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Problems

1

Year 1993 Olympiad

Level A

Problem 1. Denote by S(x) the sum of the digits of a positive integer x.Solve:

(a) x + S(x) + S(S(x)) = 1993.(b) x + S(x) + S(S(x)) + S(S(S(x))) = 1993.

Problem 2*. Suppose n is the sum of the squares of three positive integers.Prove that n2 is also the sum of the squares of three positive integers.

Problem 3. A red and a blue poker chip are stacked, the red one on top.Suppose one can carry out only the following operations: (a) adding twochips of the same color to the stack together, in any position; and (b) re-moving any two neighboring chips of the same color. After finitely manyoperations, is it possible to end up with only two chips left, the blue one ontop of the red one?

Problem 4. At the court of Tsar Gorokh, the royal astrologer built a clock Adaptation

remarkably similar to modern (analog) ones, with hands for hours, minutes,and seconds, all moving smoothly around the same point. He calls a momentof time lucky if the three hands of his clock, counting clockwise from thehour hand, appear in the order hours/minutes/seconds, and unlucky if theyappear in the order hours/seconds/minutes. Is the amount of lucky time ina 24-hour day more or less than the amount of unlucky time?

Remark. Tsar Gorokh (King Pea) is a character from Russian folklore. “In the ∗time of Tsar Gorokh” is a Russian idiom meaning “a very long time ago”.

Problem 5. Prove or disprove: There is a finite string of letters of thealphabet such that there are no two identical adjacent substrings, yet a pair ∗of identical adjacent substrings appears as soon as one adds any letter ofthe alphabet at the beginning or at the end of the string.

Problem 6. A circle centered at D passes through points A, B, and theexcenter O of the triangle ABC relative to side BC (that is, O is the centerof the circle tangent to BC and to the extensions of sides AB and AC).Prove that A, B, C, and D lie on a circle.

4 PROBLEMS

Level B

Problem 1. For two distinct points A and B in the plane, find the locus ofpoints C such that the triangle ABC is acute and the value of its angle at Ais intermediate among the triangle’s angles (meaning that \B ≤ \A ≤ \C ∗or \C ≤ \A ≤ \B).

Problem 2. Let x1 = 4, x2 = 6, and define xn for n ≥ 3 to be the leastnonprime greater than 2xn−1 − xn−2. Find x1000.

Problem 3. A paper triangle with angles of 20◦, 20◦, and 140◦ is cut alongone of its bisectors into two triangles; one of these triangles is also cut alongone of its bisectors, and so on. Can we obtain a triangle similar to the initialone after several cuts?

Problem 4. In Pete’s class, there are 28 students besides him. Each of the28 has a different number of friends in the class. How many friends doesPete have in this class?

Problem 5. To every pair of numbers x and y we assign a number x ∗ y.Find 1993 ∗ 1935 if it is known that

x ∗ x = 0 and x ∗ (y ∗ z) = (x ∗ y) + z for any x, y, z.

Problem 6. Given a convex quadrilateral ABMC with \BAM = 30◦,\ACM = 150◦, and AB = BC, prove that AM is the bisector of \BMC.

Level C

Problem 1. In the decimal representation of two numbers A and B, theminimal periods have lengths 6 and 12, respectively. What are the possibil-ities for the length of the minimal period of A + B?

Problem 2. The grandfather of Baron von Munchhausen built a castlewith a square floor plan. He divided the castle into 9 equal square areas,and placed the arsenal in the center square. The Baron’s father divided eachof the remaining 8 areas into 9 equal square halls and built a greenhouse ineach central hall. The Baron himself divided each of the 64 empty halls into9 equal square rooms and placed a swimming pool in each of the centralrooms. He then furnished the other rooms lavishly and connected each pairof adjacent furnished rooms by a door, locking all other doors.

The Baron boasts that he can tour all his furnished rooms, visiting eachexactly once and returning to the starting point. Can this be true?

Problem 3. From any point on either bank of a river one can reach theother bank by swimming a distance of no more than 1 km.

(a) Is it always possible to pilot a boat along the whole length of the riverwhile remaining within 700 m of both banks?

(b) * Same question with 800m.

YEAR 1993 OLYMPIAD 5

Remark. Both the answer and the degree of difficulty of the problem depend on adaptation

what additional assumptions are made. Naturally the boat is to be considered apoint. The original problem had a note saying so, and also that “the river joinstwo round lakes, each 10 km in radius, and the river banks consist of straight linesegments and arcs of circle.” But in spite of this precision, ambiguities remain.Can there be islands in the river? And does “within 700m”refer to the straight-line distance or the swimming distance?(See figure.) Warning: Part (b) is surprisingly difficult, unlessislands are allowed, in which case both parts are easy.

Problem 4. Given real numbers a and b, define pn = [2{an+b}], where {x}denotes the fractional part of x and [x] the integer part.

(a) Can all possible quadruples of 0s and 1s occur as substrings of the ∗sequence p0, p1, p2, . . . , if we are allowed to vary a and b?

(b) Can all possible 5-uples of 0s and 1s occur?

Problem 5. In a botanical classifier, a plant is identified by 100 features.Each feature can either be present or absent. A classifier is considered tobe good if any two plants have less than half of their features in common.Prove that a good classifier cannot describe more than 50 plants.

Problem 6. On the side AB of a triangle ABC, a square is constructedoutwards; let its center be O. Let M and N be the midpoints of AC andBC, and let the lengths of these sides be a and b. Find the maximum of thesum OM + ON as the angle ACB varies.

Level D

Problem 1. Knowing that tanα + tanβ = p and cot α + cotβ = q, findtan(α + β).

Problem 2. The unit square is divided into finitely many smaller squares,not necessarily of the same size. Consider the small squares that overlap(possibly at a corner) with the main diagonal. Is it possible for the sum oftheir perimeters to exceed 1993?

Problem 3. We are given n points in the plane, no three of which lie on aline. Through each pair of points a line is drawn. What is the least possiblenumber of pairwise nonparallel lines among these lines?

Problem 4. We start with a number of boxes, each with some marbles inthem. At each step, we select a number k and divide the marbles in eachbox into groups of size k with a remainder of less than k; we then removeall but one marble from each group, leaving the remainders intact.

Is it possible to ensure that in 5 steps each box is left with a singlemarble, if initially each box has at most (a) 460 marbles, (b) 461 marbles?

Problem 5. (a) It is known that the domain of a function f is the segment[−1, 1], and f(f(x)) = −x for all x; also, the graph of f is the union of

6 PROBLEMS

finitely many points and straight line segments (with or without endpoints). ∗Draw a possible graph for f .

(b) Is it possible to draw the graph of f if the domain of f is (−1, 1)?the whole real line?

Problem 6*. A fly lives inside a regular tetrahedron with edge a. Whatis the shortest length of a flight the fly could make to visit every face andreturn to the initial spot?

Year 1994 Olympiad

Level A

Problem 1. A local business gets apple and grape juices in identical jugsand makes a mixed drink, which it packages in same-size bottles. One jug ∗of apple juice used to be enough for exactly 6 bottles of the mix, and onejug of grape juice for exactly 10 bottles of the mix. Then the recipe waschanged, and one jug of apple juice is now sufficient for exactly 5 bottles ofthe mix. How many bottles of mix is one jug of the grape juice good fornow? (The drink is not diluted with water.)

Problem 2. A student multiplying two three-digit numbers noticed that ifhe wrote the two numbers next to one another, the resulting six-digit numberwould be seven times greater than the product. Find the two factors.

Problem 3. In a triangle ABC, let P and Q be the bases of the perpen-diculars dropped from B to the bisectors of A and C. Prove that PQ ‖ AC.

Problem 4. Four grasshoppers sit at the vertices of a square. Every nowand then one of them hops over another, landing at a point symmetric, withrespect to the jumped-over grasshopper, to where it jumped from. Provethat at no time can the grasshoppers occupy the vertices of a bigger squarethan the original one.

Problem 5. The royal astrologer considers a moment in time favorable ifthe hour, the minute and the second hands of the clock all lie on the sameside of some diameter of the clock face. All other times are consideredunfavorable. The hands all turn smoothly around the same point. Is theamount of favorable time in a 24-hour day more or less than the amount ofunfavorable time?

Problem 6. Two people play a game on a piece of graph paper, 19 × 94squares in size. Each, in turn, colors a square of any desired size, so long asits edges coincide with lines of the grid and no part of it has been coloredyet. The player who colors the last square wins. Who is the winner underoptimal play, and what is the winning strategy?

8 PROBLEMS

Level B

Problem 1. Is there a nonconvex pentagon no two of whose five diagonalsintersect, other than at a vertex?

Problem 2. Sue starts with a line segment of length k, and Leo with one oflength l. First Sue divides her segment into three parts, then Leo divides hisinto three parts. If it is possible to build two triangles from the six segmentsobtained, Leo wins; otherwise Sue wins. Depending on the ratio k/l, whichplayer can be sure to win, and what should the winning strategy be?

Problem 3. Prove that the equation

x2 + y2 + z2 = x3 + y3 + z3

has infinitely many integer solutions.

Problem 4. Two circles intersect at the points A and B. Tangents aredrawn to both circles through A. The tangents intersect the circles at thepoints M and N . The lines BM and BN intersect the circles again at thepoints P (on BM) and Q (on BN). Prove that the segments MP and NQhave the same length.

Problem 5*. Dropping one of the digits of a certain positive integer leaves ∗another number that divides the first. The dropped digit is not the leftmostone. What is the highest possible value for the first number, assuming itdoes not end in 0?

Problem 6. In a variation on the game of Battleship, ten ships must beplaced in a 10 × 10 square of graph paper: one ship has dimensions 1 × 4,two are 1 × 3, three are 1 × 2, and four are 1 × 1. The ships cannot toucheven at a corner, but they can be placed at the sides of the square.

(a) Prove that if the ships are placed from largest to smallest, there isalways room for all of them, regardless of where earlier ones were placed.

(b)* Show by an example that if the ships are laid down from smallestto largest, this may not be the case.

Level C

Problem 1. A student multiplying two seven-digit numbers noticed thatif she wrote the two numbers next to one another, the resulting 14-digitnumber would be three times greater than the product. Find the two factors.

Problem 2. An infinite sequence of numbers xn, where n runs over positiveintegers, is defined by the condition

xn+1 = 1 − |1 − 2xn|,where 0 ≤ x1 ≤ 1.

Prove that the sequence is eventually periodic if and only if x1 is rational.


Problem 3. Each of the 1994 members of the Parliament of Dunces hasslapped exactly one colleague in the face. Prove that it is possible to drawfrom this Parliament a 665-member committee none of whom have slappedone another.

Problem 4. Let D be a point on side BC of triangle ABC. Circles areinscribed in the triangles ABD and ACD, and a common outer tangent tothe circles (distinct from BC) is drawn. Let its intersection with AD becalled K. Prove that the length of AK does not depend on the position ofD on BC.

Problem 5. Consider an arbitrary polygon, not necessarily convex.(a) Is there always a chord of the polygon that divides it into two pieces

of equal area?(b) Prove that the polygon can be divided by a chord into pieces whose

areas are each at least one-third the total area of the polygon.(By a chord of a polygon we mean a line segment whose endpoints

belong to the polygon’s perimeter while the segment itself lies entirely onthe polygon, including its perimeter.)

Problem 6*. Can a polynomial P (x) have a negative coefficient if all thepowers P n(x), for n > 1, only have positive coefficients?

Level D

Problem 1. Devise a polyhedron such that no three of its faces all havethe same number of edges.

Problem 2. See Problem 2 of Level C.

Problem 3. A spherical cherry of radius r is dropped into a goblet whoseaxial cross-section is the graph of the function y = x4. What is the largestvalue of r that allows the cherry to touch the very bottom of the goblet? (Inother words, what is the radius of the largest circle contained in the regiony ≥ x4 and touching the origin?)

Problem 4. A convex polyhedron has nine vertices, one of which is A. Thetranslations that send A into each of the other vertices form eight congruentpolyhedra. Prove that at least two of these eight polyhedra have an interiorpoint in their intersection.

Problem 5*. The extensions of the sides AB and CD of a convex quadri-lateral ABCD intersect at a point P ; the extensions of the sides BC andAD intersect at Q. Consider three pairs of bisectors: those of the outerangles of the quadrilateral at vertices A and C, those of the outer angles atvertices B and D, and those of the outer angles at vertices P and Q of thetriangles QAB and PBC. Prove that if each of these three pairs of linesintersects, the intersection points are collinear.

10 PROBLEMS

Problem 6. Prove that for any k > 1, there exists a power of 2 such thatamong its last k digits, there are at least as many 9s as other digits. (Forexample: 212 = . . . 96, 253 = . . . 992.)

Year 1995 Olympiad

Level A

Problem 1. Isaac Newton spent a groat a day for a loaf of bread and a ∗tankard of ale. When prices went up by 20%, he started buying half a loafof bread and the same ale for a groat. Will a groat be enough to buy the ∗ale if prices rise by 20% again?

Problem 2. Prove that all the numbers of the form 10017, 100117, 1001117,and so on are divisible by 53.

Problem 3. Consider a convex quadrilateral and a point O inside it suchthat \AOB = \COD = 120◦, AO = OB and CO = OD. Let K, L and Mbe the midpoints of the sides AB, BC and CD, respectively. Prove that (a)KL = LM ; (b) the triangle KLM is equilateral.

Problem 4. To manufacture a closed box of volume at least 1995 cubicunits in the shape of a parallelepiped we have (a) 962, (b) 960, (c) 958square units of material. Is the material sufficient?

Problem 5. Roads lead from a city to several nearby villages; there is nodirect communication between villages. A truck loaded with goods to bedelivered to all the villages starts out from the city. The amount of fuelspent on a leg of the trip is proportional to the current load weight and thedistance. Suppose the load to be delivered to each village weighs, in some ∗units, the same as the distance from the city to the item’s destination. Provethat the fuel cost does not depend on the order in which the deliveries aremade.

Problem 6. A straight line cuts a triangle AKN off of the regular hexagonABCDEF so that AK + AN = AB. Find the sum of the angles at whichthe segment KN is seen from the vertices of the hexagon, i.e., find

\KAN + \KBN + \KCN + \KDN + \KEN + \KFN.

Level B

Problem 1. Prove that if we insert any number of digits 3 between the 0sin 12008, we get a number divisible by 19.

12 PROBLEMS

Problem 2. Consider an equilateral triangle ABC. For an arbitrary pointP inside the triangle, consider the intersection A′ of the line AP with BCand the intersection C ′ of CP with BA. Find the locus of points P forwhich the segments AA′ and CC ′ have equal length.

Problem 3. A strip is a rectangle of size 1 × k, where k is some positive ∗integer. For what integers n can one cut a 1995× n rectangle into strips allof different lengths?

Problem 4. Suppose ab = cd, where a, b, c, d are positive integers. Can ∗a + b + c + d be a prime?

Problem 5. We start with four identical right triangles. In one move we cancut one of the triangles along the altitude perpendicular to the hypotenuseinto two triangles. Prove that, after any number of moves, there are twoidentical triangles among the whole lot.

Problem 6. A team of geologists took 80 cans of food on an expedition.The weights of the cans were all different and were listed precisely in aninventory. After a while the labels became unreadable and only the cookknew which can was which. She boasted that she could prove the identityof all the cans simultaneously using only the inventory and a few weighingoperations. Her balance has two pans and a pointer that shows the differencein weight between the contents of the pans.

(a) Show how the cook can prove her claim using four weighings.(b) Can she do it in only three?

Level C

Problem 1. The number sinα is known. What is the largest number ofpossible values for (a) sin α

2 ? (b)* sin α3 ?

Problem 2. See Problem 58.9.2. REF 58.9.2

Problem 3. The diagonals of a trapezoid ABCD meet at a point K. Lettwo circles be constructed, each having one of the lateral sides of the trape-zoid as a diameter. Supposing that K lies outside both circles, prove thatthe tangents from K to these circles have equal lengths.

Problem 4. See Problem 58.9.5. REF 58.9.5

Problem 5*. Prove that if a, b and c are integers such that a/b+ b/c+ c/aand a/c + c/b + b/a are also integers, then |a| = |b| = |c|.Problem 6. A panel has a number of buttons and a number of light bulbs.Each button is connected to some of the lights, and pressing the button flipsthe state of the lights it’s connected with from on to off or vice versa. Some ∗of the lights are on at the start.

It is known that, for any set of lights, there is a button that flips an oddnumber of lights from this set. Prove that one can switch off all the lightsby pressing buttons.


Level D

Problem 1. Prove that

|x| + |y| + |z| ≤ |x + y − z| + |x − y + z| + |−x + y + z|for all real numbers x, y, z.

Problem 2. Is it possible to color the edges of an n-angled prism with threecolors so that all three colors occur among the edges of any face and amongthe edges meeting at any vertex? Answer for (a) n = 1995 and (b) n = 1996.

Problem 3. In a triangle ABC, consider the median AA1, the bisectorAA2, and a point K on AA1 such that KA2 ‖ AC. Prove that AA2 ⊥ KC.

Problem 4*. (a) Divide the interval [−1, 1] into black and white intervalsso that the integral of any linear function over intervals of one color is thesame for both colors.

(b) Same problem with quadratic trinomials instead of linear functions.

Problem 5. Find the largest value of n such that there are two-sided infinitestrings A and B satisfying the following conditions: ∗

• Any subtring of B of length at most n is contained in A.• A is periodic with minimal period 1995, but B is not (it may be ape-

riodic or have a different period). ∗The strings can contain arbitrary symbols.

Problem 6*. Prove that there exist infinitely many nonprime values of nsuch that 3n−1 − 2n−1 is divisible by n.

Problem 7. Is there a polyhedron and a point outside it such that fromthis point none of the polyhedron’s vertices is visible?

Year 1996 Olympiad

Level A

Problem 1. If a + b2/a = b + a2/b, is it true that a = b?

Problem 2. Ten iron weights are arranged around a circle. A small bronzeball is placed between each pair of neighboring weights. The mass of eachball equals the difference of the masses of the neighboring weights. Provethat the balls can be placed on the two pans of a balance so as to bring itinto equilibrium.

Problem 3. Each intersection of a square grid is home to a gardener, andthere are flowers planted everywhere. Each flower must be looked after bythe three gardeners closest to it. Draw the area that the gardener at theorigin must look after.

Problem 4. The side BC of an equilateral triangle ABC is divided intothree equal parts by the points K and L, and point M divides the side ACin the ratio AM : MC = 1 : 2. Prove that the sum of angles AKM and ALMis 30◦.

Problem 5. A rook is at the corner of an n × n chessboard. It makesn2 successive moves (of any length), alternating between horizontal andvertical motion. For what values of n can this sequence of moves be chosenso that the rook stops on all the squares of the board and finally returns toits original position? (Note that each move must be followed by one in aperpendicular direction.)

Problem 6. (a) Eight students were given eight problems to work on. Itturns out that each problem was solved by five students. Prove that thereare two students such that each problem was solved by at least one of them.

(b) If each problem was solved by four students, show by a counterex-ample that two students satisfying the condition in part (a) need not exist.

Level B

Problem 1. Prove that any convex polygon has at most 35 angles less than170◦.

16 PROBLEMS

Problem 2. Suppose the numbers a, b, and c satisfy the inequalities

|a − b| ≥ |c|, |b − c| ≥ |a|, |c − a| ≥ |b|.

Prove that one of these numbers is the sum of the other two.

Problem 3. Consider two points A and B on the circle circumscribed toa triangle ABC (the circumcircle), and assume the tangents to the circlethrough A and B meet at a point M . Choose a point N on the side BC sothat the line MN is parallel to the side AC. Prove that AN = NC.

Problem 4. The integers from 1 to n are written in a row. The samenumbers are written under them in a different order. Is it possible to writethe second row so that all the sums of pairs of corresponding numbers inthe two rows are perfect squares when (a) n = 9; (b) n = 11; (c) n = 1996?

Problem 5. Two points A and B on a circle divide it into two arcs. Con-sider all the chords joining a point on one arc AB to a point on the other.What is the locus of midpoints of these chords?

Remark. The statement is ambiguous on whether the endpoints A and B shouldbe regarded as part of the arcs. For definiteness, assume they are not.

Problem 6. Ali Baba and a thief must share a trove of 100 gold coins.They arrange them into 10 piles of 10 coins each. Ali Baba takes four cups,places them next to four piles of his choice, and transfers from each of thesepiles into the corresponding cup some number of coins —at least one, butnot the entire pile. The number of moved coins need not be the same forthe four piles.

After that, the thief must permute the cups without leaving them all inplace, and then move the coins from each cup onto the pile the cup is nextto now.

This process is repeated: Ali Baba again places the empty cups near anyfour of the ten piles, and so on. At any time, Ali Baba can take three pilesof his choice and leave; his opponent gets all the rest.

What is the greatest number of coins that Ali Baba can walk away withif the thief also works to get as many coins as possible?

Level C

Problem 1. The positive numbers a, b, and c satisfy the equation a2 +b2−ab = c2. Prove that (a − c)(b − c) ≤ 0.

Problem 2. A hundred points are marked on a piece of paper in a 10×10 Adaptation

square array. How many lines not parallel to either side of the square arerequired if we want lines going through all 100 points?


Problem 3. Divide side BC of an equilateral triangle ABCinto n equal parts with points P1, P2, . . . , Pn−1 as in thefigure, so BP1 = P1P2 = · · · = Pn−1C. Choose a pointM on the side AC so that AM = BP1. Prove that\AP1M+\AP2M+· · ·+\APn−1M = 30◦ if (a) n = 3;(b) n is an arbitrary positive integer.

A

P1 P2 Pn−1B C

M

Problem 4. A rook is on a corner square of an m × n chessboard. Twoplayers take turns moving it, either straight across or up and down, anynumber of squares each time; but the rook is not allowed to land on, oreven cross, a square on which it has already landed or through which it haspassed. The player who cannot make a move loses. Which player can forcea to win: the one who plays first or the second one? Describe a winningstrategy.

Problem 5. Two laws are in force in a certain country:

(1) A person may play basketball only if he/she is taller than most ofhis/her neighbors.

(2) A person gets free bus rides only if she/he is shorter than most ofher/his neighbors.

By a person’s “neighbor” is meant anyone whose house (regarded as a point)lies inside a certain circle whose center is the person’s house and whose radiusis chosen by the person — and a different choice of radius is allowed for thepurposes of the first law and the second law!

Is it possible that 90% of people or more are allowed to play basketballand that 90% or more are allowed free travel?

Problem 6. Prove that for any polynomial P (x) of degree n with positiveinteger coefficients, there exists an integer k such that P (k), P (k + 1), . . . ,P (k+1996) are composite numbers if (a) n = 1; (b) n is an arbitrary positiveinteger.

Level D

Problem 1. See Problem 1 on page 16.

Problem 2. Find a polynomial with integer coefficients having5√

2+√

3 +5√

2−√

3 as a root.

Problem 3. A point is chosen on each of eight evenly spaced parallel planes.Can these points be the vertices of a cube?

Problem 4. Prove that there exist infinitely many positive integers n withthe property that n is the sum of two perfect squares, but n − 1 and n + 1are not.

Problem 5. A point X lies outside two disjoint circles ω1 and ω2, andthe tangent segments drawn from X to ω1 and ω2 are equal. Prove that theintersection of the diagonals of the quadrilateral formed by the four tangency

18 PROBLEMS

points coincides with the intersection of the two internal common tangentsof ω1 and ω2.

Problem 6. The 2n possible strings of length n containing only the numbers1 and −1 form a table with 2n rows and n columns. Suppose that a numberof entries in the table are replaced by 0. Prove that one can select certainrows so that their sum (defined as the string obtained by adding each columnseparately) is a string of zeros.

Year 1997 Olympiad

Level A

Problem 1. Chess pieces are placed on the squares of a chessboard insuch a way that each row has at least one piece and the number of squaresoccupied in each row is different for every row. Prove that one can chooseeight occupied squares so that every row and every column contains exactlyone of the chosen squares.

Problem 2. The walk from an observation station to the top of MountStromboli takes 4 hours along the road plus 4 hours along a path. Thereare two craters at the top. The first erupts for 1 hour, then lies quiet for17 hours, then erupts for 1 hour again, and so on. The second crater eruptsfor 1 hour, lies quiet for 9 hours, erupts for 1 hour and so on. It is dangerousto walk either along the path or along the road during eruptions of the firstcrater, but when the second crater erupts, only the path is dangerous.

Professor Garibaldi, a volcanologist, sees both craters start eruptingsimultaneously, exactly at noon. Will Garibaldi ever be able to walk up tothe top of the volcano and back without endangering his life?

Problem 3. Points M and N are chosen inside an acute angle XOY so that\XON = \Y OM . Points Q and P are chosen on the line segments OXand OY , respectively, so that \NQO = \MQX and \NPO = \MPY .Prove that the paths MPN and MQN have the same length.

Problem 4. Prove that there exists a positive integer that yields a com-posite number whenever any three adjacent digits are replaced by arbitrarydigits. Is there a 1997-digit number with this property?

Problem 5. In a rhombus ABCD, the angle B measures 40◦, a point E isthe midpoint of BC, and F is the foot of the perpendicular dropped from Ato DE. Find the measure of angle DFC.

Problem 6. A banker found out that one of a number of seemingly identicalcoins is counterfeit (lighter than the others). He asked an expert to find thelighter coin. To make matters interesting, the expert decided to do this

20 PROBLEMS

using only a two-pan balance, without weights, and to use each coin in atmost two weighings.

What is the greatest number of coins such that the expert can be sureof being able to determine the counterfeit coin after n weighings?

Level B

Problem 1. The length of one side of a triangle is one-third the sum ofthe lengths of the other two. Prove that the angle opposite this side is thesmallest angle of the triangle.

Problem 2. You have nine pieces of cheese, all of different weights. Can youalways cut one of them into two parts so that the ten pieces thus obtained ← In the original, the

solution assumes theweights are all different.

The translation I got tried

to relax that assumption.

But I think that doesn’twork since the problem

insists that one piece must

be *divided* into two andthis would not be possible

with weights 1,1,...,1,8. So

I’m clarifying that the

weights have to bedifferent.

can be divided into two groups that weigh the same, each with five pieces?

Problem 3. A convex hexagon AC1BA1CB1 satisfies the relations AB1 =AC1, BC1 = BA1, CA1 = CB1, and \A + \B + \C = \A1 + \B1 + \C1.

Prove that the area of the triangle ABC is half that of the hexagon.

Problem 4. Imagine n metro trains running in the same direction at equalintervals along a circular line. Stations A, B, and C are located along the

Adaptation

line in this order, and the distances from A to B, from B to C and from Cto A are all the same.

Ira and Lyosha enter stations A and B, respectively, at the same time,planning to take the next train. It is known that if they enter their stationsat the moment when Roma, the conductor of one of the trains, is goingthrough an underground portion of the line, then Ira will get on a trainbefore Lyosha; at any other time Lyosha will be the first to board, or theywill both board simultaneously. What portion of the line runs underground?

Problem 5. Two round-robin tournaments had the same 2n participants.(In a round-robin tournament every participant plays against every otheronce.) A win is worth one point, a tie half a point, and a loss zero.

Prove that if the scores of each player in the two tournaments differ byat least n points, they must differ by exactly n points.

Problem 6. Suppose that 1 + x + x2 + · · · + xn = F (x)G(x), where Fand G are polynomials whose coefficients are only zeros and ones. Provethat one of the polynomials F (x) and G(x) is representable in the form(1 + x + x2 + · · · + xk)T (x), where k > 0 and T is also a polynomial withcoefficients 0 and 1.

Level C

Problem 1. Is there a convex solid, other than a ball, whose orthogonalprojections onto three pairwise perpendicular planes are disks?


Problem 2. Prove that the minimum perimeter of a quadrilateral withgiven diagonal lengths and angle between diagonals is attained by a paral-lelogram.

Problem 3. (a) As the perimeter of a quadrilateral is traced clockwise,each of its sides is extended by its length. The endpoints of the extensionsturn out to be vertices of a square. Prove that the original quadrilateral isa square.

(b) Prove that if a similar construction applied to an n-gon yields aregular n-gon, the original n-gon is also regular.

Problem 4. Given real numbers a1 ≤ a2 ≤ a3 and b1 ≤ b2 ≤ b3 such that

a1 + a2 + a3 = b1 + b2 + b3

anda1a2 + a2a3 + a1a3 = b1b2 + b2b3 + b1b3,

prove that if a1 ≤ b1, then a3 ≤ b3.

Problem 5. In a round-robin tournament everyone plays everyone else; awin scores 1 point and a loss 0. (We assume there are no ties.) Supposewe compute a “superscore” for each player by adding up the scores of theopponents who lost to that player. If the superscores of all players are thesame and there are more than two players, prove that the scores of theplayers are also all the same.

Problem 6. Consider the powers of five (1, 5, 25, 125, 625, . . . ) and take thesequence of their first digits: 1, 5, 2, 1, 6, . . . Prove that any segment ofthis sequence written in reverse order occurs in the sequence of first digitsof powers of two (1, 2, 4, 8, 1, 3, 6, 1, . . . ).

Level D

Problem 1. Points C ′, A′, and B′ are taken on the sides AB,BC, and CA of a triangle ABC, respectively (see figure).Prove that the triangle A′B′C ′ has area

AB′ · BC ′ · CA′ + AC ′ · CB′ · BA′

4R,

where R is the circumradius of the triangle ABC.A

B

CB′

C′A′

Problem 2. Calculate the integral∫ π/2

0cos2(cosx) + sin2(sinx) dx.

Problem 3. Three functions,

f1(x) = x +1

x, f2(x) = x2, and f3(x) = (x − 1)2,

are written on the blackboard. You are allowed to add, subtract, and mul-tiply these functions (and so also square them, cube them, . . . ), multiply

22 PROBLEMS

them by an arbitrary number, add an arbitrary number to them, and per-form the same operations with the expressions thus obtained. Construct thefunction 1/x by means of these operations. Prove that if we erase any of thefunctions f1, f2, f3 from the blackboard, the function 1/x will no longer beconstructible.

Problem 4. Is it possible to cut a regular tetrahedron with a unit edge intoregular tetrahedrons and octahedrons with edges of length less than 1/100?

Problem 5. Let a, b, and c be positive numbers such that abc = 1. Provethe inequality

1

1 + a + b+

1

1 + b + c+

1

1 + c + a≤ 1. (1)

Problem 6. A disk of radius 1 and a finite number of (infinite) strips whosewidths add up to 100 are given on the plane. Prove that each of the stripscan be translated so that together they cover the disk completely.

Year 1998 Olympiad

Level A

Problem 1. Are there positive integers x, y, and z that satisfy the equation28x + 30y + 31z = 365?

Problem 2. Is it possible to find eight positive integers such that none ofthem is divisible by any other, but their squares are all divisible by each ofthe eight initial numbers?

Problem 3. The dialgonals AC and BD of a parallelogram ABCD meetat a point O. Point M lies on the line AB and \AMO = \MAD. Provethat M is equidistant from C and D.

Problem 4. Some of the numbers a1, a2, . . . , a200 are written in blue, therest in red. If we erase every red number, the remaining numbers will be allthe positive integers from 1 to 100, written in ascending order. If we eraseevery blue number, the remaining numbers will be all positive integers from100 to 1 written in descending order. Prove that the set {a1, a2, . . . , a100}coincides with the set of integers from 1 to 100.

Problem 5. Several guests are seated around a round table. Some of themare acquainted with each other; each acquaintance is mutual. All the ac-quaintances of each person (including the person himself or herself) areseated at equal intervals. For different people these intervals may be differ-ent. It is known that any two guests have at least one common acquaintanceat the table. Prove that everybody is acquainted with everybody else.

Problem 6. A red square is covered by 100 white squares. All 101 squaresare congruent and the sides of each white square are parallel to the sides ofthe red square. Is it always possible to remove one of the white squares sothat the remaining white squares will still completely cover the red square?

Level B

Problem 1. Is 49 + 610 + 320 a prime number?

24 PROBLEMS

Problem 2. Two altitudes, AD and CE, are drawn in an acute triangleABC. Then a square ACPQ and two rectangles CDMN and AEKL aredrawn, with AL = AB and CN = CB. Prove that the area of ACPQ isthe sum of the areas of AEKL and CDMN .

Problem 3. Each of the inhabitants of a village either always tells the truthor always lies. The villagers formed a circle, and a traveler asked each ofthem whether his or her neighbor to the right is honest or lies. Based on theiranswers, the traveler managed to determine the exact fraction of truthfulpeople among the villagers. Your task is to find this fraction without havingheard the answers.

Problem 4. A network of military bases in Mathistan is connected byroads, each road starting and ending at a base. A set of roads is saidto be essential if, when these roads are closed, there are bases no longerconnected by roads. An essential set is said to be strategic if it does notcontain a smaller essential set.

Let two different strategic sets V and W be given. Prove that the set ofroads that belong to exactly one of V and W is essential.

Problem 5. A point O lies inside a rhombus ABCD. The angle DABmeasures 110◦, while \AOD and \BOC measure 80◦ and 100◦, respectively.What values can the measure of angle AOB take?

Problem 6. A number of distinct points are marked on the interval [0, 1].Each of these points is either halfway between two other marked points(not necessarily its neighbors) or halfway between a marked point and anendpoint of the interval. Prove that all the marked points are rational.

Level C

Problem 1. Let a, b, and c be nonnegative integers such that 28a + 30b +31c = 365. Prove that a + b + c = 12.

Problem 2. A square of side 1 is cut up into rectangles. One side of eachrectangle is marked. Prove that the sum of lengths of all the marked sidesis at least 1.

Problem 3. A road of length 1 km is entirely lit by a certain number ofstreet lamps, each of which lights a segment of road 1m long. Suppose thatturning off any of the lamps causes the road no longer to be lit in its entirety.What is the highest possible number of lamps?

Problem 4. Is there a positive integer divisible by 1998 whose digits addup to less than 27?

Problem 5. An equilateral triangle ABC is cut out of plywood and placedon the floor. Three nails are hammered into the floor, one touching each sideof the triangle, so that it is impossible to turn the triangle without lifting it


off the floor. The first nail divides the side AB in the ratio 1 : 3, measuringfrom A, another one divides BC in the ratio 2 : 1, measuring from B. Inwhat ratio does the third nail divide the side AC?

Problem 6. The positive integers from 1 to n are arbitrarily arranged ina sequence (the number n is fixed). An arrangement is said to be bad if wecan select 10 of its numbers (not necessarily consecutive) so that they forma decreasing sequence. All other arrangements are said to be good. Provethat the number of good arrangements does not exceed 81n.

Level D

Problem 1. Three numbers x, y, z satisfy the equation

x + y + z − 2(xy + yz + xz) + 4xyz = 12 .

Prove that at least one of them is equal to 12 .

Problem 2. It is known that a continuous function f :

(1) is defined on the entire real axis;(2) has a derivative at each point (thus, the graph of f has a unique tangent

at each point);(3) is such that its graph does not contain points one of whose coordinates

is rational and the other one irrational.

Does it follow that the graph of f is a straight line?

Problem 3. Two medians, AK and BL, are drawn in a scalene triangleABC. The angles \BAK and \CBL measure 30◦. Find the angles of thetriangle ABC.

Problem 4. Find all the solutions of 3x + 4y = 5z in positive integers.

Problem 5. Is it possible to make a closed chain of 61 identical coherentlyrotating cogwheels in space in such a way that the angles between meshedcogwheels are at least 150◦? Assume that:

(1) the cogwheels are disks (just for simplicity);(2) two cogwheels are meshed if the corresponding circles have a common

tangent at their common point;(3) the angle between two meshed cogwheels is the angle between the radii

of their circles drawn at the point of tangency;(4) the first cogwheel must be meshed with the second, the second with

the third, . . . , the 61st with the first; no other pair of cogwheels shouldhave a common point.

Problem 6. See Problem 6 on page 25.

Answers

Year 1993 Olympiad

Level A

1. (a) No solutions. (b) x = 1963. 3. No. 4. The amounts of luckyand unlucky time are the same. 5. There is always such a string.

Level B

1. The locus is given by the shaded domains and solidA Bcurves in the figure. (The points on the dotted curves do not

belong to the locus.) 2. x1000 = 12 · 1000 · 1003 = 501500.

3. No. 4. Pete has 14 friends. 5. 58.

Level C

1. 4 or 12. 2. Yes. 3. 700 m is not always possible, but 800 m is(if islands are not allowed). Here we’re considering the swimming distance ∗rather than the straight-line distance. 4. (a) Yes. (b) No. 6. Themaximum equals 1

2(1 +√

2)(a + b) and is attained when \ACB = 135◦.

Level D

1. pq/(q−p), unless p = q, in which case the answer iseither 0 (if p = 0) or undefined (if p 6= 0). 2. Yes.3. n lines if n > 2; one line if n = 2. 4. (a) Yes.(b) No. 5. (a) One solution is shown in the figure.(b) No. 6.

4√10

a.

1

2

−1

2

1

2

−1

2

−1

1

−1

1

0

Year 1994 Olympiad

Level A

1. Fifteen bottles. 2. 143 and 143. 5. There is more favorable timethan unfavorable time. 6. The first player wins.

Level B

1. Yes. 2. If k > l, Sue wins; otherwise Leo wins. 5. The number180625 becomes 17 times smaller if 8 is stricken out.

Level C

1. 3333334 and 1666667. 5. (a) Not always. 6. Yes.

Level D

3. r = 34

3√

2.

Year 1995 Olympiad

Level A

1. Yes. 4. Yes in all three cases. 6. 240◦. ∗Level B

2. The locus is the union of the altitude of △ABC fromvertex B with an arc of circle AC of measure 120◦ lying in ∗△ABC; see figure. 3. For n ≤ 998 and for n ≥ 3989.4. No. 6. (b) No, she can’t. A C

B

∗Level C

1. (a) 4; (b) 3.

Level D

2. (a) yes; (b) no. 5. n = 1994. 7. Yes.

Year 1996 Olympiad

Level A

1. Yes, it is true. 3. 5. For even n.

Level B

4. (a) Yes; (b) no; (c) yes. 5. Let O be the centerof the circle. Consider the circles with diameters AOand BO (see figure). The desired locus is the interior ofthese two circles minus their intersection. 6. 72 coins.

A B

O

Level C

2. 18. 4. For m = n = 1 player 2 wins. In all other cases, player 1 wins.5. Yes, it is possible.

Level D

2. x5 − 5x3 + 5x − 4. 3. Yes, they can.

Year 1997 Olympiad

Level A

2. Yes; Garibaldi can leave at the beginning of the 38th hour, say. 5. 110◦.6. 2n2 + 1 coins.

Level B

2. Yes, it is always possible. 4. If n is divisible by 3, there are no under-ground portions; if n = 3k + 1 or n = 3k + 2 for k ∈ Z, two thirds or onethird, respectively, of the track runs underground.

Level C

1. Yes, such a solid exists.

Level D

2. π/2. 3. 1/x = f1(x) − 12

(f2(x) − f3(x) + 1

). 4. Yes.

Year 1998 Olympiad

Level A

1. Yes. For example, we can take x = 1, y = 4, and z = 7. 2. Yes.6. No.

Level B

1. No. 3.12 . 5. 80◦ or 100◦.

Level C

3. 1998. 4. No. 5. In the ratio 5 : 7, measuring from vertex A.

Level D

2. No. 3. \B = 120◦, \C = arccos 54√

7, \A = arccos 2√

7. 3. Unique

solution: x = y = z = 2. 5. Yes.

Year 1993 Olympiad

Level A

1. (a) Use divisibility by 3. (b) Estimate x from below and find the remain-der after division of x by 9.2. Rewrite the expression (a2 + b2 + c2)2 in a different form.3. Consider all pairs of chips with a red chip above a blue chip.4. Favorable and unfavorable times are interchanged by reflection.5. Use induction on the number of letters in the alphabet.6. Prove that \ADB = \ACB.

Level B

2. Write down xn − xn−1 for n small.3. From what triangles can a triangle with the initial angles be obtained?4. Consider the most friendly and the least friendly of Pete’s classmates.6. Consider the points symmetric to B with respect to AM .

Level C

1. The digits of A + B must repeat in blocks of 12.4. Draw the points {an + b} on the circle of unit circumference.5. Estimate the total number of distinctions between all possible pairs ofplants with respect to all features.

Level D

4. Let there be 1, 2, . . . , n stones in the boxes. Express the maximal numberof stones after one move in terms of n and k. Investigate for which k thisnumber is minimal.5. The graph of such a function would be left unchanged by a 90◦ rotationaround the origin.6. Let P and Q be the midpoints of the sides KL and MN of a spacequadrilateral KLMN . Then

PQ ≤ 12(KN + LM).

Year 1994 Olympiad

Level A

1. Express the volume of the bottle in terms of the volume of the jug.2. If x and y are the three-digit numbers to be found, then 7xy = 1000x+y.3. Extend the perpendiculars until they intersect the line AC.4. The grasshoppers can only hop to points in a square grid.5. After an integer number of hours, the second hand and the minute handreturn to their positions.6. Use the axial symmetry of the rectangle.

Level B

2. Leo should try to make isosceles triangles.3. Set z = −x.4. It suffices to prove that the triangles AQN and AMP are congruent.Prove that they are similar and that the arcs AQ and MA are equal.6. (b) To prove that it is possible to place the next 1 × 3 ship, draw 8auxiliary 1 × 3 ships in such a way that a 1 × 4 or 1 × 3 ship would touchnot more than two of these auxiliary ships.

Level C

1. Let x and y be the 7-digit numbers required. Then 3xy = 107x + y.2. (a) Observe how the denominators of the terms of the sequence change.(b) Fix n. One can write xn+k = bkxn + ak. What is |bk|?3. Prove that some member of Parliament was slapped at most once.4. AK = 1

2(AB + AC − BC).5. (b) Draw a chord and move it so it remains parallel to itself. (If the chordgoes through a concave vertex of the polygon, it may split the polygon intomore than two parts; do not forget to account for this case.)6. Note that the condition “for all n > 1” can be replaced by “for n = 2and n = 3”. If P has a term with a zero coefficient while all the coefficientsof P 2 and P 3 are positive, it is possible to “jiggle” P a bit so it satisfies thecondition of the problem.

46 HINTS

Level D

1. Try to construct a polyhedron with 6 faces.3. What is the smallest value of r for which the circle of radius r with center(0, r) intersects the curve y = x4 other than at the origin?4. Show that the polyhedron obtained by applying a homothety with centerA and factor 2 contains the eight translated polyhedra.5. Use Fact 25.6. It is enough to find m such that 2m is just a bit less than a power of 10.For this, it suffices to find n such that 2n + 1 is divisible by a sufficientlylarge power of 5.

Year 1995 Olympiad

Level A

2. Consider the difference between consecutive numbers.3. Consider a 120◦ rotation about the point O.5. You can either compare the fuel costs for deliveries having almost the ∗same route, or deduce an explicit expression for the cost for any delivery.6. Rotate all the angles of interest so their vertices coincide with A.

Level B

1. Consider the difference between consecutive numbers.2. For a fixed A′, there exist precisely two points C ′ such that AA′ = CC ′.3. Start with a similar problem for the 5 × n rectangle.5. The order in which we cut the triangles does not matter.6. (a) The cook should first put the 27 lightest cans on the left pan.(b) Use the pigeonhole principle (Fact 1). ∗Level C

1. (b) Express sin 3x in terms of sinx.3. Use the tangent-secant theorem (power of a point) for an appropriatesecant.5. Let p be a prime divisor of a. Reduce the fractions to the least commondenominator, and consider the highest powers of p dividing the numeratorand denominator.6. Use induction. (Assume it’s been proved that all the bulbs, except anarbitrary fixed one, can be switched off.)

Level D

1. |a + b| ≤ |a| + |b|.3. Complete the triangle BAC to a parallelogram and use similarity.4. If f is a polynomial of degree n, then f(x) − f(x−d) is a polynomial ofdegree n − 1, for any fixed d 6= 0.6. Look for n of the form 3b − 2b.

Year 1996 Olympiad

Level A

1. A nonzero factor can be canceled from both sides of an equation.2. Let mi be the mass of the ith weight; then (m1−m2)+(m2−m3)+ · · ·+(m9 − m10) + (m10 − m1) = 0.3. Consider a flower and find out which gardeners look after it.6. (a) Consider two cases: (1) there is a student who has solved six problems,and (2) each student has solved at most five problems.

Level B

1. Use the formula for the sum of angles of a convex polygon.3. Use Fact 15 (page 105).4. (c) Reduce the problem to a similar one with a smaller n.6. First, prove that the thief can always ensure that there are no piles of lessthan four coins. Then show that Ali Baba can ensure that there are sevenpiles of at most four coins each.

Level C

2. Consider the 36 unit squares adjacent to the sides of the big square.3. Express the angles in question in terms of angles at the points A and M .5. Consider a country with 10 inhabitants whose houses are placed along astraght line.6. If k divides x − y, then k divides P (x) − P (y).

Level D

2. Compute5√

2 +√

3 · 5√

2 −√

3.3. Write the equations of parallel planes drawn through the vertices of theunit cube in such a way that the distances between any two neighboringplanes are the same.5. Use the law of sines and Fact 17 (page 105).

Year 1997 Olympiad

Level A

2. Start the walk right after the first crater erupts and at such a time thatfour hours later the second crater has just calmed down.3. Consider the reflections of M in the lines OX and OY .5. Extend the lines DE and AB until they meet.6. Consider the analogous problem under the condition that each coin canbe used in only one weighing, rather than two.

Level B

1. The smallest angle of a triangle is opposite the shortest side.2. Cut the largest piece.4. If the distance between two points is a multiple of the interval betweentrains, then either both points are underground or both are above ground.5. Divide the participants of the tournament into two groups: those whosescore in the second tournament is greater than in the first tournament, andthose whose score in the first tournament is greater than in the secondtournament.6. Suppose that the coefficient of x in the polynomial F (x) is equal to one.Consider the smallest m such that the coefficient of xm in F (x) is zero. Thentake k = m − 1.

Level C

2. Translate the quadrilateral ABCD by the vector−→AC .

3. Suppose that the construction in question applied to a polygon A1 . . . An

yields a polygon B1 . . . Bn. Prove that the polygon A1 . . . An can be recov-ered from B1 . . . Bn.4. Use the relations between the coefficients of a cubic polynomial and itsroots.5. Assuming that the statement is not true, consider the participants withthe greatest number of wins and those with the smallest number of wins.

52 HINTS

6. Prove that for any k there exists a power of 2 whose decimal representationstarts with a 1 followed by k zeros.

Level D

2. Write the integral as the sum of two integrals and apply the substitutiony = π/2 − x to one of them.3. Applying the allowed operations to polynomials, we can obtain only apolynomial; therefore, we can’t do without f1. To prove that f2 is indis-pensable, consider the derivatives of functions at x = 1. The indispensabilityof f3 is proved by means of complex numbers.5. Set a = x3, b = y3, and c = z3.

Year 1998 Olympiad

Level A

1. Notice that 365 is the number of days in a year.2. Use prime factorization.3. Join the midpoints of the sides AB and CD of the parallelogram.5. If a person has acquaintances sitting next to each other, this person isacquainted with all the guests.

Level B

1. (a + b)2 = a2 + 2ab + b2.3. Imagine that all truthful villagers become liars, and all liars are reformed.4. If we close all the roads in a strategic set, the bases break up into exactlytwo disconnected sets such that no road joining two bases in one of thesetwo sets belongs to the strategic set of roads.5. Prove that there are exactly two such points O.6. This problem requires a certain knowledge of linear algebra (namely, theGaussian elimination algorithm); see Field2Elem and also Fact 25. CITE Field2Elem

Level C

1. Assume that the statement is not true and derive a contradiction.2. Write out the sum of the areas of all the rectangles.3. If two segments of a road lighted by two lamps overlap, then these lampsare neighboring.4. Invent a criterion for divisibility by 999.5. Suppose there is one nail hammered on a side of the triangle. Find all thepoints in the plane such that the triangle can be rotated clockwise aroundany of these points and all the centers of possible counterclockwise rotations.6. If an arrangement is good, the numbers can be colored in nine colors sothat all the numbers of each of these colors form an increasing sequence.

54 HINTS

Level D

1. Factor the difference between the left- and right-hand sides of the givenequation.2. Define the function for x > 1 by one formula and for x ≤ 1 by anotherformula.4. Analyze the parity of the variables.

Solutions

page 95

Year 1993 Olympiad

Level A

Problem 1. (a) Since x, S(x) and S(S(x)) have the same remainder upondivision by 3 (by Fact 6), the sum x+S(x)+S(S(x)) is divisible by 3. Since1993 is not divisible by 3, there is no solution.

(b) Clearly x < 1993. It is easy to see that 1899, 1989 and 999 have thelargest sum of digits among the numbers from 1 to 1993. Thus S(x) ≤ 27.Further, S(S(x)) ≤ S(19) = 10 and S(S(S(x))) ≤ 9. The equation to besolved implies that

x = 1993 − S(x) − S(S(x)) − S(S(S(x))) ≥ 1993 − 27 − 10 − 9 = 1947.

As in part (a), we know that x, S(x), S(S(x)) and S(S(S(x))) leave thesame remainders when divided by 9; let’s call this remainder r. Then 4r hasremainder 4 when divided by 9 (because that’s the remainder of 1993). Inother words, 4r − 4 is divisible by 9, which implies that r − 1 is divisible by9 (since 4 and 9 are coprime; see Fact 9). So r = 1.

Now list the numbers from 1947 through 1993 that have remainder 1upon division by 9: they are 1954, 1963, 1972, 1981 and 1990. We canverify directly that only 1963 satisfies the equation.

Problem 2. Let n = a2 + b2 + c2. Expand the square and rearrange: page 96

(a2+b2+c2)2 = a4+b4+c4+2a2b2+2a2c2+2b2c2

= (a4+b4+c4+2a2b2−2a2c2−2b2c2)+(2ac)2+(2bc)2

= (a2+b2−c2)2+(2ac)2+(2bc)2.

We can assume that a ≥ b ≥ c > 0, so a2 + b2 − c2 > 0. Thus we haveexpressed n2 as the sum of squares of three positive integers.

Remarks. 1. Try to prove a similar statement for the sum of four and more squares.

2. For the sum of two squares the analogous statement is not true: (12 + 12)2 = 4cannot be represented as the sum of two squares of positive integers, althoughan analogous identity is true:

(a2 + b2)2 = (a2 − b2)2 + (2ab)2.

58 SOLUTIONS

For more on sums of two squares, see the solution to Problem 59.11.4 on page 4 REF 59.11.4

and Remark 2 thereto.

3. A famous theorem of Lagrange says that any positive integer can be representedas the sum of four integer squares. (An integer square, or simply a square, meansthe same as the square of an integer. This includes 0 as well as the squares ofpositive integers.)

4. The number 7 is not a sum of three squares. It turns out that a positive integer ∗is not representable as a sum of three squares if and only if it is of the form(8k + 7) · 4m.

5. Suppose that instead of specific numbers we consider sums of squares of ar-bitrary variables. An identity going back at least to the third-century Greekmathematician Diophantus of Alexandria says that

(a2 + b2)(c2 + d2) = (ac + bd)2 + (ad − bc)2,

which of course can also be written as

(a21 + a2

2)(b21 + b2

2) = (a1b1 + a2b2) + (a1b2 − a2b1)2.

This elegant equality was extended by Euler in the eighteenth century to thecase of four pairs of variables:

(a21+a2

2+a23+a2

4)(b21+b2

2+b23+b2

4)

= (a1b1−a2b2−a3b3−a4b4)2+(a1b2+a2b1+a3b4−a4b3)

2

+(a1b3+a3b1+a4b2−a2b4)2+(a1b4+a4b1+a2b3−a3b2)

2.

A similar identity also exists representing

(a21 + a2

2 + · · · + a2k)(b2

1 + b22 + · · · + b2

k) (1)

as a sum of squares of polynomials when k = 8. However, for no other k —apart from the already discussed k = 2, 4, 8 and the trivial case k = 1— is theproduct (1) writable as a sum of squares of polynomials! The proof of this factis far from elementary; to intrigue the reader, we mention only that the identity page 97

for k = 2 is related to complex numbers, for k = 4 to quaternions, and fork = 8 to Cayley numbers, also called octonions. More about this in the bookCantor-Solodovnikov. CITE

Cantor-Solodovnikov

Problem 3. First solution. Consider all pairs of chips in the stack, notnecessarily adjacent. There are four possibilities for such pairs, in terms ofthe ordering of colors: RR, RB (meaning a red chip above a bluechip), BR and BB. For instance, in the stack RBRB of the figure,there are three RB pairs (first and second chips from the top, firstand last, third and last).

BRBR

The reader can check that the parity (see Fact 23) of the number of RBpairs cannot change under our operations. For example, suppose we inserttwo red chips at a spot that has k blue chips below it. It is easy to see thatthis adds exactly 2k new RB pairs, so the parity remains the same. Theother operations — insertion of two blue chips and removal of two blue ortwo red chips— can be analyzed similarly.


Now, in the initial position, there is exactly one RB pair. In the desiredfinal position, there are none. Since 1 and 0 have different parities, it’s notpossible to reach one state from the other.

Remark. The parity of the number of RB pairs in this solution is an invariant(see Fact 2), since it remains unchanged no matter what happens to the stack,under the rules of the problem. One could alternatively consider the numberof red chips below which there is an odd number of blue chips; the parity ofthis number is also invariant. (Check it!)

The next solution also relies on an invariant, but of a different type: atransformation of the real line into itself. (A transformation is a map that is ∗ removed “Though

more abstract... finalstate.”

I’m leaving “Recall...”here since it’sbackground materialand not part of thesolution proper.

one-to-one and onto, and therefore has an inverse.)Recall that two maps from a set to itself can be composed, that is, applied

successively, producing a third map. If f : R → R and g : R → R are maps ofthe real line, the composition fg (also written f ◦ g) is the map obtained byapplying g, then f :

(fg)(x) = f(g(x)).

The composition of any map f with the identity (the map Id such that Id(x) = xfor all x) is of course f again. The composition of several maps depends on ∗the order in which they are applied, but not on the grouping: in symbols,(fg)h = f(gh), because both of these maps take x to f(g(h(x))).

page 100

Second solution. Consider two transformations of the real line, r and b,defined by r(x) = 1 − x (this is a reflection in the point 1

2 of the line) andb(x) = −1 − x (a reflection in the point −1

2).Obviously, composing r with r brings each point of the line back to itself,

and so gives the identity map; in algebraic notation (see preceding remark),rr = Id. Similarly, bb = Id. ∗

The composition of two reflections of the line is, in general, a translation.For instance, applying b, then r, has the overall effect of a translation to theright by 2:

(rb)(x) = r(b(x)) = 1 − (−1 − x) = x + 2. (1)

Now let’s associate to red chips R the reflection r, and to blue chips Bthe reflection b. For any stack of chips, imagine reading off the colors fromthe bottom of the stack up and applying consecutively the reflection corre-sponding to each color. The overall effect of all these reflections r and b, inthe given order, is again a map of the real line.

The initial stack in the problem, RB, corresponds to applying b then r;this, according to equation (1), is the translation by 2. But if we considerthe reverse stack BR, the composition gives a different result altogether:

b(r(x)) = −1 − (1 − x) = x − 2.

The composition of transformations is not, in general, commutative!Now the key observation is this: two adjacent chips of the same color

cancel each other out from the point of view of composition: their combined

60 SOLUTIONS

effect, as we have seen, is the identity. This means that inserting or remov-ing two adjacent chips of the same color has no effect on the overall maprepresented by the stack of chips; for example, rb = bbrb = brrbrb. There-fore no sequence of such insertions of removals can lead from the stack RBto the stack BR, as they represent different overall maps.

Remarks. 1. (Follow-up on the second solution, for more advanced readers familiarwith group theory language.) The set of compositions is the group of transfor-mations of the real line generated by r and b, and these generators satisfy therelations r2 = 1 and b2 = 1.

One can also define an abstract group with generators r and b— these lettersnow being regarded just as symbols— subject to the relations r2 = 1 and b2 = 1,meaning that whenever bb or rr appear in a product of generators, they can beerased. By interpreting multiplication as composition we obtain a map from theabstract group to the group of transformations; it can be checked that this is anisomorphism (that is, the group of transformations has no additional relationsbeyond those in the abstract group). Mathematicians call such an isomorphism ∗a faithful representation of the abstract group in R. This abstract group can beshown to be isomorphic to Z o Z/2, the semidirect product of Z and Z/2.

2. The problem we’ve solved (twice!)comes up in a proof of an amusing and geo-metrically intuitive fact: Two continuous paths inside a square, each of whichjoins a pair of opposite vertices, must intersect. We sketch the proof in the casewhere the paths are polygonal. (The reader acquainted with continuity mighttry to prove that this implies the general case, using two facts from topology:the image of a continuous path [0, 1] → R2 is compact; and if two compact setsdon’t intersect, there is some minimum positive distance between points in one ∗set and points in the other.)

So let’s consider two nonintersecting polygonal paths as described, one blueand one red, inside a square. We make the square’s sides vertical and horizontal.Now move a vertical line continuously from the left edge of the square to theright edge, keeping track of the intersections of this line with the red and bluepaths. The pattern of red and blue intersection points corresponds to the stackof chips in the problem. (Since the paths don’t intersect we can assume thatnone of the polygonal segments is vertical: just jiggle a vertex a tiny bit if it is.Therefore the intersection of either path with a vertical line is a finite set.)

The intersection points always appear and disappear in pairsof a single color, as in the problem. (See the diagram on the rightfor an appearing pair and a disappearing one.) Therefore if on theleft edge of the square we have the red dot above the blue dot, wecannot have the inverse situation on the right edge, contradictingthe assumption that the paths join opposite corners.

3. The geometric result just proved is closely related to the Jordan curve theorem,which states that every closed plane curve without self-intersections divides theplane into two parts, an inside and an outside. This intuitively clear statementis difficult to prove, since the curve may have no smooth pieces. But in the caseof polygonal closed curves an argument very similar to the one above providesa proof.


page 101

Problem 4. The key idea is that the mirror image of a lucky pattern ofhands is unlucky, and vice versa.

Consider the position of the hands at two distinct times: T seconds be-fore noon and T seconds after noon. The patterns formed by the hands ofthe clock at those moments are mirror images of each other, with respectto the axis of symmetry formed by a vertical diameter. (Why?) For exam-ple, here is a clock showing 1 hour, 15 minutes, and 22 seconds after noon(01:15:22 PM) and another showing the same amount of time before noon(10:44:38 AM):

12

9

6

12

3

6

Now, a moment’s thought shows that reflection interchanges lucky andunlucky patterns of hands. Thus, to each lucky moment before noon therecorresponds an unlucky moment after noon. An interval of lucky moments inthe morning is matched by an interval of unlucky moments in the afternoonor evening, and both intervals have the same length. Similarly, an intervalof unlucky moments before noon is matched by an equal interval of luckymoments after noon. Hence the total amount of lucky time in the day equalsthe total amount of unlucky time.

∗Remarks. 1. It’s not enough to establish a one-to-one correspondence between

lucky and unlucky moments; we need to consider the length of the intervals.This question is discussed further in Remark 2 after the solution to Problem57.8.5, which deals with a similar situtation (see page 83). REF 57.8.5

2. It is also not enough to argue that each lucky pattern of hands has a mirrorimage that is an unlucky pattern of hands, because the mirror image mightconceivably not correspond to an actual moment in time. Some combinationsof hand positions simply cannot occur on a properly functioning clock. (Findexamples!) So one must establish a correspondence between actual intervals oftime.

Problem 5. Consider the sequence of strings

A, ABA, ABACABA, ABACABADABACABA, . . .

each string being obtained from the previous one by writing it twice and page 102

inserting the first unused letter between the two copies.When we run out of letters, the last string will provide an affirmative

answer to the problem. We will prove this by complete induction (Fact 24).For convenience, we will denote the n-th string of the sequence by Zn, soZ1 = A, Z2 = ABA, etc. We also denote the n-th letter of the alphabet byxn, so x1 = A, x2 = B, . . . , x26 = Z. Then the sequence of strings is defined

62 SOLUTIONS

by the properties

Z1 = x1 and Zn+1 = Znxn+1Zn.

The “multiplication” on the right-hand side simply means that we are con-catenating (writing one after the other) the string denoted by Zn, the letterxn+1, and again the string Zn, is this order.

Obviously this process stops when we run out of letters, that is, aftern = 26. We claim that for any n ≤ 26, (a) the string Zn doesn’t haveidentical adjacent substrings, but (b) a pair of identical adjacent substringsappears as soon as one writes any one of the first n letters of the alphabeteither at the beginning or at the end of Zn.

Base of the induction. For n = 1, the statement is obvious.

Induction step. Suppose (a) and (b) are true for Z1, . . . , Zn−1, andconsider the n-th string, Zn = Zn−1xnZn−1.

Suppose Zn has two identical adjacent substrings. They cannot containthe central letter xn, since there is only one copy of it. Therefore they lieboth to the left or both to the right of the central letter; that is, they’reidentical adjacent substrings of Zn−1. But such substrings cannot exist, bythe induction assumption. This proves statement (a) for Zn.

To prove (b), again we proceed by cases. Suppose we write after Zn oneof the first n letters of the alphabet. If the letter we wrote is the centralletter xn, the result is two copies of Zn−1xn. If we wrote any other letter xk,the string now ends with Zn−1xk, with k < n. But the induction assump-tion says that Zn−1xk contains two identical adjacent substrings somewhere.Either way, we’ve found the desired identical substrings. The argument alsoapplies if we write a letter to the left of Zn instead of to the right.

Having proved the induction, we now just take n = 26, so all the letters ∗ I prefer to leave the50 million here foresthetic reasons, butI’ve put it inparentheses.

of the alphabet are allowed. (The word Zn thus constructed has 226 − 1letters, or more than 50 million!)

Remark. This construction is important in combinatorics and the theory of semi-groups. Try to prove that in any infinite sequence of letters of the alphabet,there must be somewhere a string in the pattern of Zn, for any n. By a stringin the pattern of Zn we mean one that is obtained from Zn by replacing eachletter xi by a fixed nonempty string Xi; for instance “abracadabra” is a stringin the pattern of Z2 (take X1 = abra and X2 = cad).

Problem 6. The circle tangent to BC and to the extensions of sides ABand AC is an excircle of the triangle ABC. Its center O is the intersection of page 103

the bisectors of the angle A and the exterior angle at vertex B; see Fact 16.

A

B

C

D

O

α/2α/2

γ

β

Let α, β, γ be the angles of △ABC. Then \BAO = α/2and \CBO = (180◦ − β)/2, and

\ABO = β + \CBO = 90◦ + β/2.


From △AOB we see that

\AOB = 180◦ − α/2 −(90◦ + β/2

)

= 90◦ − α/2 − β/2 = γ/2,

since α + β + γ = 180◦. On the other hand,we have \AOB = 1

2\ADB, because this is the angle inscribed in the circlecentered at D. Hence \ADB = γ.

Thus, \ADB = \ACB. By the converse of the theorem on inscribedangles, the points A, B, C and D lie on one circle.

Remarks. 1. The points O, C, D lie on a straight line since \AOC = β/2 and\AOD = β/2 (prove it).

2. The statement is true for the inscribed circle as well (prove it).

Level B

Problem 1. We first draw the perpendicular to ABthrough the endpoint A. Clearly, \A < 90◦ if andonly if B and C lie on the same side of this line.Applying the same argument to B, we see that thelocus of points C such that \A < 90◦ and \B < 90◦

is the strip bounded by the perpendiculars to AB atboth A and B. (See figure on the right.)

A B

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

A B

Next we construct the circle with diameter AB.From Fact 14 we know that a point C is outside thiscircle if and only if it satisfies \ACB < 90◦. Takingthe intersection with the strip already found, we seethat the locus of points C such that the ABC is acuteis the shaded set to the left. page 104

Finally we study the condition that the angle Ais intermediate between the others, that is, either

\B ≤ \A ≤ \C or \C ≤ \A ≤ \B. (1)

Since the greater angle of a triangle subtends the longer side, the condition\B ≤ \A ≤ \C is equivalent to

AC ≤ BC ≤ AB.

The points equidistant from A and B are those on the perpendicular bisectorL of the segment AB; therefore AC ≤ BC if and only if C is in the half-planedetermined by L and containing A.

At the same time, BC ≤ AB if and only if C lies in the circle of centerin B and radius AB. Thus, the locus of points satisfying AC ≤ BC ≤ AB

64 SOLUTIONS

is the shaded set in diagram (a) below:

A B

(a)

A B

(b)

A B

(c)

Similarly, the condition \C ≤ \A ≤ \B is equivalent to AB ≤ BC ≤AC, and the corresponding locus is depicted in diagram (b) of the previouspage. The union of the sets in (a) and (b), shown in diagram (c), is thereforethe locus of points C satisfying (1). There remains to draw the intersectionof this locus and the one calculated immediately before (1).

Problem 2. We look at the first several terms of the sequence, in searchof patterns. We find x3 = 9, the least nonprime greater than 2x2 − x1 = 8;then x4 = 14 is the least nonprime greater than 2x3−x2 = 12 (because 13 is page 105

prime). Continuing we obtain x5 = 20, x6 = 27, and so on, where each time— after that exceptional 13— the least nonprime greater than 2xn−1−xn−2

appears to be just 2xn−1 − xn−2 + 1, a composite number already.So we will conjecture that, for n > 4, the number 2xn−1 − xn−2 + 1 is

composite and so equals xn. If this is so, we can write, for n > 4,

xn − xn−1 = xn−1 − xn−2 + 1.

That is, the differences between successive elements of the series increase ∗by one each time, forming an arithmetic progression: x4 = 14 = x3 + 5,x5 = 20 = x4 + 6, x6 = 27 = x5 + 7, . . . , or, in compact form,

xn = xn−1 + n + 1 for n ≥ 4.

If the differences form an arithmetic progression, the numbers xn themselvescan be calculated by summing up the progression. This would give ∗

xn = x3 +5+6+7+ · · ·+(n+1)

= 2+3+4+5+6+7+ · · ·+(n+1)︸︷︷︸n terms

= 12n

(2+(n+1)

)= 1

2n(n+3).

(Here we broke x3 down as 2 + 3 + 4 to make the calculation less messy.) ∗Conjecturally, then, we conclude that

xn = 12n(n + 3) for n ≥ 4. (1)

There remains to prove that this is really the answer. We do it by complete ∗induction.

Base of the induction. For n = 4, the equality (1) is true.


Induction step. Let (1) be true for x4, . . . , xn. We need to prove thatxn+1 = 1

2(n+1)(n+4). We have

2xn − xn−1 = 2 · 12n(n + 3) − 1

2(n − 1)(n + 2) = 12(n + 1)(n + 4) − 1.

Thus, xn+1 will be equal to 12(n+1)(n+4) if this number is composite. This

is indeed so: if n is even, 12(n+1)(n+4) has a factor 1

2(n+4) > 1, and if n is

odd, it has a factor 12(n+1) > 1.

Now we just have to apply formula (1) to n = 1000, obtaining x1000 =501500.

Problem 3. Assume that at some step we get a triangle similar to theinitial one. All its angles are multiples of 20◦.

Lemma. All angles of the preceding triangle, and, more generally, of allpreceding triangles are multiples of 20◦.

Proof. Let the triangle with angles α, β, γ be a child of thetriangle with angles α1, β1, γ1, obtained by cutting alongan angle bisector. Arrange the labeling so that α1 = 2αand β = β1, as in the figure. Since the sum of theangles of any triangle is 180◦, we deduce that

α α

γ1γβ=β1

γ1 = 180◦ − α1 − β1 = (α + β + γ) − 2α − β = γ − α.

Clearly, then, if α, β, γ are multiples of 20◦, so are α1, β1, γ1, the anglesof the parent triangle (see Fact 5). But then the angles of the grandparenttriangle are also multiples of 20◦, and so on. ˜

This, however, cannot be true even after the first cut of the initial trian-gle: if we start with an angle of 20◦, we get an angle of 10◦, and if we startwith the angle of 140◦ we get an angle of 70◦. The contradiction shows thatit is impossible to get a triangle similar to the initial one.

Remarks. 1. For any positive integer n, it is possible to find an initial triangle sothe construction leads to a similar triangle after n appropriate bisector cuts,and no sooner.

2. For a similar situation, see Problem 64.9.4. REF 64.9.4

Problem 4. A classmate of Pete’s can have between 0 and 28 friends.Of these 29 possibilities, we know that 28 occur. Thus, either there is aclassmate who has 28 friends, or there is a classmate who has no friends.But if a classmate is friends with everyone, then everyone has at least onefriend. So it’s not possible for both 0 and 28 to occur: either the tally offriends is 1, . . . , 28 for the various classmates, or it is 0, . . . , 27. page 107

Denote the classmate of Pete’s with the most friends by A and the onewith the least friends by B. In the first case just considered, A is everybody’sfriend, while B has only one friend, A. In the second case B has no friends,while A is friends with everyone except B. In either case, A is a friend ofPete’s, and B is not.

66 SOLUTIONS

Now let’s send A and B to another class. Then Pete is left with 26classmates and everybody has one fewer friend in this class than before.Thus each classmate still has a different number of friends in this class.

We again send the classmate with most friends and the one with leastfriends to another class. We can keep doing this until we have sent away 14pairs of classmates. Each pair included exactly one friend of Pete’s, so Petehad 14 friends in his class.

Remarks. 1. Several ideas work together toward the solution: friendship is assumedto be a symmetric relation; it’s useful to look at extreme cases; and we are ableto apply inductive descent.

2. There is one very short, but wrong solution: Let x be the number of Pete’sfriends. Now replace all friendships by nonfriendships, and all nonfriendshipsby friendships. Then Pete’s classmates will again each have a different numberof friends, so the conditions of the problem are still satisfied, meaning that Petewill again have x friends. But at the same time, we know that Pete now has28 − x friends (his nonfriends in the original situation). Therefore

x = 28 − x.

Where is the mistake? (Hint: Have we shown that the problem has a uniquesolution?) Although wrong, this argument can point the way to the answer.

3. Solve the same problem if Pete has 27 classmates.

4. This is an extension of a problem you may have heard before: Prove that inany group of more than one person, there are two people with the same numberof friends within the group (the number can be zero).

Problem 5. In the second identity set y = z. Then we get

(x ∗ y) + y = x ∗ (y ∗ y) = x ∗ 0.

Thus, x ∗ y = x ∗ 0− y. It remains to compute x ∗ 0. For this, set x = y = zin the second identity; we get

x ∗ 0 = x ∗ (x ∗ x) = (x ∗ x) + x = 0 + x = x.

Thus, x ∗ y = x ∗ 0 − y = x − y, so 1993 ∗ 1935 = 1993 − 1935 = 58.

Remark. Check that with x ∗ y = x − y, both identities are indeed verified.page 108

Problem 6. First solution. (See figure.) Let B′ be thereflection of B in the line AM . Since AB = AB′ and\BAB′ = 2\BAM = 60◦, the triangle ABB′ isequilateral. Hence, the points A, C, B′ lie ona circle of center B. Since the inscribed angle is ∗ “the” → “a”;

removed “of radiusAB”

half the central angle, it follows that \ACB′ = 30◦. Then,since \MCA = 150◦, the points C, B′, M lie on the same line. Now, byconstruction, AM is the bisector of \BMB′, and so also of \BMC.

A

B

M

C

B′

60◦

30◦


Second solution. Through the points A, C and M , draw acircle with center O (see figure). Since \ACM = 150◦,the arc AM has measure 60◦, so the triangle MAOis equilateral. Point B lies on a symmetry axis of△MAO, since \BAM = 30◦. It follows that

\AMB = \AOB = 12\AOC = \AMC,

where the second equality is a corollary of thecongruence △ABO = △CBO. A

B

M

C

O

Level C

Problem 1. The least period length of a decimal divides any other periodlength of that decimal; see Fact 4. (We regard any terminating decimal ashaving minimal period length 1).

Lemma. If k is a (not necessarily minimal) period length for each of twodecimals P and Q, then k is also a period length for P + Q and P − Q.

page 109

Proof. Recall (Fact 13) that a recurring decimal P with period k can bewritten in the form

P =X

10l(10k − 1),

where X is an integer. Similarly, we can write

Q =Y

10m(10k − 1),

where Y is an integer. Without loss of generality we may assume that l ≥ m.We obtain

P ± Q =X ± 10l−mY

10l(10k − 1),

where again the numerator is an integer, so the decimals corresponding toP + Q and P − Q are recur with periods of length k. ˜

Now we can solve the problem. We know that A has least period 6 andB has least period 12. The lemma implies that 12 is a period length ofA+B, so the divisors of 12 are the candidates for the least period of A+B.But 6 cannot be a period length of A+B, otherwise B = (A+B)−A wouldhave a period of length 6, contradicting the assumption. Hence the leastperiod of A + B cannot be a divisor of 6.

Two possibilities remain for the least period of A + B: 12 and 4. Bothoptions are possible:

A = 0.(000001), B = 0.(000000000001), A + B = 0.(000001000002);

A = 0.(000001), B = 0.(011100110110), A + B = 0.(0111).

68 SOLUTIONS

Remarks. 1. How would someone come up with this last example? By workingbackwards! Take any decimal of least period 4 and call it A + B. Now subtractany decimal B of least period 6. The result has least period 12. (Why?)

2. We can find all possible least periods of a sum of two decimals. Let m, n, k bethe least periods of the decimals P , Q and P + Q, respectively. Then k dividesthe least common multiple lcm(m,n) of m and n, by the lemma; but at thesame time m divides lcm(n, k), and n divides lcm(m, k). Let p1, . . . , ps be allthe prime divisors of lcm(m,n), and write

m = pα1

1 . . . pαs

s , n = pβ1

1 . . . pβs

s ,

where the exponent are allowed to be 0; see Fact 10. The preceding argumentsimply that k = pγ1

1 . . . pγs

s , where

γi =

{max(αi, βi) if αi 6= βi,

any number from 0 through αi if αi = βi.

It can be shown that the opposite is also true: any such k can be a least period page 110

of a sum of two decimals whose least periods are m and n, respectively.

Problem 2. We can certainly tour the eight rooms in any ∗ make a tour of →tour (to compensatefor next addition)

hall, going clockwise, say. We can also combine tours oftwo adjacent halls (top diagram) by using two of the threedoors that separate the two halls (middle diagram).

Now the key observation is that we can always add ahall to our tour, so long as it is shares a wall with a hallalready on the tour. For example, the bottom diagram addsa third hall to the first two. We can continue adding onehall at a time, until the tour includes all the halls of thecastle.

Problem 3. The answer to (a) is that it is notalways possible; see figure on the right for a

counterexample. In it, AC = 1000 m, AB > 1400 m,CD = 1 m. The segment AB divides the river intotwo parts, and boat going down the river end to endmust cross this segment at some point. The distance from anypoint of AB to one of the banks exceeds 700m.

A

B

C

D

Part (b) turned out to be unexpectedly difficult, unlessislands are allowed, in which case a counterexample is nothard to find (see figure on the right; the river has width justunder 1 km).

The intention of the authors of the problem was to not allow islands,and we (the authors of the book) were at first unable to solve it under thisconstraint. A contest was announced, and a solution was found through thejoint efforts of A. Akopyan, V. Kleptsyn, M. Prokhorova, and the authors.It turned out to be more of a research-level mathematics problem than anOlympiad problem! page 111


The solution is given in full in Appendix B, but the main steps— eachof which requires a whole argument in itself —are these:

• Show that no disc of radius 750m and center in the river (that is, not ∗in a lake) lies entirely on water.

• Deduce that if a point in the river is not within 750 m of one bank, itlies within 750 m of the other bank.

• Use the (possibly very complicated) boundary of the set of points thatlie within 750 m of the left bank to prove that there is a path thatremains within 800 m of both banks.

page 112

Problem 4. Since we’re taking fractional parts of numbers, it’s a good ideato visualize the real line “rolled up” into a circle of unit length, so numberswith the same fractional part correspond to the same point

0

ba+b

2a+b

1

2

on the circle. (See figure on the right, and compare the re-mark after the solution of Problem 60.10.6.) It is clear that REF 60.10.6

pn = 0 if xn = {an + b} lies on the upper semicircle [0, 12),

and pn = 1 if xn lies on the lower semicircle. Furthermore,xn = {an + b} is the point on the circle obtained from {b} by n consecutiverotations through the arc {a}.

(a) A rational value of a leads to a periodic pattern for the sequence pn.(Why?) Trying out b = 0 and the simplest possible fractions for a, we seethat all 4-tuples of 0s and 1s can occur:

• The “bigon” with a = 12 gives the sequence 010101. . . , so 0101 can

occur. (We only list 4-tuples starting with 0, since we know the com-plementary ones can be obtained by choosing b = 1

2 instead of b = 0,leading to the replacement of pn by 1 − pn.)

• The equilateral triangle with a = 13 gives the sequence 001001. . . , so

we get 0010 and 0100.• The square with a = 1

4 gives 0011 and 0110.

• The octagon with a = 18 gives 0000, 0001, 0011 again, and 0111.

(b) First solution. We will prove that the string 00010 cannot be realizedfor any a and b. (In the second solution below we give a general conditionnecessary for a string to be realizable.)

Consider three consecutive terms of the sequence: xn, xn+1 and xn+2.If the points xn and xn+2 are diametrically opposed, each point is obtainedfrom the previous one by a 90◦ rotation, and we’re in the situation of thesquare in part (a); the sequence pn cannot contain three zeros in a row. ∗

If the points xn and xn+2 are not diametrically opposed, they divide thecircle into two distinct arcs, and either xn+1 lies on the longer arc, as in

70 SOLUTIONS

diagram (a), or xn+1 lies on the shorter arc, as in (b).

(a)

xn

xn+1

xn+2 (b)

xn

xn+1xn+2

Suppose xn+1 lies on the longer arc. Then any other three consecutivepoints xm, xm+1 and xm+2 are similarly situated, being obtained from thepoints xn, xn+1 and xn+2 by the same rotation. This means that three ∗such points cannot appear in the upper half-circle; that is, the substring 000cannot appear in the sequence pn.

Now suppose instead that xn+1 lies on the shorter arc xnxn+2. Then anyother three consecutive points xm, xm+1 and xm+2 are similarly situated.In particular, if xm and xm+2 belong to the upper half-circle, so does xm+1;that is, the substring 010 cannot appear in the sequence pn.

We have considered all the scenarios and proved that the string 00010cannot be encountered.

Second solution. Let’s look at runs of consecutive 0s and 1s in the sequencep0, p1, p2, . . . . We claim that all such runs have either the same length orlengths differing by 1, except that the first run may be short (it can start inthe middle, so to speak). In particular, the string 00010 can never occur forany a and b, because it would mean the sequence has a run of 0s of lengthat least 3, but also a run of 1s of length 1.

The reason the runs have almost uniform length is that the spacingbetween xn and xn+1 around the circle is the same for all n. Indeed, supposefirst that 0 < a ≤ 1

2 and let i ≥ 1 be the integer uniquely defined by

i ≤ 1

2a< i+1, or equivalently, ia ≤ 1

2 < (i+1)a. (1)

We now show that a change from 0 to 1 must be followed by a changefrom 1 to 0 after exactly i or i+1 entries. More formally, suppose pn = 0and pn+1 = 1; that is, xn belongs to the upper semicircle [0, 1

2), while ∗xn+1 = xn + a belongs to the lower semicircle [12 , 1). The third inequality in(1) gives a < 2a < · · · < ia ≤ 1

2 . It follows that xn+1, xn+2, . . . , xn+i all liein [12 , 1), giving i consecutive 1s. If we had pn+i+1 = pn+i+2 = 1, there would ∗be i + 1 consecutive intervals of length a inside the lower semicircle: fromxn to xn+1, then to xn+2, and so on up to xn+i+2. But this is impossible bythe last inequality in (1). Therefore the run of 1s stops at length i or i + 1. ∗

A completely analogous argument shows that a change from l to 0 mustbe followed by a change from 0 to 1 after exactly i or i+1 entries. This provesour claim for a ∈ (0, 1

2 ]. If a ∈ (12 , 1) we replace a by 1−a, which corresponds

to a rotation through the same angle but in the opposite direction. Weneedn’t consider values of a outside [0, 1) because only the fractional partof a matters. Finally, for a = 0, there is only one run, of infinite length. Soour claim is proved in all cases.


Remarks. 1. This problem is relevant in symbolic dynamics; see Smale-horseshoe. CITE Smale-horseshoe

2. The original Olympiad problem said “. . . the sequence determined by some aand b”, which is perhaps ambiguous: the question might be whether all possiblequadruples of 0s and 1s can occur for some fixed choice of a and b. The second ∗solution to part (b) shows that the answer is no: any values of a and b thatallow the string 0000 cannot allow the string 0100.

Problem 5. Let m be the number of plants in a certain good classifier. Letus estimate the total number S of distinctions between all pairs of plantswith respect to all features. There are 1

2m(m−1) pairs of plants, and eachpair differs in at least 51 features, so

S ≥ 51 · 12m(m−1).

page 113There is another way to look at S. Let mi be the number of plants

having feature i. The number of pairs of plants that can be distinguished bymeans of feature i is (m−mi)mi. Summing over all the features, we obtainthe total number S of distinctions:

S =100∑

i=1

(m−mi)mi.

Now, the arithmetic mean of m−mi and mi is 12m, so the inequality between

the arithmetic and geometric means (Fact 26) gives (m − mi)mi ≤ 14m2.

Therefore S ≤ 100 · 14m2 = 25m2. Combining this with the earlier bound,

we obtain

51 · 12m(m−1) ≤ S ≤ 25m2. (1)

Subtracting 25m2 from the first and last expressions and simplifying we get12m(m−51) ≤ 0. Hence m ≤ 51.

It remains to prove that m 6= 51. If m = 51, we have a strict inequality

mi(m − mi) < 14m2

(since we have an integer on the left and a fraction on the right). That meansthe second inequality in (1) is strict, implying m < 51. This contradictionimplies that a good classifier cannot describe more than 50 plants.

Remarks. 1. One might be tempted to guess that a good classifier can describe 50plants, but with a bit more work one can show that this is far from true. Whatwe do is extend the classifier with one extra feature, which we declare presentif and only if an even number of the original 100 features were present. Thisnew classifier has 101 features and any pair differs by at least 52 of them: if apair differs in 51 of the original features, it must also differ in the new feature.(Why?)

Now the same arguments as above yield

52 · 12m(m−1) ≤ S ≤ 101 · 1

4m2,

which leads to m ≤ 34. Thus, the new classifier (and hence the initial one) candescribe at most 34 plants.

72 SOLUTIONS

2. This problem is related to error correcting codes. Replace plants by messagesand descriptions of features by length-n strings of bits (0s and 1s). The classi-fier— which is now a collection of m strings— is said to be a code of length n.The minimum number d of differences between two sequences in the code is thecode distance; in our problem d = 51.

If we take a message in the code and distort it arbitrarily by flipping nomore than 1

2 (d − 1) positions, we are still able to recover the original message,simply by selecting the message that shares the most bits with the distortedone. (There will be at most 1

2 (d−1) differences, whereas the comparison with

any other string will give at least d − 12 (d−1) > 1

2 (d−1) differences.) This iswhy the code is said to be error-correcting. page 114

One important open problem of code theory is finding the maximal size(number of different messages) of a length-n error-correcting code with codedistance d, for arbitrary n and d. A famous result in this direction is thePlotkin–Levenshtein theorem, which establishes an upper bound (called thePlotkin bound) in the case d > 1

2n, and provides certain natural conditions thatguarantee the bound can be achieved. In our problem we have n = 100 andd = 51, so Plotkin’s bound applies, and its value is 34. This bound is achievable:there does exist a code with 34 messages.

Problem 6. Let the remaining vertices of the square onside AB be D and E, so the square is ABDE, and set γ =\ACB. (See figure on the right.) Applying the intercepttheorem to the triangle ADC we see that CD = 2OM ;similarly, CE = 2ON . Therefore it suffices to find themaximum of CD + CE = 2(OM + ON).

First solution. On side BC of △ABC, construct asquare CBD′E′ outwards. (See figure below.) TrianglesABD′ and DBC have two sides and the included angle equal, so CD = AD′.

γ

A B

C

DE

NM

O

A B

C

DE

D′

E′

c

c

a

a√

2

b

In the triangle ACD′, two sides are known: AC = b and CD′ = a√

2.Moreover, \ACD′ = γ + 45◦. The side AD′ attainsits maximal value when the triangle degenerates intoa segment, so

max(CD) = max(AD′) = b + a√

2

for γ = 135◦. Similarly, we have max(CE) = a+b√

2,again for γ = 135◦. Thus, each of OM and ONattains its maximum when γ = 135◦, and so doestheir sum:

max(OM + ON) =1 +

√2

2(a + b).

Second solution. Let \CAB = α, \ABC = β, c = AB, d = CD, e = CE.The law of cosines for △ABC gives page 115

c2 = a2 + b2 − 2ab cos γ.


Next we apply the law of cosines to △AEC:

e2 = b2 + c2 − 2bc cos(90◦ + α) = b2 + c2 + 2bc sin α,

since cos(90◦ + α) = − sin α. Substituting c2 from the first formula into thesecond, we get

e2 = 2b2 + a2 − 2ab cos γ + 2bc sin α.

The law of sines for △ABC implies that sin α = (a/c) sin γ. Therefore

e2 = 2b2 + a2 + 2ab (sin γ − cos γ).

Similarly, d2 = 2a2 + b2 + 2ab (sin γ − cos γ).Hence both e and d attain their maximum values when sin γ−cos γ does,

which is to say, when γ = 135◦. Hence the maximum of e+d is also attainedfor γ = 135◦, and it equals 1

2(1 +√

2)(a + b).

Level D

Problem 1. If tan(α + β) is defined, then

tan (α + β) =tanα + tanβ

1 − tanα · tanβ=

p

1 − tanα · tan β. (1)

The product of tangents is related with p and q as follows:

q =1

tanα+

1

tanβ=

tanα + tanβ

tanα · tanβ=

p

tanα · tanβ. (2)

We deduce from (2) that either p and q are both zero or they are bothnonzero; therefore we have only these cases to consider:

1. If p = 0 = q, then (1) implies tan(α + β) = 0. We have to verify herethat the denominator of (1) does not vanish. Indeed, since p = 0, wehave

tanα = − tanβ, so 1 − tanα · tanβ = 1 + tan2 α > 0.

2. If p 6= 0, q 6= 0 and p 6= q, then (2) implies tanα · tanβ = p/q, so (1)implies tan(α + β) = pq/(q − p).

3. If p 6= 0, q 6= 0 but p = q, then tan(α + β) is not defined.

Problem 2. We will construct a subdivision into squares satisfying thecondition of the problem. We first divide the unit square into four equalsquares. The little squares that intersect the main di-agonal only at a vertex will be called level-1 squares.We subdivide each of the remaining squares into fourequal squares of side 1

4 . The little squares of side 14 page 116

that intersect the main diagonal only at a vertex willbe called level-2 squares. We continue in this way (seefigure) until we have 500 levels of squares.

12

2

3

3

3

3

1

2

1

4

1

8 1

2

2

3

3

3

3

There are 2k squares at level k, each with side 2−k. Hence the totalperimeter of all level-k squares is 4, and the total perimeter of all squaresintersecting the diagonal is 4 · 500 > 1993.

74 SOLUTIONS

Remarks. 1. A stronger result is in fact true: The unit squarecan be partitioned into squares in such a way that the sumof perimeters of squares intersecting the main diagonal ina segment exceeds any given number. The construction isa modification of the previous one; we make the under-the-diagonal level-1 square have side 3

5 , say, instead of 12 , while

the complement is subdivided into two squares of side 25 along the diagonal,

plus an irregular area that we leave aside, knowing that it can be subdividedinto tiny squares (since 3

5 is a rational number).We then repeat the procedure on the two new squares along the diagonal to

get level-2 squares, and so on. All level-k squares now have side (25 )k, instead

of (12 )k as before, and there are 2k−1 of them, instead of 2k. So now the total

perimeter of level-k squares is 2 · (45 )k; that is, it decreases in a geometric pro-

gression instead of being the same for all levels. But the ratio of the progressioncan be made arbitrarily close to 1, by replacing the number 3

5 by some rationalnumber very close to 1

2 . So the sum of perimeters can be made as large asdesired.

2. This problem arose during a lecture of the illustrious mathematician N. N. Luzin,when he wanted to shorten the proof of a theorem of Cauchy (Luzin loved toimprovise). Luzin conjectured: Fix a curve of a bounded length in the unit squareand consider a partition of the square into little squares. The total perimeterof the little squares that intersect the curve is bounded by a constant dependingon the curve only. A. N. Kolmogorov, who was to become just as famous amathematician, was at the lecture and soon constructed a counterexample.

Problem 3. For n = 2 the answer is obviously 1. So assume n ≥ 3.

The desired number is at most n, because we can exhibit an arrangementof points generating only n pairwise nonparallel lines: the vertices of aregular n-gon,

We prove this by showing that there are as many nonparallel lines asthere are axes of symmetry of the polygon. To each side and each diagonal,we assign an axis of symmetry: the perpendicular bisector of this side ordiagonal. Two sides or diagonals have the same perpendicular bisector ifand only if they are parallel. Therefore we just need to count the axes ofsymmetry of a regular n-gon.

For n odd, each axis of symmetry passes through a unique vertex. Hence,the total number of axes of symmetry is n. page 117

For n even, an axis of symmetry passes either through a pair of oppositevertices or through the midpoints of opposite sides. There are n/2 axes ofeither type. Hence, the total number of axes of symmetry is again n.

Now we prove the converse: The desired number is at least n. That is,for any arrangement of n points, no three of which lie on a line, we canalways find n pairwise nonparallel lines.

It is easy to find n − 1 lines: just take a point and consider all the linesfrom it to the other points. It is a bit harder to construct the n-th line.

One method is to introduce cartesian coordinates on the plane. Amongthe n points, take one, say O, having highest y-coordinate, and move the


origin there. Among the other n−1 points, choose A such that the ray OA ∗makes the biggest possible angle with the positive x-axis, andB such that the ray OB makes the least possible angle withthe positive x-axis. All the rays connecting O with theother points lie inside the angle AOB, by our choice ∗of A and B (see figure). Thus they must intersect thesegment AB and hence cannot be parallel to it. Nowwe just take AB as the n-th line.

A B

O

Remarks. 1. The points A, O and B are adjacent vertices of the convex hull of then points, that is to say, the smallest convex set containing all the ∗points. (It can be shown that the convex hull of a finite set ∗is a polygon whose vertices are contained in the set. Theconvex hull of the seven points in the preceding figurehas five vertices; see figure on the right.)

2. The given noncollinearity condition cannot be replaced by the weaker one thatthe points do not all lie on a single line. For example, for the set consistingof the vertices of a regular 2k-gon and its center, there are only 2k nonparallellines.

Problem 4. Clearly, the worst-case scenario is when all marble populationsoccur at the start; that is, we have n boxes—where n = 460 in part (a)and n = 461 in part (b) —and there is a box with j marbles for everyj = 1, 2, . . . , n. So from now on we assume this is the situation.

We start with the observation that a box having m = qk + r marbles(0 ≤ r < k) before a step with group size k with be left with q + r marblesafter that step; see Fact 7.

Lemma 1. After the first step, with group size k, there is a number f(n, k)such that the marble populations are exactly all the numbers in the range1, 2, . . . , f(n,k), and no others.

page 118

Proof. Let f(n, k) be the highest marble population in a box at the end ofthe step. We show by reverse induction on j (Fact 24) that there is a boxwith exactly j marbles, for all j = 1, . . . , f(n, k).

Suppose this is true for some j. Then there exist numbers m (startingpopulation), q (quotient) and r (remainder) such that 1 ≤ m ≤ n, 0 ≤ r < k,m = qk + r, and j = q + r.

If r > 0, we look at the box that started off with m−1 marbles. If r = 0,we take the one that had m − k marbles. Either way, we have found a boxholding exactly j − 1 marbles at the end of the step. This completes theinduction step and proves the lemma. ˜

This lemma effectively reduces the situation at the end of the first stepto the original situation, with a smaller number of boxes (we just ignoreboxes with duplicate populations). There remains to select k so the newlargest population f(n, k) is as low as possible.

76 SOLUTIONS

Lemma 2. The largest population f(n, k) after the first step is given by

f(n, k) =[n+1

k

]+ k − 2. (1)

Proof. The function “population of a box at the endof the first step” grows by 1 when its first argument(the initial population) grows by 1, except when theargument is 1 less than a multiple of k, in which casethe function drops by k−2 (see figure, where k = 5).Thus the maximum of the function is always achieved for an argument valuethat is 1 less than a multiple of k, and that is as large as possible under thiscondition. So let Q = [(n+1)/k] be the quotient of the division of n+1 by k.The box that started off with Qk − 1 = (Q−1)k + (k−1) marbles achievesthe maximum, and its new population is Q + k − 2 = [(n+1)/k] + k − 2marbles. This proves (1). ˜

Lemma 3. For a given value of n, the function f(n, k) achieves its mini-mum when k =

[√n+1

]+ 1.

Proof. To find the value of k that minimizes f(n, k), we first write

f(n, k) =[n+1

k+ k

]− 2.

The function inside the brackets decreases in the interval (0,√

n+1 ) andincreases in the interval (

√n+1, n]. Since [x] does not decrease, the function

f(n, k) attains its maximum either at k =[√

n+1]

or at k =[√

n+1]+ 1.

It remains to show that we always have

f(n, [√

n+1 ] + 1) ≤ f(n, [√

n+1 ]).

Let [√

n+1 ] = s. Then

s2 ≤ n+1 < (s+1)2. (2)

Thanks to (1) it suffices to prove that[n+1s+1

]<

[n+1

s

].

Equation (2) implies that page 119

[n+1s+1

]≤ s and

[n+1

s

]≥ s.

Therefore, it suffices to prove it is not possible for both sides to equal s. But[n+1

s

]= s =⇒ n+1

s< s+1 =⇒ n+1

s+1< s =⇒

[n+1s+1

]< s.

The lemma is proved. ˜

We are now ready to find the minimum number of steps required for anystarting value of n. We simply apply repeatedly the function

g(n) = f(n, [√

n+1 ] + 1),


which represents the maximum population after one step, with an optimalchoice of k. We verify by direct computation that, after five applications,

g(g(g(g(g(460))))) = 1 but g(g(g(g(g(461))))) = 2.

The sequences of steps for n = 460 and n = 461 are as follows:

n step k n

460 1 22 461

40 2 7 41

10 3 4 11

4 4 3 5

2 5 2 3

1 2

To recapitulate, we spell out the details of the argument for n = 461.The three lemmas say that if the boxes start with all populations from 1through 461, then after step 1 there remain all populations from 1 throughg(461) = f(461, 22) = 41; after step 2 there remain all populations from 1through g(41) = 11; after step 3, all populations from 1 through 5; after step4, all populations from 1 through 3; and after step 5, populations 1 and 2.Thus, for n = 461 it is not always possible to be left with a single marble ineach box.

Remarks. 1. Instead of proving that

f(n, [√

n+1 ] + 1) ≤ f(n, [√

n+1 ]),

one could just check case by case.

2. This problem originated in computer programming. A text containing one ormore blank spaces between words had to be processed so as to leave preciselyone blank between words. The programmers solved the problem by recursivelyapplying the operation of selecting a positive integer k and replacing every group ∗of k consecutive blanks by a single blank.

page 120

Problem 5. (a) A possible solution is shown in the figure. The function is ∗defined by

f(x) =

−12 − x for − 1 ≤ x < −1

2 ,

x − 12 for − 1

2 ≤ x < 0,

0 for x = 0,

x + 12 for 0 < x ≤ 1

2 ,12 − x for 1

2 < x ≤ 1.

1

2

−1

2

1

2

−1

2

−1

1

−1

1

0

(b) We will show that a function satisfying the conditions of the problemcannot exist on the interval (−1, 1). The case of the function defined on thewhole real axis is analogous.

Step 1. Suppose, to the contrary, that such a function f(x) exists. Itsgraph is mapped to itself under a clockwise 90◦ rotation: if (x, y) is a point

78 SOLUTIONS

on the graph, we have y = f(x), so f(y) = −x by assumption, and the point(y,−x) also belongs to the graph; but this is precisely the image of (x, y)under the specified rotation.

By applying the 90◦ rotation repeatedly we see that the graph maps toitself also under 180◦ and 270◦ rotations.

Step 2. This implies that the coordinate axes cannot intercept the graphexcept at the origin: any other intersection would imply three more inter-sections (obtained by 90◦, 180◦ and 270◦ rotations), and in particular therewould be two distinct intersections of the graph with the y-axis, which isimpossible.

Step 3. Now consider the portion of the graph that lies within the openfirst quadrant {(x, y) : x > 0, y > 0}. Recalling the assumption that thegraph is a union of finitely many points and line segments, we can write thisintersection as

L1 ∪ L2 ∪ · · · ∪ Ln ∪ P1 ∪ P2 ∪ · · · ∪ Pm,

where the Lk are line segments and the Pj are points. We may assume thatthe line segments Lk are pairwise disjoint and open (meaning the endpoints page 121

are excluded) and that the points Pj are distinct and do not belong to anyof the line segments Lk.

For each k, let Jk be the line segment obtained from Lk by a clockwise90◦ rotation. Next, for each j, let Qj be the point obtained from Pj bythe same rotation. All these points and segments lie on the open fourthquadrant, and in fact

J1 ∪ J2 ∪ · · · ∪ Jn ∪ Q1 ∪ Q2 ∪ · · · ∪ Qm

is precisely the intersection of the graph with the fourth quadrant. (Why?)Since we already know the graph does not intersect the positive x-axis,

we conclude that the intersection of the graph with the half-plane x > 0consists of 2n line segments (the Lk and Jk) and 2m points (the Pj and Qj).None of the line segments can be vertical. Thus the projections of all theseline segments and points on the x-axis partition the interval (0, 1) into 2nopen intervals and 2m points. But it is impossible to divide an open intervalinto an even number of subintervals using an even number of points! Wehave reached a contradiction.

Problem 6. Clearly a shortest flight touches each faceof the tetrahedron exactly once. Let the tetrahedronhave vertices ABCD, and let the shortest flight bethe space quadrilateral EFGH , where E ∈ △ABC,F ∈ △BCD, G ∈ △ABD, and H ∈ △ACD(see figure). Our job is to find the smallest possibleperimeter for the quadrilateral EFGH .

A

B

C

D

E

FG

H

Draw the symmetry plane of the tetrahedron containing CD; note thatit is perpendicular to AB. Let E1F1G1H1 be the reflection of EFGH inthis plane (so E1 and G1 lie on the same faces as E and G, respectively,


while F1 lies on the same face as H , and H1 lies on the same face as F ).The quadrilaterals EFGH and E1F1G1H1 have the same perimeter. page 122

Lemma. In any space quadrilateral, the distance between the midpoints oftwo opposite edges is less than or equal to the mean of thelengths of the remaining edges.

Proof. Let KLMN be the quadrilateral and let P andQ be the midpoints of KL and MN , respectively (seefigure). Denote by R the midpoint of the diagonalLN . We have PR = 1

2KN and RQ = 12LM . Hence

PQ ≤ PR + RQ = 12(KN + LM). ˜

K

L

M

N

P

Q

R

Denote the midpoints of the segments EE1, FH1, GG1, and HF1 byE2, F2, G2, and H2, respectively. These points also lie on the faces of thetetrahedron. By the lemma, the perimeter of E2F2G2H2 is no greater thanthat of EFGH . Moreover, E2 and G2 lie on the symmetry plane of thetetrahedron containing CD; that is, they lie respectively on the mediansCT and DT of the faces ABC and ABD, where T is the midpoint of AB.

Next we repeat this symmetrization operation for another symmetryplane. That is, we reflect E2F2G2H2 in the symmetry plane of ABCDthat contains AB, obtaining a quadrilateral E3F3G3H3, and then we takethe midpoints of the segments E2G3, F2F3, G2E3, and H2H3,obtaining a quadrilateral E4F4G4H4, whose vertices all belongto one of the two planes of symmetry of ABCD consideredso far (one through AB and the other through CD).

Specifically, the vertices E4 and G4 lie on CT and DT ,while F4 and H4 lie on the medians AS and BS of thefaces BCD and ACD, where S is the midpoint of CD. page 123

(See figure on the right.)

A

B

C

D

S

T

Again, the perimeter of E4F4G4H4 is no longer than that of E2F2G2H2,which as we know is no longer than that of EFGH . Hence, the perimeterof EFGH is at least 4d, where d is the distance between CT and BS.

It remains to construct a path of length 4d and find d. Let the commonperpendicular to CT and BS intersect CT at E0 and BS at F0. Let G0 bethe reflection of E0 in the plane ABS. It follows from symmetry that F0G0 isthe common perpendicular to BS and DT . Similarly we construct the pointH0 such that G0H0 is the common perpendicular to DT and AS and H0E0

is the common perpendiculars to AS and CT . The perimeter of E0F0G0H0

is 4d. We also have to prove that the bases of our common perpendicularslie on the faces of the tetrahedron, rather than on their extensions. Thiswill be checked below (we still have to calculate d).

Draw the plane through AB perpendicular to CT and takethe projection of the tetrahedron on this plane. We obtaintriangle ABD′, in which AB = a and D′T = a

√2/3, by

80 SOLUTIONS

the formula for the length of the altitude of the regulartetrahedron. (See figure on the right.) A B

D′

S′

T

The projection sends S to S′, the midpoint of D′T . Hence, d equals thedistance between T and the line BS′, because the common perpendicular isparallel to the plane of projection. It is also obvious that the base of theperpendicular dropped from T onto the line BS′ lies on the segment BS′

and not on its extension; hence F0 lies on the segment BS. We similarlyprove that the remaining vertices of the quadrilateral lie on the medians,not on their extensions.

In the right triangle BTS′, the legs BT = 12a and TS′ = 1

2a√

23 are

known. Hence

BS′ = 12a

√53 , d =

BT · TS′

BS′=

a√10

.

Remarks. 1. The solution implies that there are three suitable paths. (Why?)

2. An analogous problem on the plane is known: A beetle crawls inside a trianglewith sides a, b, c. What is the shortest length of a path that visits each sideand returns to the initial point? page 124

It the case of an acute triangle the answer is the path joining the basesof the altitudes; this is known as Fagnano’s problem. (See Coxeter, Chapter 4, CITE Coxeter

§ 5.) For a right or obtuse triangle the path degenerates into an altitude traveledtwice; see Problem 70 in ShklyarskiChentsovYaglom. CITE Shklyars-

kiChentsovYaglom

Year 1994 Olympiad

Level A

Problem 1. First solution. A bottle of mix used to require 16 of a jug of

apple juice and 110 of a jug of grape juice. Hence, the volume of a bottle is

equal to 16 + 1

10 = 415 of the volume of a jug.

After the change in the recipe, a bottle of drink requires 15 of a jug of a

apple juice and 1x of a jug of grape juice. Hence, the volume of the bottle is

equal to1

5+

1

x=

4

15

of the volume of the jug. So we have an equation1x

=415

− 15

=115

. Thus,x = 15.

Second solution. To make 30 bottles of the drink according to the old recipe,one needed 5 jugs of apple juice and 3 of grape juice: a total of 8 jugs. Afterthe change, 30 bottles require 6 jugs of apple juice, and hence 8−6 = 2 jugsof grape juice. Now one jug of grape juice serves 30/2 = 15 bottles.

Problem 2. Let x and y be the three-digit numbers to be found. If wewrite three zeros after x, we get 1000x. If instead we write the digits of y, page 125

we get 1000x + y. See Fact 11. Thus we have the equation

7xy = 1000x + y. (1)

First solution. Divide both sides of (1) by x:

7y = 1000 +y

x.

The number y/x is positive and smaller than 10, since y ≤ 999 and x ≥ 100.Therefore

1000 < 7y < 1010.

Dividing this inequality by 7 we get

14267 < y < 1442

7 .

82 SOLUTIONS

Since y is an integer, y is equal to either 143 or to 144. Let y = 143.Substituting this value of y into (1) we deduce that

7x · 143 = 1000x + 143.

Solving this equation we find that x = 143. If y = 144, a similar equationyields x = 18, which has too few digits.

Second solution. Rewrite (1) in the form

1000x = (7x − 1)y.

It is not difficult to see that x and 7x − 1 have no common divisors exceptfor 1 and −1. Indeed, if d is a common divisor of x and 7x − 1, then ddivides 7x, and hence it divides 1 = 7x− (7x− 1); see Fact 5. But 1 is onlydivisible by 1 and −1.

Thus 7x−1 divides the product 1000x and is relatively prime to x. Thisimplies that 7x − 1 divides 1000 (see Fact 9). Since

7x − 1 ≥ 7 · 100 − 1 = 699 >1000

2,

the only possibility is 7x − 1 = 1000. Hence x = 143. Substituting back in(1), we find y = 143.

Remark. Compare with Problem 57.10.1. REF 57.10.1

page 126

Problem 3. Extend the segments BQ and BP so theyintersect the line AC at A′ and C ′, respectively. In thetriangle ABC ′, the segment AP is both an altitude anda bisector (see figure). Hence, this triangle is isosceleswith AB = AC ′, and AP is a median too: BP = C ′P .Similarly, the segment CQ is a median of △CBA′,that is, BQ = QA′. Hence PQ is a midline of△A′BC ′, i.e., PQ ‖ A′C ′. But then PQ ‖ AC. A

B

C

PQ

A′ C′

Problem 4. Let the initial configuration be a 1×1 square on a piece of graphpaper. Whenever a grasshopper jumps, it must land on a crossing of the grid,because the jump is symmetric with respect to a crossing and starts from acrossing (see figure; we use the fact that a square grid is centrally symmetricwith respect to any of its crossing). Another way to see thisis to choose coordinates so that one of the corners of theinitial square is the origin; then the x- and y-coordinates ofeach grasshopper must remain whole numbers, because thedifference in each coordinate between the two grasshoppersinvolved in a jump is replaced by its negative.

In particular, since the distance between two grasshoppers cannot beless than 1, at no time can their positions coincide with the corners of asmaller square than the original one.

Now suppose the grasshoppers could land on the corners of a largersquare than the original one. The reverse of a legal jump is also a legal


jump, so by performing the jumps backwards, we’d be in a situation wheregrasshoppers starting at the vertices of a (larger) square ended up occupyingthe vertices of a smaller square. But we know this is impossible.

Remark. If the grasshoppers had started at the vertices of some other parallelogram,they would be destined to always jump to crossings of the slanted grid made ofcopies of the initial parallelogram. The set of crossings of such a grid is calleda lattice.

page 127

Problem 5. We start with some obvious remarks.

• Any two hands of the clock define an unfavorable sectorbetween their extensions, in the sense that if the thirdhand is in that sector, the moment is unfavorable. Thesize of this sector does not exceed 180◦ (see figure).

• After a whole number of hours, the minute and second hands will returnto the exact same position they occupy now.

• After six hours, any unfavorable moment is replaced by a favorable one(the hour hand rotates through 180◦ and moves from an unfavorablesector to a favorable one).

• The converse is not true: there are favorable moments, six hours afterwhich the time is again favorable. For example, any time between3:00:00 and 3:00:05 is favorable, but so are the times six hours later,from 9:00:00 to 9:00:05.

Now imagine the day divided into favorable and unfavorable intervals oftime. The operation of adding six hours maps all unfavorable intervals tofavorable ones of the same length (though the latter might be part of largerfavorable intervals). So the total length of favorable intervals is at least asmuch as that of unfavorable intervals. The same operation of adding sixhours also maps some favorable intervals to favorable intervals (consider the5-second interval discussed in the previous paragraph). Thus the favorableintervals add up to strictly more time than the unfavorable ones.

Remarks. 1. A similar situation is considered in Problem 56.8.4. REF 56.8.4

2. The original question asked “what is there more of, favorable time or unfavorabletime?” Obviously what we’re interested in is the total length of the intervals offavorable and unfavorable time, not whether there are more moments of eachkind of time. There are infinitely many moments of each of the two kinds, andthey can be placed in one-to-one correspondence, just as the intervals [0, 1] and[0, 2] can be placed in one-to-one correspondence through the map x 7→ 2x.

The question of when it is possible to establish a one-to-one correspondencebetween two infinite sets is interesting and deep. (For finite sets, the answer issimple: a one-to-one correspondence exists if and only if the sets have the samenumber of elements.) Here are some example results; for details, see Settheory: CITE Settheory

(a) There is a one-to-one correspondence between the set of all integers andthe set of positive integers.

(b) There is a one-to-one correspondence between the set of all rational num-bers and that of positive integers.

84 SOLUTIONS

(c) There is no one-to-one correspondence between the set of all rational num-bers and that of all irrational numbers.

(d) There is a one-to-one correspondence between the points of a line segmentand the points of a square.

Problem 6. The first player colors the 18×18 square adjacent to the longerside of the rectangle so that it is equidistant from the right and left edges ofthe rectangle (see figure for an analogous situation with a 7×14 board). Therest of the rectangle is imagined to be divided into congruent halves. Nowthe second player’s every move can be coun-tered by the first with a symmetric move; thesecond player would only be able to preventthis by coloring a square that straddles thesymmetry axis, which is impossible.

Level B

Problem 1. Let D and E be points inside a triangle ABC.Draw all 10 segments between pairs of points; they canbe divided into two non-self-intersecting polygonal linesof five segments each (see figure). Either of these twopolygonal lines can be taken as the desired pentagon;the other polygonal line is made of the diagonals ofthe pentagon. A

B

C

D

E

Problem 2. First case. If k > l, Sue wins, by securing a part longerthan the sum of all the other parts. For example, she can cut k into partsof length

l + 23(k − l), 1

6(k − l), 16(k − l).

(See figure below.) The largest of these, l+ 23(k− l), cannot serve as a side of page 129

any triangle: by the triangle inequality the sum of the lengths of the othertwo sides should be greater than this, but the sum of the lengths of all othersegments is only l + 1

3(k − l).

k

l + 2

3(k−l) 1

6(k−l) 1

6(k−l)

l

Second case. If k ≤ l, Leo wins. Let Sue divide her segment into partsof length k1, k2, k3, where k1 ≥ k2 ≥ k3. Then Leo divides his segment lso that his largest part is of the same length as Sue’s largest part, and theremaining two are of the same size: k1,

12(l − k1) and 1

2(l − k1). (See figurebelow.) Since k1 ≥ k2, it is possible to construct two isosceles triangles from


the resulting segments, of lengths

k1, k1, k2,l − k1

2,l − k1

2, k3.

Indeed, from segments of lengths a, a, b it is possible to construct an isoscelestriangle if and only if b < 2a. Obviously k2 < 2k1. On the other hand,

2 · l − k1

2= l − k1 > k3,

since k1 + k3 < k ≤ l.

k

l

k1 k2 k3

k11

2(l−k1)

1

2(l−k1)

k2

k1 k1

k3

1

2(l−k1)

1

2(l−k1)

Problem 3. Here is an infinite set of solutions: page 130

x = k(2k2 + 1), y = 2k2 + 1, z = −k(2k2 + 1), for k = 0, 1, 2, . . . .

Verification is straightforward.How can this answer be guessed? Since we only have to find infinitely

many solutions, not all of them, we can get rid of one variable and two cubesby taking z = −x. Then we get the equation 2x2 + y2 = y3, which we canrewrite as x2 = 1

2(y−1)y2 or

x = ±√

y − 1

2y.

So if (y−1)/2 is a perfect square, x is an integer. Thus we take (y−1)/2 = k2,which gives

y = 2k2 + 1, x = k(2k2 + 1).

Remark. There certainly exist other solutions, e.g.,

x = 12k(k2 − 1) + 1, y = − 1

2k(k2 − 1) + 1, z = −k2 + 1 for any k ∈ Z.

Problem 4. First solution. Let the circles be disposedas in the figure, that is, with the center of each outsidethe other. Other configurations are treated similarly;see the remark after the second solution.

It suffices to prove that the triangles AQNand AMP are congruent. For this, we provethat they are similar and that the correspond-ing sides AQ and AM are equal.

By the theorem on inscribed angles, \P =\N and \Q = \M , so △AQN is similar to△AMP .

A

B

M

N

P

Q

To prove that the segments AQ and AM are congruent, we prove thatthe corresponding arcs, measured counterclockwise, are the same. The first

86 SOLUTIONS

of these arcs has measure twice the inscribed angle ABQ; see Fact 15. Sim-ilarly, the arc AM equals twice the angle MAN . Thus, it suffices to provethat \ABQ = \MAN .

Now, the angle ABQ is an exterior angle of the triangle ABN , and hence\ABQ = \ANB + \BAN . We see that \MAN = \BAM + \BAN ; but\ANB = \BAM , since each of them equals half the arc AB on the circlethrough A, B and N . Hence

\MAN = \ANB + \BAN = \ABQ.

The problem is solved.

Second solution. By the tangent-secant theorem (power of a point), we have

MP =AM2

MB, NQ =

AN2

NB.

Thus it suffices to prove that page 131

AM2

AN2=

MB

NB. (1)

Let’s show first that the triangles AMB and NAB aresimilar. We have \AMB = \BAN , since they bothhave half the measure of the arc AB on thecircle ABM (see figure). By the same token,\MAB = \NAB. This proves the similarityof AMB and NAB, which in turn implies that

AM

AN=

AB

NB, and

AM

AN=

MB

AB.

Multiplying these equalities we get (1).

A

B

M

N

P

Q

Remark. If the center of one of the circles lies inside the other circle, the picturewill be different. If the center of each of the circles is inside the other circle, weget still another picture. In these cases, in the first solution, instead of provingthat \ABQ = \MAN , we have to prove that \ABQ + \MAN = 180◦. Thesecond solution remains valid in either case. The reader is encouraged to workout the details.

Problem 5. Let x be the dropped digit, a the part of the number to theleft of x, and c the part of the number to the right of x. Let c contain ndigits. Then the original number equals a · 10n+1 + x · 10n + c; see Fact 11on page 104. After x is struck out it becomes a · 10n + c.

(These formulas encompass the case n = 0, with the convention c = 0,but we can dispose of this case quickly. Indeed, in this case 10a + x isdivisible by a. It follows that x is divisible by a, and hence a ≤ x ≤ 9.This would mean the original number has only two digits. We are going toconstruct a six-digit number below, so the case c = 0 is irrelevant.)

Now consider the ratio between the two numbers:

r =a · 10n+1 + x · 10n + c

a · 10n + c, where c < 10n.


Subtracting 10 from both sides of this equation and simplifying we reach

r − 10 =x · 10n − 9c

a · 10n + c≤ x

a≤ 9

a≤ 9. (1)

Set l = r − 10. Multiplying both sides of the equality in (1) by a · 10n + cand manipulating we get

(x − la) · 10n = (l + 9)c. (2)

If l ≤ 0, the left-hand side is positive, so l + 9 > 0. Thus, page 132

−8 ≤ l ≤ 9.

Moreover, l 6= 0; otherwise the original number ends in 0.

Lemma. a < 10.

Proof. Consider the two cases: l > 0 and l < 0.If l > 0, equality (2) implies that x − la > 0, and hence

a <xl≤ 9

l≤ 9.

If l < 0, then

x − la =(l + 9)c

10n<

9 · 10n

10n= 9,

implying −la < 9, so a < 9. ˜

The lemma implies that the original number has n + 2 digits. The nextstep is to find the greatest possible value of n.

By assumption c does not end in 0. Thus c cannot be divisible by both2 and 5 (see Fact 10 on page 104).

Suppose first that c is not divisible by 2. Consider the right-hand sideof (2). Since 1 ≤ l + 9 ≤ 18, the number l + 9 can be divisible by 24 = 16,but it cannot be divisible by 25. Hence n ≤ 4. If n = 4, then l + 9 = 16,and (2) becomes

(x − 7a) · 54 = c.

Since x is a one-digit number, it follows that a = 1, and hence x = 8 orx = 9. If x = 9, then c ends with a 0 and hence does not fit. If x = 8, thenc = 625 and the final answer is

180625 = 10625 · 17.If, on the other hand, c is not a multiple of 5, then l + 9 can be divisible

by 5 but not by 52, so n ≤ 1, and the number will certainly not be thelargest possible.

Problem 6. For part (b), it is easy to give an example where page 133

the first few ships are arranged so that there is no room forthe last one; see figure on the right.

In part (a), there is a hidden difficulty: many students assumed it’senough to prove that one can place the last 1× 1 ship, but in fact one mustshow that each ship in sequence can be placed.

88 SOLUTIONS

The 1× 4 ship can be placed. To prove that each of the two 1× 3 shipscan be placed, too, we mark eight “phantom” 1 × 3 ships, all in the samedirection and leaving two cells in between them (see first diagram below).If a 1 × 4 and at most one 1 × 3 ship have been placed, each of them cantouch or overlap with at most two phantom ships; therefore some phantomship will be untouched and can be replaced by an actual 1 × 3 ship.

The reasoning for the 1 × 2 ships is similar; it involves twelve parallel1×2 phantom ships, again with two cells between them (see middle diagram).Suppose the 1 × 4 ship, two 1 × 3 ships and no more than two 1 × 2 shipshave been laid down. Each touches at most two phantom ships, so at leastone phantom ship will be untouched and can be replaced by an actual ship.

For the final step of the proof we consider sixteen 1×1 phantom ships (seerightmost diagram above). Each of the six long ships already placed touchesat most two phantom ships, while each of the (at most three) 1 × 1 shipsalready placed touches one phantom ship. Thus at most fifteen phantomships are affected, and one is left over, showing that the last 1 × 1 ship canbe placed.

Remark. It is interesting to find the maximal number of identical ships of, forexample, size 1 × 4, that one can place.

Level C

Problem 1. We offered two detailed solutions to the very similar problem page 134

57.8.2. Here we give the corresponding arguments in slightly abbreviated REF 57.8.2

form.First solution. Let x and y be the numbers to be found. Then 3xy =

107x + y, or 3y = 107 + y/x. Since 0 ≤ y/x ≤ 10, we have 107 < 3y <107 + 10, and therefore 33333331

3 < y < 333333623 . Now one can either

consider the three cases y = 3333334, y = 3333335, and y = 3333336, orobserve that

y

x≤ 3333336

1000000< 4,

and hence 107 + 1 ≤ 3y < 107 + 4. Only one number in this interval isdivisible by 3, namely, 107 +2. So 3y = 107 +2, and hence y = 3333334 andx = 1666667.

Second solution. Since

107x = (3x − 1)y (1)


and since x and 3x−1 have no common divisors, 107 is a multiple of 3x−1.But 3x − 1 ≥ 3 · 106 − 1, so either 3x − 1 = 5 · 106 or 3x − 1 = 107. Only3x − 1 = 5 · 106 fits, so x = 1666667. The relation (1) implies y = 3333334.

Problem 2. We first show that all the xn lie in the interval [0, 1]. Indeed,

0 ≤ xn ≤ 1 =⇒ −1 ≤ 1− 2xn ≤ 1 =⇒ 0 ≤ |1− 2xn| ≤ 1 =⇒ 0 ≤ xn+1 ≤ 1.

Rationality implies periodicity. If xn is rational, then so is xn+1 and itsdenominator is no greater than that of xn (assuming both are irreducible,of course). Indeed, let xn = pn/qn be irreducible. Then

xn+1 = 1 −∣∣∣∣qn−2pn

qn

∣∣∣∣ =qn − |qn−2pn|

qn.

If this fraction is irreducible, its denominator is the same as that of xn; if itis reducible, its denominator decreases after simplification.

Thus, all the terms of the sequence are rational numbers between 0 and1. But there are only finitely many such irreducible fractions with denomi- page 135

nators not greater than a given q. Therefore some term of the sequence willreappear and at that point the sequence will become periodic.

Periodicity implies rationality. First solution. Eliminating the absolutevalue sign in the equation xn+1 = 1−|1−2xn|, we see that either xn+1 = 2xn

or xn+1 = 2− 2xn. Therefore xn+1 = a1 +2b1xn, where a1 is an integer andb1 = ±1. Similarly, xn+2 = a2 + 4b2xn. Continuing this process we get

xn+k = ak + 2kbkxn,

where ak is an integer and bk = ±1.Suppose that xn+k = xn. Then xn is a solution of a linear equation with

integer coefficients:

ak + 2kbkxn = xn.

This equation has a unique solution, since 2kbk = ±2k 6= 1. The solution isrational, so xn is rational. But then xn−1 is rational, xn−2 is rational, andso on. Finally, x1 is a rational number.

Second solution. Write x1 in binary notation (Fact 12 on page 104):

xn = 0.a1a2a3 . . . .

If a1 = 0, the binary representation of xn+1 is obtained from the binaryrepresentation of xn by a shift (verify this):

xn+1 = 0.a2a3a4 . . .

If a1 = 1, the binary representation of xn+1 is obtained from the binaryrepresentation of xn by a shift and the simultaneous interchange of 0 with 1,a process we will call inversion. Let us prove that if xn is periodic, then an

is also periodic. If xn+k = xn and an even number of inversions took placebetween them, then an+k = an. If an odd number of inversions occurredbetween xn and xn+k, we have an+2k = 1 − an+k = an. Either way the

90 SOLUTIONS

sequence an is periodic. But periodic binary representations, like decimalones, correspond to rational numbers (see Fact 13 on page 104).

Remarks. 1. The function y = f(x) = 1−|1−2x| is linear on each of the segments[0, 1

2

]and

[12 , 1

]:

y =

{2x for 0 ≤ x ≤ 1

2 ,

2 − 2x for 12 ≤ x ≤ 1.

Similarly, the function page 136

y = fn(x) = f(f . . . f(x) . . . )︸︷︷︸n times

sending x1 to xn is linear on each segment [k/2n, (k+1)/2n]. The graphs of thefirst three of these functions are as follows:

y

xy = f(x)

y

xy = f1(x)

y

xy = f2(x)

Because of the shape of its graph, the function f is called the tent map. It isan important object in symbolic dynamics.

2. For every T = 2, 3, . . . , there exists at least one point with period T : for instance,the x-coordinate of the last intersection point of the segment {(x, y) | y = x,0 ≤ x ≤ 1} with the graph of the function y = fT (x). Explicitly, this value isx1 = 2T /(2T + 1).

Contemplate the question: How many periodic trajectories are there foreach period T? (Or, which is almost the same, how many points x are there forwhich x = fT (x) and x 6= fk(x) if k < T?)

Problem 3. First solution. Two members of Parliament will be said to beenemies if one of them has slapped the other. The statement of the problemis the particular case n = 665 of the following result:

Lemma. If there are M ≥ 3n − 2 members of Parliament and each hasslapped exactly one colleague, it is possible to make up a parliamentary com-mittee with n members, none of whom is enemies with another.

Base of the induction. For n = 1, the statement is obvious.

Induction step. Since there are as many members as there were slaps,the pigeonhole principle (Fact 1) implies that there is a member who wasslapped at most once. This member, therefore, has at most two enemies (oneslapee and perhaps one slapper). We draft this member into the committee,and set aside his or her enemies, reducing the number of available membersby at most 3. Thus there are at least M−3 ≥ 3(n−1)−3 members. Thereforethe conditions of the lemma are satisfied with n replaced by n − 1, and by page 137

the inductive hypothesis one can select a committee of n−1 members of this


reduced parliament containing no enemies. With the one previously chosenwe get n committee members none of which are enemies.

Second solution. Consider a directed graph Γ (see fact 3 on page 101), whosevertices are members and whose edges correspond to slaps, the source beingthe slapper and the target the slapee.

Define a directed tree as any directed graph whose vertices can be par-titioned into levels satisfying the following properties: there is one vertex oflevel 1, called the root, with no edges emanating from it; and every vertexof level j > 1 has one outgoing edge, which goes into a vertex of level j − 1.

Lemma. Each connected component of Γ consists of disjoint directed treeswhose roots are linked together in an oriented cycle.

1

1

1 1

1

2

3

34 3 2

Proof. Starting from any vertex, we follow directed edges until we return toa vertex already visited, which must happen since the graph is finite. Hence,each connected component has a cycle. We assign the vertices in these cycleslevel 1. Among the remaining vertices, we assign level 2 to those from whichthere is an edge leading to a vertex of level 1. Vertices of level 3 will bethose from which there is an edge leading to a vertex of level 2, and so on.More formally, the level of a vertex not in a cycle is 1 plus the number ofedges that separate that vertex from a cycle. (This number is well-definedsince the path leading from a vertex to a cycle is unique.) ˜

Next we assign one of three colors to each vertex, so that no edge con-nects vertices of the same color. We start with the cycles. In cycles of even page 138

length, we color the vertices alternatingly red (R) and green (G). In cyclesof odd length, we paint one vertex blue (B), then alternatingly paint theother vertices R and G.

R

G

R G

B

G

R

RRGR

We next color the trees according to their roots. If the root is R or B,even levels are colored G and odd levels starting from 3 are colored R. If theroot is G, even levels are colored R and odd levels are colored G.

92 SOLUTIONS

Clearly, at least one of the colors is used in at least one third of vertices.The committee is then declared to consist of all members of that color.

Remarks. 1. If each cycle contains an even number of nodes, two colors suffice.

2. Give an example where it is impossible to select a committee with 668 members(out of 1994).

3. The lemma in the first solution can be generalized: If there are M members ofParliament and each has slapped exactly k colleagues, it is possible to make upa parliamentary committee with [M/(2k + 1)] members containing no enemies.

Supply a proof; you can follow closely the first solution given above.

Problem 4. Let’s try to guess the length of the segment AK. Considerthe limit case as the point D tends to C. Then one of the circle shrinks toa point and the other becomes the circle inscribed in △ABC; the point Kmerges with P , the tangency point of this circle with side AC.

Now, there is a formula for the length of thesegment between a vertex A of a triangle ABC andthe tangency point of the incircle to a sidecontaining A: it reads

AB + AC − BC

2.

Let us prove that AK is given by this formula,not only in the limit case but in general.

A

B CDE

PB′

F

Q

KM

NC′

Denote the tangency points as in the figure. Observe that AB′ = AP ,AC ′ = AQ, KM = KP , KN = KQ, since these are pairs of tangent page 139

segments drawn from a single point. Moreover, MN = EF since these aresegments of common tangents to two circles (see Fact 17). We can now write

AB + AC − BC = AB′ + AC ′ − EF = AB′ + AC ′ − MN

= AP + AQ − MN = AP + AQ − (MK + KN)

= (AP − KP ) + (AQ − KQ) = 2AK.

Problem 5. (a) The answer is negative. For example, take the polygonin the figure, consisting of three identical square roomsconnected by thin bent corridors. We will prove that nochord splits the polygon into two pieces of equal area.

Let the area of the polygon be S, and let the areaof each room be 0.3S, so the total area of the corri-dors is 0.1S. If the chord intersects only a corridor,then two rooms are on one side of it, and their areais greater than 0.5S.

0.3S

0.3S

0.3S

If the chord intersects one of the rooms, it cannot intersect the “hub” ofthe corridors, and so again there are two rooms on one side of the chord.

(b) First solution. The idea is to move the ends of a chord through thepolygon in such a way that the area of the smallest piece is maximized. Fornotational simplicity we assume the total area is 1.


Select the direction of the chord so it is not parallel to any side ordiagonal of the polygon; let this be the vertical direction. Then the chordcannot ever pass through more than one vertex of the polygon.

The chord divides the polygon into exactly two pieces if no vertex liesin the interior of the chord, and into exactly three pieces otherwise. page 140

Choose any vertical chord with no vertex in its interior. It divides thepolygong into two pieces. If the smallest piece has area at least 1

3 , theproblem is solved. Otherwise, we start moving the chord either right or left,so as to make the area of the smallest piece grow.

As long as the chord does not touch a concave vertex, the area growscontinuously (see Remark to the solution of Problem 63.11.4). However, REF 63.11.4

when the chord reaches a concave vertex, one of its endpoints may jumpsuddenly to another part of the perimeter.

What happens when the chord encounters such a troublesome vertex?(Note that the chord cannot meet two vertices simultaneously.)

Consider the situation at the exact moment when three regions exist.Let a be the area of the region where the chord had been so far, and b, c,with b ≥ c, the areas of the other two regions. Note that b > 1

3 , since a < 13

(otherwise we’d be done).There are three possible configurations, shown below, and in all of them

we push the chord slightly into the b region, thus combining the two regionsof smallest area. In one case, shown in the rightmost diagram, this involvesa reversal in the direction of movement of the chord.

b

c

a

c

a

b

b

a

c

If b < 23 we are done, because the combined piece has area just above

a + c = 1 − b ≥ 13 , while the other has area just below b, which as we know

exceeds 13 .

If b ≥ 23 , we continue moving the chord into the area b; the area of the

smallest piece has jumped from a to a + c, but remains less than 13 , and we

repeat the process.Now, a troublesome vertex can only be crossed once, because the recently

swept region, of area a, keeps growing throughout — that is, at each time itcontains all earlier versions of itself —whereas the “recently swept regions”available for a given troublesome vertex are all disjoint. (These are theregions of area a, b, c in the figure.) The number of vertices being finite, weeventually run out of troublesome vertices, meaning that no obstacle remainsfor the smaller piece to attain area 1

3 . (Compare Fact 2 on page 100.)

94 SOLUTIONS

Second solution. Any polygon can be partitioned into triangles whose sidesare sides or internal diagonals of the polygon. (We will not prove this fact,so this solution is only a sketch.)

Consider such a triangulation. Each of the internal edges of the trian-gulation splits the polygon into two pieces. Let AB be an edge for which page 141

the area of the smaller piece is maximal. Let ABC be a triangle of thetriangulation adjacent to AB on the side of the larger piece.

Denote by SAB, SBC and SAC the areas of the pieces adjacent to thecorresponding sides of the triangle ABC and not containing it. (If a side isan edge of the polygon, the corresponding area is defined to be 0.) Then

SAB + SBC + SAC + SABC = 1,

where, for simplicity, we have defined the area of the polygon to be 1. Wemay assume that SBC ≥ SAC . Further, from the maximality conditionimposed on the chord one can deduce that

SAB ≥ SBC ≥ SAC .

These inequalities imply that SAB + SABC ≥ 13 . If SAB ≥ 1

3 , then theproblem is solved; otherwise let X be a point on side AC such that SAB +SABX = 1

3 . The chord passing through B and X satisfies the conditions ofthe problem. (Check it!)

Remarks. 1. A variation of part (a) is “Can a chord always be found that dividesthe polygon into some number of pieces of equal area?” The answer is still no;it is not difficult to adapt the solution given above to this case. For example,let the areas of the rooms in the figure on page 92 be 0.3S, 0.33S and 0.36S.Prove that such a polygon cannot be cut by a single chord into any number ofpieces of equal area.

2. A polygonal annulus is a region whose boundary consists of two polygonal curves,instead of one. There are polygonal annuli that cannot be divided by a chordinto two polygons whose areas are more than a third the total area. Can youcome up with an example? Also, explain what breaks down in the proof givenin the first solution to (b) above.

3. The second solution of part (b) shows that it is always possible to select thedesired chord so that it contains a vertex.

Problem 6. We call a polynomial anxn+an−1xn−1+· · ·+a1x+a0 positive if

all its coefficients are positive (ai > 0 for i = 0, . . . , n). Clearly, the productof positive polynomials is positive. Therefore if the square and cube of apolynomial are positive, so is any higher power, because any higher poweris a product of squares and cubes.

We first find a nonpositive polynomial whose square and cube are posi-tive. Experimenting with 0 and 1 as coefficients—a process made especiallyeasy using the shorthand discussed in Remark 2 below—we find that thepolynomial f(x) = x4 + x3 + x + 1, although it has a 0 coefficient, has apositive square and cube.

Now we need a polynomial with a negative coefficient. For this, we page 142


“jiggle” the polynomial f a bit (mathematicians say “perturb”). That is,we replace the 0 coefficient by a small negative coefficient, defining

g(x) = f(x) − εx2, (1)

where ε > 0 is very small. Then the coefficients of g2 and g3 are close tothose of f2 and f3, and so must still be positive.

So the answer to the question is affirmative, and an example is given byx4 + x3 − εx2 + x + 1 for any ε positive and small enough. One can checkexplicitly that ε = 1

10 will work.

Remarks. 1. To be rigorous, the closeness argument must use the fact that thecoefficients of the square of a polynomial F (and also those of the cube) dependcontinuously on the coefficients of F . (For this notion, see for instance theremark to the solution of Problem 63.11.4.) Continuity follows from the fact REF 63.11.4

that the coefficients of F 2 are themselves given by polynomial expressions inthe coefficients of F : if F (x) = a4x

4 + a3x3 + a2x

2 + a1x + a0, the coefficientsof F 2 are a2

4, 2a3a4, a23 + 2a2a4, and so on. Therefore a small change in the ai

causes only a small change in the coefficients of F 2.

2. The multiplication of polynomials can be sped up by writing just the coefficients,particularly if they’re simple. For instance, the multiplication of x2 +1 by x+1can be represented by the schematic calculation

1 0 1+ 1 0 1

1 1 1 1

where we interpret the result as x3 + x2 + x + 1. Also note that the “hole” inx2+1— that single 0 between 1’s— gets filled when we multiply this polynomialby x + 1. Using this idea it’s not hard to come up with the polynomial in thesolution, which “fills its own hole” when multiplied by itself:

1 1 0 1 1+ 1 1 0 1 1+ 1 1 0 1 1+ 1 1 0 1 1

1 2 1 2 4 2 1 2 1

Incidentally, the square of the integer 11011 is 121242121. How does this relateto the previous calculation? Something similar happens with the cube of 11,which is 1331— these are the coefficients of (x + 1)3.

How far can you push this observation? What happens if you square thepolynomial x4 + 2x3 + 2x + 1 and the number 12021?

Level D

Problem 1. The figure shows an example with 6 faces:two triangles, two quadrilaterals, and two pentagons.Such a polyhedron can be obtained from a tetrahedronby clipping off two corners.

page 143

96 SOLUTIONS

Remarks. 1. This is essentially the simplest example. Indeed, let the polyhedronhave n faces. Since no more than two faces have the same number of edges,there are at least

[12 (n+1)

]types of faces (triangles, quadrilaterals, and so on).

Since n ≥ 4, this means there are faces with 4 or more edges. But such a facehas at least 4 neighbors, so n ≥ 5. Therefore there are faces with 5 or moreedges. But then n ≥ 6.

Another solution, with more than the minimal number of faces, can beobtained from a cube by clipping off two neighboring corners.

2. Solve the following related problem from the 1973 Moscow Mathematical Olym-piad: Prove that any convex polyhedron has two faces with the same number ofedges. (Hint: Euler’s formula.)

3. Prove or disprove: Any polyhedron with 10n faces has n faces with the samenumber of edges.

Problem 3. We first solve an auxiliary problem: to find the smallest circlewith center on the y axis, tangent to the x axis, and intersecting the graph

of the function y = x4 somewhere beside the origin(see figure). In other words, what is the least r forwhich the system

x2 + (y − r)2 = r2, (1)

y = x4. (2)

has a nonzero solution? It is intuitively clear that thisproblem is equivalent to the initial one, and we shallprove this equivalence rigorously later.

y

x

y = x4

Substituting (2) into (1), simplifying and dividing by x2 (assuming x >0) we get

x6 − 2rx2 + 1 = 0

whence we deduce that

r(x) =12

(x4 +

1x2

).

The smallest value r0 is the minimum value of r(x) for x > 0. We have

r′(x) = 2x3 − 1x3 .

On the semiaxis x > 0 we see that r′(x) < 0 for x < x0 = 1/ 6√

2, r′(x0) = 0,and r′(x) > 0 for x > x0. Hence r′(x) decreases for x < x0 = 1/ 6

√2, it

attains its minimum at x0, and it increases for x > x0.Thus, the smallest value of r for which the circle has a common point

with the curve y = x4 (away from the origin) is page 144

r0 = r(x0) =3 3√

2

4.

It remains to show that the same r0 solves the initial problem as well.First, let’s check that the circle of radius r0 with center at (0, r0) lies above


the graph of y = x4. Indeed,

r(x) ≥ r0 for any x 6= 0. (3)

Substituting the expression for r(x) in (3) we get

x6 − 2r0x2 + 1 ≥ 0 for any x 6= 0.

Multiplying both sides of this inequality by x2 and replacing x4 by y we get

x2 + (y − r0)2 ≥ r2

0 for any x and y = x4.

But this means that the cherry of this radius can lie in the goblet.Finally we verify that if r > r0 the cherry will not touch the bottom of

the goblet. Indeed, in this case,

x60 − 2rx2

0 + 1 < 0,

so for y0 = x40, we have

x20 + (y0 − r)2 < r2.

This means that the circle of radius r tangent to the x-axis at the originintersects the graph of the function y = x4; so the cherry will not touch thebottom.

Remarks. 1. It is not difficult to see that, in cross-section, the biggest cherry istangent to the goblet at the bottom and at two more points. In space, the setof points where the cherry is tangent to the goblet is the union of a point anda circle.

2. A similar problem can be solved for the goblet whose vertical cross-section isof the form y = |x|a, for any real a > 0. For the x-coordinate of the tangencypoint of the thickest cherry with the goblet we have an equation

x2(a−1) =a − 2

a.

(a) For a > 2 there is a nonzero solution; the situation is similar to that ofa = 4.

(b) For a = 2, we get x = 0 and r = 12 , which is the curvature radius of the

graph at the origin; so the biggest cherry touches the goblet only once, atthe bottom.

(c) For 0 < a < 2 no cherry can touch the bottom of the goblet.

Problem 4. The homothety with center A and scale factor 2 sends theinitial polyhedron M into a polyhedron M ′ whose volume is 8 times greaterthan that of M . Let us prove that all 8 translated polyhedralie inside M ′.

Let B be the translate of the vertex A, let X be anarbitrary point of M , and let Y be the image of the pointX under the same translation (see figure). We must prove page 145

that Y belongs to M ′. A X

B Y

K

Since M is convex, the segment BX is entirely contained in M ; in par-ticular, its midpoint K belongs to M . The quadrilateral ABY X is a par-allelogram, so Y is obtained from K under a homothety with the center atpoint A and factor 2; hence Y belongs to M ′.

98 SOLUTIONS

Next we observe that points close to A do not belong to any of thetranslated polyhedra. Indeed, let us turn M so that A lies above all theother vertices of M . Then there exists a plane that passes below A but aboveall the other vertices of M . This plane cuts off of M a small polyhedronN containing A. Clearly, N does not intersect any of the 8 translatedpolyhedra.

If all the translated polyhedra had disjoint interiors, the volume of M ′

would be at least the sum of volumes of the translated polyhedra plus thevolume of N . But the sum of volumes of the translated polyhedra is equalto the volume of M ′. This is a contradiction.

Remarks. 1. It is easy to generalize the result to n-dimensional space.

2. At the Thirteenth International Mathematical Olympiad, held in the town ofZilina, Czechoslovakia, in 1971, a weaker formulation was offered: Prove that atleast two of the nine (not eight) polyhedra intersect.

Problem 5. It is difficult to illustrate the problem (the intersection pointslie far from each other), so let’s think. First, recall that the points of abisector are equidistant from the sides of the angle it bisects.

For each of the four lines containing a side of the quadrilateral ABCD,define a function fi, where i = 1, 2, 3, 4, equal to the directed distance to thisline: if the point lies on the same side of the line as the quadrilateral, we takethe usual distance, otherwise we take minus the distance. The functions fi

depend linearly on the coordinates of the points (see Fact 25), i.e., page 146

fi(x, y) = aix + biy + ci.

A point lies on the bisector of an outer angle of the quadrilateral if andonly if the sum of the values of the two functions corresponding to the legsof the angle vanishes.

For each of the two intersection points of the bisectors spoken about inthe formulation of the problem, we see that the sum of all four functionsvanishes.

But the sum of linear functions is a linear function, and the set of pointswhere a nonconstant linear function vanishes is a straight line. The sum ofthe fi does not vanish identically since it is positive inside the quadrilateral.Hence the intersection points of the bisectors lie on one line.

Remarks. 1. A spatial generalization of the result of the problem can be obtainedusing the same idea.

2. Here is another problem that can be solved using a similar idea: in a triangle,find the locus of the points the sum of whose distances to two sides equalsthe distance to the third. The intersection of a bisector with the opposite sidesatisfies this condition; therefore so does any point on a line segment connectingtwo of these intersections. (Why? Be careful.)

3. In school, a linear function is any polynomial of degree at most 1. In linearalgebra, a linear function is any polynomial of degree at most 1 that does nothave a constant term. Functions with a constant term are said to be affine.


Problem 6. When does a power of 2 have a string of 9s among its last fewdigits? When it’s a bit less than a number divisible by a high power of 10.For example, 212 + 4 is divisible by 100, 253 + 8 is divisible by 1000.

Let’s first try to find numbers of the form 2n+1 divisible by a high powerof 5. Then we will multiply those numbers by the corresponding power of2 and obtain numbers of the form 2k(2n + 1) divisible by a high power of10. Simplifying and subtracting the smaller summand we get the requiredpower of 2. page 147

Lemma 1. For all k ≥ 1, the number 22·5k−1

+ 1 is divisible by 5k.

Proof. We use induction on k (see Fact 24 on page 24). The base of induc-tion (k = 1) is obvious. To prove the induction step, we write

22·5k−1

+ 1 = 45k−1

+ 1.

Let a = 45k−1

. By the induction hypothesis a + 1 is divisible by 5k. Then

45k

+ 1 = a5 + 1 = (a + 1)(a4 − a3 + a2 − a + 1).

Since a + 1 is divisible by 5k, it suffices to prove that the second factor isdivisible by 5. Indeed, a is of the form 5m − 1, so all the summands of thesecond factor give residue 1 after division by 5 (see Fact 7), and their sumis divisible by 5. The lemma is proved. ˜

Thus, the number 2k(22·5k−1

+1) ends with at least k zeros. It is easy tosee that 2k has at most k/2 digits, if k > 1. Hence, among the last k digitsof the number

22·5k−1+k,

no more than k/2 digits can differ from 9.

Remarks. 1. Similar arguments prove that the numbers

40, 41, 42, . . . , 45k−1

have different residues after division by 5k. Try to prove a more general factoften encountered in number theory, called Hensel’s lemma:

If x − 1 is divisible by pk, where p > 2 is a prime and k > 0, but is notdivisible by pk+1, then xn − 1 is divisible by pk+r if and only if n is divisibleby pr.

2. At the Leningrad Mathematical Olympiad in 1981, a similar problem was posed:Is there a positive integer power of 5, whose last 100 digits contain at least 30zeros in a row?

Year 1995 Olympiad

Level A

Problem 1. Let a and b be the initial prices (in groats) of ale and bread, ∗respectively. Then 1 = a + b. After the first rise, the prices become 1.2a for ∗ale and 1.2b for bread, so

1 = 1.2a + 12 · 1.2b = 1.2a + 0.6b.

Subtracting this equation from 1=a+b gives 0=0.2a−0.4b, which simplifies ∗to a = 2b. Substituting into 1 = a + b we get 3b = 1, so b = 1

3 and a = 23 . page 148

The price of ale after the second rise is (1.2a) · 1.2 = 0.8 · 1.2 = 0.96. ∗Thus, Isaac can still buy his ale with a groat. ∗Problem 2. We work by induction (see Fact 24 on page 109).

Base of the induction. Since 10017 = 53 · 189, the claim is true forthe first element of the sequence.

Induction step. Suppose a number in the sequence is divisible by 53, ∗and look at the difference between this number and the next:

1001 . . . 1︸︷︷︸k times

7 − 1001 . . . 1︸︷︷︸k−1 times

7 (1)

This difference equals (1001−100) ·10k = 901 ·10k, because the last k digits ∗cancel out; it is therefore divisible by 53, since 901 = 53 · 17. Now, if the ∗number on the right in (1) is a multiple of 53 and the difference is too, so isthe number on the left (Fact 5, page 102). This proves the induction step. ∗Remarks. 1. One can show that the quotients upon division by 53 are of the form

18 . . . 89.

2. See problem 58.9.1 for a similar question. REF 58.9.1

Problem 3. (a) The segment KL is a midline ofthe triangle ABC, so its length is half that of AC. ∗Similarly, LM has half the length of BD. ButAC is taken to BD under a 120◦ rotation aboutthe point O (clockwise in the case of the figure).Hence AC = BD, so KL = 1

2AC = 12BD = LM .

A

B

C

D

O

K

L

M

120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦120◦

∗

102 SOLUTIONS

(b) Again by the midline theorem, the vectors−→AC and

−→KL are parallel, as ∗

are−−→LM and

−−→BD. The angle between

−→AC and

−−→BD is 120◦, since a 120◦ ∗

rotation takes one to the other. Therefore the angle between−→KL and

−−→LM page 149∗is 120◦, and the angle

−→LK and

−−→LM is the complement of that, or 60◦. Using ∗

part (a), we deduce that the triangle KLM is equilateral.

Problem 4. Among parallelipipeds of fixed volume, the one with the leastarea is the cube; see Fact 26 on page 111. Therefore the minimum amountof material needed is

6 · ( 3√

1995)2. (1)

This is a messy number, so let’s try to find a solution in integers. The nearestcubes to 1995 are 1728 = 123 and 2197 = 133; moreover, a 12× 12× 13 boxdoesn’t have enough volume (1872). A 12× 13× 13 box does (2028), but itsarea, 2 · (12 · 13 + 12 · 13 + 13 · 13) = 962, only gives an answer to part (a).

To do better, we can try adding 1 to one dimension while subtracting 1from another (see also Remark 3 below). Indeed, the 11 × 13 × 14 box has ∗volume 2002, and area 958, so it works for (a), (b) and (c).

Remarks. 1. One can check that this is the best possible answer in integers: 957units are not enough. Allowing real numbers, the minimum area is 950.853, thevalue of (1).

2. The two-dimensional counterpart of the result we used is that the rectangle ofleast perimenter for a given area is a square. Derive this from the inequalitybetween the geometric and arithmetic means. ∗

3. Because the dimensions are roughly the same, adding 1 to one dimension whilesubtracting 1 from another changes the volume and area very little (recall that(x − 1)(x + 1) is “almost equal” to x2). In the case of 12 × 13 × 13 we have a ∗1.5% excess margin in the volume— we can go down from 2028 to 1995— whilethe area only needs to go down by about 0.5%, from 962 to 958. So the strategyworks. ∗

Problem 5. First solution. It suffices to prove that the fuel cost does notchange under an interchange of two consecutive deliveries. Indeed, by ap-plying such transpositions we can reorder the deliveries in any way we want.(This statement, which is easily proved by induction on the number of vil-lages, is an important theorem in permutation theory; see Chapter 2 inTranspositions.) CITE Transpositions

page 150Suppose that at some point the bus visits village M and then visits Nnext, delivering goods of respective weights m and n. By assumption, theone-way distances to M and N are m and n as well. If we change the routejust by interchanging N and M , this obviously does not affect the amountof fuel spent on other legs of the trip. More subtly, it does not affect theportion of the fuel bill due to the goods delivered to any other villages. Thisis a consequence of the fact that the cost per mile is proportional to the total ∗ removed “ This is

clear ... that comelater”

weight, and so can be written as a sum of costs for each package separately.Apart from the packages delivered to M and N , all packages on the trucktravel the same miles under either route (before or after the interchange).


Thus we just have to compare the cost of carrying goods of weights mand n to the villages M and N . In the first case, a load of m + n willbe carried over a distance m (to village M), then a load n will be carried ∗over a distance m + n — back to the city and out to N . Hence, the cost isproportional to ∗

(m + n)m + n(m + n) = (m + n)2.

In the second case the cost is proportional to (m + n)n + m(m + n), which ∗is the same.

Second solution. Let the weights of the goods be a1, a2, . . . , an in the orderthey are delivered. The fuel cost is proportional to

a1(a1 + 2a2 + 2a3 + · · · + 2an) + a2(a2 + 2a3 + · · · + 2an)

+ · · · + an−1(an−1 + 2an) + a2n.

Expanding the products and reordering we get

a21 + a2

2 + · · · + a2n + 2a1a2 + 2a1a3 + · · · + 2a1an + 2a2a3 + · · · + 2an−1an,

which equals (a1 + a2 + a3 + · · ·+ an)2. This expression does not depend onthe order of enumeration.

Problem 6. Let N lie on AB and K on AF , as in the figure. Observethat FK = AN . Select points P on BC, R on CD, S on DE and T onEF so that FK = AN = BP = CR = DS = ET .Then \KBN = \TAK, \KCN = \SAT , \KDN =\RAS, \KEN = \PAR, \KFN = \NAP . Thus

\KAN+\KBN+\KCN+\KDN+\KEN+\KFN

= \KAN+\TAK+\SAT+\RAS+\PAR+\NAP

= \KAN+\KAN = 120◦+120◦ = 240◦.A

B

C

D

E

F

K

N

PR

S

T

Level B

Problem 1. We work by induction (see Fact 24 on page 109).

Base of the induction. Since 12008 = 19 · 632, the claim is true forthe first element of the sequence. page 151

Induction step. We need to show that if one of our numbers is divisibleby 19, so is the next. To this end it is enough to show that the differencebetween two consecutive numbers is divisible by 19 (see Fact 5 on page 102). ∗But

1203 . . . 3︸︷︷︸n times

08 − 1203 . . . 3︸︷︷︸n−1 times

08 = 1083 · 10n = 19 · 57 · 10n.

Remarks. 1. It is not difficult to show that the quotients upon division by 19 areof the form 63 . . . 32.

2. See problem 58.8.2 for a similar question. REF 58.8.2

104 SOLUTIONS

Problem 2. At first let A′ be any point on BC, and drawthe segment AA′. We claim that, among the segmentsfrom C to points on the side AB, there are at most twowhose length equals AA′: namely, the segments CC1

and CC2 such that

\C1CA = \A′AC and \C2CB = \A′AC.A C

B

A′C1

C2

P1P2

(The two coincide if AC1 = AC2 is the bisector.) ∗Indeed, CC1 = AA′ because △ACC1 and △CAA′ are congruent (case ASA).Similarly, CC2 = CC1 by the congruence of △ACC1 and △BCC2. Thatthese are the only two points on AB whose distance to C equals AA′ followsfrom the fact that a circle intersects a line in at most two points.

Now we observe that the desired locus is made up of all intersections ofAA′ with CC1 and CC2, as A′ varies. page 152∗Now recall that an isosceles triangle has a symmetry axis that containsthe altitude, bisector, and median from the corresponding vertex, and isalso the same as the perpendicular bisector of the opposite side. Let P1

be the intersection of AA′ with CC1. It lies on the altitude of the triangleABC dropped from B: indeed, the triangle AP1C is isosceles, since theangles at its base are equal. Therefore AP1 = CP1, that is, P1 lies onthe perpendicular bisector of AC, which is also the altitude in question.Conversely, all points on this altitude satisfy the condition of the problem,by symmetry.

The locus of P2, the intersection of AA′ and CC2, is less obvious. Weknow that the triangles A′AC and C2CB are congruent, and rotating thelatter by 120◦ about C (clockwise in the figure above) aligns it with theformer. Thus CC2 rotated by 120◦ becomes parallel to AA′; that is, theangle AP2C is 120◦. This can also be seen as follows:

\AP2C = 180◦ − \A′AC − \C2CA

= 180◦ − \A′AC − (60◦ − \A′AC) = 120◦.

The locus of points (inside the triangle ABC) that see thesegment AC under an angle of 120◦ is an arc with endpointsA and C and measure 2 · (180◦ − 120◦) = 120◦. (Since the ∗sides of an equilateral triangle are seen from its center at120◦, this arc intersects the other part of the locus —thealtitude —at the center of the triangle. This correspondsto C1 = C2 being the midpoint of AB and A′ being themidpoint of BC.) Reversing the argument, we see thatall points on this arc satisfy the condition of the problem.(Check it!) A C

B

∗Problem 3. Suppose first that n ≤ 1995, so no strip can be longer than1995. Since all strips have different, integer lengths, the most area they cancover is 1+2+· · ·+1995 = 1

2 ·1995·(1995+1), or 1995·998 (see Remark aftersolution). So the maximum possible value of n (not exceeding 1995) is 998. page 153


Now, one can show that n can be 998, by exhibiting adecomposition of the 1995×998 rectangle into strips of differ-ent lengths. We follow the pattern of the figure on the right(which was drawn with longest side 9 instead of 1995). Thatis, one row of the rectangle is occupied by a single strip of length m = 1995,the next by strips of lengths 1 and m − 1, and so on until the last row, ∗which is occupied by strips of lengths (m − 1)/2 and (m + 1)/2. Note thatm = 1995 is odd.

It is also obvious that any value of n less than 998 will work; startingfrom the 1995 × 998 rectangle, delete some rows until only n are left.

Inspired by this particular case, we prove:

Lemma. An M × N rectangle, with M ≥ N , can be divided into strips allof different lengths if and only if M ≥ 2N − 1.

Proof. We follow the technique used above for M = 1995. The maximumarea that can be covered is 1

2 M(M+1) (the sum of integers from 1 to M),

so MN ≤ 12 M(M+1), or N ≤ 1

2(M+1). There remains to show that forall such values of N the rectangle can be covered. The case of M odd isverbatim the same as for M = 1995. Now take the case that M is even. ∗Then N ≤ (M +1)/2 means that N ≤ M/2. If N = M/2 we can cover theM×N rectangle by strips of lengths M , 1, M−1, 2, M−2, . . . , M/2−1,M/2+1. Any rectange with fewer rows (that is, N < M/2) can be coveredby deleting some rows from the previous example. ˜

We now solve the problem in the case where 1995 is the shorter di- ∗ deleted spurious“are”

mension (n > 1995). We simply take N = 1995 in the lemma and obtainM ≥ 2 · 1995 − 1 = 3989. Thus the final answer is “n ≤ 998 or n ≥ 3989”.

page 154

Remarks. 1. We used the equality 1 + 2 + · · · + n = 12n(n+1). This formula can

easily be proved by induction. We take this occasion to give another elegantproof, whose underlying idea is often used to find the sum of any arithmeticsequence.

Set X = 1 + 2 + · · · + n and compute the sum of all numbers in the table

1 2 3 . . . n−2 n−1 n

n n−1 n−2 . . . 3 2 1

Since the sum of the elements in each column is equal to n + 1 and there are ncolumns, the sum is equal to n(n + 1). On the other hand, the sum of elementsin each row is X, so 2X = n(n + 1), implying our statement. ∗

2. One can show a priori that, in the situation of the problem, the sides of thestrips should be parallel to the sides of the given rectangle. Indeed, assumeotherwise, and let A be a vertex of a slanted strip closest to the boundary ofthe original rectangle. Consider the sides of all the strips containing A on theirboundary. The angle between such consecutive sides is right, which means thatall rectangles having A on their boundary are slanted. But this contradictsminimality.

106 SOLUTIONS

Problem 4. First solution. The hypothesis implies that the number

a + b + c + d = a + b + c +ab

c=

(a + c)(b + c)

c

is an integer. Hence the fraction is reducible. Since both factors in thenumerator are greater than in the denominator, after division by c each ofthe factors turns into a number greater than 1. Thus, a + b + c + d is theproduct of two factors, each greater than 1, so it cannot be prime.

Remark. We used a nontrivial statement: If the fraction (xy)/z is an integer, thenz can be represented as a product of integers ts so that x is divisible by t and yis divisible by s.

If you master the second solution, you will see how to prove this statementrigorously.

Second solution. We know that a, b, c, d are positive integers and thatab = cd. We claim that in this situation there exist positive integers u, v,w, z such that a = uv, b = wz, c = uw, d = vz. ∗

Indeed, consider the prime factorizations of a, b and c. Since c divides ab,all its prime factors are present among the prime factors of a and b combined;more precisely, a factor appearing m times in the prime factorization of cmust be accounted for at least m times in the prime factorizations of a andb taken together.

Let p1, . . . , pn be all the prime factors of a and b, and write

a = pk1

1 pk2

2 . . . pkn

n , b = pl11 pl2

2 . . . plnn , c = pm1

1 pm2

2 . . . pmn

n ,

where some of the numbers ki, li, mi can equal zero. As we observed above, ∗page 155mi ≤ ki + li for i = 1, . . . , n. Therefore we can write mi = ti + si, where

ti ≤ ki and si ≤ li. Set

u = pt11 pt2

2 . . . ptnn and w = ps1

1 ps2

2 . . . psn

n .

Then c = uw. Besides, u divides a and w divides b. Hence a = uv andb = wz, where v and z are integer. Finally,

d =ab

c=

uvwz

uw= vz,

proving our claim.Now note that

a + b + c + d = uv + wz + uw + vz = (u + z)(v + w).

Both factors are greater than 1, so a + b + c + d is not a prime.

Problem 5. First solution. The general idea is this: At some point, three atleast of the starting triangles must be cut. After that there will be two groupsof three identical triangles each. Again, from each of these two trios, twotriangles must be cut. Two children coming from one group willbe identical with one another and with two children fromthe other group (see shaded triangles in the figure).We’re back to having four identical triangles!


But we must be careful about the argument: what if the cuts are madefirst in one triangle and its children, leaving the other starting triangles forlater? Perhaps then we might not get the trios of the previous paragraph?

To help organize our thoughts, let’s call a cut in an initial triangle a ∗∗level-1 cut. Any cut of a triangle obtained from a level-1 cut will be calleda level-2 cut, and so on.

Suppose we have a recipe —a sequence of cuts—that takes us from fourequal triangles to all different triangles. The key observation is that theorder of the cuts doesn’t matter, so long as the recipe still makes sense. Inparticular, the end result of a recipe can be obtained via a different recipeconsisting of a reordering of the same cuts: all level-one cuts made first,then all level-two cuts, etc.

Now start with a recipe having as few steps (cuts) as possible. Afterthe reordering, which leaves the number of steps unchanged, there must bethree level-one cuts right at the beginning; otherwise two starting triangleswill never be touched. After all level-one cuts are made, there must befour level-two cuts; otherwise there will be two first-generation children leftuntouched at the end.

These four level-two cuts produce four identical triangles, as discussed ∗(see figure on previous page). We discard all other triangles apart from these SL: make sure figure

position remains the samefour.Now, if we apply to these triangles all the cuts in the recipe that still

make sense (skipping steps that involve discarded triangles or their descen-dents), we end up with a subset of the triangles obtained from the full recipe,and of course these must be all different. But then we’ve found a shorter page 156

recipe to go from four identical triangles to all different triangles! (At least7 steps shorter.) But we had assumed our starting recipe was as short aspossible, so we have reached a contradiction.

Second solution. We will look for an invariant of the problem (see Fact 2 onpage 100) that we can use to our advantage. An obvious choice is the area.

Let each starting triangle have area 1 and let p and q = 1−p, be theareas of the triangles resulting from the first cut. We can assume p < q, ∗the case of an isoceles triangle being trivial. By induction, the area of anytriangle is pmqn, where m and n indicate respectively how many times thesmaller or the larger of the two children was chosen after a cut. Triangleswith the same pair (m,n) have the same area and, since they also have thesame shape, must be congruent.

Obviously, the total area of the triangles is 4 throughout the process.Now suppose that after some sequence of cuts, no congruent triangles are

left; we will try to obtain a contradiction. Each final triangle correspondsto a different pair (m,n). If we sum the areas of all (m,n)-triangles for0 ≤ m < ∞ and 0 ≤ n < ∞, whether or not they occur, we should obtain a ∗∗number strictly greater than 4, since, by assumption, the final triangles areamong this set, but are finite in number.

108 SOLUTIONS

Using the formula for the sum of a geometric series, we have

4 <

∞∑

m=0

∞∑

n=0

pmqn =

( ∞∑

m=0

pm

)·( ∞∑

n=0

qn

)=

1

1−p

1

1−q. (1)

It would be nice if the rightmost expression were at most 4, because then ∗we’d have our contradiction. But in fact this only occurs if p = q = 1

2 . Too ∗bad we don’t control p and q. . .

But wait! We’re free to devise any invariant we want. We can choose anew invariant to imitate the case p = q = 1

2 , whatever the actual values ofthese numbers. Let the “wealth” of a triangle be divided equally between itstwo children whenever a cut is made (and let it be 1 for the initial triangles).

The wealth of a triangle corresponding to the pair (m,n) is then 2−(n+m).Again, the total wealth is invariant and equal to 4. Now the partial sumthat replaces (1) is

4 <∞∑

m=0

∞∑

n=0

(12

)m+n=

( ∞∑

m=0

(12

)m)·( ∞∑

n=0

(12

)n)

=1

1/2

1

1/2= 4,

and we are done.

Remark. The following problem posed by Fields Medal winner Maxim Kontsevichis solved similarly: A bacterium sits at the (0, 0) square of an infinite squaregrid. After one minute it divides into two bacteria, which go on living in thesquares immediately above and to the right of the initial square. One of thesesplits after one minute, again occupying the squares immediately above and tothe right of where the parent was. The splits continue, always one bacterium ata time; moreover each square can fit only one bacterium at any given moment.Prove that there will always be at least one bacterium in some square (x, y) page 157

with x + y ≤ 5, x ≥ 0, y ≥ 0.

Problem 6. (a) The cook introduced, for her convenience, an imaginaryweightless can, so the number of cans became divisible by 3. She thennumbered the cans 1 through 81 in order of increasing weight.

To prove that she numbered the cans correctly, she first divides theminto three groups: cans 1 to 27, cans 28 to 54, and cans 55 to 81. She placesthe first group on the left pan and the third group on the right pan. Thebalance shows the greatest possible difference in weight between groups of27 cans, something that can be confirmed from the inventory. No othergrouping of 27 cans on each pan can give that readout.

Next the cook divides each group into three piles of 9 cans each. Sheplaces the lightest piles from each group (cans 1–9, 28–36 and 55–63) onthe left pan and the heaviest piles from each group (cans 19–27, 46–54 and73–81) on the right pan. This yields the greatest difference in weight betweensets of 27 cans of which 9 are taken from each group. This again is a numberthat can be calculated from the inventory.


Thus, after the second weighing there are 9 piles ordered according totheir weight; each pile has 9 cans and the geologists believe that each pilehas the composition claimed by cook.

In the third weighing, the cook divides each pile into 3 trios. She placesthe lightest trio from each pile on the left pan and the heaviest three canson the right pan. Again the readout is the maximum difference that can beachieved between sets of 27 cans of which 3 are taken from each pile. Thusthe makeup of each trio has been proved.

In the fourth weighing the cook places the lightest can from each trio onthe left pan and the heaviest on the right pan. Again because the readoutis the maximum possible, the identity of each can has been established.

(b) Three weighings are not enough, whatever method the cook uses. Indeed,the first weighing defines three sets of cans: those on the left pan, those onthe right pan and those that didn’t take part in the weighing. At least oneof these sets contains 27 or more cans (since 26 ·3 < 80); let’s call it X. Thesecond weighing defines three subsets of X: one containing those elementsof X that were placed on the left pan, and so on. One of these subsetscontains nine or more cans; call it Y . The third and last weighing definesthree subsets of Y accordingly, at least one of which has three or more cans.The cans in this last set simply cannot be distinguished on the basis of theweighings.

Level C page 158

Problem 1. (a) First we show that sinα/2 cannot takeα

(π−α)/2

α/2π−α ∗

more than four different values. Indeed, if sinα = sinβ,then either β = α + 2πk or β = π − α + 2πk, where kis an integer. Hence either β/2 = α/2 + πk or β/2 =(π − α)/2 + πk, respectively. The corresponding points on the unit circleare therefore α

2,

α2

+ π,π2− α

2,

3π2

− α2

.

(They are shown in the figure as the endpoints of the thin diameters.) Someof them may coincide, but there are no more than four in total.

It remains to give an example when the values of sine at these four pointsare pairwise different. For example, let sinα =

√3

2 ; then the points on thecircle corresponding to α/2 are π/6, π/3, 7π/6 and 4π/3. The values of thesine at these points are 1

2 ,√

32 , −1

2 and −√

32 .

(b) If sin α = 0, then α = kπ, where k is an integer. Hence sin α/3 can beequal to sin 0 = 0, sin π/3 =

√3

2 and sin 4π/3 = −√

32 .

Let us show that sin α/3 cannot take more than three different values.We can argue as in part (a) but we will give another proof.

Let sin α/3 = t. Recall the formula sin 3β = −4 sin3 β + 3 sin β. We get

sin α = sin(3 · α

3

)= −4t3 + 3t.

110 SOLUTIONS

It remains to recall that every cubic polynomial has no more than threedifferent roots, see Fact 20).

Remarks. 1. The algebraic approach is available for part (a) too, but it’s not asneat as part (b) because cosines must come in too. Recall the identity cos 2θ =

cos2 θ − sin2 θ = 1 − 2 sin2 θ; it gives sin α/2 = ±√

(1− cos α)/2. Now, givensin α, there are two opposite choices for cos α (why?). So there are, in general,four choices for sinα/2,

2. More generally, if n is any odd integer, sinnφ is a polynomial of degree |n| in ∗sin φ, while cos nφ is a polynomial of degree |n| in cos φ for any integer n. In ∗both cases the polynomial contains only exponents of the same parity as n. SeeRemark to Problem 59112 on page 132. REF 59112

3. If sinnφ is given, the highest number of values that sinφ can take is n for nodd, and 2n for n even. Prove this yourself.

4. If cos nφ is given, then cos φ can assume no more than n different values. Provethis too.

page 159

Problem 3. Let AB and CD be the sides ∗of the trapezoid. Denote by M and N,respectively, the second intersections ofthe lines AC and BD with the circles (seefigure). If the straight line AC is tangent tothe circle with diameter AB, we set M = A, and similarly, if the straightline BD is tangent to the circle with diameter CD, we set N = D.

A

B

K

C

D

M N

Recall, that, by the secant-tangent theorem, the squares of the lengths ofthe tangents from point K to the circle are equal to KM ·KA and KN ·KD.Hence, we have to prove that

KM · KA = KN · KD. (1)

Since \AMB is subtended by a diameter, we see that \AMB = 90◦,see Fact 14. (This is true also if M = A since the tangent is perpendicularto the radius whose end point is the tangent point.) Set φ = \AKB. In theright triangle KMB we have KM = KB cosφ. Similarly, KN = KC cos φ.Substituting the expressions so obtained into (1), we see that we only have toprove that KB ·KA = KC ·KD. This is a known property of any trapezoid,a consequence of the similarity of the triangles AKD and CKB.

Variant solution. Here is a more geometric proof of (1). It involves showingthat the points A, M , N , D lie on a single circle.

Since \BMC = 90◦ and \BNC = 90◦, the points B, M , N and C lieon a single circle with diameter BC. This implies that \CMN = \CBN =\BDA, since BC ‖ AD.

But then \AMN + \NDA = 180◦. Therefore the points A, M , N andD also lie on a single circle. Thus KM ·KA = KN ·KD (power of a point).

This solution, however, is essentially diagram-based, and fails if, forexample, one or both points M or N lie on the extension of the diagonal.


These cases are not difficult to consider separately, or one can use orientedangles. The case where M = A and N = D is more difficult.

We can overcome the case M = A in either of two ways. One way isto make a correct interpretation of all statements for the case of coincidingpoints. For example, the statement that “ A, M , N and D lie on a singlecircle” should be understood as follows: the line AK is tangent to the circlethrough points A, N and D.

The second way is as follows: since M = A, the lines AC and AB are per-pendicular. We may assume that the lines AC and CD are not perpendicular ∗ ∗since, if they were, ABCD would be a parallelogram, not a trapezoid (andfurthermore the conclusion obviously holds for parallelograms). Clearly, thelines AB and BD cannot be perpendicular. So it suffices to interchange Awith B and C with D.

Problem 5. First solution. Assume to the contrary that not all three num-bers have the same absolute value. Also assume that gcd(a, b, c) = 1; if thatis not so, substitute for the three original numbers their quotients by theirgcd, something that does not disturb either the hypotheses or the conclusion.

Now one of the three numbers is distinct from ±1; let it be divisible bythe prime p. At least one of the other two numbers is not divisible by p;otherwise p would divide the gcd. Assume that p does not divide c.

Given an integer x, denote by k(x) the power of p in the prime factoriza-tion of x; see Fact 10 on page 104. Thus k(c) = 0. Without loss of generalitywe assume that k(a) ≥ k(b).

We havea

b+

b

c+

c

a=

a2c + b2a + c2b

abc. (1)

In the denominator of this fraction, p appears with power k(a)+k(b). Hence page 161

the numerator should be divisible by pk(a)+k(b). We will show that this isimpossible.

The summand a2c contains p to the power 2k(a) ≥ k(a) + k(b), and b2acontains p to the power k(a)+2k(b) ≥ k(a)+k(b); hence a2c+b2a is divisible

by pk(a)+k(b). But c2b only contains p to the power k(b) < k(a) + k(b), so

c2b is not divisible by pk(a)+k(b). The sum (a2c+ b2a)+ c2b, therefore, is not

divisible by pk(a)+k(b) (see Fact 5). This is a contradiction.

Second solution. Set: x = a/b, y = b/c, z = c/a. Then xyz = 1 and

xy + yz + zx =xy + yz + zx

xyz=

1

x+

1

y+

1

z=

a

c+

c

b+

b

a

is an integer. Consider the polynomial

P (t) = t3 − (x+y+z)t2 + (xy+yz+zx)t − xyz.

We have shown that all the coefficients of this polynomial in t are integers.Since the highest coefficient is equal to 1, all rational roots of this polynomialare actually integers (see Remark). But as we recognize from Viete’s formulafor cubic polynomials (Fact 20 on page 108), or can check by expanding

112 SOLUTIONS

P (t) = (t−x)(t−y)(t−z), the roots are precisely x, y and z. Thus, x, y,and z are integers; since xyz = 1, we have x = ±1, y = ±1 and z = ±1.Equivalently, |a| = |b| = |c|.Remark. In the second solution we used a particular case (an = 1) of the following:

Theorem. Let f(x) = a0+a1x+· · ·+anxn be a polynomial with integer coefficients.Any irreducible fraction r/q that is a root of f(x) satisfies r |a0 and q |an.

Proof. The condition f(r/q) = 0 is equivalent to

a0qn + a1rq

n−1 + · · · + anrn = 0. (2)

Suppose q does not divide an. Then there is a prime p and an integer t > 0 suchthat pt divides q but not an. Since r and q are relatively prime, p does not divide r.So the term anrn is not divisible by pt, whereas every other term on the left-handside of (2) is so divisible. But this is impossible (see Fact 5 on page 102).

The proof that r divides a0 is analogous, interchanging the roles of q and r. ˜page 162

Problem 6. First solution. Observe first that the result does not dependon the order in which the buttons are pushed. We will work by inductionon the number n of light bulbs. For n = 1, the statement is obvious.

Let the statement be true for n−1 lights; we must prove it for n lights.Consider the i-th light. By the induction hypothesis we can switch off

all lights, except perhaps this one. Denote the set of buttons we have topush for this by Si.

If, for some set Si, all the lights go out, we are done. Otherwise, aftereach Si only the i-th light is glowing.

What happens if we first push all the buttons from Si and then pushall the buttons from Sj? We invite the reader to check that this will flipthe state of exactly two lights, the i-th and j-th ones, leaving the remaininglights in the same state as at the beginning.

This means we can switch off any two lights, and hence any even numberof lights. Thus if initially an even number of lights is on, we are done.

Assume that initially an odd number of lights is on. By assumptionthere is a button that flips an odd number of lights. Push it; now then thenumber of lit bulbs is even, and we can switch them off in pairs as in theprevious case.

Second solution. This solution uses linear algebra over the field F2 with twoelements, and so falls outside the high-school curriculum. But the readermay be familiar with the usual linear algebra over real numbers, and thesame notions apply with minor changes, which we point out along the way.

Arithmetic in F2 = {0, 1} is modulo 2: the addition and multiplicationtables are the obvious ones, except that 1+1 = 0. (See Fact 25 and the firsttwo chapters of Field2elem.) CITE Field2elem

Let the lights be numbered from 1 through n. To every state of thepanel we assign a row vector x = (x1, . . . , xn), where xi = 1 if the i-th lightis on and xi = 0 otherwise. Such vectors are elements of the n-dimensionalspace Fn

2 , analogous to Rn.


We also assign a vector to each button: b = (b1, . . . , bn), where bi = 1if and only if that button flips the i-th light. Pushing the button changesthe state from x to x + b. Pushing several buttons in sequence amounts toadding the sum of the corresponding b’s to the initial vector x. If that sumequals the initial vector, the final vector x +

∑b equals 0, meaning that page 163

all the lights are off. Our problem, therefore, is to prove that the vectorscoming from buttons span the whole n-dimensional space.

To a set of lights we assign a linear functional (i.e., a linear function ∗Fn

2 → F2), given by(x1, . . . , xn) 7→

∑xi,

where the sum runs over the indices i of the lights in the given set. BecauseF2 has only one nonzero element, any linear functional is obtained is thisway, for some set of lights.

The functional defined by a certain set of lights vanishes at the vectorcoming from a button if and only if this button flips an even number oflights in the set. Therefore the assumption that for any (nonempty) set oflights there is a button flipping an odd number of lights from this set canbe recast in the language of linear algebra as follows: No nonzero functionalvanishes on all vectors coming from buttons. ∗ removed “(A

nonzero functional...)”But this implies what we wish to show, for the following reason. It

is a fact of linear algebra that given a vector subspace and a point in thecomplement of this subspace, there is a linear functional on the whole spacethat vanishes on the given subspace but not at the chosen point. So if thevectors coming from buttons don’t span the whole space, there is a linearfunctional that vanishes on all vectors coming from buttons, but not on some page 164

vector outside the span; and this is contrary to the assumption.

Remark. Call a set of lights parity-invariant if any button changes the state of ∗an even number of lights in the set. The problem stipulates that there are noparity-invariant sets. Consider the case where parity-invariant sets are allowed. ∗ shortened to “case” for

better layout∗∗ deleted “(Why?)”

If the set of initially glowing lights has an odd number of lights in commonwith a parity-invariant set, it’s impossible to switch off all the lights.

∗Prove, by induction or using linear algebra, that if the set of initially glow-ing lights has an even number of lights in common with any parity-invariant ∗set, then it is possible to switch off all the lights.

Level D

Problem 1. Since the absolute value of a sum does not exceed the sum ofabsolute values (see Remark), we have

|x+y−z|+ |x−y+z| ≥ |(x+y−z)+(x−y+z)| = 2|x|.We similarly get

|x−y+z|+ |−x+y+z| ≥ 2|z| and |−x+y+z|+ |x+y−z| ≥ 2|y|.Adding up the three inequalities and dividing the sum by 2, we obtain thedesired result.

114 SOLUTIONS

Remark. Although the inequality

|x + y| ≤ |x| + |y| (1)

can easily be proved case by case, we give an elegant proof. Since both sidesof the inequality are nonnegative, we can square them and get an equivalentinequality. In other words, it suffices to prove that |x + y|2 ≤ (|x| + |y|)2.Expanding squares and using the fact that |a|2 = a2 for every a, we see thatthis is equivalent to x2 + 2xy + y2 ≤ x2 + 2|x| |y| + y2, which is obvious.

Inequality (1) is true for vectors x and y in n-dimensional space as well.The same proof works, with slight modifications. On the plane, (1) is equivalentto the triangle inequality.

Problem 2. The vertex condition can be reformulated as follows: Everyvertex has exactly one incident edge of each color.

(a) Since 1995 is divisible by 3, we can color the bottomedges red, green and blue in turn. We then color the match-ing top edges with the same colors. Now each vertical edgeis adjacent to edges of two colors; we use the remainingcolor for the vertical edge.

br g b r

r

g

gbb

rg b

r

Clearly, the colors of the lateral edges repeat with period three; hence,on each lateral face, the lateral edges are of different colors, and the top andbottom edges have the third color, as required.

(b) Suppose for a contradiction that the prism is colored as required. Theremust be three consecutive edges of different colors on the top face; other-wise the perimeter of that face is colored alternately with only two colors,contrary to assumption.

Take such a run of three edges; let their colors be red, green, blue,proceeding counterclockwise as seen from above. (The other cases are com- ∗pletely similar.) The vertical edges going down from the two intermediatevertices clearly have their colors determined (blue and red). Therefore thebottom edges between and adjacent to these vertical edges also have theircolors determined, matching the upper ones. (Why?) Further, the nextvertical edges, at both ends of the run, must be green, and so the next edgeson the top face in both directions are also determined. We now have a runof five edges: blue, red, green, blue, red, as in the figure above. page 165

We conclude by induction that the top and bottom edges are coloredwith period 3. But 1996 is not divisible by 3, so this is not possible. ∗ comma → .

Problem 3. First solution. We extend the median AA1 by its length andcomplete the triangle to a parallelogram ABDC, asin the figure.

A B

K

C D

A1

A2Recall that a bisector divides the opposite sideof a triangle in a ratio proportional to the adjacentsides:

A2B

A2C=

AB

AC.


Because A2K is parallel to AC and BD, it divides the transversals AD andBC into proportional segments (this is sometimes called the “generalizedintercept theorem”: see Remark 1 below), Thus the point K divides AD inthe ratio

KD

KA=

A2B

A2C=

AB

AC=

CD

AC.

Hence, K divides AD in the same ratio as the bisector of \ACD does,by the same property of the bisector already mentioned. Therefor CK isthe bisector of \ACD! Thus, AA2 and CK are perpendicular being the ∗bisectors of the inner one-sided angles at parallel lines.

The latter property is proved as follows. Let R be the point of intersec-tion of these bisectors. Then

\RAC + \RCA = 12(\BAC + \DCA) = 1

2 · 180◦ = 90◦.

Second solution. We use a notable property of trapezoids (see Remark 2below): the midpoints of the bases, the intersection of the diagonals and theintersection of the lines containing the nonparallel sides are all collinear.

A B

K

C

A1

A2

B1

C1

R

We work as follows (see figure). We first extend the segment CK to theintersection with line AB at point C1. In the trapezoid AKA2C, the

midpoint B1 of the base AC, the intersection of the diagonalsR and the intersection A1 of the extensions of the nonpar-

allel sides are collinear; therefore R lies on the midline page 166

A1B1 of the triangle ABC, and we get CR = RC1.Thus, in the triangle CAC1, the segment ARis simultaneously a median and a bisector, andhence it is also an altitude.

Remarks. 1. The generalized intercept theorem. Let A1, A2, A3 be pointson a straight line and B1, B2, B3 points on another line. If the segments A1B1,A2B2 and A3B3 are parallel (or exactly one is a point while the other two areparallel), we have

A1A2

A2A3=

B1B2

B2B3.

The proof is easy. We may assume that A2B2 is not a point. Through A2

draw a line parallel to the line containing B1, B2, B3. Let this line intersect thelines A1B1 and A3B3 at points C1 and C3, respectively. Then A2C1B1B2 is aparallelogram, so A2C1 = B2B1. Analogously, A2C3 = B2B3. It remains to usethe similarity of the triangles A1A2C1 and A3A2C3 (which is the usual intercepttheorem).

2. A remarkable property of the trapezoid. In a trapezoid, the midpointsM,N of the bases, the intersection P of the diagonals and the intersection Q ofthe lines containing the nonparallel sides are all collinear.

The simplest proof uses homothety. There exists a homothety with centerP sending one base of the trapezoid onto the other base, and therefore sendingM to N . Therefore P lies on the line MN .

Similarly, there exists a homothety with center Q sending one base of thetrapezoid onto the other. So Q also lies on the line MN .

116 SOLUTIONS

Problem 4. We prove by the induction on n that it is possible to divideany interval into white and black intervals so that the sum of the integralsof any polynomial of degree ≤ n over white intervals is equal to that overblack intervals.

Base of the induction. For n = 0, the polynomial is a constant. Theintegral is the product of this constant by the length of the interval. So wedivide the interval in half.

Induction step. Denote the midpoint of the interval [a, b] by c. Bythe induction hypothesis, we can divide [a, c] into black and white intervalsso that the sum of integrals of any polynomial of degree ≤ n− 1 over whiteintervals is equal to that over black intervals. Let B1, . . . , Br be the blackintervals and W1, . . . ,Ws be the white ones, numbered left to right. Wedenote the integral of the function f(x) over Bi by

∫

Bi

f(x) dx.

We similarly denote the integral of the function f(x) over Wi. Now trans- page 167

late each interval Bi to the right by c, color the result white and denoteit by Ws+i. Similarly, translate Wi to the right by c, color it black anddenote it by Br+i. Clearly, the intervals B1, . . . , Br+s together with theintervals W1, . . . ,Wr+s form a division of the interval [a, b] into black andwhite intervals. The figure illustrates the cases n = 1 and n = 2.

n = 1

B1 B2W1 W2

a c b

n = 2

B1 B2 B3 B4W1 W2 W3 W4

a c b

We claim that the division of [a, b] thus obtained has the desired propertyfor all polynomials of degree ≤ n. To prove this, we first use the change ofvariables y = x − c, dy = dx to check that

∫

Wr+i

f(x) dx =

∫

Bi

f(y+c) dy,

(see 28 on page 112). Then, since the integral of a difference equals thedifference of the integrals, we have

∫

Wr+i

f(x) dx −∫

Bi

f(x) dx =

∫

Bi

(f(x+c) − f(x)) dx.

Analogously,∫

Wi

f(x) dx −∫

Bs+i

f(x) dx = −∫

Wi

(f(x+c) − f(x)) dx.


Adding up these equalities for all i, we get

r+s∑

i=1

∫

Wi

f(x) dx −r+s∑

i=1

∫

Bi

f(x) dx

=

r∑

i=1

∫

Bi

(f(x+c) − f(x)) dx −s∑

i=1

∫

Wi

(f(x+c) − f(x)) dx. (1)

Now let f(x) be a polynomial of degree ≤ n. Then f(x + c) − f(x) isa polynomial of degree at most n − 1; see Fact 22 on page 109. By theinduction hypothesis the right-hand side of (1) vanishes, so the left-hand page 168

side also vanishes.

Remarks. 1. Consider the subdivision of the unit interval into subintervals oflength 2−n−1 according to the procedure above. Letting “+” stand for a black ∗interval and “−” for a white one, the resulting sequence of signs is the n-th termof the Morse sequence

+, +−, +−−+, +−−+−++−, +−−+−++−−++−+−−+, . . .

where from each term A we obtain the next by writing A twice and then inter-changing + and − everywhere in the second copy.

The n-th term of the Morse sequence can also be constructed as follows:For every k = 0, . . . , 2n − 1, the k-th sign in the sequence is the parity of the ∗number of 1’s in the binary representation of k (see Fact 12 on page 104).

One can show that in the strings of the Morse sequence no substring appearsthree times in a row.

2. The expression

Dcf(x) = f(x + c) − f(x)

is called the first difference of the function f with step c. The second differenceis the first difference of the first difference, and so on.

Before computers, first and higher differences were used for tabulating func-tions. For example, the function sinx can be approximated by a polynomial ofdegree n, and this can be used to compute a table whose k-th column containsthe values of the k-th differences of this polynomial. One starts filling out the ∗n-th column, which is constant for a degree-n polynomial. Then, given the k-thcolumn, it’s easy to fill out the (k−1)-st, and so work backward to the functionitself. Thus, one can tabulate any function using summation only, so long as itcan be approximated by polynomials.

Problem 5. We first show that n ≥ 1994. Let A have period 1000 . . . 00(1994 zeros), and let B be periodic as well, with period 1000 . . . 000 (1995zeros). Any piece of B of length at most 1994 occurs in A, and the otherconditions are also satified.

We next show that n < 1995 by proving that, for any A of period 1995,if all substrings of B of length 1995 are found in A, then B = A, apart fromreindexing.

Indeed, assume that A and B satisfy the hypotheses. We first show thatB repeats with periodicity 1995 (that is, 1995 is a period of B). Otherwise,

118 SOLUTIONS

B would contain a substring of length 1996 with different first and lastsymbols:

B = . . . x . . . . . . . . .︸︷︷︸1994 symbols

y . . . ,

where x 6= y. Let Z be the substring between x and y, and assume x occursin Z exactly k times. Consider the two strings xZ and Zy. Both occur inA, since they have length 1995. But their length coincides with the period page 169

of A, so xZ must be a cyclic permutation of Zy. In particular, x shouldappear the same number of times in xZ as in Zy. But this is clearly not thecase. Therefore B repeats with periodicity 1995.

Now let’s show that any shorter periodicity exists in B. Take a sub- ∗string of B of length 1995. By assumption, this substring occurs somewherein A. But we already know that B and A are completely determined by anysubstring of length equal to 1995; therefore B and A coincide as claimed,and 1995 is the minimal period of B.

Remark. The proof also implies that if a two-sided infinite string has minimal periodn, then any two identical segments of length n−1 are offset by an integer number ∗of periods.

Problem 6. The difference of powers suggests using the fact that, if r is amultiple of s, then as−bs divides ar−br, where a, b are variables or numbers. ∗(See Fact 8 on page 103.) ∗ removed “This is

easily...”Thus, if s is a divisor of n−1, the difference 3n−1 − 2n−1 is divisible by

3s − 2s. We want 3n−1 − 2n−1 to be divisible by n, so we look for n in theform 3s − 2s and such that n−1 is divisible by s. (We also need n itself tobe composite, but this comes free with our choice of n, so long as s is notprime.) ∗

We need 3s − 2s − 1 to be divisible by s. Let us take s = 2t; then thesecond summand is divisible by s (why?), and we just need 3s−1 = 32t−1 tobe divisible by s = 2t. Let us prove that this is always the case by induction.

For t = 1 this is clear: 9 − 1 divides 2. Let it be true for some t ≥ 1.Then for t + 1, we have

32t+1− 1 = (32t

+ 1)(32t− 1).

The first factor is even, and the second is divisible by 2t by the inductionassumption; hence the product is divisible by 2t+1 and we are done.

Remarks. 1. Why does the statement of the problem specify that n must be com-posite? It’s because otherwise Fermat’s little theorem (see [Chapter 3, § 6]MTF, ∗

CITE [for example) makes the problem easy. The theorem says: If p is prime and a isnot divisible by p, then ap−1−1 is divisible by p. Use it to show that 3n−1−2n−1 page 170

is divisible by n for n prime and distinct from 2 and 3.

2. There are infinitely many composite numbers n such that 2n−1 − 1 is divisible ∗by n.


Problem 7. Let’s arrange six sticks (long thin parallelepipeds)in the pattern shown in the figure, resting on the faces ofan imaginary central cube. The sticks almost touch theirneighbors, but not quite.

From the center of the construction, no vertices arevisible, because either end of any stick is hidden by a face ∗ removed “which lies

much closer”of another stick. Looking in any direction, we see onlyportions of faces and edges. ∗ removed 2nd figure

Next we make five bridges between the sticks, to connect them all into ∗ removed “at least”a polyhedron. This creates new vertices, but we can contrive to hide themfrom the center as well. For instance, a bridge might be made by weldinga thin, straight wire (with a square cross section) from one end of a stickto one end of another. The new vertices are very close to old ones, so theyremain invisible.

Remark. In two dimensions such an example is impossible: For any polygon anda point outside it, at least one of the polygon’s vertices is seen from the point(although, perhaps, no side is seen completely).

Back in three dimensions, one can guarantee that at least one vertex isvisible if there is a plane that separates the eye from the polyhedron.

Year 1996 Olympiad

Level A

Problem 1. Reducing each side of the equation to a common denomina-tor — for instance, a + b2/a = (a2 + b2)/a — we obtain

a2 + b2

a=

a2 + b2

b.

From the initial equation it is clear that a 6= 0 and b 6= 0. Therefore, page 171

a2 + b2 > 0 (see the remark below). So we can eliminate the factor a2 + b2

from both sides to obtain 1/a = 1/b. Hence a = b.

Remark. The following obvious statements are often used in olympiad problemsolving (and in mathematics generally): (1) the square of a nonzero number ispositive; (2) the sum of the squares of several numbers is nonnegative. If sucha sum is zero, then each of the numbers is zero.

Problem 2. Denote by mi the masses of the weights and by xi the massesof the balls. The sum (m1−m2)+(m2−m3)+ · · ·+(m9−m10)+(m10−m1)equals 0, because each mi occurs exactly twice, with opposite signs, leadingto complete cancellation.

The mass xi of the i-th ball is the absolute value of the differencemi − mi+1, so the preceding condition can be rewritten as

±x1 ± x2 ± · · · ± x9 ± x10 = 0,

where some signs of the xi are pluses and the others are minuses. Choosethe balls whose masses enter into this sum with a plus sign and put themon the left pan of the balance. Put the other balls on the right pan. Thebalance will then be in equilibrium.

Problem 3. Consider a grid square with a flower init. Divide it into four equal quadrants (small squares).We can assume without loss of generality that thesides of the quadrants have unit length. Suppose ourflower is in the upper left quadrant, as in the diagram;the other cases are all similar. Denote the gardenersat the vertices of the square by A, B, C, D.

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1A1

A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2A2 B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1B1

B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2B2

122 SOLUTIONS

We will prove that the flower will be looked after by gardeners A, B, andC. We divide all the gardeners into four groups, each group being locatedin one of the four shaded sectors in the figure. page 172

Gardener A is closer to the flower than B. This follows from thePythagorean Theorem, because the vertical separation from the flower isequal for both gardeners and the horizontal separation is less for A than forB (being less than 1 for the former and greater than 1 for the latter).

Similarly, gardener A is closer to the flower than C, and gardeners B andC are closer than D. Therefore, D does not look after the flower. Neitherdoes any other gardener from the D sector: they are all even farther fromthe flower than D (consider the vertical and horizontal distances again).

Now consider the B sector. Clearly, all its gardeners (apart from B) arefurther from the flower than B, and so even further than A. We will showthey are also further than C, and so do not look after the flower. Take,for instance, gardener B1. The vertical separation from B1 to the floweris greater than 2, and the horizontal separation is greater than 1. So thisgardener is farther from the flower than C.

It is exactly analogous to show that no gardener in the C sector, apartfrom C, takes care of the flower.

Finally, consider the A sector. We must show that A1 and A2, thenearest gardeners apart from A, are further from the flower than B and Care. Consider A1 (the proof for A2 is similar). To see that A1 is furtherthan B, just look at horizontal separations. To prove that A1 is further thanC, consider the perpendicular bisector to the segment A1C. Its points areequidistant from A1 and C. Points below the bisector are closer to C thanto A1. Clearly, the flower is below the bisector; therefore, it is closer to C.

Thus, the gardeners that look after the flower are A, B, and C.Now consider the gardener at the origin, say X. We already know that

X does not look after flowers unless they’re in one of the four squares (of theoriginal grid) touching the origin. And we need only find the intersection ofX’s area of responsibility with one of these squares: the pieces in the other page 173

three squares are found by symmetry.For instance, consider the square XY ZT in the figure. Again, we divide

it into quadrants. As we have seen, gardener X is one of those who looksafter the three shaded squares. Flowers in the fourth quadrantare looked after by Y , Z, and T . Hence, gardener X looks afterthe flowers growing in these three small squares and in the nineother squares obtained from them by reflection. X T

Y Z

Problem 4. Without loss of generality, assume K is closerto B than L, as in the figure. Then MKC is an equilateraltriangle (since MC = KC and \MCK = 60◦). Hence,AB is parallel to MK (since \MKC = \ABC = 60◦)and the angles AKM and BAK are equal (being alter-nate interior angles).

A

B CK L

M


Notice that the angles BAK and CAL are equal by symmetry (or by thecongruence of △BAK and △CAL, which follows from the SAS property).Hence \AKM = \CAL and

\AKM + \ALM = \CAL + \ALM = \LMC.

The last equation follows from the fact that \LMC, being an exterior angleof triangle AML, equals the sum of its two opposite interior angles.

Thus, it remains to prove that \LMC = 30◦. But ML is a median inthe equilateral triangle MKC; hence ML is an angle bisector of this triangle.

Remark. Compare with the solution to Problem 3 on page 130.

Problem 5. We can assume that the rook startsfrom the upper left corner. A solution for even n isillustrated on the right. The rook visits first all thesquares on the top two rows, then the squares on thethird and fourth rows, and so on.

We now show that the task is impossible for oddn. Consider any row apart from the top one. Whenthe rook hits this row, its next move must be to an-other square of the same row, and then to another row. Thus, the squaresof the chosen row are paired up by the moves within that row. (Note that page 174

the rook visits each square exactly once, since the number of moves is n2.)Therefore any row contains an even number of squares.

In other words, after its first visit to a row, the rook will have visitedtwo of its squares. After the second visit, there will be four such squares. Ifn is odd, a moment would come when the rook must visit the last unvisitedsquare on a row, and then it would be unable to move. (See also Fact 23 onpage 109.)

Problem 6. (a) Let N be the greatest number of problems solved by asingle student. So, for instance, if N = 8 one student solved all problemsand we’re done; if N = 7 we take a student who solved 7 problems and onewho solved the remaining problem, and again we’re done.

Consider the student/problem pairs that resulted in solutions. Thereare at least 40 such pairs (exactly 40 if we interpret “solved by five stu-dents” more strictly; either interpretation works). By the pigeonhole prin-ciple (page 100), at least one student solved 5 or more problems. So we’releft with two cases for N :

N = 6, so there is a student who solved exactly six problems. Each of theremaining two problems was solved by 5 among the remaining 7 stu-dents, and since 7 is less than 5 × 2, there must be some student whosolved both problems (pigeonhole principle again).

N = 5, that is, each student solved at most five problems. Since there were atleast 40 solutions, each student must have solved exactly five problems,and each problem must have been solved by exactly five students.

124 SOLUTIONS

So suppose student 1 solved problems 1–5. We claim there is astudent among the other seven who solved problems 6, 7, and 8. Ifthis were not true, each of the seven would have solved at least threeof the first five problems. Together with student 1, this would give5 + 7 × 3 = 26 solutions of problems 1–5. But this cannot be, becauseeach problem was solved by exactly five students. Therefore our claimis true.

(Another way to think of this last paragraph is as follows. Eachproblem was left unsolved by three students; in the case of problems 6,7, and 8, these are student 1 and two more. Therefore, all in all, thereare at most seven students— including student 1— who failed to solve page 175

one or more problems among the last three. That leaves at least onestudent who solved all three.)

(b) We need to fill an 8 × 8 table (rows representing students, columnsrepresenting problems) so that each column has four or more entries and notwo rows complete each other to form the full set {1, 2, . . . , 8}. We may aswell assume that each column has exactly four entries.

Any solution is likely to involve some trial and error, but we use what welearned in part (a) to direct the search. In particular, we keep the definitionof N . The cases N = 8, N = 7 and N = 6 won’t help us, because our earlierarguments show that in these cases, there are two students who solved allproblems between themselves. (In the case N = 6 the relevant inequality isnow 7 < 4 × 2.)

Next we try for a counterexample assuming N = 5. Naturally, we startby filling row 1 in positions 1–5. By assumption, in the subtable consistingof the last three columns and seven rows, there are 3× 4 = 12 entries. Eachof these rows has at most two entries in columns 6–8; otherwise it wouldcomplete row 1. It is reasonable to place exactly two entries on each of rows3–8, leaving row 2 with nothing in columns 6–8; thisway we can make row 2 a duplicate of row 1 withoutlosing any freedom.

The simplest way to choose two of the numbers 6,7, 8 for each of the last six rows so that each numberis chosen four times is to group the rows in pairs, as inthe figure: rows 3 and 4 skip column 6 and get entriesin columns 7 and 8; the next two rows skip column 7instead; and the last two rows skip column 8.

1

2

1 2 3 4 5

3

4

5

6

7

8

6 7 8

students

pro

ble

ms

We have now filled everything in, except for the subtable consiting ofcolumns 1–5, rows 3–7. This is similar to the original task, but with twoentries per column (out of six), so it’s easy to solve. Also, we must take carethat no two rows that complete each other in the first five columns shouldalso complete each other in the last three. This is best achieved by keepingthe pairing of rows already introduced; we divide the first five columns intothree chunks (of lengths 2,2,1 or 3,1,1) and fill one chunk per pair of rows,


like this:

or

Level B

Problem 1. First solution. Suppose the statement is not true. Then a page 176

certain n-gon has at least 36 angles less than 170◦, and the remaining n−36angles less than 180◦ (convexity). Therefore, the sum of all angles of thispolygon is less than 36 ·170◦+(n− 36) ·180◦. But it is well known that thissum is equal to (n − 2)180◦. This yields the inequality

(n − 2) · 180 < 36 · 170 + (n − 36) · 180.

After removing the parentheses and an obvious simplification, we obtain theinequality 180 · 34 < 170 · 36, which is false.

Second solution. If an interior angle of a polygon is less than 170◦, thecorresponding exterior angle is greater than 10◦. If we had 36 or more suchangles, their sum would be greater than 360◦. But the sum of exterior anglesof a convex polygon is 360◦. This contradiction completes the proof.

Problem 2. First solution. First assume that one of the numbers is zero.For instance, let a = 0 (the other cases are similar). Then we have theinequalities |b| ≥ |c| and |c| ≥ |b|, whence |b| = |c|, that is, b = c or b = −c.In the first case, we have b = a + c; in the second case, a = b + c, and we’redone.

Now suppose that none of the numbers a, b, and c is zero. Withoutloss of generality we can assume that a is the greatest of the three numbersin absolute value (that is, |a| ≥ |b| and |a| ≥ |c|). We can also assumethat a > 0 (if it isn’t, we replace all three variables by their negatives, anoperation that affects neither the problem’s hypotheses nor the conclusion).Then |a| = a, |a−b| = a−b, |a−c| = a−c.

Under these assumptions, the inequality |b − c| ≥ |a| implies that b andc have different signs. (Why?)

Consider the two possible cases.(1) b > 0, c < 0. Then |b| = b, |c| = −c, |b − c| = b − c, and the

given inequalities take the form a − b ≥ −c, b − c ≥ a, a − c ≥ b. The firstinequality implies that b ≤ a + c; the second one, that b ≥ a + c; therefore,b = a + c.

(2) b < 0, c > 0. Then, as in the previous case, we obtain the inequalitiesa−b ≥ c, c−b ≥ a, a−c ≥ −b. Therefore, in this case we have simultaneous

126 SOLUTIONS

inequalities c ≥ a + b and c ≤ a + b, that is, c = a + b. Thus, the statementis proved in both cases.

Second solution. Square both sides of the inequality |a − b| ≥ |c| and carryall terms to the left-hand side. We get (a− b)2 − c2 ≥ 0. The left-hand sidecan be factored as a difference of squares: (a − b − c)(a − b + c) ≥ 0, or,which is the same,

(a − b − c)(b − c − a) ≤ 0.

By a similar argument, the products (b−c−a)(c−a−b) and (c−a−b)(a−b−c)are also nonpositive.

Multiplying the three products, we find that page 177

(a − b − c)2(b − c − a)2(c − a − b)2 ≤ 0.

But if a product of nonnegative numbers is not positive, it must be zero.Hence at least one of the numbers is zero and we’re done.

Problem 3. The angles BNM and NCA

A

B

C

M Nare equal, because the lines MN and ACare parallel. The inscribed angle NCA isequal to half the measure of its interceptedarc AB

⌢. The angle BAM between a tangent

and a chord is also equal to half the measureof the arc AB

⌢which it intercepts (Fact 15 on

page 105), so \NCA = \BAM . Hence the anglesBNM and BAM are equal, and the quadrilateralAMBN is cyclic. We have the following chain of equations for angles:

\NCA = \BAM = \MBA = \MNA = \NAC.

The second equation follows from the fact that AMB is an isosceles triangle(the tangent segments drawn to a circle from an external point are equal),the third one from the fact that AMBN is a cyclic quadrilateral, and thelast one from the equality of alternate interior angles of parallel lines.

It follows that \NCA = \NAC; hence ANC is an isosceles triangle,i.e., AN = NC.

Remark. The attentive reader may have noticed that we used the fact that \BCAis acute. Check that otherwise there is no point N on the side BC such thatMN ‖ AC.

Problem 4. (a) The sum of the number 9 and the number beneath it liesbetween 10 and 18. Since there is only one perfect square in this interval,the number under 9 must be 7. Similarly, the number 7 must be writtenabove the number 9 on the second row. In the same manner, it can be shownthat the numbers under 4, 5, and 6 must be 5, 4, and 3, respectively. Now page 178

it is not difficult to complete the answer:

1 2 3 4 5 6 7 8 9

8 2 6 5 4 3 9 1 7


(b) Using the same reasoning we see that the only number that can bewritten under 11 is 5. But the number under 4 must also be 5, which isimpossible.

(c) The idea is to reduce the problem to a similar one for smaller n. Thefirst square past 1996 is 2025 = 452, which equals 1996 + 29. So if undereach number k from 29 through 1996 we write the number 2025 − k, thesum-is-a-square condition is satisfied for all columns after the first 28. Thisreduces the problem to the case n = 28.

Repeating the same strategy we write under each of the numbers k =21, 22, . . . , 28 its counterpart 49−k, reducing the problem to the case n = 20.Then we write 36−k under each k = 16, 17, 18, 19, 20, reducing the problemto n = 15. Finally, under each k = 1, 2, . . . , 15 we write 16 − k.

Remark. When do you run into trouble with this strategy? Try to prove that theproblem can be solved whenever n > 11.

Problem 5. We begin with two simple statements:

(1) The segment joining the midpoint of a chord to the circle’s center isperpendicular to the chord.

(2) The condition that the endpoints of a chord lie on different arcs ABis equivalent to the chord intersecting the segment AB (in an interior point).

Thus the problem can be reworded as follows: Given three points A,B, and O with AO = BO, find the locus of all points M such that theperpendicular to MO going through M crosses the open segment AB.

We now prove that the perpendicular to MO at M intersects the opensegment AB if and only if exactly one of the two angles OMA and OMB isobtuse. Indeed, the perpendicular intersects AB if and only ifA and B lie in opposite half-planes with respect to thisperpendicular; this condition can also be expressed bysaying that exactly one of the two points (A or B) liesin the same half-plane as the point O, and the other liesin the half-plane opposite O. But a point P lies in thehalf-plane opposite to O if and only if the angle PMOis obtuse (as illustrated by point A in the figure on theright), and P lies in same half-plane as O if it is acute page 179

(point B in the figure).

A

B

OM

The locus of points M such that \AMO is the interior of the circle withdiameter AO, and the locus of points M such that \BMOexceeds 90◦ is the interior of the circle with diameter BO(see Fact 14 on page 105). This means that the locus inquestion consists of the points lying inside one, but notboth, circles with respective diameters AO and BO. Inother words, it is the union of the interior of these circlesminus their intersection —the shaded region in the figure.

A B

O

128 SOLUTIONS

Problem 6. Let us show that Ali Baba can always arrange for seven pilesto end up with at most four coins, and the thief can ensure that no pile evercontains less than four coins. That will mean that Ali Baba can walk awaywith 100 − 7 · 4 = 72 coins.

We start by proving that the thief can ensure there are at least four coinsin each pile after each exchange. This is certainly true at the beginning.Suppose it is still true at a certain point and some coins are then placed inthe cups. If there are two cups with the same number of coins, the thief cansimply swap them, leaving the situation unchanged. If the number of coinsis different for each cup, then the two cups with most coins will contain atleast three and four coins, respectively, and the thief can swap these cups.Then the new piles will again consist of at least four coins each.

Now we show that Ali Baba can always reduce seven piles to four coinsor less.

Suppose there are four piles with more than four coins each. Let x1 ≥ page 180

x2 ≥ x3 ≥ x4 ≥ 5 be the numbers of coins in these piles. We’ll show thatAli Baba can work with these piles alone until one of them has four coins orless. He chooses the partitions

x1 = y1 + 1, x2 = y2 + 2, x3 = y3 + 3, x4 = y4 + 4,

and puts 1, 2, 3, and 4 coins from these piles into the respective cups. Afterthe thief rearranges the cups, the new piles will consist of

y1 + z1, y2 + z2, y3 + z3, y4 + z4 (1)

coins, where z1, z2, z3, z4 is a certain permutation of the numbers 1, 2, 3, 4.Let x′1, . . . , x

′4 be the four numbers in (1), reordered if necessary to preserve

their nonincreasing order. There are three possibilities:

(1) If the first cup was moved, the tallest pile got even taller (sincez1 > 1). In other words, x′1 > x1.

(2) If the first cup stayed put and the second was moved, the secondtallest pile grew; it either surpassed the first and we have x′1 > x1 (sincewe renumbered the piles), or this did not happen and we have x′1 = x1 andx′2 > x2.

(3) If the first two cups stayed put, the last two were interchanged; inthis case x′1 = x1, x′2 = x2, and x′3 > x3.

Ali Baba then repeats this process so long as these same four piles stillhave more than four coins each. We must show that this cannot go onforever. Conclusions (1)–(3) can be rephrased as follows: on each applicationof the process, x1 can only go up, never down; if x1 doesn’t go up, x2 canonly go up, never down; if neither x1 and x2 go up, x3 must go up. Thismeans that no triple (x1, x2, x3) can ever occur more than once. (To convinceyourself of this, consider the parallel situation where x1, x2, x3 represent thedigits of a decimal number and obey the same rules: the number as a wholemust increase at each step.) But there are only finitely many triples, so


eventually it must be impossible to apply the process again, indicating thatsome pile now has four coins or less.

Considering now all the piles, the net effect so far is that the number ofpiles with four coins or less has increased. As long as Ali Baba can repeatthe whole procedure, this number will keep increasing. It only stops whenthere are at least seven piles with four coins or less.

Level C page 181

Problem 1. First solution. We can assume that a ≥ b (the case a ≤ b istreated similarly). Then b2 ≤ ab, a2 ≥ ab, and therefore,

a2 ≥ a2 + b2 − ab ≥ b2,

whence a ≥ c ≥ b. It follows that the first factor in the product (a−c)(b−c)is nonnegative and the second one is nonpositive. But then the product isnonpositive, completing the proof.

Second solution. Consider a triangle with a 60◦ anglebetween two sides of lengths a and b. By the law ofcosines, the length of the third side is

b a

c

60 ◦

√a2 + b2 − 2ab cos 60◦ = c.

Since the greatest angle of any triangle is at least 60◦ and the smallestangle is at most 60◦, the angle opposite side c is of intermediate value inour triangle. Since a greater angle in a triangle lies opposite a longer side,either a ≤ c ≤ b or b ≤ c ≤ a. Hence one of the two factors a − c andb − c is nonnegative and the other is nonpositive. Therefore, their productis nonpositive.

Problem 2. Draw all 17 lines parallel to a diagonal ofthe square and passing through at least two of the markedpoints (see figure). This takes care of all points except twocorners of the array, which require an 18th line.

To see that fewer lines will not suffice, consider thecenters of the 36 unit squares around the edges of thearray. Clearly, any line not parallel to the edges of the array crosses at most page 182

two of these points, so we need at least 18 lines to account for all of them(pigeonhole principle, see page 100).

Problem 3. Since an exterior angle of a triangle is equal to the sum ofits two opposite interior angles, the triangle APkM yields the equation\PkMC = \PkAC + \APkM , or \APkM = \PkMC − \PkAC. Addingtogether all these equations, we see that the sum in question is equal to

(\P1MC + · · · + \Pn−1MC) − (\P1AC + · · · + \Pn−1AC).

130 SOLUTIONS

Suppose that n is odd; the complementary case is similar. Reflection aboutthe altitude of the equilateral triangle P1MC shows that the angles PkMCand Pn+1−kMC add up to 60◦ (see figure on the next page).

Then the terms in the first sum fall into (n − 3)/2 pairs,each of which adds up to 60◦, and two unpaired terms:\P1MC = 60◦ and \P(n+1)/2MC = 30◦. Hence the firstsum is equal to 30◦ ·n. The terms of the second sum fallinto (n− 1)/2 pairs with a total of 60◦ each. Hence thesecond sum equals 30◦ · (n− 1). The statement of theproblem follows immediately.

A

P1 Pk Pn+1−kB C

M

Remark. Compare with Problem 4 on page 15.

Problem 4. The case m = n = 1 is obvious: there’s no room for even asingle move. For definiteness, assume that the board consists of n rows andm columns, with m ≥ n, and that the rook starts from the upper left corner.

It turns out that the winning strategy for the first player is to make eachmove the longest possible. To prove this, we’ll assume that there are boardson which this strategy does not win and refute this conjecture by derivinga contradiction. Among all such boards, we can take one with smallestpossible area. Since on an 1 × n board the first player obviously wins, wecan assume that both dimensions are greater than 1: m ≥ n ≥ 2. page 183

The first player moves the rook along the entire longer side, the top row.The second one is forced to move down. Consider three cases:

(a) The second player makes a one-square move. We’re nowin the starting situation of the same game played on anm × (n−1) board (see figure). Since m ≥ 2, this is notthe 1 × 1 board.

(b) The second player moves all the way down. Then the firstmoves all the way across, per the strategy. If m = n = 2,the first player wins immediately. If not, we’re in thesituation we’d be in after the initial move if the gamewere being played on an (m−1) × (n−1) board.

(c) The second player moves by k squares, where 1 < k < n−1. Thefirst player moves all the way across, per the strategy. At this point,whether the next move by the second player is up or down, we’re in the in the original, the

black-edged rectangle istoo long.

situation we’d be in after the first player’s initialmove if the game were being played on a smallerboard: either an (m−1) × (n−1) board (thick solidline in the figure; note that this cannot be the 1× 1board since n ≥ 4) or an m × (n−k) board (doublethin line).

In every case, we have reduced the situation to onethat is encountered, if the strategy is followed, during a game played ona board of lesser area (and distinct from 1 × 1). But, by our least-area


assumption, on such a board the first-player strategy wins. Hence it winson the m × n board as well, and we arrive at a contradiction.

Problem 5. We answer the question with an example. Suppose there are 10inhabitants in the country, and that their houses are placed along a straight page 184

line in ascending order of their owners’ heights. Suppose the intervals be-tween the houses, in kilometers, are 1, 2, 3, 4, 5, 4, 3, 2, and 1:

Then everyone except the tallest person can travel free of charge. Indeed,the five shortest persons can choose circles of enormous radius, so each willbe shorter than 5 of their 9 neighbors. The rest of them must choose a circleenclosing just one neighbor.

On the other hand, everyone except the shortest person will be allowedto play basketball, if the five tallest choose a large circle and all the restchoose a circle enclosing just one neighbor.

Problem 6. It is readily seen that P (N) > P (M) for N > M > 0, becausethe coefficients are positive. We also have P (N) > 1 for N > 0.

Next we claim that if k divides x− y, then k divides P (x)− P (y). Thisfollows by induction on the number of monomials in P . For one monomialit is obvious since

xn − yn = (x − y)(xn−1 + xn−2y + · · · + yn−1).

But if the statement is true for two polynomials, it’s also true for their sum.Now let’s set

A = P (1)P (2) . . . P (1996).

Since P (k) divides A, it also divides P (A + k) − P (k), for k = 1, . . . , 1996.Hence P (k) divides P (A + k). But P (k) > 1 and P (A + k) > P (k). Itfollows that P (A+k) is a composite number for k = 1, . . . , 1996, completingthe proof.

Level D page 185

Problem 2. Introduce the symbols a =5√

2 +√

3 and b =5√

2 −√

3, sothe desired root is x = a + b. By direct substitution we see that a5 + b5 = 4and ab = 1, so

a +1a

= x and a5 +1a5 = 4.

132 SOLUTIONS

Our goal is to relate the expression on the right to powers of x. By expandingthe binomial powers we see that

x5 =(a +

1a

)5= a5 +

1a5 + 5

(a3 +

1a3

)+ 10

(a +

1a

),

x3 =(a +

1a

)3= a3 +

1a3 + 3

(a +

1a

).

Substituting the value of a3 +1a3 from the second equation into the first, we

obtain(a +

1a

)5=

(a5 +

1a5

)+ 5

((a +

1a

)3− 3

(a +

1a

))+ 10

(a +

1a

).

Hencex5 = 4 + 5(x3 − 3x) + 10x

orx5 − 5x3 + 5x − 4 = 0.

Remark. It is possible to express an +1

ancan be expressed in terms of a +

1

afor

any n:

an +1

an= Pn

(a +

1

a

).

The polynomials Pn are related to the so-called Chebyshev polynomials Cn bythe relation Cn(x) = 1

2Pn(2x). The Chebyshev polynomials are defined by theformula

cos(nx) = Cn(cos x).

The link between these formulas follows from the relation cosx = 12 (eix +e−ix),

where i =√−1; see Zorich. See also Remark 2 to Problem 2 (page 110). CITE Zorich

Problem 3. Any set of evenly spaced parallel planes is equivalent to anyother under a similarity, and any cube is also similar to any other. So thequestion can be rephrased thus: Can the vertices of the unit cube [0, 1]3 lieon evenly spaced parallel planes?

Now, a family of parallel planes is made up of the constant sets of alinear function ax+by+cz; and the spacing between two such planes isproportional to the difference between the values of the function on each.We wish to find a, b, c so that the values of the function on the vertices of page 186

the unit cube are evenly spaced. These values are 0, a, b, c, a+b, b+c,a+c, and a+b+c. A moment’s thought will show that our condition will besatisfied if, for example, a = 1, b = 2 and c = 4. In other words, the planesdefined by x+2y+4z = k, for k = 0, . . . , 7, contain each one vertex of theunit cube.

Problem 4. This problem and many similar ones can be solved using mod-ular arithmetic, which is the consideration of remainders upon division byvarious integers. Choosing the best integer by which to divide in order todraw useful conclusions (the modulus) usually involves some trial and error,but where squares are involved, it’s often useful to work modulo 4. The re-mainder of a perfect square upon division by 4 is either 0 or 1 (see Remark


1 after the solution). So sums of two squares can only give 0 + 0, 0 + 1 or1 + 1 modulo 4; numbers that give a remainder of 3 modulo 4 cannot besums of two squares.

Similarly, the remainders of a square modulo 9 are 0, 1, 4, or 7 (seeRemark 1); considering the possible sums of these remainders we find thatsums of two squares can only give 0, 1, 2, 4, 5, 7, 8 modulo 9. (Note that4 + 7 ≡ 2 modulo 9, since 11 = 9 + 2.)

Here are two of many possible solutions based on these facts:

First solution. The sum n = (36k + 2)2 + 42, where k is an integer, givesthe same as 22 + 42 modulo 4 and also modulo 9, since 4 and 9 both divide36. Therefore:

• n =≡ 0 modulo 4, which means that n−1 ≡ 3 modulo 4. So n − 1cannot be a sum of two squares.

• n =≡ 0 modulo 9, which means that n+1 ≡ 3 modulo 9. So n + 1cannot be a sum of two squares.

Since k can take any value and n grows with k if k ≥ 0, we have foundinfinitely many values of n as desired.

Second solution. The sum n = 9k + 1, where k > 0 is an integer, gives 2modulo 4 and gives 1 modulo 9. Therefore:

• n + 1 cannot be the sum of two squares since n+1 ≡ 3 modulo 4.• n−1 gives 1 modulo 4 and 0 modulo 9, which are both allowed. So this

path doesn’t look promising, but now we look at arithmetic modulo 3.If n − 1 = 9k can be written in the form 9k = a2 + b2, there are twopossibilities: either a and b are both non-multiples of 3, so a2 and b2

each have remainder 1 modulo 3, which disagrees with their sum beinga multiple of 3; or a and b are both multiples of 3, and we can divide by9 to obtain a similar equation, 9k−1 = (a/3)2+(b/3)2. In the latter casewe just apply the argument again to the new equation, and concludethe proof by reverse induction (see Fact 24). Note that if k − 1 = 0 wecannot apply the argument, but then there is obviously no solution.

page 187

Remarks. 1. It is easy to find what remainders are allowed for a square, for anymodulus q. For suppose that n gives r modulo q; that is, we have an equationn = kp + r, with k an integer. Then n2 = k2q2 + 2kqr + r2; that is, n2 has thesame remainder as r2, modulo q. So all we need to do is look at all values or rand take their squares modulo q.

• q = 3: We have 02 = 0, 12 = 1, 22 = 4 ≡ 1 modulo 3. So the possibilitiesare 0 and 1. The square of any non-multiple of 3 leaves a remainder of 1modulo 3.

• q = 4: We have 02 = 0, 12 = 1, 22 = 4 ≡ 0, 32 = 9 ≡ 1 modulo 4. So thepossibilities are again 0 and 1.

• q = 9: We can save time by noting that the remainders 5 to 8 can berelated to their “complements”; that is, 9− t is the same (modulo 9) as −t,so (9−t)2 is also the same (modulo 9) as (−t)2 = t2. Thus we only need

134 SOLUTIONS

to consider the squares of 0, 1, 2, 3, 4. They give 02 = 0, 12 = 1, 22 = 4,32 ≡ 0 and 42 ≡ 7 modulo 9.

• Also useful: the square of an odd number always yields a remainder of 1when divided by 8. (Check it!)

2. A famous criterion for a positive integer n to be representable as the sum of twosquares requires that any prime factor of n of the form 4k + 3 must appear inthe prime factorization of n with an even power [§ 4–5]NumbRepresSumSquar. CITE [

See also Remark 2 on page 57.

Problem 5. Denote by O1 and O2 the centers of the circles and by r1 andr2 their radii, respectively.

Consider the point D at which a common interior tangent to the circlesmeets the segment O1O2. Then (see Fact 17)

DO1

DO2=

r1

r2.

Let AC be a diagonal of the quadrilateral in question and let S be thepoint at which it meets O1O2 (see figure).

A

B

C

D=S

X

O1 O2

By the law of sines, we have

SO1

sin \O1AS=

r1

sin \O1SA,

SO2

sin \O2CS=

r2

sin \O2SC. (1)

Consider the circle with center X passing through A and C. It is not hard page 188

to see that the lines AO1 and CO2 are tangent to this circle. Expressingthe angles between these tangents and the chord AC in terms of interceptedarcs (see Fact 15), we obtain the equation

\O1AS + \O2CS = 12(

⌢AC +

⌢CA) = π

(where the second arc contains the point B in the figure and the firstdoesn’t); hence sin \O1AS = sin \O2CS. The angles O1SA and O2SC,being vertical angles, are equal. Therefore, by (1), we have

SO1

SO2=

r1

r2.

Since there is only one point dividing a line segment in a given ratio, thepoints D and S coincide. A similar reasoning is true for the second diagonal.


This means that the diagonals of the quadrilateral and the common internaltangents cross the line of centers at the same point, and we’re done.

Remark. This problem implies the following remarkable statement: If a quadrilat-eral ABCD is circumscribed around a circle, the intersection point of the linesjoining the opposite points of contact coincides with the intersection point ofits diagonals. Indeed, take as X the incenter of the quadrilateral. Let ω1 be thecircle centered at A and passing through the points at which the incircle of thequadrilateral touches AB and AD. Denote by ω2 a similar circle with centerC. Applying the statement of the problem to the configuration thus obtained,we see that the lines joining the opposite points of contact of the incircle meeton AC and, for similar reasons, on BD at the same time.

Problem 6. First solution. We will choose certain rows of the modifiedarray one by one, and keep track of the sum S = S(m) of the m rows chosenso far. The first row we take is the modified version of the row (1, 1, . . . , 1)of the original array. The string S(1) simply coincides with this row and,obviously, consists of zeros and ones. If it has only zeros, this row aloneyields the desired set. Otherwise, we find in the original array the rowobtained from S(1) by replacing 1 by −1 and 0 by 1. The correspondingrwo in the modified array row is chosen to be our second string. Then S(2),the sum of S(1) and the chosen row, consists of 0s and 1s again.

Suppose that we have already chosen k rows. If the sum S(k) coincideswith one of the previous sums S(m), where m < k, then the sum of the page 189

k−m rows indexed from m+1 to k is the zero string and we’re done. If thesum S(k) does not coincide with any of the previous sums, then we take upnext the row in the modified array corresponding to the row in the originalarray obtained from S(k) by replacing 1 by −1 and 0 by 1. This row couldnot have been chosen before, because different sums S(m) specify differentrows to be chosen in the array, and the sum S(k) has never occurred before.

If at a certain step we obtain a sum that has occurred before, then, aswe have already seen, the problem will be solved; if not, we will eventuallyhave exhausted all the rows of the array. That is, we’ll have 2n differentsums S(k). Since the number of different strings of 0s and 1s of length nis also equal to 2n, each of these strings will coincide with one of the sumsS(k). In particular, for a certain k, the string S(k) will consist of zeros, andthe desired set is the set of the first k chosen rows.

Second solution. Denote by ai the rows of the initial array and by bi the rowsof the modified array, where i = 1, 2, . . . , 2n. We will construct yet a thirdarray, with the rows ci = ai − 2bi. In other words, any two correspondingelements of ci and ai either coincide (if the element of ai was replaced byzero) or are opposite (in all the other places). In particular, the new arrayconsists of ±1. Therefore, for any i there exists a j(i) such that ci = aj(i).Now let us consider the sequence ik specified by the recurrence relationik+1 = j(ik). (The first term of the sequence is chosen at will.) Since thissequence can take only finitely many values, it must contain equal terms.

136 SOLUTIONS

Suppose that ik = il for certain k and l, k < l, and all the terms of thesequence with numbers less than l are different. Then

bik +bik+1+ · · ·+bil−1

= 12(aik−cik)+ 1

2(aik+1−cik+1

)+ · · ·+ 12(ail−1

−cil−1)

= 12(aik−aik+1

+aik+1−aik+2

+ · · ·+ail−1−ail)

= 12(aik−ail) = 0,

which completes the proof.

Year 1997 Olympiadpage 190

Level A

Problem 1. The numbers of pieces on each row ranges from one to eight.Since these numbers are different for different rows, there is a row with ex-actly one piece, another with two, and so on. We number the rows accordingto the number of pieces in them.

On row 1, we mark its single piece. On row 2 there are two pieces, sowe can choose one that’s not in the same column as the first marked piece.We mark this second piece. Similarly, we can mark at least one of the threepieces on row 3 in a column not already occupied by a marked piece, and soon. See also Fact 1.

Problem 2. The round trip along the road and the path takes 16 hours.Therefore, no trouble will come from the first crater if the walk starts rightafter that crater has finished an eruption.

The round trip along the path takes 8 hours. This means that the walkon the path will be safe from the second crater if it starts immediately afterthat crater has finished an eruption.

Therefore the whole trip will be safe if the first crater has just finishedan eruption at the time Professor Garibaldi sets out, and the second craterhas just settled down 4 hours later, by the time the volcanologist reachesthe path.

Let us find such a moment. The first crater erupts during the 1st, 19th,37th hours. The second erupts during the 1st, 11th, 21st, 31st, 41st hours.Therefore, if Professor Garibaldi sets out at the beginning of the 38th hour,reaching the path at the beginning of the 42nd hour, everything will be fine.

Remark. We have solved the problem by listing possibilities. However, we couldset up a Diophantine equation. The first crater erupts during hours of the form18x + 1, where x is a positive integer, and the second crater during hours ofthe form 10y + 1. We want to have a 4-hour interval between eruptions; thisleads to the equation

10y − 18x = 4.

The smallest positive integer solution to this equation is y = 4, x = 2.page 191

138 SOLUTIONS

Problem 3. Let L and K be the reflections of M in the lines OX and OY ,respectively (see diagram below and to the left). Then K, P , and N lieon the same line, and NK = NP + PK = NP + PM . Indeed, segmentMK is perpendicular to the line OY , and if A is their intersection point,then MA = AK (by the definition of reflection). The right triangles MAPand KAP have equal legs; therefore, they are congruent and \KPA =\MPY = \NPO. In addition, we have PK = PM .

A

O

X Y

N

M

PQL

K

O

X Y

N

M

PQL

K

Similarly, points N , Q, and L lie on the same line, and NL = NQ+QL =NQ + QM . It remains to prove that NL = NK. We do this by showingthat the triangles KON and LON are congruent. Indeed, since K and Mare symmetric about the line OY , we have KO = MO (see diagram aboveand to the right).

Likewise, MO = LO, and hence LO = KO. Further, the triangles KONand LON have a common side, and

\KON = \KOP + \PON = \POM + \PON

= \QON + \PON = \XOY.

One shows similarly that \LON = \XOY . Thus, \KON = \LON,and the triangles KON and LON are congruent by the SAS property, whichcompletes the solution.

Remark. Given two points M and N on the same side of a line OY , the shortestpath from M to N touching the line is the path MPN considered in our prob-lem. This property, which the ancient Greeks knew and could demonstrate —it appears as a theorem in Heron’s Catoptrics — is one way to explain the factthat a ray of light reflected in a mirror makes the same angle coming in andgoing out (because of the physical principle that light “likes” to take the short-est path). Or, put in a whimsical way: A fireman at point N has to put out afire at point M having filled his pail in a river, represented by the straight line page 192

OY . To what point on the river should he go in order to make as short a runas possible?

Now the initial problem can be interpreted as follows: if the river is theangle XOY and the angles NQO and MQX are equal, it doesn’t matter towhich side of the angle the fireman runs.


Problem 4. Any number ending in 0 is composite. Therefore if we choosea number ending in 0, we only have to worry about changing the last triple Rewritten to give reader a

chance to understand how

one might come up withsuch an idea.

of digits, because changing any other triple will leave it a composite number.Next, if we choose our number N to end in 000, the condition “replacing

the last triple by anything gives a composite number” is the same as thecondition that N +1, N +2, . . . , N +999 are all composite. Now, if N is amultiple of all numbers from 1 to 999— for instance, N = 999!, it is clearthat N +k is a multiple of k for k = 1, . . . , 999. Thus N +2, . . . , N +999 areall composite. But what to do about N +1?

One idea is to make N even bigger, and look at the second hundrednumbers after N , instead of the first hundred. For instance, if N = 2000!,then N +1000, N +1001, N +1002, . . . , N +1999 are all composite, and soN +1000 satisfies the conditions of the “Prove that” part of the problem.

But 2000! has more than 1997 digits, because it’s greater than 10001000.To find an N with 1997 digits, we note that many factors are superfluous.We can start by dropping all the factors from 1 to 999. We keep the factor1000 since we want N to end in 000, but we can drop all other even factors,because, N being even, N +1002, N +1004, . . . , N +1998 are also even andtherefore composite! So now N is the product of 501 numbers under 2000,and we see that it has less than 1997 digits from an easy estimate:

2000501 = 2501 · 1000501 < 10501/3 · 10501×3 = 101670.

(Here the inequality comes from the fact that 23 < 10.)To conclude we extend N with zeros to the right so it acquires exactly

1997 digits, and we finally add 1000 according to the earlier plan; this givesa 1997-digit number satisfying all the required conditions.

Remark. You may have encountered the idea in the second paragraph when solvinga simpler problem: Prove that there exist 1000 consecutive composite numbers.

page 193

Problem 5. First solution. Let the lines DE and ABintersect at G. The triangles DEC and BEG arecongruent, by ASA (see figure). Thus BG =CD = BA, which means that A, G, and Clie on a circle with center B; moreover AGis a diameter of this circle. Since \AFG = 90◦, thepoint F must lie on the same circle, by the 90◦ case of the inscribed angletheorem (Fact 14).

A B

CD

EF

G

Again by the inscribed angle theorem, we have \GFC = 12\GBC =

12(180◦ − 40◦) = 70◦. It follows that \DFC = 180◦ − \GFC = 110◦.

Second solution. We can do without the inscribed angle theorem. To seethat F lies on the circle, note that BF = BA = BG, because BF is themedian of the right triangle AFG, and hence equals half the hypotenuse AG.

Now consider the sum of the angles of the quadrilateral ABCF . Wehave

\BCF + \CFA + \FAB + \ABC = 360◦.

140 SOLUTIONS

But \BCF + \FAB = \CFA because the triangles CBF and FBA areisosceles. Therefore we get 2\CFA + 40◦ = 360◦, or \CFA = 160◦. Itfollows that \CFD = 360◦ − \AFC − \AFD = 360◦ − 160◦ − 90◦ = 110◦.

Problem 6. First, let’s solve a simpler problem: suppose each coin can beused in only one weighing, rather than two. What is the greatest numberof coins such that the expert can be sure of being able to determine thecounterfeit coin after k weighings?

If two coins are placed on the same pan of the balance in one of theweighings, it will be impossible to identify which of them is counterfeit (as-suming one of them is), because they cannot take part in a second weighing.Therefore, each weighing involves one coin on each pan.

If the balance is not in equilibrium, the counterfeit coin is obviouslythe lighter of the two. If it is in equilibrium, the number of questionable page 194

coins decreases by two. It follows that k weighings allow us to identify thecounterfeit coin from among 2k + 1 coins.

Returning to the original problem, let’s denote by f(n) the unknownnumber of coins. Suppose that in the first weighing, there were s coins oneither pan. If the balance did not come to equilibrium, then we must lookfor the counterfeit coin among s coins each of which can be used only once,and we are permitted to make n − 1 weighings. As we have seen above,s ≤ 2(n − 1) + 1 = 2n − 1.

If the balance is in equilibrium, then we obtain the original problem forthe f(n)−2s coins that were not weighed the first time, with n−1 weighingsallowed; thus,

f(n) − 2s ≤ f(n − 1).

Hencef(n) ≤ f(n − 1) + 2s ≤ f(n − 1) + 2(2n − 1).

It follows that

f(n) ≤ 2(2n − 1) + 2(2n − 3) + · · · + 2 · 3 + f(1).

It is easy to see that f(1) = 3; therefore, by the formula for the sum ofarithmetic progression, we have f(n) ≤ 2n2 + 1.

On the other hand, if we have 2n2 + 1 coins and each time we take the Translation said “See also

Fact “ref{fact:A7}”, butno such label exists.

Which fact is meant?

(Original has no suchreference.)

largest possible s, that is, s = 2n − 1 at the first step, s = 2n − 3 at thesecond step, and so on, then the expert will be able to find the counterfeitcoin. Thus, f(n) = 2n2 + 1.

Level B

Problem 1. Let a, b, and c be the lengths of the sides of the triangle, witha = 1

3(b + c). Since the smallest angle of a triangle is opposite the shortestside, it suffices to prove that a is the smallest of the side lengths. That is,we want to show that

a =b + c

3< b (1)


(and also that a < c, but since b and c play the same role, it’s enough toprove one inequality). Multiplying both sides of the inequality by 3 andcanceling out b, we can rewrite (1) in the form c < 2b.

By the triangle inequality, a + b > c; therefore, 13(b + c) + b > c, that is,

2b > c, which proves the inequality a < b.page 195

Problem 2. Denote by m1, m2, . . . , m9 the weights of the pieces, all inincreasing order. Put the pieces of weights m1, m3, m5, and m7 on the left,and those of weights m2, m4, m6, and m8 on the right. Then

m1 + m3 + m5 + m7 < m2 + m4 + m6 + m8.

But if we add the heaviest piece on the left, the balance tips the other way:

m1 + m3 + m5 + m7 + m9 > m2 + m4 + m6 + m8.

So moving some part of the heaviest piece to the right will achieve equality.

Problem 3. The sum of the angles of any hexagon is 720◦; this can beshown, for instance, by cutting the hexagon into two quadrilaterals. Hence

\A + \B + \C = \A1 + \B1 + \C1 = 360◦. (1)

The area of the hexagon is the sum of the areas of triangles AB1C,BCA1, AC1B, and ABC (see figure). Therefore we need to show that thearea of △ABC is equal to the sum of the areas of triangles AB1C, BCA1,and AC1B.

A

B

C

A1

B1

C1

A′

It turns out that these three triangles can be puttogether to form a copy of triangle ABC. To seethis, we first turn the triangle AB1C about point Cso that its vertex B1 goes to point A1; this is possiblesince CA1 = CB1. Suppose that the point A is takenunder this rotation to A′. Translation said “see

Fact “ref{fact:G5}”, but

no such label exists.

Original has no suchreference.

We have \A′A1B = 360◦ − \A′A1C − \BA1C =360◦−\CB1A−\CA1B = \AC1B (we have used (1)

and the fact that rotations preserve angles). Now it is clearthat triangles A′A1B and AC1B are congruent by the SAS property.

It remains to prove that triangles A′BC and ABC are congruent. Itfollows from the congruence of triangles A′A1B and AC1B that A′B = AB.So triangles A′BC and ABC are congruent by the SSS property.

page 196

Problem 4. For simplicity, we think of the trains and stations as points(or we can consider corresponding points in each train and each station).

Clearly, if n is divisible by 3, Ira and Lyosha always leave simultaneously,so there are no underground portions. Suppose that n is not divisible by 3.Then, no matter when they come to their stations, either Ira or Lyosha willtake the train first.

Denote by l the length of the interval between trains. If a certain pointX is underground, any point Y whose distance from X is a multiple of l is

142 SOLUTIONS

underground as well. Indeed, if Ira enters the station when Roma is at X,then Ira will catch a train first. But when Roma is at Y , the arrangementof trains is the same as when Roma is at X, so in this case Ira will also bethe first to catch a train, hence point Y is also underground.

Thus, the underground portions occur in a periodical pattern of periodl, and it’s enough to study the situation within one interval of length l. (Seefigure; “hairy” portions of the line lie underground.)

Consider the moment when a train leaves station B.Suppose the train Ira will take—the nearest train uplinefrom A— is x units away from A at this moment. Thenthe entire interval between this train and point A liesunderground. Indeed, if the train Roma is conducting isin this interval, it will pick up Ira first, because Lyoshahas just missed a train. B

AC

x

Let’s show that the interval of length l − x just down the line from Ais not underground. Indeed, when a train arrives at station A, the nearesttrain upline from B will be at a distance l − x from B. It follows that ifRoma’s train is in the interval of length l−x mentioned above, then Lyoshawill get on a train first, because Ira has just missed a train.

Thus, in an interval of length l, a portion of length x lies undergroundand the rest, of length l − x, above ground. Since the pattern is periodic,the ratio of underground to above-ground portions along the entire road is page 197

the same, x to l − x.We still have to find x. Clearly, the value of x is equal to the remainder

of the division of the distance BA by l. If D is the total length of the track,the distance BA is 2

3D and l is equal to D/n, so x = {2n/3}D, wherethe braces denote the fractional part of a number. Thus, if n yields theremainder 1 upon division by 3, then x = 2D/3, and if the remainder is 2,then x = D/3.

Problem 5. We divide the participants of the tournament into two groups:those whose second score was higher than the first and those whose firstscore was higher. At least one of these groups has n players or more. Wecan assume, for instance, that the first group does so, and that it consistsof x players, where x ≥ n.

Suppose that the total score of the first group in the second tournamentis D points more than in the first. Then, by assumption,

D ≥ xn. (1)

These x players increased their score at the expense of the other 2n − xplayers, since matches inside one group always contribute the same totalnumber of points, 1

2x(x − 1), to the intra-group total score. Each matchwith a player from the other group can add at most one point to the totalscore; hence

D ≤ x(2n − x). (2)


Comparing inequalities (1) and (2), we obtain 2n − x ≥ n. If inequality (1)and/or (2) were strict, we would have 2n−x > n, contrary to the assumptionx ≥ n. Therefore, x = n and D = n · n, and so the players from the firstgroup increased their personal scores by exactly n points each.

The same argument applies to the second group, because it also consistsof n players. page 198

Remark. It follows from the solution that the situation in the problem can onlyoccur if the participants fall into into two groups of n players each, such thateveryone in the first group beat everyone in the second group in the first tour-nament, and the second group beat everyone in the first group in the secondtournament.

Problem 6. Suppose that

F (x) = a0 + a1x + a2x2 + · · · and G(x) = b0 + b1x + b2x

2 + · · · .

By assumption,

(a0 + a1x + a2x2 + · · · )(b0 + b1x + b2x

2 + · · · ) = 1 + x + x2 + · · · + xn.

The constant term of a product of polynomials is equal to the product oftheir constant terms, so a0 ·b0 = 1. Hence a0 = 1 and b0 = 1. The coefficientof the first power of x in the product is a0b1+a1b0. Since it equals 1, we haveeither a1 = 1, b1 = 0 or a1 = 0, b1 = 1. Assume without loss of generalitythat a1 = 1 and b1 = 0.

If all the coefficients ai of F (x) are equal to 1, the statement of theproblem is true. If at least one of them is zero, there is some m for which

a0 = a1 = · · · = am−1 = 1, am = 0.

We want to prove that the polynomial F (x) can be represented in the form(1 + x + x2 + · · · + xm−1)T (x), where T is also a 0-1 polynomial.

If some coefficient bl is equal to one, where 1 ≤ l < m, we immediatelysee that the coefficient of xl greater than 1, because the term xl of theproduct can be obtained by multiplying both a0 by blx

l or al by b0xl. Hence

bl = 0 for 1 ≤ l < m. Now it is clear that bm = 1: otherwise the coefficientof xm would be zero.

The sequence of coefficients a0, a1, . . . can be viewed as a sequence ofalternating intervals: first a interval of ones, then a interval of zeros, thenagain ones and so on. Consider one of these intervals: ar = · · · = ar+s−1 = 1with ar−1 = 0, ar+s = 0. The length s of this interval cannot be greaterthan m: otherwise the term xr+m of the product could be obtained both bymultiplying arx

r by bmxm and by multiplying ar+mxr+m by b0. page 199

We prove that the length of such a interval cannot be less than m.Suppose that this is not true and consider the leftmost interval ar, . . . , ar+s−1

of ones surrounded by zeros whose length s is less than m.Let’s examine the computation of the term xr+s in the product F (x) ·

G(x). Since ar+s = 0, it must appear as the product of some term of theform auxu and some term bvx

v. (Of course, u + v = r + s.)

144 SOLUTIONS

It is not difficult to see that if u 6= 0, then au−1 = 0; otherwise wecould obtain xr+s−1 in two ways: au−1x

u−1 · bvxv = ar+s−1x

r+s−1 · 1. Thecase u = 0 is similar. Thus there must be a interval of ones of the formau, . . . , au+m−1. (Its length is m, because it lies to the left of the intervalar, . . . , ar+s−1.)

But the number r + m − v = r + m − (r + s − u) = u + (m − s) lieson the interval u, . . . , u + m − 1, whence ar+m−v = 1, and we arrive at acontradiction: arx

r · bmxm = ar+m−vxr+m−v · bvx

v.Consequently, all intervals of ones have the same length m. Long division

(see Fact 19) convinces us that the polynomial F can be represented as theproduct of the polynomial 1 + x + · · · + xm−1 by a polynomial all of whosecoefficients are zeros or ones.

Remark. It is easier to understand this solution— and especially to come up withit— by drawing intervals on a line and studying what happens when they areshifted. Geometrically speaking, the problem can be formulated as follows:

Suppose a interval of length L can be covered by translates of a certainunion S of disjoint little intervals contained in L (so each point of the intervalis covered by one of the little intervals and these little intervals do not overlap,but only touch one another at their endpoints). Then all the intervals in Shave the same length l, and, of course, the quotient L/l is an integer.

See how simple and beautiful the solution becomes when translated intogeometric language!

Level C

Problem 1. First solution. We take cartesian coordinates in space and callα, β, and γ the coordinate planes, with equations x = 0, y = 0, and y = 0,respectively. Consider the ball B defined by page 200

x2 + y2 + z2 ≤ 1.

Its projection on the plane α is a disk of radius 1 with center at the origin.The set of points whose projections hit this disk is a cylinder, denoted by C1;it is defined by the inequality Original said x2 + y2 ≤ 1,

in conflict with the

definition of α.y2 + z2 ≤ 1.

In a similar way we define the cylinders C2 and C3 as the sets of pointswhose projections onto the planes β and γ fall inside the respective unitdisks centered at the origin.

Let C be the intersection of C1, C2, and C3. We will show that Csatisfies the conditions of the problem. We know that C is convex, beingthe intersection of convex sets.

Let’s show that the projections of C on the planes α, β, and γ are disks.For instance, consider the plane α. The projection of C on α lies in theunit disk of α, by construction. But the unit disk in α is itself contained inC, because it projects onto the two other planes inside the respective unit


disks (the projections are diameters). Thus the projection of the solid C onα both contains and is contained in the unit disk; therefore, they coincide.

Next, C is not a ball. Indeed, the point (a, a, a) lies in C1, C2, and C3

if 2a2 < 1, and it lies outside of the unit ball B centered at the origin if3a2 > 1; clearly there are values of a satisfying both conditions. One lastthing needs to be checked, though: could C be some other ball? The answeris no, because a ball having a different size or center than the unit ball Bcentered at the origin would have a different projection than B (and hence page 201

than C) in at least two coordinate planes.

Second solution. Consider a ball and its projections on the three planes,which are circles. Take any point A on the sphere (the boundary of theball) that does not project to the circumference of any of the three circles;in effect A can be any point of the sphere not lying on a coordinate plane.

Now take any plane intersecting the ball but so close to A that theintersection is entirely contained in the same octant of space as A. (Forinstance, if the three coordinates of A are positive, the octant it defines isthe set of points whose three coordinates are positive.) If we lop off the littlespherical cap on one side of this plane, what’s left of the ball is no longera ball, but it has the same projections on the coordinate planes as the ball!This set is convex, being the intersection of a ball and a half-space.

Problem 2. Translate the quadrilateral ABCD bythe vector

−→AC (see figure). The image is a quadrilat-

eral A′B′C ′D′, where A′ = C, and the quadrilateralBB′D′D is a parallelogram, because the segmentsBD and B′D′ are parallel and congruent. Let A0,B0, C0, and D0 be the midpoints of segments BD,BB′, B′D′, and D′D, respectively.

AB

CD B′

C′D′

A0

B0

C0D0

We claim that A0B0C0D0 is a parallelogram whose diagonals form thesame angle and have the same lengths as the diagonals of the quadrilateralABCD. The statement about the lengths follows from the fact that thesegments A0B0 and C0D0 are the midlines of triangles B′BD and B′D′D,respectively. The statement about the angles is true because the segmentsB0D0 and BD are parallel and congruent, and so are A0C0 and AC.

Therefore, it remains to prove that the perimeter of quadrilateral ABCDis not less than the perimeter of parallelogram A0B0C0D0. But the perime-ter of the parallelogram is equal to B′D +BD′ (by the property of triangle’smidlines). By the triangle inequality, BC + CD′ ≥ BD′ and B′C + CD ≥B′D. The proof is completed by adding these inequalities.

Problem 3. We go directly to part (b), since (a) is a particular case of it.Obviously, our construction applied to a regular polygon A1A2 . . . An yieldsa regular polygon B1B2 . . . Bn. (To justify this rigorously, notice that thepolygon A1A2 . . . An maps to itself under a 2π/n rotation; hence the polygonB1B2 . . . Bn also maps to itself under the same rotation.) page 202

146 SOLUTIONS

Since all regular n-gons are similar, any of them can be obtained by theextension procedure from some other regular n-gon. It remains to prove thatthe polygon A1A2 . . . An is uniquely determined by the polygon B1B2 . . . Bn.

First solution. We prove by induction (see Fact 24) a more general state-ment: Let A1A2 . . . An be a polygon. Let B1 be the point on the ray A1A2

beyond the point A2 such that A1B1/A1A2 = α1; let B2 be the point on theray A2A3 beyond the point A3 such that A2B2/A2A3 = α2; and so on. Thenthe points A1A2 . . . An are uniquely determined by the polygon B1B2 . . . Bn

and the coefficients α1, α2, . . . , αn.

Base of the induction (n = 3). Let B′ be the intersection point ofthe line A1A2 with the segment B2B3 (see figure below and to the left). Wecan find the ratio B2B

′/B′B3 using Menelaus’ theorem (see remark below),because we know the ratios A2B2/A2A3 and A3B3/A3A1. Therefore, thepoint B′, and hence the line A1A2, are uniquely determined. Similarly, wecan determine the lines A2A3 and A3A1; therefore the triangle A1A2A3 isunique.

A1

A2

A3

B1

B2B3 B′

A1

A2

A3

An

B1

B2

B3

Bn−1

Bn

B′

page 203

Induction step. Let the line AnA2 intersect the segment B1Bn at B′

(see figure above and to the right). By Menelaus’ theorem, ifwe know theratios AnBn/AnA1 and A1B1/A1A2, we can find the ratio B1B

′/B′Bn. Thisdetermines B′ uniquely. Another application of Menelaus’ theorem allows usto find the ratio AnB′/AnA2. The induction conjecture applied to the poly-gons A2A3 . . . An and B′B2 . . . Bn−1 shows that the points A2, A3, . . . , An

are determined uniquely. Now it is not difficult to determine the point A1.

Remark. Menelaus’ Theorem. Let A′, B′, C ′ be points on the sides BC, AC,AB of a triangle ABC or on their extensions. The points A′, B′ and C ′ arecollinear if and only if

BA′

CA′· CB′

AB′· AC ′

BC ′= 1 (1)

and either all the points lie on the side extensions or only one of them does.We prove the implication we actually used above: collinearity implies (1).

Suppose that the points A′, B′, and C ′ lie on the same line. Projection to aperpendicular line maps these points to the same point. Denote this point byP and the projections of A, B, and C by A0, B0, and C0. Then

BA′

CA′· CB′

AB′· AC ′

BC ′=

B0P

C0P· C0P

A0P· A0P

B0P= 1.


Second solution. Suppose the polygons B1B2 . . . Bn and A1A2 . . . An are re-lated as in the problem. Then A2 is the midpoint of A1 and B1, a factthat we can express by the equation A2 = 1

2A1 + 12B1. Similarly we have

A3 = 12A2 + 1

2B2 = 14A1 + 1

4B1 + 12B2. We interpret these sums as “weighted

averages” of points on the plane, which make sense as long as the weightsadd up to 1. (Such averages are also called affine combinations.)

Continuing in this way we eventually get

An = 12An−1 + 1

2Bn−1 = 14An−2 + 1

4Bn−2 + 12Bn−1 = · · ·

=1

2n−1 A1 +1

2n−1B1 +

1

2n−2B2 + · · · + 1

4Bn−2 +

12Bn−1

and finally, since A1 is also the midopint of An and Bn,

A1 =12An +

12Bn = · · · =

12n A1 +

1

2nB1 +

1

2n−1B2 + · · · + 1

4Bn−1 +

12Bn.

In this last equation we have A1 on one side and a weighted averageof A1, B1, . . . , Bn on the other. This implies that A1 is a weighted averageof B1, . . . , Bn alone, with known coefficients. (Formally we can treat theequation A1 = 2−nA1+2−nB1+ · · ·+ 1

2Bn as if A1 and the Bi were numbers,subtracting 2−nA1 from both sides and then multiplying by the right numberto adjust the weights back to 1.) And since now A1 is uniquely determinedby the Bi, so are A2, A3, . . . , An, as we wished to show.

Remark. A slick way of phrasing the same argument in a different language is this:Place weights of masses 1, 2, . . . , 2n at the points B1, B2, . . . , Bn. Then A1 is the(uniquely determined) center of mass of this system, for the following reason: page 204

Let’s modify the system by placing an additional weight of mass 1 at A1.Now the center of mass of the weights at A1 and B1 is at A2, so we can replacethe unit weights at A1 and B1 by a double weight at A2. Next we replacethe weights at A2 and B2 by a weight of mass 4 at A3. We continue in thisway until we end up with a single weight, located at A1, which is therefore thecenter of mass of the modify system. Going back to the original system involvessubtracting the auxiliary weight at A1; but adding or removing weights at thecenter of mass cannot change the center of mass.

Third solution. The homothety (scaling) with center B1 and scale factor12 takes the point A1 to A2. Similarly, the homothety with center B2 andfactor 1

2 takes A2 to A3, and so on. Finally, the homothety with center Bn

and factor 12 takes An to A1.

Thus the composition of these n homotheties fixes the point A1. Buta composition of homotheties is also a homothety, and a homothety leavesexactly one point fixed (unless it fixes every point; but this cannot be thecase because our composition has scale factor 1

2 · 12 · 1

2 · · · = 12n ). Hence the

point A1 is defined uniquely.

Problem 4. The expressions in the problem occur in the relations betweena cubic polynomial and its roots. Let’s recall these relations, which are

148 SOLUTIONS

the degree-three case of Vieta’s formulas (see also Fact 20). A polynomialP (x) = (x−a1)(x−a2)(x−a3) becomes, upon expanding the product,

P (x) = x3 − (a1+a2+a3)x2 + (a1a2+a1a3+a2a3)x − a1a2a3.

A similar expression can be written for Q(x) = (x−b1)(x−b2)(x−b3). Theequalities in the problem mean that these polynomials differ only in theconstant term. Therefore, the graph of one polynomial is obtained from thegraph of the other by a vertical translation.

For x ≤ b1, we have Q(x) ≤ 0, because each of the three factors in theexpression for Q(x) is negative or zero, and there are an odd number offactors. Thus, Q(a1) ≤ 0 and P (a1) = 0. Therefore the graph of y = Q(x)is obtained from the graph of y = P (x) by a downward translation, or thetwo coincide. In particular, Q(a3) ≤ P (a3) = 0. But for x > b3 we haveQ(x) > 0. Consequently, a3 ≤ b3.

Problem 5. Suppose that the scores of competitors are not the same.Let M be the highest score and m the lowest score. We call players whosescore is M leaders, and those whose score is m trailers.

The superscore of a leader is at least Mm, being the sum of M numbers,each of which is at least m. For an analogous reason, the superscore of atrailer is at most Mm. page 205

Therefore, if the superscores are all the same, leaders and trailers allhave a superscore of Mm. Several stunning conclusions follow:

• All the wins of a leader were against a trailer. Otherwise, the leader’ssuperscore would be strictly more than Mm.

• There is only one leader. For if there were more, one would have beatthe other, contradicting the previous item.

• All the wins of a trailer were against a leader. Otherwise, the trailer’ssuperscore would be strictly less than Mm.

• There is only one trailer. Otherwise, contradiction with the previousitem.

• M = 1 and m = 0, since there was only one match between a leaderand a trailer, and only such matches can contribute to M and m, as aconsequence of the “All the wins” statements.

• There were only two players in the tournament. For a third playerwould have beat both the leader and the trailer (again because of the“All the wins” statements). Therefore there would be a score greaterthan M , contrary to assumption.

Hence, if there are more than two players and the superscores are all thesame, the scores must also be all the same.

Remark. For the solution to another problem involving tournament scores, seeProblem 5 on page 142.

Problem 6. It is enough to show that any initial segment of the sequenceof first digits of powers of 5 occurs in reverse in the sequence of first digitsof powers of 2.


This would be very easy if the sequence of powers of 2 were extended tothe left with the first nonzero digit of each negative power. That’s becausethe first nonzero digit of 1/2n is the same as the leftmost digit of 5n.

So the problem is to find among positive powers of 2 the same sequenceof starting digits as among negative powers of 2. If some power of 2 wereequal to some power of 10 this would be easy, because the sequence of firstdigits would be periodic (if 2i =10j with i, j > 0, the decimal representationof 2n+i would be the same as that of 2n followed by j zeros).

This is too much to hope for (why?), but it points us in the right direc-tion: Since we only need to find matches for sequences of finite length, itsuffices to show that for any k there exists a power of two, x = 2n, whosedecimal representation is of the form

100 . . . 0︸︷︷︸k zeros

∗, (1)

where ∗ stands for an arbitrary strings of digits. If this is so, we will have page 206

2n−1 = 2n/2 = 500 . . . 0∗, then 2n−2 = 250 . . . 0∗, then 2n−3 = 1250 . . . 0∗,etc. To be more precise, write x = 10N + y, with y < 10N−k; then

2n−l =x

2l=

10N + y

2l= 10N−l5l +

y

2l;

so the first digit of 2n−l coincides with the first digit of 5l for l < k.The existence of a power of two of the form (1) can be proved using a

standard fact about irrational numbers (see Remark 1 below). But we givehere an alternate, explicit construction.

Since there are finitely many k+1-tuples of digits and infinitely manypowers of 2, we can certainly find two powers of 2 sharing the same firstk +1 digits, say 2a and 2b, with a > b. We take the quotient 2a−b and applythe following observation:

Lemma. Suppose A and B, with A > B, are positive integers that have the I’ve tightened theargument to account for

(a) the case when B has

no “next digit” – which

cannot be discounted apriori, even among powers

of 2; and (b) the case

when p− q < k, whose

non-occurrence amongpowers of 2 requires a bit

of justification.

same first k+1 digits. Consider the first place, counting from the left, wherethey have different digits. If A has the larger digit there, or if B only hask+1 digits, the fraction A/B has an infinite decimal representation beginningwith a one and k zeros. If B has the larger digit, the decimal representationof A/B begins with k nines.

Proof. Denote by D the number formed by the first k+1 digits of A and B.Suppose that A is a p-digit number and B is a q-digit number, with p ≥ q.Then A = 10p−k−1D + α and B = 10q−k−1D + β, where α < 10p−k−1,β < 10q−k−1. We have

∣∣∣AB

− 10p−q∣∣∣ =

∣∣∣∣10p−k−1D + α

10q−k−1D + β− 10p−q

∣∣∣∣ =|α − 10p−qβ|

10q−k−1D + β

<10p−k−1

10q−1= 10p−q−k. (2)

150 SOLUTIONS

If A has the larger digit in the first different place or if B only has k+1 digits(in which case β = 0), then α ≥ 10p−qβ, which means that all expressionsin (2) enclosed in absolute value bars are nonnegative, and the inequalitiesabove can be rewritten as

10p−q ≤ A/B < 10p−q + 10p−q−k.

This means that A/B begins with a one and k zeros (some of which may lie page 207

to right of the decimal point, if p − q − k < 0).The case where B has the larger digit in the first different place is treated

similarly. ˜

Our quotient 2a−b, with a > b, has to be an integer, and it cannot be apower of 10 (and if it could, this would only help, as already seen). So thek zeros or nines are all in the integer decimal representation of 2a−b.

So far we have shown that for a given k we can find a power of 2 startingeither with a 1 and k zeros or with k nines. In the first case, the problemis solved. In the second case, let N be the number of digits of a power of2 starting with k nines; note that N > k, since a power of 2 cannot end in9. Applying the italicized statement with N instead of k we obtain a newpower of 2 that either begins with a 1 and N zeros (and then the problem issolved) or begins with N nines, which is more nines than the earlier powerof two. In the latter case, by the lemma, the quotient of this power of 2 bythe earlier one is a power of 2 beginning with a 1 and k−1 zeros. So againin this case the problem is solved (for k−1 rather than k, but this does notmatter since we could just start with a larger k).

Remarks. 1. Finding a power of 2 of the form (1) is the same as finding i, j > 0such that 0 ≤ 2i − 10j < 10j−k, or equivalently

1 ≤ 2i

10j< 1 +

1

10k. (3)

Taking base-10 logarithms, these inequalities become 0 ≤ i log10 2 − j < ε,where we denote by ε the number log10(1 + 10−k). Finally, this inequality canbe rewritten in the form {i log10 2} < ε, where the braces denote the fractionalpart of a number. page 208

So we must look at the sequence of fractional parts ai = {i log10 2}. It isconvenient to think of the fractional part of a number as a point on a circle ofunit length. (This circle can also be seen as the interval [0, 1] with endpointsidentified. See the solution to Problem 4 on page 69 for a similar situation.)On the circle, each point ai is separated from the next one, ai+1, by an arc ofthe same length α, always in the same direction. Here α = log10 2, but thediscussion below holds for any α.

If α is a rational number, written in lowest terms as p/q, the points ai forma periodic sequence of period q. (Why?) On the circle they occupy the verticesof a regular q-gon. We already know that this cannot happen in our case, sincethe ≤ in (3) cannot be an equality. Thus log10 2 is irrational.


Now, for α irrational it is an easily proved theorem (see below) that thesequence ai = {iα} will eventually land arbitrarily close to any point on thecircle. In particular, for α = log10 2, there is some i such that ai = {iα} < ε.

2. Theorem (Density of irrational orbits). If α > 0 is irrational, the pointsai = {iα}, for i ≥ 1, form an everywhere dense sequence in the circle of unitlength or in the interval [0, 1]. That is, given a point x ∈ [0, 1] and a positivenumber ε, no matter how small, there exists some i such that |iα − x| ≤ ε.

Proof. If α is irrational, all the ai are distinct. (Why?) Take an integer N with I’ve turned the “proofsketch” into a proof since

it does not take that much

space and it’s a good

example of rigorousargumentation with

fractional parts.

1/N < ε, and divide the circle into intervals [0,1/N), [1/N,2/N), . . . , [(N−1)/N,1).Since the ai are all distinct and there are infinitely many of them, we can find twoin the same interval, say am and an, with n > m. Set k = n − m.

Assume first that an > am. Then ak = {kα} = {nα − mα} = {an − am} =an −am < 1/N . From the sequence of numbers 0, ak, 2ak, . . . , take the largest thatdoes not exceed x, say jak. Then x − jak < (j+1)ak − jak = 1/N < ε. Thus wehave found a point in the sequence ai that’s within the required distance from x.(Note that jak = ajk since jak ≤ x < 1.)

The case an < am can be turned into the previous case by replacing α by 1−αand x by {1 − x}; this replaces every ai by 1 − ai. (Why?) ˜

Level D

Problem 1. One formula for the area of a triangle ABC ia AB·BC ·CA/4R,where R is the circumradius. So what we must show is that the ratio of theareas of triangles A′B′C ′ and ABC is equal to

AB′ · BC ′ · CA′ + AC ′ · CB′ · BA′

AB · BC · CA.

Define x = AB′/AB, y = BC ′/BC, z = CA′/CA. Then it is easy to computethe ratios of the areas of the triangles AB′C ′, A′BC ′, and A′B′C to the areaof the triangle ABC. They are equal to x(1 − y), y(1 − z), and z(1 − x).Therefore, the solution reduces to checking the identity

1 − x(1 − y) − y(1 − z) − z(1 − x) = xyz + (1 − x)(1 − y)(1 − z).

Multiply out and see!

Problem 2. We have∫ π/2

0cos2(cos x) + sin2(sinx) dx =

∫ π/2

0cos2(cosx) dx +

∫ π/2

0sin2(sinx) dx.

Perform the substitution (see Fact 28) y =π2− x in the second integral; page 209

then dy = −dx, and

∫ π/2

0sin2(sinx) dx =

∫ 0

π/2sin2

(sin

(π2− y

))(−dy) =

∫ π/2

0sin2(cos y) dy.

152 SOLUTIONS

Since cos2(cosx) + sin2(cos x) = 1, we have

∫ π/2

0cos2(cosx) + sin2(sinx) dx =

∫ π/2

0cos2(cos x) dx +

∫ π/2

0sin2(cos x) dx

=

∫ π/2

0dx =

π2.

Problem 3. Notice that f2(x)−f3(x) = 2x−1; we can add 1 and multiplythe expression 2x thus obtained by frac12. This yields the function x.Subtracting it from f1(x), we obtain

1

x= f1(x) − 1

2(f2(x) − f3(x) + 1).

We turn to the proofs of impossibility. Since we are not allowed to divide,it is impossible to express 1/x only in terms of the functions f2 and f3: wecannot obtain x in the denominator. In other words, the addition andmultiplication of polynomials yields only polynomials, whereas the function1/x is not a polynomial.

The proof that we cannot do without the function f2 is more interesting.Let’s compute the derivatives of f1 and f3 at the point x = 1. Both thesederivatives are equal to zero. If two functions have zero derivatives at thepoint 1, then the derivatives of both their sum and product is also zero. Thelatter follows from the computation

(fg)′(1) = f ′(1)g(1) + f(1)g′(1) = 0.

Thus, all the functions that can be obtained by our operations from thefunctions f1 and f3 have zero derivatives at point 1. And the derivative of1/x at this point is nonzero.

Only one Olympiad competitor managed to solve this part of the problem.He considered the series expansions of f1 and f3 in terms of x−1 and pointedout that both expansions start with (x− 1)2, while that of 1/x has a linearterm. This is equivalent to the argument above.

It remains to prove that we cannot do without the function f3. We willgive three proofs.

First solution. Any function obtained from f1 and f2 will clearly be of page 210

the form f(x)/xn, where n ≥ 0 and f(x) is a polynomial. Such functionsare called Laurent polynomials. (Warning: a Laurent polynomials may notbe a polynomial!)

By multiplying top and bottom by x if necessary, any Laurent polyno-mial can be written in the form f(x)/xn with n even. Let A be the setof Laurent polynomials f(x)/x2k such that the polynomial f(x) − f(−x) isdivisible by x2 + 1.

We know that f1 ∈ A, because f1(x) = f(x)/x2, where f(x) = x(x2+1),and so f(x)− f(−x) = 2x(x2 + 1) is divisible by x2 + 1. It is even easier tocheck that f2 ∈ A: in this case f(x) = x and f(x)−f(−x) = 0. On the other


hand, 1/x is not in A, because for f(x) = x the function f(x)− f(−x) = 2xis not divisible by x2 + 1.

It remains to show that the sum and the product of Laurent polynomialsin A belong to A as well. The statement about the sum is left as an exerciseto the reader. Consider two Laurent polynomials f(x)/x2k and g(x)/x2l inthe set A. Then

f(x)

x2k· g(x)

x2l=

f(x)g(x)

x2(k+l),

and so f(x)g(x)−f(−x)g(−x) = f(x)(g(x)−g(−x)

)+g(−x)

(f(x)−f(−x)

)

is divisible by x2 + 1.

Remark. An alert reader may have noticed that the definition of A is a bit fuzzy, inthat membership in A might depend on how an element is expressed. That is,could we have f1(x)/x2k1 = f2(x)/x2k2 , yet f1(x)−f1(−x) is divisible by x2+1,while f2(x)− f2(−x) is not? The answer is no, because f1 and f2 are obtainedfrom one another only by multiplication by a power of x, and therefore thesame is true about f1(x) − f1(−x) and f2(x) − f2(−x). But the factorizationof a polynomial is unique (see Fact 21), so xnf(x) is divisible by x2 + 1 if andonly if f(x) is.

Second solution. (Uses complex numbers —see Remark 1 below.) The func-tions f1, f2 and 1/x are defined for all nonzero complex numbers. Let’s look page 211

at their values at the point x = i =√−1. We have f1(i) = i + 1

i = i− i = 0

and f2(i) = i2 = −1, which are real numbers. Therefore, the value of anyfunction obtained from f1 and f2 at the point i will be real. But the valueof 1/x at the same point is 1/i = −i, which cannot be expressed in terms ofreal numbers via multiplication and addition.

Remarks. 1. Complex numbers are extremely important in mathematics, physicsand most other sciences. They are a generalization of the ordinary (“real”)numbers, just as real numbers are a generalization of the integers.

To work with complex numbers we include among our numbers one whosesquare is −1. It is denoted i, or

√−1. Then, for addition and multiplication to

make sense, we need to include also all the numbers of the form a + bi, where aand b are real. (For b = 0 this formula yields ordinary real numbers a, showingthat real numbers are a particular case of complex numbers.)

Complex numbers can be added, subtracted, multiplied, and divided (if thedivisor is not zero). These operations satisfy the same properties— commuta-tivity, associativity, distributivity— as the corresponding ones on real numbers.The sum of two complex numbers a + bi and c + di is the complex number

(a+c) + (b+d)i.

The product can be found using the properties listed above, including the defin-ing property i2 = −1:

(a+bi)(c+di) = ac + adi + bci + bdi2 = (ac−bd) + (ad+bc)i.

Every polynomial has a root among the complex numbers. This is knownas the fundamental theorem of algebra. Complex numbers have many otherbeautiful properties and applications. Googling the words “complex numbersintroduction” brings up several good, well-explained sites.

154 SOLUTIONS

2. Our second solution works if we interpret “arbitrary number” in the statementof the problem to mean “arbitrary real number”. But if we’re using complexnumbers, it’s natural to ask: what if we allow multiplication by, and additionof, complex numbers? The answer is still that 1/x cannot be expressed in termsof f1 and f2. Here’s one reason: f1(i) = f1(−i) and f2(i) = f2(−i). Therefore,any function obtained from f1 and f2 takes the same value at the points i and−i; but this is not true of the function 1/x.

3. There is a deep connection between this problem and the vast and beautiful areaof mathematics known as algebraic geometry. We will say some words about it, I expanded this a bit to

give the reader some

chance to understand

what’s going on.

even though it goes far beyond the high-school curriculum.Consider all the functions that can be obtained from the single function x

by the allowed operations, including multiplication by and addition of a complexnumber. We can get only polynomials, and we get all polynomials in this way.They form a set denoted by C[x], where C stands for the complex numbers. Thisset is a ring — more exactly, a commutative and associative ring with identity—since it has an addition and a multiplication operation with these properties.

If we start from the function x2 instead of x, we obtain the ring C[x2] ofpolynomials with even-degree terms only. This is a subring of C[x], and it isdistinct from C[x] since x (for example) is in C[x] but not in C[x2].

Next, consider all functions that can be obtained from x and 1/x by theallowed operations. We get another ring, the ring of all Laurent polynomials;we denote it by C[x, x−1]. Any function in this ring is nicely behaved in C∗, thecomplex plane minus the origin, just as any polynomial is nicely behaved in allof C. Note that C[x] is a subring of C[x, x−1], and the two do not coincide.

More generally, take some finite number of Laurent polynomials, f1, . . . , fn,to serve as seeds. The functions that can be obtained from these generators(seeds) by the allowed operations form a subring of the ring of all Laurentpolynomials; this subring— let’s call it A— may or may not be all of C[x, x−1].

It can be shown that A is the ring of nicely behaved functions (algebraicfunctions) on some space X called an algebraic curve. The full ring of Laurentpolynomials is the ring of algebraic functions on C∗ = C \ 0.

The inclusion A ⊂ C[x, x−1] corresponds to a algebraic map (or morphism)of curves φ : C∗ → X. The ring A equals all of C[x, x−1] if and only if this mapis an isomorphism, that is, it has an inverse and the inverse is also an algebraicmap.

There are three reasons why φ might not be an isomorphism: φ is notsurjective, φ is not injective, or the derivative of φ at certain points is 0.

If A = C[x] is the ring of functions generated by f2 and f3 in the problem,the first reason takes place: A is the set of algebraic functions on the spaceX = C, and φ : C∗ → C is the inclusion, which is not surjective because 0 isnot in its image.

If we start from f1 and f2, the curve X can be taken as the set of (y, z) ∈ C2

such that y2z−(1+z)2 = 0, and the map φ takes x ∈ C∗ to the pair (y, z) definedby y = x + 1/x and z = x2 — that is, y = f1(x) and z = f2(x). (Note thatthe polynomial defining X vanishes identically when these values of y and z aresubstituted.) Since the points i and −i in C∗ have the same image in the curveX (check it out!), the map φ is not surjective and so is not an isomorphism. Wesay that X is a singular curve with an ordinary double point. page 212


Finally, suppose we start from f1 and f3; this time φ takes x ∈ C∗ to thepair (y, z) ∈ C2 defined by y = x + 1/x and z = (x − 1)2. (The curve X hasa complicated equation, y2z − (y + z)2 + 4y − 4 = 0.) The map φ is not anisomorphism because its derivative vanishes at the point 1. We say the algebraiccurve X has a cusp, or pinch point.

Problem 4. Cutting the regular tetrahedron along the planes parallel toits faces and passing through the midpoints of its edges, we partition it intoa regular octahedron and four regular tetrahedrons. Each small tetrahedronis obtained from the big one by a homothety with scale factor 1

2 and centerat a vertex of the big tetrahedron (see figure on the left).

It is a little harder to visualize how to cut a regular octahedron. Asin the case of the tetrahedron, let us shrink the octahedron in half fixingone of its vertices. Repeating this for all six vertices, we obtain six smalloctahedrons. After they are carved out of the big octahedron, we find thatthe remainder consists of eight regular tetrahedrons adjoining the faces ofthe big octahedron. One vertex of each of these tetrahedrons is the center ofthe initial octahedron, and all the others are midpoints of the octahedron’sedges (see figure above and to the right).

After the first step (at which only the tetrahedron is sliced) we obtainan octahedron and tetrahedrons with edges of length 1

2 ; after the secondstep (at which four tetrahedrons and one octahedron are cut) we obtainoctahedrons and tetrahedrons with edges of length 1

4 , and so on. After theseventh step the edge lengths of tetrahedrons and octahedrons has become1

128 , which is less than 1100 .

page 213

Problem 5. Set a = x3, b = y3, c = z3. Then xyz = 3√

abc = 1.For any x and y we have the inequality x2 − xy + y2 ≥ xy (expand the

square (x−y)2). Multiplying both sides of this inequality by x + y, whichis legal since because x > 0 and y > 0, we obtain x3 + y3 ≥ (x + y)xy. Itfollows that

1

1 + a + b=

xyz

xyz + x3 + y3≤ xyz

xyz + (x+y)xy=

z

x + y + z.

156 SOLUTIONS

Similar inequalities can be obtained for1

1+b+cand

11+c+a

. Now just addtogether the three inequalities:

1

1+a+b+

1

1+b+c+

1

1+c+a≤ z

x+y+z+

x

x+y+z+

y

x+y+z= 1.

Problem 6. We start by combining parallel strips so no two directions arethe same. We then fix a point O and assign to each strip either of thetwo vectors at O whose length is the strip’s width and whose direction isperpendicular to the strip’s boundary.

Divide the plane into 12 sectors (angles) of measure 30◦ with vertex O.For each pair of opposite sectors, choose the vectors lying within the twosectors, including their boundaries, and find the sum of their lengths. Thisyields six sums, at least one of which must be 100

6 or more, since their totalis 100.

Choose a pair of opposite sectors whose corresponding sum is at least1006 . Replacing some of these vectors by their opposites, if needed, we can

ensure that all of them fall within the same 30◦ sector.A sum of vectors does not depend on the order of summands. Take the

vectors in the clockwise order of their directions — so the direction of each isobtained from the previous one’s by a clockwiserotation through at most 30◦—and draw themin this order so the head of each vector is thetail of next. This gives a convex polygonal lineOO1O2 . . . On, as in the figure. The length ofthis broken line, as we have mentioned above, isat least 100

6 . O

O1

O2

On−1

On

M

page 214

The distance O0On, where O0 is another name for O, cannot be less than1006 · cos 30◦. That’s because all the component vectors lie within 30◦ of the

sum vector; that is, the length of the projection of each segment Oi−1Oi,where i = 1, . . . , n, onto the line O0On is at least Oi−1Oi · cos 30◦.

Now translate each strip so its edges pass through the endpoints of thecorresponding segment Oi−1Oi, We claim that the polygon MOO1O2 . . . On,where M is the intersection point of the perpendiculars to segments OO1 andOn−1On drawn through the points O and On, respectively, will be completelycovered by the strips.

Indeed, consider any point X in the polygon MOO1O2 . . . On. A simple,but somewhat incomplete, argument is to consider the point Y on the brokenline OO1O2 . . . On closest to X. If Y lies on OiOi+1, then XY ⊥ OiOi+1,and hence X lies in the strip perpendicular to the segment OiOi+1.

Let’s give a longer, but more rigorous, reasoning. Suppose the stripsdo not cover the point X. The angle XOO1 is not obtuse. (Why?) Noris the angle XO1O acute; if it were, the stripperpendicular to OO1 would cover X. It followsthat \XO1O ≥ 90◦, and, since \OO1O2 < 180◦,

O

O1

O2X

\XO1O2 = \OO1O2 − \XO1O < 90◦.


Similarly, one shows that the angle XO2O3 is acute. Continuing byinduction, we finally conclude that if point X is not covered by any of thestrips perpendicular to segments OO1, . . . , On−2On−1, then all the anglesXO1O2, XO2O3, . . . , XOn−1On are acute. But then X lies in the stripperpendicular to the segment On−1On: a contradiction. page 215

Now recall that the angle between the vectors OO1 and On−1On does notexceed 30◦. Therefore, the angles MOOn and MOnO of the triangle MOOn

are at least 60◦. Therefore this triangle contains theequilateral triangle constructed on the side OOn

(see figure). The inradius of a regular triangle withside length a is equal to a/(2

√3). It remains to

check the inequality

60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦60 ◦

O

On

M

1006 · cos 30◦

2√

3> 1,

which is amply satisfied (any number above 24 would do in place of 100).

Year 1998 Olympiad

Level A

Problem 1. One solution is given by the number of months in a nonleapyear that have 28, 30 and 31 days, respectively: 28 · 1 + 30 · 4 + 31 · 7 = 365.Another solution is x = 2, y = 1, z = 9.

Remark. Compare with Problem 1 on page 24.

Problem 2. Let p1, p2, . . . , p8 be different prime numbers. We claim thatthe numbers n1 = p2

1p2 . . . p8, n2 = p1p22 . . . p8, . . . , n8 = p1p2 . . . p2

8 satisfythe desired conditions. Indeed, each of the primes p1, p2, . . . , p8 appears inthe prime factorization of n2

i with a power of at least two. Therefore, n2i is

divisible by each of the nj . To show that ni is not divisible by nj for i 6= j,just note that pj appears in the prime factorization of nj with power 2, butin that of ni with power 1 only. (Here of course we’re using the fact thatthe prime factorization of an integer is unique: Fact 10 on page 104).

page 216

Problem 3. First, given the parallelogram ABCD,there is at most one point M on the line AB with I reorganized this proof to

avoid relying on thefigures in the analysis of

cases, and especiailly to

deal cleanly with the

exceptional cases.

\AMO = \MAD, for the following reason. Since Oand D are on the same side of the line AB, saying that \AMO = \MADis equivalent to saying that MO is parallel to the line AD′ obtained fromAD by reflection in AB (see figure above). But only one parallel to AD′

can go through O. Hence the line MO is unique, and so is M .

D O

MA B

D′

Next, a point is equidistant from C and D if and only if it lies on theperpendicular bisector of the segment CD. We draw this line; it intersectsCD at its midpoint P , and it intersects the line AB atsome point N (see figure). We must show that N = M .

The line OP is a midline of △BDC, because O is themidpoint of the diagonals. Hence OP ‖ BC ‖ AD.Also, the line through O parallel to PN forms arectangle PP ′N ′N , and O is the midpoint of theside P ′N ′, being equidistant from oppositesides of the parallelogram.

O

A B N(=M)N ′

CD PP ′

160 SOLUTIONS

Then, if N is closer to B than to A, as in the figure, we have thefollowing series of equalities: \NAD equals the exterior angle at vertex D ofthe parallelogram (corresponding angles), which equals \DPO (alternatinginternal angles), which equals \ANO (congruence of △OPP ′ and △ONN ′,from the equalities P ′O = ON ′ and P ′P = N ′N). Thus \NAD = \ANO,and N satisfies the defining property of the point M . Since M is unique,we have N = M .

If N is closer to A than to B, we have instead \NAD = \ADC =\CPO = \ANO, so again N must be equal to M . The same figure asbefore illustrates this situation, with the sole change we that switch A withB and C with D.

There are two exceptional cases: When the initial parallelogram is arectangle, the lines PN and P ′N ′ coincide; then \NAD = \ANO becauseboth are right angles. And when N = A, the angles NAD and ANO areundefined; there is, strictly speaking, no point M as in the statement, butstill the point N is characterized by the condition in the first paragraph ofthis solution.

Problem 4. Suppose that k among the numbers a1, a2, . . . , a100 are blueand 100−k are red. Since the blue numbers are written in increasing order,these k numbers are the numbers from 1 through k. Similarly, the 100 − kred numbers are the numbers 100, 99, . . . , k + 1. Therefore, all the numbersfrom 1 to 100 occur among a1, a2, . . . , a100.

Problem 5. If a person X has acquaintances sitting next to each other—in particular, if X is acquainted with either of X’s neighbors —then X isacquainted with all the guests. Let us prove that there is such a guest X. page 217

Consider any two neighbors. If they are acquainted, we have found our X.If they are not, a common acquaintance of theirs can serve as X, since thisperson has two acquaintances who are neighbors.

Having found X, we now observe that X’s neighbors are also acquaintedwith everybody, because they are acquainted with X, who is seated next tothem. The neighbors of these neighbors are also acquainted with everybody,and so on around the table.

Problem 6. Not necessarily. See the figure for thecorresponding counterexample using eight squaresinstead of 100. The example can be describedexplicitly as follows:

A

1

B C

9

D

Denote by A,B,C,D the vertices of the redsquare. Divide the diagonal AC into 100 equalsegments and denote their endpoints, in order, by thenumbers 1, 2, . . . , 101, with point A denoted by 1 andpoint C by 101. For each pair of points k and k + 1(k = 1, 2, . . . , 100), there exist exactly two squares of the given size whosesides are parallel to the sides of the red square and pass through points kand k +1: one of these squares contains B and doesn’t contain D, while the


other contains D but not B. If k is odd, we let the k-th white square be thesquare that contains B, and if k is even, the square that contains D. The100 white squares thus selected completely cover the red square; but if weremove the square whose sides pass through points k and k+1, the segmentof the diagonal between points k and k + 1 will not be covered. (Comparewith Problem 3 on page 24.)

Level B

Problem 1. We have

49 + 610 + 320 = (29)2 + 2 · 29 · 310 + (310)2 = (29 + 310)2.page 218

Remark. One might be tempted to tackle this problem by calculating the remain-ders of the given number upon division by various primes, in search of a smalldivisor. But this will go nowhere, because 29 + 310 is a prime!

Problem 2. Draw the third altitude BS and extend itto meet PQ at T (see figure). Then the rectanglesASTQ and AEKL are of equal area. Indeed,by the similarity of the right triangles ABS andAEC, we have AE/AC = AS/AB, which impliesAE · AB = AS · AC, which implies AE · AL = AS · AQ.We prove in the same way that CSTP and CDMN haveequal area.

A

B

C

DE

K

L

M

N

PQ

S

T

Remark. The equation AE · AB = AS · AC can also be derived from the fact thatits sides are both equal to the power of the point A with respect to the circlepassing through the points B, E, S, and C.

Problem 3. Let x be the fraction of truthful villagers. Imagine that thetruth tellers become liars and the liars become truth tellers. Then the trav-eler will receive exactly the same answers! Indeed, the honesty of each vil-lager flips, but so does the honesty of the neighbor about whom the villageris informing; therefore, the answer remains the same.

Now, after the change, the fraction of honest villagers is 1 − x ratherthan x. Thus, the traveler cannot distinguish between a circle in which theportion of truthful people is x and one in which this portion is 1 − x. Thismeans the traveler can determine x only if x = 1 − x, or if x = 1

2 .

Remark. Number the villagers in clockwise order and set xi = 1 if the i-th villageris a liar, and xi = 0 otherwise. Then the answer of the i-th villager can beexpressed as the value of xi + xi+1, where the addition is performed modulo 2(that is, 0 + 0 = 0, 0 + 1 = 1, 1 + 1 = 0). Therefore, the information receivedby the traveler can be viewed as a system of simultaneous linear equations overthe field with two elements. See Field2elem,SystemLinearEq. CITE

Field2elem,SystemLinearEq

page 219Problem 4. By thinking of the bases as the nodes of a graph whose edgesare the roads (see Fact 3), we see that the set of all bases can be divided

162 SOLUTIONS

into a certain number of connected components. Two bases belong to thesame connected component if and only if it’s possible to go from one to theother using the available roads.

By definition, a set of roads is essential if and only if after its removalthere are at least two connected components.

Initially, there is only connected component. Otherwise the empty setwould be an essential set. It would therefore be the only strategic set. Butthe problem asserts the existence of two different strategic sets.

Lemma. Closing all the roads in a strategic set causes the bases to breakup into exactly two connected components. No road joining two bases withineither of these components can belong to the strategic set.

Proof. Let S be a strategic set and consider the connected componentsobtained by the removal of S. If a road b ∈ S joins two bases in the sameconnected component, restoring it will not change the number of connectedcomponents; that is, the set S \ b still breaks up the bases into more thanone component, so it’s still essential, and S is not strategic.

Now we show that the removal of S can only create two connected com-ponents. Let a be any road in S; as we have seen, it joins two differentconnected components, say X and Y . If we restore the road a, the com-ponents X and Y merge, but all other components are unaffected. So ifthere is a third component the set S \ a is still essential; that is, S is notstrategic. ˜

Let the connected components arising from removing the strategic set Vbe called A and B, and let the ones arising from removing W be C and D.

Set K = A ∩ C, L = A ∩ D, M = B ∩ C, andN =B∩D (see figure). The sets K, L, M , N are pair-wise disjoint and their union is the set of all bases. page 220

We claim that at most one of them is empty. Indeed,if K = L = ?, then A is empty, which is impossible.If K = N = ?, then A = D and B = C; but this con-tradicts the assumption that V and W are differentstrategic sets. The remaining cases are similar.

K

L

M

N

A B

C

D

By removing the strategic set V , we disconnect K from M , K from N ,L from M , and L from N , but K remains connected to L and M to N . Byremoving the strategic set W , we disconnect K from L, K from N , L fromM , M from N , while K remains connected to M and L to N .

Thus, the set of roads that belong to exactly one of V and W are all theroads joining K and L, K and M , L and N , M and N . When this set ofroads is removed, the bases split into at least two (nonempty) sets, K ∪ Nand L∪M , which are not connected to each other. In other words, we havean essential set.

Remark. This set is not necessarily strategic: for instance, sets K and N might benonempty, with no roads from K to N .


Problem 5. The locus of points O such that \AOD = 80◦

and O lies on the same side of AD as B is an arc of circlepassing through A and D, and the locus of points O forwhich \BOC = 100◦ with point O lying on the sameside of BC as A is an arc of circle passing throughB and C (see (see the figure). The point O must lieon both these two circles. This means that there areat most two such points.

A

B

C

D

O2

O1

Let’s locate the two possibilities for O. The first, O1,can be taken on the diagonal AC so that \BO1C = 100◦.Then, obviously, \AO1B = 80◦, and, by symmetry with respect to AC, wehave \AO1D = \AO1B = 80◦. Similarly, the second point, O2, can be page 221

taken on the diagonal BD so that \BO2C = 100◦. In this case, \AO2D =80◦ and \AO2B = 100◦.

The two points must be distinct, since they lie on different diagonalsand do not coincide with the intersection point P of the diagonals, Also,both points lie inside the rhombus: for instance, O2 lies on the diagonalBD rather than on its extension. To see this, consider the triangle BPC.Since \BO1P = 80◦ > 55◦ = \BCP , the point O1 lies inside the rhombus.Similarly, \CBP = 35◦ < 80◦ < \CO2P ; hence O2 lies on the diagonalBD rather than on its extension. Therefore both values \AOB = 80◦ and\AOB = 100◦, occur.

Problem 6. Let x1, . . . , xn be the coordinates of the marked points, whichwe will treat as unknowns. We assume x1 < x2 < · · · < xn. We also set The draft translation of

the first solution was

different from the original.It was somewhat garbled,

so I made adaptations for

clarity.

Also I’m factoring out the

initial portion of the twosolutions, which is

common.

x0 = 0 and xn+1 = 1.By assumption, for each i = 1, . . . , n, we have

xi = 12(ai + bi), (1)

where ai < xi < bi and ai, bi belong to the set {x0, . . . , xn+1}. Generalizingthe notation, what we have are n equations of the form

xi = β0x0 + β1x1 + · · · + βnxn + βn+1xn+1. (2)

Each coefficient βj on the right-hand side is rational and nonnegative, sinceit equals 0 or 1

2 ; and at least one xj , with j > i, has a strictly positivecoefficient, because bi in (1) equals xj for some j > i.

Important: Though we always use the symbol βj for the coefficient of xj ,the number it represents can of course vary across equations, and along thecourse of the solution.

First solution. On the right-hand side of (2) we can omit the term inx0 (since x0 = 0) and the term in xi, whose coefficient is 0 by construction.We therefore write

xi = β1x1 + · · · + βixi + · · · + βn+1xn+1, (3)

where the hat indicates that the term underneath it is absent.

164 SOLUTIONS

We now simplify these n equations by substitution. The first equationgives x1 in terms of x2, . . . , xn+1; we take this value of x1 and substitute itwherever x1 appears on the right-hand side of the other n−1 equations. Wethus get n − 1 equations involving only x2, . . . , xn+1. We claim that theyhave the same properties as the original set of equations, namely: the left-hand sides are respectively x2, . . . , xn; the right-hand sides have rational,nonnegative coefficients; and in the equation having xi on the left, at leastone variable xj , with j > i, has a positive coefficient.

That the new coefficients are rational and nonnegative is obvious, sincethey are products (or sums of products) of the old ones. To show the posi-tivity condition, notice that a positive coefficient of xj , where j > i, eitherremains unchanged or is increased by a positive multiple of the coefficientof xj in the first equation.

There is one way in which the new equations can differ from the old onesin (3): the i-th equation (for some values of i) might gain a xi term on theright, if it originally involved x1 and the value of x1 given by first equationinvolves xi. To make this i-th equation comply with the scheme in (3), wemove the new term αxi to the left and divide by 1−α. We must show thatthis is not a division by 0: but if 1 − α were 0, our equation at this stagewould be 0 = β0x0 + · · ·+βn+1xn+1, where all the products on the right arenonnegative and at least one is positive, and this is an impossibility.

For a similar reason the division by 1 − α preserves the nonnegativityof the coefficients: indeed, 1−α cannot be negative, otherwise the negativeproduct (1−α)xi would equal a positive sum. And of course the coefficientsremain rational, and at least one remains positive (for j > i). This reducesthe system of equations to another one with the same structure as (3), butwith one fewer equation and one fewer variable.

We then repeat this procedure, eliminating x2 to get a system of n − 2equations of the form (3) involving only the unknowns x3, . . . , xn, togetherwith the constants x0 and xn+1. We continue eliminating variables one byone, until we finally reach a single equation xn = βn+1xn+1, which is arational number (recall that xn+1 = 1). Then back-substitution gives thevalues of xn−1, xn−2, and so on back to x1, always with rational numbers.

Remarks. 1. What we did was in essence to use Gaussian elimination, a generalmethod for solving systems of linear equations (see also Fact 25). I reinstated the remarks,

which were absent from

the draft translation. (Did

the authors decide tomake changes between the

original and the draft

translation stage?)

2. The fact that we could always divide by 1−α in applying the method is essential,

page 223

as it ensures that our system has a unique solution. A system with rationalcoefficients admitting more than one solution has irrational solutions as well asrational ones; for example the system

{x1 = 2x2 − 2

x2 = 12x1 + 1

has the solution x1 = 2√

2, x2 =√

2 + 1 among many others.


On the other hand, if the solution is unique, it can be found by Gaussianelimination, which only involves addition, subtraction, multiplication, and di-vision— operations that cannot turn rational numbers into irrational ones (seeFact 25 on page 110). This is the basis for the second approach to the problem,discussed next.

Second solution. (Above grade level.) The system of simultaneous linearequations (2) is rational ; that is, it has rational coefficients and rationalconstant terms. If a rational system of linear equations has a unique solution,the solution is rational (see last pargraph of Remark 2 above). Thus we willhave solved the problem if we can show that the solution of (2) is unique.

Let x1, x2, . . . , xn and y1, y2, . . . , yn be two different solutions. Then thenumbers t1 = x1 − y1, t2 = x2 − y2, . . . , tn = xn − yn form a nonzero solutionto the homogeneous system corresponding to (2), that is, the system in whichall constant terms βn+1xn+1 are replaced by 0.

But then the numbers ti have the following property: each of themeither lies halfway between two others or it lies halfway between 0 andanother number tj . This is only possible if all the numbers are equal tozero. Indeed, choose among the ti the one with the greatest absolute value.If it isn’t zero, it can’t be the midpoint of an interval between two othernumbers.

Having shown that all the ti are 0, we see that our system has a uniquesolution, as needed.

Level C

Problem 1. Suppose a + b + c ≤ 11. Then

28a + 30b + 31c ≤ 31(a + b + c) ≤ 11 · 31 = 341,

which is less than 365. So this cannot happen. Similarly, if a + b + c ≥ 14, page 224

we have

28a + 30b + 31c ≥ 28(a + b + c) ≥ 28 · 14 = 392

which is also impossible since 369 > 365.There remains to prove that a + b + c cannot be equal to 13. Suppose

that a + b + c = 13. The possibility a = 13, b = c = 0 does not satisfy theassumption:

28 · 13 + 30 · 0 + 31 · 0 = 364 6= 365.

The last case to refute is a + b + c = 13, a < 13. In this case, we haveb + c = 13 − a > 0 and

28a + 30b + 31c = 28(a + b + c) + 2b + 3c ≥ 28 · 13 + 2(b + c).

The first term is equal to 364 and the second one is at least 2. It followsthat the sum is at least 366 and can’t be equal to 365.

Remark. Compare Problem 98.8.1. REF 98.8.1

166 SOLUTIONS

Problem 2. First solution. Denote by ai the marked side of the i-th rec-tangle and by bi a perpendicular side. The area of the initial square is equalto the sum of the areas of the rectangles:

∑aibi = 1. But bi ≤ 1 for all i;

therefore,∑

ai ≥ 1. at the end of 2nd sol.,

changed “is 1” to “is at

least 1” because a pointmay presumably be on the

boundary of two

rectangles, leading todouble counting. (Or does

razrezali exclude that

possibility?)

Is it worth mentioning the

possiblitity that with

infinitely many rectangles,

the sides needn’t beparallel to the square?

Second solution. We project all the marked segments onto one side of thesquare. If it is completely covered by the projections, then their total lengthis at least 1. If there is a point on the side that is not covered by theprojections, we draw the perpendicular to the side through that point. Thisperpendicular is covered by rectangles with a marked side parallel to it —otherwise the foot of the perpendicular would lie in the projection of themarked side. The total length of marked sides of the rectangles covering theperpendicular is at least 1.

Problem 3. We number the lamps in order along the road. If the segments

omitted “by natural

numbers” (the use of

ordinals already implies

that).

lit by the n-th and (n+2)-nd lamps overlap, the (n+1)-st lamp can be turnedoff. Therefore, segments with different odd numbers are disjoint. It is im-possible to place more than 999 disjoint (closed) segments of length 1m ona segment of length 1000 m. Hence there are at most 1998 lamps, because ifthere were 1999 or more, there would be at least 1000 odd-numbered lamps. page 225

Next we show how to arrange 1998 lamps so none of them can be turnedoff. Let the centers of the segments they light form a uniform sequencewhose endpoints are 1

2 m apart from the endpoints of the road. The interval

between two neighboring centers is 9991997 m. Hence the distance between the

n-th and (n+2)-nd lamps is 19981997 m, leaving a gap of 1

1997 m between thesegments they light. This gap is lit only by the (n+1)-st lamp. Therefore,none of the lamps can be turned off.

Remark. Compare Problem 98.8.6. REF 98.8.6

Problem 4. A number that is divisible by 1998 is also divisible by 999. Wewill show that the sum of digits of a positive number divisible by 999 is at ∗least 27.

As is well-known (or see Fact 6 on page 102), a number is divisible by 9if and only if the sum of its digits in decimal notation is divisible by 9. Thecriterion for divisibility by 999 can be formulated similarly:

Split the decimal representation of the number into groups of three digitsfrom right to left. (We call these groups “triples”, even though the leftmostmay have only one or two digits.) Add these triples together as numbers.The initial number is divisible by 999 if and only if the sum thus obtained isdivisible by 999.

For instance, 97902 is divisible by 999, since the number 97 +902 = 999is divisible by 999.

One can view these triples as the “digits” of the number in 1000-arynotation. This criterion is proved just like the criterion for divisibility by 9;we leave this task to the reader.


Now take a number divisible by 999, split it into triples of digits, andcompute the sum of the triples. If the new number is greater than 1000,repeat the operation, and keep doit it until obtaining a number smallerthan 1000. This will necessarily happen, because each operation decreasesthe number. (Proof: if a1, . . . , ak are nonnegative integers, with ak 6= 0 andk ≥ 1, then a0 + 1000a1 + · · · + 1000kak > a0 + a1 + · · · + ak.) page 226

The only positive integer smaller than 1000 and divisible by 999 is 999itself, whose digits add up to 27. Thus, it remains to show that our opera-tions never increase the sum of digits: this will mean that the sum of digitsof the initial number could not be less than 27.

Obviously, when a number is split into triples, the total sum of theirdigits is the same as that of the number. Let us show that when the triplesare added, the sum of digits does not increase. Denote by S(X) the sum ofdigits of X. It follows from the algorithm of digitwise addition that

S(X + Y ) = S(X) + S(Y ) − 9P (X,Y ),

where P (X,Y ) is the number of carries performed when we add X and Y .This means that S(X + Y ) ≤ S(X) + S(Y ) and we’re done.

Problem 5. We start with the “problem of one nail.” Suppose that wehave one nail hammered in the floor at a point M on the side AC. Fix apoint O; it will be the center of the hypothetical rotation. Is it possible torotate the triangle around the point O by a small angle, and if it is, then inwhat direction?

Suppose that the triangle is positioned as in the figure.Draw the perpendicular to AC at M . It divides the planeinto two half-planes. If O lies in the same half-plane asC, then we can rotate the triangle around O counter-clockwise, and if it lies in the other half-plane, thena clockwise rotation is possible.

A

B

C M

O

Now imagine that O lies right on the perpendicular.Then, if O and B are on the same side of the line AC(which is true whenever O lies inside the triangle), we’ll be unable to rotatethe triangle in either direction. If O and B lie on different sides of the lineAC, the triangle can be rotated in both directions. page 227

A

B

C

−++ −+−

++−

+−−+−+

−−+

+++

Returning to the three-nail problem, draw theperpendiculars to the three sides of the trianglegoing through the nails. We claim that if theydo not meet at the same point, the triangle canbe rotated. Indeed, in this situation the per-pendiculars partition the plane into sevenregions. To each region we assign a stringof three plus or minus signs. The first sign for a region is + if the nail on ABdoes not prevent the triangle from rotating clockwise around points in thisregion, and so on; see figure on the right. Different regions will have differentstrings of signs. There are eight possible strings and seven regions. Hence,

168 SOLUTIONS

at least one of the strings will be either +++ or −−−. In the first case, wecan rotate the triangle clockwise, and in the second, counterclockwise.

If instead the perpendiculars meet at a single point, sayR, let D, E, and F be the positions of the nails on the sidesAB, BC, and AC, respectively (see figure). Then thenails at D and E block the rotation around any centerinside the angle DRE (and its vertical angle), so dothe nails at E and F for the angle ERF and F,D forthe angle FRD. It follows that the only possibleposition of the center we are left with is R, theintersection of the perpendiculars. But this point lies inside the triangle,and, as already seen, each of the nails blocks the rotation around it. Thus,rotation is impossible if and only if the perpendiculars to the sides drawnfrom the corresponding nails meet at one point.

A B

C

D

E

FH

K

R

It remains to determine the position F of the third nail on the side AC.Denote by K the intersection of AC with the perpendicular to AB throughD; similarly, let H be the intersection of AC with the perpendicular to BCthrough E. Then page 228

AK =AD

cos 60◦= 1

2AC and CH =CE

cos 60◦= 2

3AC.

The triangle HRK is isosceles, since the angles HKR and KHR each mea-sure 30◦. Hence HF = FK and F divides the side AC in the ratio 5:7,measuring from A.

Problem 6. Step 1. We prove that the numbers in a good arrangementcan be colored with nine colors, denoted by 1, 2, . . . , 9, so that numbers ofthe same color form an increasing sequence. We do this by going throughthe numbers from left to right, assigning to each the lowest possible colorsuch that the number most recently assigned this color is smaller than thecurrent number.

Suppose that nine colors are not enough to complete such a coloring.This means that, at a certain moment, we are unable to give the currentnumber, a10, any color from 1 to 9; in particular we must have encounteredearlier some number a9 > a10 that received color 9. Next, a9 couldn’t getcolor 8 because of an earlier number a8 > a9 that got this color. Extendingthis reasoning down to color 1, we obtain 10 numbers in decreasing order.

Step 2. An arrangement of the numbers from 1 to n colored as describedabove is completely determined by the color of each number from 1 to n andthe color of each place in the string, because the numbers on the places ofthe same color increase. There are 9n ways to color the numbers from 1 ton using 9 colors. The number of colorings of the n places using 9 colors isthe same. Thus, the number of good arrangements is at most 81n.

Remark. There are exactly n! permutations of the numbers from 1 to n, and anever smaller fraction of them is “good” as n increases. This follows from thebound in the problem together with Stirling’s formula, a result from calculus


that says that n! is approximately equal to√

2πn(n/e)n. More precisely,

limn→∞

81n

n!= lim

n→∞

81n

√2πn(n/e)n

≤ limn→∞

81n

(n/e)n= lim

n→∞

(81e/n)n = 0.

Here e ≈ 2.71828 is the base of natural logs. See also Fact 4 on page 101.

Level D

Problem 1. We have

x+ y + z − 2(xy + yz + xz) + 4xyz − 12 = 1

2 (2x− 1)(2y− 1)(2z− 1). (1)

If the left-hand side of the equation is zero, then at least one of the factors page 229

in the right-hand side is zero. Therefore, one of the numbers x, y, z is 12 .

Remark. How would one come up with this factorization? We offer two suggestions.

(a) The original polynomial involves the so-called symmetric functions of thevariables x, y, z (x+y+z, xy+yz+xz, and xyz); these bring to mind Vieta’sformula (Fact??) in the cubic case. To be able to make the polynomial fit theformula we would like to get rid of the coefficients 2 and 4, which we can do byscaling the variables via the substitutions x = X/2, y = Y/2, and z = Z/2:

x+y+z−2(xy+yz+xz)+4xyz

= 12 ((X+Y +Z)−(XY +Y Z+XZ)+XY Z). (2)

So let the symmetric functions on the right be the coefficients of a cubicpolynomial in the dummy variable t:

−t3 +(X +Y +Z)t− (XY +Y Z +XZ)t2 +XY Z = −(t−X)(t−Y )(t−Z),

where the equality is exactly Vieta’s formula. Substituting t = 1 gives

−(1 − X)(1 − Y )(1 − Z) = −1 + 2 × right-hand side of (2).

Going back to the original variables we get

−(1 − 2x)(1 − 2y)(1 − 2z) = −1 + 2 × left-hand side of (2),

which is the same as (1).

(b) A general result (see Fact 19) says that if a polynomial in one or more vari-ables takes the value 0 identically when x is replaced by a, this polynomial isdivisible by x − a.

Denoting the polynomial on the left-hand side of (1) by P (x, y, z), we seethat P (1

2 , y, z) — the polynomial in y and z obtained by replacing x by 12 —

is identically 0. Hence P (x, y, z) is divisible by x − 12 . Similarly, P (x, y, z) is

divisible by y− 12 and z− 1

2 . Therefore, P (x, y, z) is equal to (x− 12 )(y− 1

2 )(z− 12 )

up to a constant factor (see Fact 21).

Problem 2. Here is a counterexample:

f(x) =

{2 − x if x ≤ 1;

1/x if x > 1.

The graph of this function is shown on the right. x

f↼x↽

The first property is obvious, and the second is true because the functions2 − x and 1/x have the same derivative at x = 1.

170 SOLUTIONS

Let us prove the third property. Obviously, if x is rational, then f(x) isalso rational. Let y = f(x) be rational. Then either x = 2−y or x = 1/y. Inany case, x is rational. Therefore, rational values of x correspond to rationalvalues of f(x) and vice versa.

Problem 3. Denote by M the centroid of the tri-angle ABC (the centroid is the intersection of themedians). Consider the equilateral triangle APQsuch that AK is its median and P lies on the lineAB. Since the centroid divides the medians inthe ratio 2:1, the point M is the centroid, andthus, the center, of triangle APQ. It follows page 230

that \KPM = 30◦ = \KBM . ThereforeM , K, P , and B lie on a circle. Since \MKP is a right angle, MP is adiameter of this circle. This circle meets AP at two points, one of which isP and the other one, B, is the midpoint of AP .

A

B

P

K C

ML

Q60◦

30◦

30◦

30◦

Let the triangle APQ have side length 2a. We have AB = a and AK =√3a. The triangle BKP is equilateral, BK = a = KC, and BC = 2a.

Further, \ABK = 120◦, whence by the law of cosines AC2 = AB2 +BC2−2AB · BC cos 120◦ = a2 + 4a2 + 2a2 = 7a2. Finally, by the law of cosineswe find cos \ACB = (4 + 7 − 1)/(2 · 2

√7) = 5/(4

√7) and cos \CAB =

(7 + 1 − 4)/(2√

7) = 2/√

7.

Problem 4. Both sides of the equation must yield the same remainderupon division by 3. The remainder of the left-hand side is 1 (see Fact 7);therefore, z is an even number (see Remark 1). Similarly, the left-handside has remainder 1 upon division by 4; therefore, x is even, too. Thus,4y = 5z − 3x = 52z0 − 32x0 , i.e., 22y = (5z0 − 3x0)(5z0 + 3x0). Therefore,5z0 − 3x0 = 2k and 5z0 + 3x0 = 2l, where k and l are nonnegative integersand k + l = 2y. It follows that 5z0 = 1

2(2k + 2l) and 3x0 = 12(2l − 2k).

The number 2l−1 − 2k−1 = 3x0 is odd; hence k = 1. Then 2k = 2 and3x0 = 2l−1 − 1. Consequently, l−1 is even: l−1 = 2s. It follows that3x0 = (2s − 1)(2s + 1) is the product of two powers of 3 that differ by 2; the page 231

only possibility for these two numbers is 1 and 3. Then s = 1, l = 3, and2y = 4. Thusj x = y = z = 2 is the unique solution to our equation.

Remark. It is not difficult to prove by induction that the remainder of 5z upondivision by 3 is 1 if z is even and 2 if z is odd.

In fact, the remainders of an upon division by b, for fixed a and b, alwaysform a periodic sequence (Fact 4). This can be derived from the last statementin Fact 7.

Also it should be mentioned that if b is prime, the period of this sequenceis a divisor of p − 1. This statement is equivalent to Fermat’s Little Theorem;see the remark Problem 95.11.6. REF 95.11.6

Problem 5. On the face of it, it may seem that the desired constructionis impossible: if the first cogwheel rotates clockwise, then the second onerotates counterclockwise, the third rotates clockwise, and so on. So the 61st


cogwheel must rotate clockwise, which does not agree with the rotation ofthe first cogwheel.

However we are in space, not on the plane. And the direction of rotationdepends on the side from which we look at the cogwheel. Here is one possiblesolution:

x

y

z

1

7

1314

3738

61

page 232

Let us describe the gear in the figure: cogwheels 13 through 61 and 1 liein the xy-plane; all other cogwheels have their centers in the xz-plane, andlie in planes perpendicular to the xz-plane. Cogwheels 13, 15, 17, . . . , 61rotate clockwise if we look at the xy-plane from above, while cogwheels 14,16, . . . , 60, 1 rotate counterclockwise. The numbers of cogwheels undergothe “parity change” in the direction of motion occurs along the chain ofcogwheels from 1 to 13; these two cogwheels at the endpoints of this chainrotate in opposite directions in the xy-plane.

Remark. Experienced readers will have noticed that thissolution in essence uses the Mobius strip. It would the draft translation was

rather garbled here. I

hope this is what you

mean. (Actually the

drawing shown, with aruled M. strip, cannot

contain any circles, but I

guess the claim is

substantially correct.)

have been possible to draw the cogs on a Mobius striphaving the traditional shape shown in the figure, butit would have been harder to justify the correctness ofthe answer.

The phenomenon of parity change is connected with the fact that theMobius strip is a nonorientable, or one-sided, surface.

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