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Solutions For PUMaC 2006
Taotao Liu, Aaron Potechin, Nathan Savir, Jian Shen, Jeffrey Tang
December 14, 2006
These are the official solutions for the 2006 Princeton University Mathematics Competition tests.c© 2006 PUMaC.
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Contents
1 Individual Tests 51.1 Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2.1 Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.2 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.3 Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.2.4 Advanced Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.2.5 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2 Team Test 232.1 Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3 Power Test 293.1 Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.2.1 Characteristic functions of intervals . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2.2 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2.3 Rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.2.4 Almost-refinability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4 Relay Test 374.1 Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4.2.1 Round 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.2.2 Round 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
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Chapter 1
Individual Tests
1.1 Rules
1. Five different Individual Tests will be offered in the following topics: Algebra, Geometry, NumberTheory, Calculus, and Advanced Topics. The Advanced Topics Test will include topics such as in-equalities, sequences and series, combinatorics, probability, and game theory.
2. Each student will take only one Individual Test.
3. No more than two students from each team may take the same Individual Test.
4. Half-teams (teams of 5) must have one student take each individual test.
5. The Individual Test that each student is taking must be declared on the registration form by the LateRegistration Deadline (November 22, 2006), and may be changed up to the beginning of the TeamTest on the day of the competition.
6. If a student takes an Individual Test that he or she is not signed up for, his or her results will bedisqualified for that test.
7. All Individual Tests will consist of 10 short-answer questions, and will last 60 minutes.
8. The first five problems will be worth 2 points each and the last five problems will be worth 3 pointseach.
9. During the Individual Test, there will be absolutely no communication between the students from thetime the questions are handed out to the time the answers are collected.
10. Proctors will give 5-minute and 1-minute warnings, and no other information.
11. At the end of the 60 minutes, students must submit completed official answer sheets to the Proctor.Copying of answers from other sheets of paper to the official answer sheet after the time limit will notbe allowed.
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1.2 Problems
1.2.1 Algebra
1. Given that x2 +5x+6 = 20, find the value of3x2 +15x+17.
Solution: Note that
3x2 +15x+17= 3(x2 +5x+6)−1 = 3(20)−1 = 59
2. Express√
7+4√
3+√
7−4√
3 in the simplest possible form.
Solution 1: Let x =(√
7+4√
3+√
7−4√
3)
. Then
x2 =(√
7+4√
3+√
7−4√
3
)2 =
(7+4
√3)
+2(√
49−48)
+(
7−4√
3)
= 16
whence x = 4 .
Solution 2: Let (a+ b√
3)2 = 7+ 4√
3 (and then also 7−4√
3), and solve the resulting system ofequations ina andb. We find that √
7+4√
3 = 2+√
3√7−4
√3 = 2−
√3
(2+√
3)+(2−√
3) = 4
3. Let r1, . . . , r5 be the roots of the polynomial x5 +5x4−79x3 +64x2 +60x+144. What is r21 + ...+ r25?
Solution: Using Vieta’s Formulas,r1 + · · ·+ r5 = 5 andr1r2 + r1r3 + r2r3 + · · ·+ r4r5 = −79. Thenr12+ · · ·+ r52 = (r1 + · · ·+ r5)2−2(r1r2 + r1r3 + · · ·+ r4r5) = 25+2(79) = 183 .
4. Find all pairs of real numbers(a,b) so that there exists a polynomial P(x) with real cofficients andP(P(x)) = x4−8x3 +ax2 +bx+40.
Solution: SupposeP(x) = x2 + rx + s. ThenP(P(x)) = x4 + 2rx3 +(r2 + r + 2s)x2 +(2rs+ r2)x+(rs+s+s2). Equating coefficients, we find
2r =−8
r2 + r +2s= a
2rs+ r2 = b
rs+s+s2 = 40
From the first equation we getr = −4. The last becomes−3s+ s2 = 40 =⇒ (s−8)(s+5) = 0. Ifs=−5 we get a = 2,b = 56 . If s= 8 we get a = 28,b =−48 . These are the only possible valuesof a andb.
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5. Find the greatest integer less than the number
1+1√2
+1√3
+ · · ·+ 1√1000000
Solution: Let Sbe the sum in question. We just note that
2(√
n+1−√
n) =2√
n+1+√
n
and clearly
2√
n+√
n−1>
1√n
>2√
n+1+√
n
Adding these inequalities fromn = 2 to 106, we obtain 1998> S−1 > 2(√
106 +1−√
2) > 1997.
Thus bSc= 1998.
6. Suppose that P(x) is a polynomial with the property that there exists another polynomial Q(x) tosatisfy P(x)Q(x) = P(x2). P(x) and Q(x) may have complex coefficients. If P(x) is a quintic withdistinct complex roots r1, ..., r5, find all possible values of|r1|+ ...+ |r5|.
Solution: Suppose thatP(x0) = 0. ThenP(x2) has roots at√
x0 and−√x0. SinceP(x) | P(x2), wemust also haveP(±√x0)= 0. Proceeding in this fashion, we find that all of the numbersx0,±
√x0,±
√±√x0, . . .
must be roots ofP(x). SinceP(x) is a quintic, that sequence must be finite. Since either 1< |√x0|<|x0| or 0< |x0|< |√x0|< 1 must be true if|x0| 6∈ {0,1}, clearly we must havex0 = 0 or |x0|= 1. Noroots are repeated, so the set of 2nth roots ofx0(6= 0) must contain either 4 or 5 elements (depend-ing on whetherx | P(x)). So |r1|+ ...+ |r5| ∈ {4,5}. To show that both are obtained, consider thepolynomials
P1(x) = x5 +x4−x3 +x2−x+1 = (x−1)(x2 +x+1)(x2−x+1)
P2(x) = x5 +x4 +x3 +x2 +x = x(x4 +x3 +x2 +x+1)
Thus, the sum in question may obtain the values{4,5} .
7. Find one complex value of x that satisfies the equation√
3x7 +x4 +2 = 0.
Solution: Think geometrically in the complex plane. Note that 1,√
3 and 2 are the sides of a righttriange. This suggests thatx4 andx7 are separated by 90 degrees, and that the magnitude ofx is 1.This suggests thatx makes an angle of 30 with the real number line, orx = cos(30)+ sin(30) · i =√
32
+i2
. It can be easily checked that this is a solution. Since the polynomial has real coefficients,
x =√
32− i
2, the complex conjugate, will also be a solution.
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8. The Lucas numbers Ln are defined recursively as follows: L0 = 2,L1 = 1,Ln = Ln−1+Ln−2 for n≥ 2.Let r = 0.21347. . ., whose digits form the pattern of the Lucas numbers. When the numbers havemultiple digits, they will ”overlap,” so r= 0.2134830. . ., not 0.213471118. . .. Express r as a rationalnumberp
q , where p and q are relatively prime.
Solution: Let Ln be thenth Lucas number. We want to findS= ∑∞n=0
Ln10n+1 .
Let φ = 1+√
52 , the golden ratio. Note that 1+φ = φ2, soφn−2+φn−1 = φn and(−φ)−n+2+(−φ)−n+1 =
(−φ)−n, and asφ0 +(−φ)0 = 2,φ1 +(−φ)−1 = 1, we conclude thatLn = φn +(−φ)−n. Then we arelooking for
S=110
∞
∑n=0
(Ln
10
)n
=110
∞
∑n=0
[(φ10
)n
+(−110φ
)n]=
110
(1
1−φ/10+
11+1/10φ
)=
110
(20
19−√
5+
20
19+√
5
)=
2(19+√
5+19−√
5)192−5
=1989
9. The curve y= x4 +2x3−11x2−13x+35 has a bitangent (a line tangent to the curve at two points).What is the equation of the bitangent?
Solution: A polynomial with two double roots hasy = 0 for a bitangent. Then let’s subtract a linearfunction from this polynomial to form one with two double roots. We want
(x+a)2(x+b)2 = (x2 +2ax+a2)(x2 +2bx+b2)
= x4 +(2a+2b)x3 +(a2 +4ab+b2)x2 +2ab(a+b)x+(a2 +b2)
So
(2a+2b) = 2 =⇒ (a+b) = 1
a2 +4ab+b2 =−11 we subtract(a+b)2 = 1 to get 2ab=−12
By inspection we see a solution isa = 3,b =−2. So thex-coefficient should be−12 and the constantshould bea2b2 = 36. Then the difference isx+1, so our bitangent isy =−x−1 .
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10. If x,y,z are real numbers and
2x+y+z≤ 66
x+2y+z≤ 60
x+y+2z≤ 70
x+2y+3z≤ 110
3x+y+2z≤ 98
2x+3y+z≤ 89
What is the maximum possible value of x+y+z?
Solution: This is an example of the linear programming problem, and there are a number of ways togo about this. Here is one solution:
We are given six inequalities. Visualize this in 3 dimensions. Each inequality represents a plane,where our point must be on one side of the plane.
Imagine that we start at a point(x,y,z) that satisfies all of the inequalities and then pick a directionto travel in. If this decreasesx+y+z, travel in the opposite direction. In this, way, we will be eitherincreasing x+y+z or keeping it constant. When we hit a plane, choose a direction parallel to that plane.When we hit another plane, the intersection will form a line, and we can travel along this line. Theonly places where we cannot do this are when we hit a point where three planes intersect, or when 3inequalities are equalities. At least one maximum will be found at one of these intersection points orat infinity.
A quick look at the inequalities shows that ifx, y, or z is infinity, x+ y+ z must be -infinity. This isclearly not a maximum. Thus, there must be at least one intersection point giving the maximum valuefor x+y+z.
Because of the symmetry, brute force is feasible here. However, we will use a slightly more efficientmethod.
Pick one intersection point to start with. The first three inequalities are a natural place to start. Solvinggivesx = 17,y = 11, andz= 21. However, this violates inequality 5, as
x+2y+3z= 102
3x+y+2z= 104
2x+3y+z= 88.
Let’s try to find the maximum value ofx+ y+ z given just inequalities 1,2,3, and 5, moving to anadjacent intersection point.
If we takex = 17−3p, y = 11+ p, z= 21+ p, this preserves the 2nd and 3rd equalities, decreases3x+ y+ 2z by 6p, and decreasesx+ y+ z by p. If we takex = 17+ p, y = 11− 3p, z = 21+ p,this preserves the 1st and 3rd equalities, increases 3x+ y+ 2z by 2p, and decreasesx+ y+ z by p.If we takex = 17+ p, y = 11+ p, z= 21−3p, this preserves the 1st and 2nd equalities, decreases3x+y+2z by 2p, and decreasesx+y+z by p.
Note thatp > 0, as otherwise one of the first three inequalities will be violated.
We want to decreasex+y+zas little as possible while decreasing 3x+y+z, so we use the first path.Settingp = 1 reduces 3x+ y+ 2z to 98, as needed and givesx = 14, y = 12, andz= 22. A quickcheck shows that this satisfies all of the inequalities. Thus, the answer is48 .
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1.2.2 Geometry
1. A,B,C,D,E, and F are points of a convex hexagon, and there is a circle such that A,B,C,D,E, and Fare all on the circle. If]ABC= 72, ]BCD= 96, ]CDE = 118, and]DEF = 104, what is]EFA?
Solution: CDEF is cyclic, meaning that]CFE = 62◦, andABCF is cyclic, meaning that]AFC =108◦. So ]EFA= 170 .
2. ABC is an equilateral triangle with side length 1. BCDE is a square. Some point F is equidistant fromA, D, and E. Find the length of AF.
Solution: Imagine shifting triangleBAC down by 1. NowB is on E andC is on D. Call the pointwhereA would beP. Clearly,EP= 1 andDP = 1. Also, AP= 1 because the triangle was shifteddownwards by 1. ThereforeF = P, and the answer is1 .
3. Find the exact value ofsin36◦.
Solution: Take a regular pentagonABCDEof side length 1. SegmentAC has length 2cos(36). Drawsegments perpenicular toAC that meetD andE. It is now clear that segmentAChas length 2cos(72)+1. By the double angle formula for cosine, cos(72) = 2cos2(36)−1. Solving the resulting quadratic
equation gives cos(36) = 1+√
54 , whence sin(36) =
√5−
√5
8.
4. There is a circle c centered about the origin of radius 1. There are circles c1, . . . ,c6, each of radius r1,such that each circle is completely inside c and is tangent to it, and c2 is tangent to c1, c3 is tangentto c2, . . ., and c1 is tangent to c6. There is a circle d which is tangent to c, c1, and c2, but does notintersect any of these circles. What is the radius of circle d? Express your answer in the forma+b
√c
d ,where a,b,c,d are integers, d is positive and as small as possible, and c is squarefree.
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Solution: The easiest way to solve this is to use inverse geometry. LetC be the center of the big circleand r be its radius. Invert the plane aboutC, keeping the big circle in place, the six little circles becomesix circles of the same size, forming a hexagon. Pick two adjacent circles plus the original circle. Thecircle we are looking for is now in the center of this triangle. The distance fromC to its center is2r
√3
3 .
Its radius is2r√
3−3r3 . The distance from the other side of the circle to C isr + 22r
√3−3r3 = 4r
√3−3r3 .
Invert the plane aboutC to restore everything to the way it started. The distance fromC to the nearside of the circle is
r2
4r√
3−3r3
=3(4
√3+3)
48−9r
Taking r minus this and dividing by 2 to get the radius (and recalling thatr = 1) gives14−6
√3
37
5. A,B, and C are vertices of a triangle, and P is a point within the triangle. If angles]BAP,]BCP, and]ABP are all30◦ and angle]ACP is45◦, what issin(]CBP)?
Solution: Let∠CBP= x,BP= a. Since∠BAP= ∠ABP, thenAP= BP= a. The length of the altitudefrom P to BC is thenasinx, makingCP= 2asinx. Since∠ACP= 45◦, the length of the altitude fromP to AC is
√2asinx. (1)
Now the sum of the angles in a triangle is 180◦, so∠CAP= 45◦−x. SinceAP= a, then the length ofthe altitude fromP to AC is asin(45◦−x). (2)
Equating(1) and(2), 1√2
cosx− 1√2
sinx =√
2sinx, whichsimplifies to cosx = 3sinx. And cos2x+
sin2x = 1, so 10sin2x = 1. x is acute, so sinx > 0. Therefore,sinx =√
1010
.
6. Given that in the diagram shown,]ACB= 65◦, ]BAC= 50◦, ]BDC = 25◦, AB= 5, and AE= 1,determine the value of BE·DE.
Solution: You see that]ABC= 180◦−]BAC−]ACB= 65◦. SoABC= ACB, and triangleABCis isoceles, withAB = AC = 5. Notice that]BDC = 25◦ = 1
2]BAC. This reminds me of a circle.Construct a circle withA as the center, andAB andAC as radii. We see from a well-known theorem
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thatD must lie on the circle. Now extendAC to meet the circle again atF , and we have Power of thePoint aboutE. SoBE ·DE = CE ·EF = 4·6 = 24 .
7. Given parallelogram ABCD, construct point F so that CF⊥BC, as shown. Also Fis placed so that]DFC = 120◦. If DF = 4 and BC= CF = 2, what is the area of the parallelogram?
Solution: ExtendCE to meet the extension ofAD at G. ThenCE⊥ BC andBC‖ AD, soCG⊥ AG.∠DFG = 180◦−∠DFC = 120◦, ∠DGF = 90◦, andDF = 4, soFG = 4cos60◦ = 2. ThenCG= 4,so[ABCD] = AD·CG= 8 .
8. Given that triangle ABC has side lengths a= 7,b= 8,c= 5, find(sin(A)+sin(B)+sin(C))·(cotA
2 +cotB2 +cotC
2
).
Solution: To evaluate the left-hand side, consider the circumcircle (whose radius is the junction ofthe perpendicular bisectors):
whereR is the circumradius. Nowm∠BCA= 12m∠BOA, and4BOD∼= 4AOD by SSS (or SAS).
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Thus,m∠BOD= 12m∠BOA= ∠C. So sinC = sin(∠BOD) = c/2
R . And simlarly,
sinA = sin(∠BOE) =a/2R
sinB = sin(∠AOF) =b/2R
sinA+sinB+sinC =1
2R(a+b+c)
To evaluate the left-hand side, consider the incircle (whose radius is the junction of the angle bisec-tors):
wherer is the radius. Now becauseBO, AO, andCO are angle bisectors, we have:
cot(A2)+cot(
B2)+cot(
C2
) =12
((cot(
A2)+cot(
B2))
+(
cot(A2)+cot(
C2
))
+(
cot(B2)+cot(
C2
)))
=12
((c−z)+z
r+
(b−z)+zr
+(a−z)+z
r
)=
12r
(a+b+c)
So
(sinA+sinB+sinC) · (cot(A2)+cot(
B2)+cot(
C2
)) =1
2R(a+b+c) · 1
2r(a+b+c)
=1
4rR(a+b+c)2
Now sinceR= abc4rs , wheres is the semiperimeter,Rr = abc
4s . And so we have
14rR
(a+b+c)2 =14· 4sabc
· (2s)2 =4s3
abc=
1007
.
9. Consider all line segments of length 4 with one end-point on the line y= x and the other end-point onthe line y= 2x. Find the equation of the locus of the midpoints of these line segments.
Solution: Consider a point(a,a) on the liney = x and a point(b,2b) on the liney = 2x. If thesetwo points are the endpoints of a line segment of length 4, then(a−b)2+(a−2b)2 = 42 (*). The
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midpoint of this line segment is(x,y) = (a+b2 , a+2b
2 ). Solving the equationsx = a+b2 andy = a+2b
2 fora andb in terms ofx andy, we geta = 4x−2y andb = 2y−2x. Substituting into (*) and simplifying,
the equation of the locus is25x2−36xy−13y2 = 4 .
10. Points P and Q are located inside square ABCD such that DP is parallel to QB and DP= QB= PQ.Determine the minimum possible value of]ADP.
Solution: Let θ = ]ADP andx = DP. For a givenθ, let’s consider the minimum possible valueof PQ− x (Ignoring the restiction thatPQ = x) If we increasex by dx, we increasePQ by 2· dx·(cos(]DPQ)). Thus, this is minimum when]DPQ= π
3 .
For a given value ofθ, let’s say we construct this minimum. Now reduceθ by da. This increasesPQ
by√
32 ·x ·da (unlessθ > π
4 which clearly is not the answer.
Thus, for the minimum angle,PQ−x is minimum whenPQ= x. If PQ−x has a negative minimumvalue, the angle can be reduced untilPQ = x. If PQ− x has a positive minimum value, then theconditions of the problem cannot be satisfied for that angle.
Now it is clear that]PDQ= ]PQD= π3 whenθ is minimum. By symmetry, it follows quickly that
θ =π12
.
1.2.3 Number Theory
1. Find the smallest positive integer that is a multiple of18and whose digits can only be4 or 7.
Solution: Let the number ben. Certainly 18| n =⇒ 9 | n, so the sum of the digits ofn is also amultiple of 9. We can do that with three digits in exactly one way (can’t do it with less than three),that is with one 4 and two 7s. But since 18| n, n is even, hencen = 774 .
2. Professor Conway collects a total of58 midterms from the two sections of his introductory linearalgebra course. He notices that the number of midterms from the smaller section is equal to theproduct of the digits of the number of midterms from his larger section. Assuming that every studenthanded in a midterm, how many students are there in the smaller section?
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Solution: The number of students in the larger section must be a two-digit number, say 10a+ b,wherea,b< 10,a> 0, andb is non-negative. We know thatab+10a+b= 58. Use Simon’s FavoriteFactoring Trick:(a+1)(b+10) = ab+10a+b+10= 68. Given the conditions ona andb, the onlysolution isa+1 = 4, b+10= 17, yieldinga = 3,b = 7, and ab= 21 .
3. Find the fifth root of 14348907.
Solution: Let y= 14348907. Consider(y/10)5 = 143.48907. Obviously 2< y/10< 3, so the leadingdigit is 2. What gives us a 7 in the last place for 5th powers? Only 7. So the answer is27 .
4. What are the last two digits of2003200520072009
, where abc
means a(bc)?
Solution We write 2003= 100x+3, since only the last two digits can affect the last two digits of thepower: (100x+3)n = 100(irrelevant)+3n. The last digits of powers of 3 repeat: 1 3 9 27 81 43 2987 61 83 49 47 41 23 69 7 21 63 89 67 1. The cycle repeats every 20 terms. Now, for the same reason2005n is like 5n (mod 20), and other thann = 0,5n (mod 20) = 5. The 5th one in the sequence (the1 is really the 0th because we started atn = 0) is 43 .
5. Find the largest integer k such that12k | 66!.
Solution: We need to count both 2s and 3s – there are[
662
]+[
664
]+[
668
]+[
6616
]+[
6632
]+[
6664
]=
33+16+8+4+2+1 = 64 2s, so 32 4s, and[
663
]+[
669
]+[
6627
]= 22+7+2 = 31 3s, so that makes
31 total 12s.
6. I have a set A containing n distinct integers. This set has the property that if a,b∈ A, then12 6 | |a+b|and12 6 | |a−b|. What is the largest possible value of n?
Solution: There are 12 possible equivalences(mod 12). Obviously we can’t have two that are thesame. If we take 0 thru 6, we have 7 different ones and no problem, but if we try to add an 8th it won’twork. This is because if we have something that’sk (mod 12), we excludek and 12−k, so if we take7 distinct ones we necessarily excluded all the rest.
7. Find the largest possible value of the expression x+y+z, x,y,z∈ Z, given that the equation10x3 +20y3 +2006xyz= 2007z3 holds.
Solution: 10x3 +20y3 +2006xyz= 2007z3. Take this equation mod 2. Clearly,z≡ 0 (mod 2). Nowtake this equation mod 4.z≡ 0 (mod 4), so this leaves 10x3 ≡ 0 (mod 4). x≡ 0 (mod 2). Nowtake this equation mod 8.z≡ 0 (mod 2) andx≡ 0 (mod 2), so this leaves 20y3 ≡ 0 (mod 8). y≡ 0(mod 2). However, ifx,y, andz are a solution, we can get another solution by dividingx, y, andz by2, and it must still be integral becausex≡ y≡ z≡ 0 (mod 2). Thus, we can continue this processforever.
This only makes sense ifx = y = z= 0, and this is the only integral solution. Therefore, the largestpossible value ofx+y+z is 0 .
8. Find all integers n (not necessarily positive) such that7n3−3n2−3n−1 is a perfect cube.
Solution: Write the given expression as(2n)3− (n+ 1)3 = m3 =⇒ (2n)3 = m3 + (n+ 1)3. ByFermat’s Last Theorem, this equation can hold in integers only if at least one of the numbers is 0. So,to make 2n = 0 we haven = 0, to makem= 0 we have 2n = n+1 =⇒ n = 1, and to maken+1 = 0we haven =−1. Our solutions are{−1,0,1}.
15
9. Consider the set of sequences{Si} that start with S0 = 12, S1 = 21, S2 = 28, and for n> 2 Sn is thesum of two (not necessarily distinct) Skn and Sjn with kn, jn < n. Find the largest integer that cannotbe found in any sequence Si .
Solution: The language in the problem boils down to: what is the largest integer that we cannotexpress as 12x+21y+28z with x,y,z≥ 0 and integer. If we consider this question in mod 12, noticethat the 12x term doesn’t matter. Once we find a linear combination of 21 and 28 that is in a paricularresidue class of mod 12, we can add multiples of 12 to get all integers greater than this in the sameequivalence class. So we are really just interested in the smallest linear combination required foreach equivalence class. Taking the maximum of these and subtracting 12 gives the largest number notattainable. So consider any constantc that we are trying to reach in mod 12 with linear combinationsof 21 and 28. We have that 21y+ 28z≡ c (mod 12) To do this we just need to make them equal inmod 3 and in mod 4 because of the Chinese Remainder Theorem. In mod 3, this equation is verysimple: z≡ c (mod 3) and in mod 4 it is also simpley≡ c (mod 4). We want to choosec so thatzandy are both maximized, so we havez= 2,y = 3 Thus our maximum is 2·28+3 ·21−12= 107 .This exact same approach generalizes so that the maximum forab,bc,ac wherea,b,c are pairwiserelatively prime positive integers, is equal to 2abc−ab−bc−ac.
10. If a1, . . . ,a12 are twelve nonzero integers such that a61 + · · ·+ a6
12 = 450697, what is the value ofa2
1 + · · ·+a212?
Solution: Since there are 6th powers, we might want to consider Fermat’s Little Theorem. 450697is 2 mod 7; from FLT and that we deduce that either 3 or 10 of theai equal 7. 76 = 117649, so therecan’t be 10 (too large). Thena1 = a2 = a3 = 7. Now we havea6
4 + ...+ a612 = 97750. The RHS is
equivalent to 6 mod 8. Sixth powers are either 1 or 0 mod 8. So we deduce there are 6 odds and 3evens among the remainingai . The RHS is equivalent to 1 mod 9. Sixth powers are either 1 or 0 mod9. From this we deduce that there are 8 multiples of 3 and 1 that isn’t. Since 66 = 46656, not all threeevens can be 6, so we find thata4 = a5 = 6, a6 = ... = a11 = 3. Subtracting, we finda12 = 2. Theanswer is then 4+6(9)+2(36)+3(49) = 277 .
1.2.4 Advanced Topics
1. What is the greatest possible number of edges in a planar graph with12 vertices? A planar graph isone that can be drawn in a plane with none of the edges crossing (they intersect only at vertices).
Solution: Euler’s formula tells us thatV −E + F = 2. Suppose we have some vertices and someedges. Consider any region (including the exterior) created by the graph. If this region touches morethan 3 vertices, then there are two vertices that can be connected without making the graph nonplanar.Suppose we go through every region and connect them all in this way. Then all regions have exactlythree vertices and three edges associated. Any particular edge must be associated with exactly tworegions, and each region is associated with three edges. So the number of regions is 2/3 the numberof edges. Then we haveV−E + 2
3E = 2, so 3(V−2) = E. If V = 12 then E = 30 .
2. 3green,4 yellow, and5 red balls are placed in a bag. (Large piles of balls of each colour are outsidethe bag.) Two balls of different colours are selected at random, andreplacedby two balls of the thirdcolour. If, at some point, there are5 green balls left in the bag, and there are at least as many yellowballs as red balls left in the bag, how many balls of each colour are left in the bag? Write your answerin the form(g,y, r), where g is the number of green balls and so on.
16
Solution: Consider this question modulo 3. At each step, we remove two balls of different colours,and add two balls of the third colour, effectively adding -1 mod 3 to the number of balls of each colour.Letting the number of green balls beG, the number of yellow balls beY, and the number of red ballsbeR, R (mod 3)≡ (Y+1) (mod 3)≡ (G+2) (mod 3). So ifG (mod 3)≡ 2, then Y (mod 3)≡ 0,andR (mod 3)≡ 1. Also notice that the total number of balls in the bag never changes. So ifG = 5,thenY + R = 7. So (Y,R) = (6,1), (Y,R) = (3,4), or (Y,R) = (0,7). We are told thatY R, so(G,Y,R) = (5,6,1). We see that this can be obtained as follows: Take a green and a red. Then wehave (2, 6, 4). Take a green and a red. Then we have (1, 8, 3). Then we take yellow+red twice and wehave (5,6,1) as desired.
3. Find the minimum value of x2 +2x+ 24x for x > 0.
Solution: Write the function asx2 +2x+ 8x + 8
x + 8x . We apply the AM-GM inequality to obtain that
the function is greater than or equal to 55√
2·83 = 5 5√
2·29 = 20 . This is obtained if we can haveequality among all the terms, which we see works whenx = 2.
4. A modern artist paints all of his paintings by dividing his 3 ft by 5 ft canvas into 21 random regions.He then colours some of the regions, and leaves some of them white. If the smallest region has areaa = 10 square inches, and the probability that any given region with area ai is left white is a
ai, then
what is the probability that any given point on the canvas is left white? (1 ft = 12 in)
Solution: Let the total area of the canvas beA = 3 f t ·5 f t = 2160in2. Suppose the regions of areaai
are calledAi . The probability that a random pointx is white is equal to the probability that the regionit is in is white. Therefore,
P{x is white}= P{x∈ Ai} ·P{Ai is white}
=ai
A· aai
=ai
2160· 10
ai
=1
216
After multiplying by the number of regions, the desired probability is21216
=772
.
5. In the diagram shown, how many pathways are there from point A to point B if you are only allowedto travel due East, Southeast, or Southwest?
17
Solution: Use the following algorithm: at pointA write the number 1. At every other vertexV, if allof the vertices from which one could travel toV have numbers, write the sum of these numbers onV.Continue until all vertices are numbered. The number onB is the number of paths fromA to B. It iseasy to verify that this method counts all paths fromA to B. We obtain the answer20 .
6. Evaluate the sum:
r
∑k=0
(r6
)(12− r6−k
)
Solution: You have a group of 12 objects, which you then split up into a group ofr objects and agroup of 12− r objects. To pick 6 objects, you must pickk objects from the group withr objects forsomek. Since we are summing overk, each possibility is counted once and only once, and the answer
is
(126
)= 924 .
7. Aaron has a coin that is slightly unbalanced. The odds of getting heads are60%. What are the oddsthat if he flips it endlessly, at some point during his flipping he has a total of three more tails thanheads?
Solution: Let P(n) be the probability that we getn more tails than heads at some point, so we’relooking for P(3). We haveP(k)P( j) = P(k+ j) because there is no distinction between startingpoints when the sequence is infinite. We also haveP(n) = 3
5P(n+1)+ 25P(n−1) from the result of
one flip. It is trivially obvious thatP(0) = 1, since we start there. ThenP(1) = 35P(1)2 + 2
5P(0), and
so 3P(1)2−5P(1)+ 2 = 0. SoP(1) = 5±√
25−246 = 5±1
6 . Obviously we want the nontrivial solution
P(1) = 23. ThenP(3) = P(1)3 =
827
.
8. Evaluate the sum:∞
∑n=0
5n+76n
18
Solution: Let S= ∑∞n=0
5n+76n . 6S= ∑∞
n=05n+76n−1 = ∑∞
n=−15(n+1)+7
6n . Subtracting, 5S= ∑∞n=0
56n +42=
51− 1
6+42= 48. So S=
485
.
9. A stick of length 10 is marked with 9 evenly spaced marks (so each is one unit apart). An ant is placedat every mark and at the endpoints, randomly facing either right or left. Suddenly, all the ants startwalking simultaneously at a rate of 1 unit per second. If two ants collide head-on, they immediatelyreverse direction (assume that turning takes no time). Ants fall off the stick as soon as they walk pastthe endpoints (so the two on the end don’t fall off immediately unless they are facing outwards). Onaverage, how long (in seconds) will it take until all of the ants fall off of the stick?
Solution: The key is to recognize that two ants colliding and reversing directions is equivalent to twoants passing each other and not affecting each other’s paths.
E(maxTwait) = n× (P(either endpoint faces inwards)+(n−1)×P(both endpoints face outwards + either of 2nd closest points face inwards)+ . . .
Sincen = 10 is even, we have that
E(maxTwait) =
( n2−1
∑i=0
34·(
14
)i
· (n− i)
)+(
12
)n
· n2
=2−n
3+
3n−13
=98991024
.
10. What is the largest possible number of vertices one can have in a graph that satisfies the followingconditions: each vertex is connected to exactly3 other vertices, and there always exists a path oflength less than or equal to2 between any two vertices?
Solution: Pick one vertex. This is connected to three distinct vertices. Each of those can be connectedto at most 2 other distinct vertices, for a total of 1 + 3 + 6 = 10. It is therefore impossible to satisfythe path-length requirement with more than 10 vertices. We verify that10 works by construction.
19
1.2.5 Calculus
1. Evaluate the limit:limx→∞
√x2 +x+1−
√x2−x−1
Solution: √x2 +x+1−
√x2−x−1 =
x2 +x+1−x2 +x+1√x2 +x+1+
√x2−x−1
limx→∞
2x+2√x2 +x+1+
√x2−x−1
≈ 2xx+x
= 1 .
2. The Three Stooges are standing at the vertices of an equilateral triangle with side length1. At thesame time Mo starts running towards Curly, Curly starts running towards Larry, and Larry startsrunning towards Mo. They all travel at the same speed s= 1 unit per second. How long, in seconds,does it take before the three of them collide?
Solution: From the symmetry in the problem we can see that the three guys will always be thevertices of an equilateral triangle. Suppose at a certain time this triangle has side lengthL. Whenthey next travel a very short distance dx, we can approximate their motion by a straight line. Thisforms a new, slightly smaller triangle with side lengthL′. We know they travel this distance in timedxs . We want to determine how quickly the distance between them decreases, i.e.,sL−L′
dx . Using Lawof Cosines, we find thatL′2 = (L−dx)2 + dx2−2Ldxcos(60) = L2−3Ldx+ 3dx2. We discard thedx2 term and findL′ =
√L2−3Ldx. Now we want
limdx→0
L−L′
dx= lim
dx→0
L−√
L2−3Ldxdx
.
Multiply by the conjugate of the numerator to get
L2−L2 +3Ldx
dx(L+√
L2−3Ldx)=
3L
L+√
L2−3Ldx.
It should now be clear that asdx→ 0, we have the closing speeds′ = s32. Then the time it takes for
them to collide is1s′ = 2
3s =23
.
3. Find all real m such that there exists a sequence a1,a2,a3, . . . where0 < ak < 1 for all k and
limn→∞
(a1a2a3 · · ·an) = m
Solution: We can findai such that the productm is any number in[0,1). Consider the sequencebn = 1− 1
(n+2)2 = (n+1)(n+3)(n+2)2 . Then
∞
∏n=0
bn =12
∞
∏n=k
bn =k+1k+2
limk→∞
k+1k+2
= 1
20
Supposem∈ [0,1). Let k be large enough that 1> k+1k+2 > m. Let a2 = bk,a3 = bk+1, . . ., and let
a1 = m(k+2)k+1 so that 0< a1 < 1 and we have the desired result. Som∈ [0,1)
4. Let a0 = 2006, a1 = 2007, and a2n = an−1+an−2 for n≥ 2. Assume that an > 0 for all n. Evaluate thelimit limn→∞ an.
Solution: If there is a limitL, it must satisfyL2 = L+L, soL = 0,2. Now, note that ifan−1 > 2 andan−2 > 2 thenan2 > 4 soan > 2. Asa0 > 2,a1 > 2, every term in our sequence is greater than 2, sothe limit must be2 .
5. Suppose x+y = 3e2. Find the maximum value of xy.
Solution: y = 3e2− x so let f (x) = xy = x3e2−x. Then ln( f (x)) = (3e2− x) ln(x) and so f ′(x) =x3e2−x
(3e2−x
x − ln(x))
. By inspection we see thatf ′(e2) = 0. Sign analysis shows that this is the only
relative extremum, and it is a maximum. So the maximum value isf (e2) = e2e2.
6. Evaluate the limit:
limn→∞
2n
∑k=n
(−1)k
k
Solution 1: ∑∞k=0
(−1)k
k is a convergent series. Therefore, for large enoughn, we know that∑∞k=n
(−1)k
k
is arbitrarily small. Therefore the limit is0 .
Solution 2: ∣∣∣∣∣ 2n
∑k=n
(−1)k
k
∣∣∣∣∣≤∣∣∣∣∣ 2n
∑k=n
1k(k+1)
∣∣∣∣∣≤∣∣∣∣∣ 2n
∑k=n
1k2
∣∣∣∣∣≤∣∣∣∣∣ 2n
∑k=n
1n2
∣∣∣∣∣≤ 1n
Clearly the RHS goes to 0 asn gets large. So the desired limit is0 .
7. Evaluate the sum:∞
∑n=1
n2
(n−1)!
Solution: Let Sk = ∑∞n=0
(n+1)k
n! . We already knowS0 = e, and we want to findS2. Note that(n+1)2
n! = n2+2n+1n! = n
(n−1)! + 2(n−1)! + 1
n! . But then = 0 term is 0, so those first two terms are just
those in the summations forS1 and 2S0. So we haveS2 = S1 +3S0. Now note thatn+1n! = 1
(n−1)! + 1n!
and we getS1 = 2S0 for the same reason. SoS2 = 5S0 = 5e .
8. Evaluate the limit:
limx→0
1x2
(cos(x)−1
x+
ln(1+x)2
)Solution: Using Taylor series for cosine and logarithm, we have
1x2
(cos(x)−1
x+
ln(1+x)2
)=
1x2
(1−x2/2+O(x4)−1
x+
x−x2/2+O(x3)2
)=
1x2(−x/2+O(x3)+x/2−x2/4+O(x3))
=−1/4+O(x)
21
So the limit asx→ 0 is −14
.
9. Suppose points A,B,C are chosen at random on the circumference of a circle of radius 1. What is theexpected value of the perimeter of triangle ABC?
Solution: On a circle, the average distance between two random points on the circle is
12π
Z 2π
02rsin
θ2
=4rπ
.
A triangle is three random points on a circle, so on average, the perimeter of the triangle will be
3· 4(1)π
=12π
10. Evaluate the sum:∞
∑n=0
1(n+1)2n
Solution: We write 2S(1/2) = ∑∞n=0
1(n+1)2n , soS(1/2) = ∑∞
n=01
(n+1)2n+1 . Write the summand asxn+1
n+1 ,
so we have a generalS(x) and observe that it is the same as
S(x) =∞
∑n=0
Z x
0
ddy
yn+1
n+1dy.
Now we exchange the order of summation and integration and take the derivative:
S(x)Z x
0
∞
∑n=0
yndy=Z x
0
11−y
dy=− ln(1−x).
We want 2S(1/2) = 2(− ln(1/2)) = 2ln(2) .
22
Chapter 2
Team Test
2.1 Rules
1. The 2006 Team Test will consist of 10 short-answer questions, and will last 30 minutes.
2. The first five problems are worth 8 points each and the last five problems are worth 12 points each.Under no circumstances will partial credit be given for an answer.
3. Proctors will give 5-minute and 1-minute warnings, and no other information.
4. At the end of the 30 minutes, teams must submit the completed official answer sheet to their Proctor.Copying of answers from the blackboard or other sheets of paper to the official answer sheet after thetime limit will not be allowed.
5. During the Team Test, any team member found communicating with anyone other than his or herteammates or Proctor will result in disqualification of the entire team from the Team Test.
2.2 Problems
1. Find the smallest positive integer n such that2n+1 and3n+1 are both squares.
Solution: We can do this by mods. If 2n+1 and 3n+1 are both perfect squares, they are restrictedin what they can be mod various numbers. 4 and 5 are the first interesting ones.
n a b0 1 11 3 02 1 33 3 2
Son≡ 0 (mod 4).n a b0 1 11 3 42 0 23 2 04 4 3
23
Perfect squares are 0, 1, or 4(mod 5), so we must haven≡ 0 (mod 5). Thus 20|n. Obviouslyn= 20doesn’t work, butn = 40 does, so that’s the answer. (One can actually check(mod 8) and find thatnecessarily 40|n.)
2. In triangle ABC, R is the midpoint of BC and CS= 3SA. If x is the area of CRS, y is the area of RBT, zis the area of ATS, and y2 = xz, then what is the value ofAT
TB? Express your answer in the forma+b√
cd ,
where a,b,c,d are integers, d is positive and as small as possible, and c is squarefree.
Solution: Let CR= BR= a, CS= 3b, AS= b, AT = kc, andBT = c. We are required to solvefor k. Now x = [CRS] = 1
2a · (3b) · sin∠C = 38 · (
12(2a) · (4b) · sin∠C) = 3
8[ABC], y = [BRT] = 12ac·
sin∠B= 12(k+1) ·(
12(2a) ·((k+1)c) ·sin∠B) = 1
2(k+1) [ABC], andz= [AST] = 12b·(kc) ·sin∠A= k
4(k+1) ·(1
2(4b) · ((k+ 1)c) · sin∠A) = k4(k+1) [ABC]. Also, y2 = xz. Substituting,( 1
2(k+1) [ABC])2 = 38[ABC] ·
k4(k+1) [ABC], which simplifies to the quadratic equation 3k2 + 3k− 8 = 0, which has positive root
k =√
105−36
.
3. Find all real solutions(x,y) to the equation y4 +2y2 +8x2 +16x4 = 24xy−8.
Solution: Write the expression asy4+2y2+8x2+16x4+4+4= 24xy. The purpose of this is to havethe same number of terms on the LHS as the sum of the powers of each variable on the LHS. We thensee if the AM-GM inequality can provide any information. Since all variables on the LHS are raisedto even powers, the terms are all positive, so the geometric mean of the LHS is6
√212y6x6 = 4xy. Then
by the AM-GM inequality, LHS≤ |RHS|. But we are given that LHS = RHS, and equality occursin the AM-GM inequality only when all terms are equal. Soy4 = 16x4 = 4, whencey = ±
√2 and
x =± 1√2. But we also need the RHS to be positive, sox andy have the same sign. Thus the solutions
are (x,y) ∈
{(√2
2,√
2
),
(−√
22
,−√
2
)}.
Alternatively, substitute 2x = z to obtainy4 + 2y2+ z4 + 2z2 = 12yz− 8. We’d like to find a nicefactorization of this, such as(y2 + a)2 + (z2 + a)2 + b(y+ cz)2. An obvious choice fora is 2, toget rid of the 8. Then we have(y2 + 2)2 +(z2 + 2)2−2y2−2z2−12yz= 0. This doesn’t seem too
24
helpful, but ifa =−2 then we have(y2−2)2 +(z2−2)2 +6y2−12yz+6z2 = 0, which is(y2−2)2 +(z2− 2)2 + 6(y− z)2 = 0. Since all of these are squares, every term must equal 0 individually, soy2 = 2,z2 = 2,y = z, which gives us the same answers as before.
4. Suppose that n> 1 and Pn(x) is a polynomial of degree n. For k= 1,2, . . . ,n we have Pn(k) = k(k+1).Also Pn(0) = 1. For all n there exists an integer m> n such that Pn(m) = Pn+2(m). Find the value ofm for n= 10.
Solution: First note that we can actually find the polynomialPn(x). This will make finding thedesired result much easier. We know thatPn(x)−x(x+1) hasn zeros at 1,2, . . . ,n. ThereforePn(x) =x(x+1)+c(x−1)(x−2) · · ·(x−n). ThenPn(0) = 1 = 0+c(−1)(−2) · · ·(−n) = c(−1)nn!, so
Pn(x) = x(x+1)+(−1)n
n!(x−1)(x−2) · · ·(x−n)
Pn(m) = Pn+2(m)
m(m+1)+(−1)n
n!(m−1)!
(m−n−1)!= m(m+1)+
(−1)n+2
(n+2)!(m−1)!
(m−n−3)!1
(m−n−1)(m−n−2)=
1(n+1)(n+2)
Clearlym= 2n+3 will make the two sides equal, regardless ofn. So whenn = 10, m= 23 .
5. How many pairs of positive integers(a,b) are there such that a< b and a,b can be the legs of a righttriangle with hypotenuse 340?
Solution: Any Pythagorean triple with even hypotenuse is a multiple of a primitive triple with oddhypotenuse. So it’s sufficient to find all triples with hypotenuses 5, 17, or 85 (the odd multiples of340). For 5 and 17 we get(3,4,5) and(8,15,17). For 85, we need to perform some calculations.All primitive Pythagorean triples can be written in the form(m2−n2,2mn,m2 +n2). If m2 +n2 = 85,then the only possible pairs are (9, 2) and (7, 6). These give the triples(36,77,85) and(13,84,85).In all, we have found4 solutions.
6. Consider the sequence1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,1, . . .
formed by writing the first power of two, followed by the first two powers of two, followed by the firstthree powers of two, and so on. Find the smallest positive integer N such that N> 100and the sumof the first N terms of this sequence is a power of two.
Solution: Let Sn be the sum of all the terms up to the first appearance of 2n−1. Then
Sn = n·1+(n−1) ·2+ ...+2·2n−2 +1·2n−1 (2.1)
2Sn = n·2+(n−1) ·22+ ...+2·2n−1 +1·2n (2.2)
Subtracting (1) from (2),Sn = −n+ 2+ 22+ ...+ 2n−1 + 2n = 2n+1−2−n. So we need the firstkpowers of 2 (which are the nextk terms of the series) to add up ton+ 2 for the sum of the seriesto be 2n+1, as required. IfNn is the number of terms in someSn, thenNn = 1+ 2+ ...+ n = n(n+1)
2 .The smallestNn > 100 occurs whenn = 14 andN14 = 105. The sum of the firstk powers of 2 is1+ 2+ 4+ ...+ 2k−1 = 2k−1. If we equate this ton−2 for n≥ 14, thenn = 29 is minimal, withN29 = 435, andk = 5. So the minimalN > 100 such that the sum of the firstN terms of the series is apower of two isN = N29+k = 440 .
25
7. S is a subset of{1,2, . . . ,100}. What is the maximum number of elements in S such that the productof any two of them is not a perfect square?
Solution: If ab= c2 for somec > 1, then eithera or b is divisible byd2 for somed > 1. To showthis, first note that there must be at least one prime factor ofc greater than 1. Letp be one such factor.Either p2|a, p2|b, or p|a,b, andc but p2 does not divide any of them. If the last case is true, sinceais not equal tob, there must be another prime factor ofc, and this process can be repeated. Sincecis finite, the process must end somewhere, and eithera or b is divisible by a square. We can have atmost one of the numbersd2n, n squarefree, in our set. Thus, if we exclude all numbers from our setthat are divisible by squares, we are guaranteed to have the maximum possible number of elements.Between 1 and 100, there are 25 numbers divisible by 4. There are 11 numbers divisible by 9, but twoof these are divisible by 36. There are 4 numbers divisible by 25, but one of these is divisible by 100.Finally, there are 2 numbers divisible by 49. This gives(100−25−11+2−4+1−2) = 61 .
8. Evaluate the sum∞
∑n=1
1n2(n+1)
Solution: Write1
n2(n+1)=
1n
(1
n(n+1)
)The inner term is a condensation of a familiar telescoping series, so let’s expand it:
1n
(1n− 1
n+1
)Distributing the1
n back in gives us1n2 − 1
n(n+1) . So we have
∞
∑n=1
1n2(n+1)
=∞
∑n=1
1n2 −
∞
∑n=1
1n(n+1)
The first sum isπ2
6 , a result of Euler which we will not show here. The second sum telescopes to 1
(look at the expansion above). So our answer is thusπ2
6−1 .
9. Suppose a,b,c are real numbers so that a+b+c= 15and ab+ac+bc= 27. Find the range of valuesthat may be obtained by the expression abc.
Solution: Vieta’s formulas may suggest to us to consider the roots of a cubic polynomial. Thatis, if p(x) = (x+ a)(x+ b)(x+ c) then p(x) = x3 + (a+ b+ c)x2 + (ab+ ac+ bc)x+ abc= x3 +15x2 + 27x+ abc. We want to find the range of possible values ofabc for which a,b,c will be real.In other words, the local maximum of the polynomial must be nonnegative and the local minimummust be nonpositive. Observe that there can only be one point where the local minimum is 0, andthere can only be one point where the local max is 0 (these are not the same point). This and thetwo given conditions are enough to determinea,b,c. Therefore, the interval between these is theone we want. If two of the variables are equal, then we have 2a+ b = 15 anda2 + 2ab= 27. Thisgivesa2 +2a(15−2a) = 27 so 3a2−30a+27= 0. This factors to 3(a−1)(a−9) = 0, soa = 1,9.Similarly a = 1 =⇒ b = 13 =⇒ abc= 13. a = 9 =⇒ b = −3 =⇒ abc= −243. Therefore,abc∈ [−243,13] .
26
10. The names of 8 people are written on slips of paper and placed in a hat. Each of the 8 people thenrandomly draw a piece of paper (without replacement). Then, the people are formed into groupssatisfying the following requirements:(i) Each person is in the same group as the person who drew his piece of paper.(ii)There are as many groups as possible while still satisfying condition (i).On average, how many groups will there be? (There might be “groups” of only one person.)
Solution: Pick one person to draw first. If he gets his own name, that will close off one group. Ifhe draws personx’s name, let personx draw next. If personx draws person one’s name, it will closeoff a group. If personx draws persony’s name, let persony draw next... etc. If a group is closed off,start a new group in the same way. Note that each time a person draws, there is exactly one draw hecan make that will close off the group. Since each group must be closed off exactly once, on average,
there are18 + 1
7 + 16 + 1
5 + 14 + 1
3 + 12 + 1
1 =761280
groups.
27
28
Chapter 3
Power Test
3.1 Rules
1. The Power Test will present an interesting definition of an advanced concept. The Power Test will last60 minutes, and students will be required to submit full solutions.
2. The Power Test will be graded out of 110 points.
3. The Power Test will have several parts. Results from previous parts may be assumed true in laterparts, even if the team fails to prove theses previous results. Students should cite these results whenthey use them. For example, From Part II, Question 1,
4. Proctors will give 5-minute and 1-minute warnings, and no other information.
5. At the end of 60 minutes, teams must submit their solutions written on official solution paper, inorder, with numbered pages within each solution (if that solution has more than one page). Collecting,copying, and/or re-ordering the solutions after the time limit will not be allowed. Solutions written onpages other than the provided official solution paper will not be accepted.
6. Solutions should be written on one side of the paper only.
3.2 Problems
A function f (x) is calledfinitely 2-refinable(we will just say “refinable” from now on) if there are constants{...,c−1,c0,c1,c2, ...} so that
f (x) = ∑j∈Z
c j f (2x− j) = . . .+c−1 f (2x+1)+c0 f (2x)+c1 f (2x−1)+ . . .
and only finitely many of the constantsc j are nonzero. For example, the functionf (x) = x is easily refinablewith c0 = 1
2 the only nonzero constant. Also,f (x) = k a constant is refinable in infinitely many differentways, so long as the sum of thec j is 1.
29
3.2.1 Characteristic functions of intervals
The function
f (x) ={
1 a≤ x < b0 otherwise
is called the characteristic function or indicator function for the interval[a,b). We often writeχ[a,b) to denotethis function.
1. Show thatχ[0,1) is refinable.
Solution: Denotef (x) = χ[0,1). Then f (2x) is 1 if x∈ [0,1/2) and 0 otherwise. Alsof (2x−1) is 1if x∈ [1/2,1) and 0 otherwise; it’s a right translate by 1/2. Thenf (2x)+ f (2x−1) = f (x), so f (x) isrefinable withc0 = c1 = 1.
2. Show thatχ[0, 32) is not refinable.
Solution: Denotef (x) = χ[0, 32). We note thatf (2x) = 1 whenx= 1/5 (say), as isf (2x+1). Suppose
thatc−1 is not zero. f (2x+1) is nonzero for somex < 0 too, so we will needc−2 =−c−1. But thenf (2x+2) is nonzero for−1≤ x <−1/2, wheref (2x+1) is zero. So we need to makec−3 =−c−2.Clearly this process will not terminate, so we won’t have a finite refinement. Soc−1 must be 0. Butthenc0 = 1, becausef (1/5) = 1. f (2x) = 1 up untilx = 3/4, andf (2x−1) is 1 when 1/2≤ x < 3/4also, soc1 must be 0. But then no other translate of the function is nonzero for 3/4≤ x < 1, so wewon’t be able to achieve the proper value of 1 on that interval. So this alternative doesn’t work either,and we conclude that there’s no finite refinement off (x).
3. For which a and b isχ[a,b) refinable?
Solution: The functionf (x) = χ[a,b) is refinable if and only ifa andb are integers. If the endpointsare integers, thenf (x) = f (2x−a)+ f (2x−b). If the endpoints are not integers, there are two possiblesituations. Ifb−a < 1, then f (2x− j) is nonzero on an interval of length less than 1/2, but we canonly translate by half-integers, so we can’t cover the entire line by translations of interval[a/2,b/2).If b−a = 1, then the translations of the intervals cover the real line, but only one translate is nonzeroat any given point, so each one that overlaps the original interval must have coefficient 1. But asa isnot an integer,a/2+ j/2 never equalsa for integersj, so the leftmost interval that overlaps[a,b) willextend too far to the left, and so we cannot makef (x) out of translations off (2x). If b−a > 1, thenwe may have overlap. But for the same reason as previously, the leftmost interval that overlaps[a,b)extends too far left. So the next interval will have the opposite coefficient to cancel out, but that oneextends too far left again, and so on, so our refinement will not be finite. If the leftmost interval hascoefficient 0, then the next farthest left interval either doesn’t covera or still extends too far left, andwe repeat this finitely many times to conclude that the function is not refinable (sinceb−a is finite).
3.2.2 Polynomials
1. Show that f(x) = x2 +1 is refinable.
Solution: f (2x− j) = (2x− j)2 +1 = 4x2−4x j + j2 +1. Try j =−1,0,1. We wantc−1(4x2 +4x+
30
2)+c0(4x2 +1)+c1(4x2−4x+2) = x2 +1, so
c−1 +c0 +c1 =14
c−1 = c1
2c−1 +c0 +2c1 = 1
Using the second equation, the first and last become
2c1 +c0 =14
4c1 +c0 = 1
2c1 =34
=⇒ c1 = c−1 =38
c0 =−12
So the function is refinable, as desired.
2. Show that all polynomials of degree 2 are refinable in infinitely many distinct ways.
Solution: This is exactly the same as the previous problem, except more general. Letf (x) =ax2 + bx+ c and let j,k,m be distinct integers. We want to find refinement coefficientsc j ,ck,cm sothat
f (x) = c j f (2x− j)+ck f (2x−k)+cm f (2x−m)a = 4c j +4ck +4cm
b =−4 jc j −4kck−4mcm
c = j2c j +k2ck +m2cm
At this point there are two approaches. The elementary approach is to solve these equations directly.We will not work this solution here because it requires lengthy algebraic manipulations, but the answeris
c j =amk+b(m+k)+4c
4( j−m)( j−k)
ck =am j+b(m+ j)+4c
4(k−m)(k− j)
cm =a jk+b( j +k)+4c
4(m−k)(m− j)
The alternative solution requires some linear algebra. We use the fact that a system of linear equationshas a unique solution if and only if the determinant of the matrix of coefficients is nonzero. So wecompute
det
∣∣∣∣∣∣4 4 4−4 j −4k −4m
j2 k2 m2
∣∣∣∣∣∣= 16(−m2k+mk2 +m2 j−m j2−k2 j +k j2)
= 16(k−m)(m− j)( j−k) 6= 0
31
since all three integers were assumed to be distinct. So the refinement coefficients are uniquely deter-mined by the coefficients of the polynomial. We have a different refinement for any three integers wechoose, so there are infinitely many distinct refinements.
3. In fact, all polynomials are refinable. Prove that for some class of polynomials, the entire class ofpolynomials is refinable. You may receive full credit for your solution even if you don’t prove the mostgeneral possible result.
Solution: Lettingn be the degree of the polynomial, we may writef (x) = ∑ni=0aixi with an 6= 0 since
the zero polynomial is trivially refinable. Substituting, we need to satisfy
f (x) = ∑j∈Z
n
∑i=0
c jai(kx− j)i .
By the binomial formula, we get
f (x) = ∑j∈Z,0≤l≤i≤n
c jai
(il
)kl xl (− j)i−l =
n
∑l=0
xl kln
∑i=l
∑j∈Z
ai
(il
)c j(− j)i−l .
Comparing with the definition off (x) and setting coefficients equal gives us:
kln
∑i=l
∑j∈Z
ai
(il
)c j(− j)i−l = al
for 0≤ l ≤ n. Making the change of summation indexm= i− l , the sum is rewritten:
kln−l
∑m=0
am+l
(m+ l
l
)∑j∈Z
c j(− j)m = al .
This is a triangularn+ 1×n+ 1 system of equations in the unknowns∑ j∈Z c j(− j)m with diagonalcoefficientsankl . Sincean,k 6= 0, a solution exists. Now, pick any set ofn+1 distinct integers, e.g. 1to n+1, to be the set ofj ’s with non-zeroc j ’s. Then our problem of findingc j ’s reduces to the matrixequationAc= b where A is then+1×n+1 matrix:
1 1 1 1− j1 − j2 . . . − jn+1
(− j1)2 (− j2)2 . . . (− jn+1)2
......
......
(− j1)n (− j2)n . . . (− jn+1)n
c is the column vector of thec j ’s, andb is the column vector of the∑ j∈Z c j(− j)m found above. ButA is the transpose of a Vandermonde matrix with distinct row vectors, so its determinant is non-zero.Thus, we can computeA−1b to find the values of thec j ’s.
3.2.3 Rational functions
1. Show that the function 1x2−1 is refinable.
32
Solution: The methods from before obviously won’t work for this kind of function. With charac-teristic functions of intervals, we considered the endpoints of the intervals. With polynomials, weconsidered equating coefficients on each power ofx. For rational functions, the obvious distinguish-ing characteristics to consider are the locations of the poles (the points where the denominator is 0). Inthis case,x2−1 = (x−1)(x+1), so we might as well decompose the function using partial fractions.We get
f (x) =1
x2−1=
12(x−1)
− 12(x+1)
f (2x) =1
4x−2− 1
4x+2
As with the characteristic functions, we see that the poles have been ”pulled twice as close together.”We can try to shift the function over left and right so that the poles of the refinement match the polesof the original function:
f (2x−1)+ f (2x+1) =(
14x−4
− 14x
)+(
14x− 1
4x+4
)f (x) = 4 f (2x−1)+4 f (2x+1)
So the function is in fact refinable, as desired.
2. Show that the function 1x2+1 is not refinable.
Solution: The difference between this and the previous problem is that this function does not havepoles (or at least, they’re not on the real line). One approach is to flagrantly factor using complexnumbers and do partial fractions again, which gives
f (x) =1
x2 +1=
12i(x− i)
− 12i(x+ i)
f (2x− j) =1
2i(2x− j− i)− 1
2i(2x− j + i)
But clearly no integerj will recreate the old denominator. To make this proof precise, observe thatthe locations of the poles are in fact atx = i andx =−i. But in the refinement, they are atx = j
2 + 12 i
andx = j2 −
12 i. These have imaginary part±1
2, regardless ofj. So we can never obtain the originalfunction by translations off (2x).
3. Show that if r(x) = p(x)q(x) , where p(x) and q(x) are polynomials with no common factors, then r(x) is
refinable only when all of the roots of q(x) are real integers.
Solution: First consider the case thatr has a complex pole, then it is not refinable. Suppose thatr hasat least one complex pole. Assume for contradiction thatr is refinable. If this is the case, then bothr(x) and∑ j r(2x− j) have exactly the same poles. Leth be the maximum of|ℑ(z)| over all poleszofr. Then all poles ofr(2x− j) will be contained in the horizontal strip|ℑ(z)| ≤ h/2. But r has a polez0 such that|ℑ(z0)| = h, which is not contained in this strip. This is a contradiction and sor is notrefinable.
We now show that ifr has a pole that does not occur at an integer, thenr is not refinable. There areonly three possible situations where the translates of the poles of∑ j c j r(2x− j) can have a non-zeroc j .
33
(a) The left-most pole ofr(2x− j) occurs in the same location as the left-most pole ofr(x).(b) All of the poles ofr(2x− j) occur at locations in between the left and right-most poles ofr(x).(c) The right-most pole ofr(2x− j) occurs in the same location as the right-most pole ofr(x).
In other words, to include translates ofr(2x) that have poles outside the region bounded by the leftand right-most poles ofr(x) would require inifitely manyc j to ensure thatr(x) and∑ j c j r(2x− j) hadexactly the same poles. From this we can assume without loss of generality that the left-most pole of∑ j c j r(2x− j) lines up with the left-most pole ofr(x), and any translate ofr(2x) that shifts the polesto the left must havec j = 0.
Supposer has at least one non-integer pole. Then there is some poleP such thatP is the left-mostnon-integer pole. SinceP is a non-integer, the mappingr(2x) will have a pole atP/2, which is notan integer or half-integer (this pole atP/2 will be the left-most pole in the functionr(2x) that is notan integer or half-integer). Thus, no translation of poles occuring at integers can establish a pole atx = P in the function∑ j c j r(2x− j). SinceP itself is non-integer, the distance between the pointsPandP/2 is not an integer or half-integer. Thus, translating a pole atx = P/2 by integers will neverresult in a pole atx = P. The only other poles that can occur left ofx = P are those that may fall inbetweenx = P/2 andx = P. If one of these poles can be integer translated to line up with the pole atx = P, it must leave a new pole at the integer translate of the pole atx = P/2. No translates ofr(2x)can cover this new pole because it is the integer translate of the left-most non-integer pole. Note alsothat there are no other non-integer poles to the left ofP. We cannot left-translate the poles on the rightside ofP due to the reasons given above. Therefore, there is no way to establish a pole atx = P inthe function∑ j c j r(2x− j), without introducing a new, uncoverable pole at the translate ofx = P/2.Hence,r(x) 6= ∑ j c j r(2x− j) andr is not refinable.
Therefore,r is refinable only when all roots ofq(x) are real integers.
4. Show that if
r(x) = ∑k
pk(x)(x−k)mk
where the sum is finite, the mk are integers, and the pk are polynomials, then r(x) is refinable only ifpk(x) is a constant for all k and mj = mk for all j ,k.
Solution: We will prove this by contradiction. If∃ j,k∈ Z such thatmj 6= mk, thenr is not refinable.In other words, every pole ofr must be of the same order andr must be a proper rational function.
Assume thatpi(x) = αi . Sincer has finitely many poles and at least one pole of positive order, it musthave a left-most pole of positive order. Let the place of occurence of this pole bex = i and the orderof the pole bemi > 0. Sincer has a pole of ordermi at x = i, the mapping∑ j c j r(2x− j) must alsohave an ordermi pole atx = i, if r is to be refinable. The translation of poles that are not of ordermi
can never establish an ordermi pole. And from the same argument as in the previous problem, wecannot use any pole occuring to the right ofx = i to establish a pole atx = i. This leaves only theoption of using translates of the contracted version of the pole atx = i itself to create a pole atx = i in∑ j c j r(2x− j), requiring the equation
ciαi
((2x− i)− i)mi=
αi
(x− i)mi.
Thus,ciαi = 2mi αi and soci = 2mi . In order to establish a pole of ordern, only translates of other polesof ordern can be used. This means that ifr has poles of varying order,r can be separated into a sum of
34
functionsrt , where eachrt is a function that has poles of all the same order. Consequently, in order forr to be refinable, all of the functionsrt must be refinable with the same refinement sequence. However,as we noted a moment ago, the left-most pole of each order,mn, will necessarily have a refinementcoefficient ofcn = 2mn. Thus, none of the functionsrt can have the same first refinement coefficientunless they have poles of the same order. Therefore, ifr has poles of varying order,r is not refinable.Moreover, sincer is assumed to have at least one pole of positive order, any refinabler cannot have apole of negative order. This means thatr cannot be reduced to the sum of a polynomial and a rationalfunction, which implies that all rational refinable functions must be proper rational functions.
3.2.4 Almost-refinability
A function f (x) is called almost-refinable if we can write
f (x) = k+∑j
c j f (2x− j)
wherek is a nonzero constant.
1. Show that if f(x) = g(x)+d, where g(x) is refinable and d is a constant, then f(x) is almost refinable.
Solution: We are given thatg(x) is refinable, so it has a refinement with refinement coefficientsc j .That is,
g(x) = ∑j
c jg(2x− j)
g(x)+d∑j
c j = ∑j
c j(g(2x− j)+d)
f (x)−d+d∑j
c j = ∑j
c j f (2x− j)
f (x) = d
(1−∑
j
c j
)+∑
j
c j f (2x− j)
So f (x) is almost refinable, as desired.
2. Show that there exists a function f(x) that is almost refinable but does not differ from a refinablefunction by a constant (that is, there is no constant d so that f(x)− d = g(x) is refinable). Yourfunction need not be defined on the entire real line.
By looking at the previous problem, we see that iff (x) is almost refinable and the sum of its refinementcoefficients isC, then we can reverse the process above and find a refinableg(x) unless∑ j c j = 0. Sowe want to find a function that satisfies
f (x) = k+∑j
c j f (2x− j)
∑j
c j = 1
It’s simplest to assume that only one refinement coefficient is nonzero. Then we only need to find asolution to the functional equation
f (x) = k+ f (2x)
35
Taking the derivative of this (assuming the function is differentiable; why not?), we getf ′(x) =2 f ′(2x), so it’s homogeneous of degree -1, like1
x . The function whose derivative is1x is ln(x). Sowe try f (x) = ln(x). Sinceln(2x) = ln(2)+ ln(x), we findk =− ln(2), and f (x) = ln(x) is an almostrefinable function which does not differ from a refinable function by a constant.
36
Chapter 4
Relay Test
4.1 Rules
1. Each team will be broken down into two sub-teams of five members each.
2. Each of the five people will receive a different problem.
3. When a person in relay positions 1, 2, 3, or 4 solves his or her problem, the answer should be writtenon an Official Relay Passing Slip, and passed to the right.
4. Every contestant is to work on his or her own problem. The only means of communication will bethrough the Official Relay Passing Slip, and the only thing that can be written on the Official RelayPassing Slip is the answer that the contestant has arrived at. This answer may be double-underlinedto avoid ambiguity (for example, to differentiate between a 6 and a 9), but no other markings may bemade on the Official Relay Passing Slip.
5. There is no limit to the number of official slips that may be passed. If, for example, a contestantrealizes that a previous answer was incorrect, he or she is allowed to pass another slip with a differentanswer.
6. Relay positions 2, 3, 4, and 5 will have problems that use the result of the previous persons question.For example, relay question 3 involves the answer to relay question 2.
7. When the person in relay position 5 gets his or her answer, it should be filled out on an Official RelayAnswer Slip. This is the only answer that will be considered for credit.
8. A Relay will last 12 minutes. Proctors will accept answers from the person in relay position 5 in the15-second windows before the 6-minute mark and before the 12-minute mark.
9. If the answer submitted at 6 minutes is correct, and no answer is handed in later, 10 points will beawarded. If an answer is handed in at 12 minutes, any answer previously handed in will be discardedby the Proctor. Correct answers submitted at 12 minutes will be awarded 5 points. Obviously, do notsubmit the same answer twice! Under no circumstances will partial credit be given for an answer.
37
4.2 Problems
4.2.1 Round 1
1. Given the triangle as shown, ABC= ADB= 90◦,AD = 4x,BD = 3x,BC= y, and CD= 9. Find y.
Solution: Using the Pythagorean Theorem on4ABD, AB = 5x. Now ∠CDB = 180◦−∠ADB =90◦ = ∠BDA and∠CBD= ∠ABC−∠ABD= 90◦−∠ABD= 180◦−∠ADB−∠ABD= ∠BAD, so4BCD4ABD (AA). Then BC
AB = CDBD, so y
5x = 93x. Therefore,y = 15 .
2. Let x be TNYWR, and let y= x3. A criminal, having escaped from prison, travelled for 10 hours
before his escape was detected. He was then pursued and gained upon at y miles per hour. When hispursuers had been 8 hours on the way, they met a train going in the opposite direction at the samerate as themselves, which had met the criminal 2 hours and 24 minutes earlier. In what time from thebeginning of the pursuit will the criminal be overtaken?
Solution: Let the criminal’s rate ber. After 18 hours, he’s gone 18r and the cops have gone 8(y+ r).We are told 2(y+ r)(2.4) was the distance between them 2.4 hours before. So(18− 2.4)r − (8−2.4)(y+ r) = 4.8(y+ r) and 10.4(y+ r) = 15.6r.104y = 52r → 2y = r. So we want(10+ t)2y = 3yt,which means 20+2t = 3t andt = 20. So the answer is20 .
3. Let z be TNYWR, and let y= z4. Evaluate
(y1
)+(y
3
)+ . . .+
(yy
), where
(mn
)is the number of distinct
teams of n people chosen from a pool of m≥ n people.
Solution: 2y−1 = 24 = 16 .
4. Let a be TNYWR. Compute the sum of the first 99 terms oflog4a− log4a2+ log4a3− . . ., wherelogxystands for the exponent that you put on x to get y. So if z= logxy, then xz = y.
Solution: log416= 2. log4an = nlog4(a). So we are looking for 2(1−2+3−·· ·+99) = 2(50) =100 .
5. Let d be TNYWR. Find the focal length of the ellipsex2
d + y2
36 = 1.
Solution:√
100−36= 8 .
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4.2.2 Round 2
1. How many ordered pairs of integers(x,y) satisfy the equations xy = 1 and x+y = 3?
Solution: If x+y = 3 thenxy = x3−x. Thenx3−x = 1 meansx =±1 or 3−x = 0, sox =∈ {−1,1,3}.So there are3 such pairs of integers.
2. Let k be TNYWR. Suppose ABC is a right triangle with integer sides and one leg of length k. Find thearea of the inscribed circle for triangle ABC.
Solution: This is a 3− 4− 5 right triangle. The area of a triangle is the semiperimeter times theinradius,sr = A. But we knows andA: 6r = 6 =⇒ r = 1. So the area of the circle isπ .
3. Let S be TNYWR, where S is the surface area of a right circular cone. Given that the slant height ofthe cone is5
6, what is the radius of the base of the cone?
Solution: The surface area of a cone isπrs+ πr2. We are given that the surface area isπ and theslant height is5
6. Soπ = πr 56 +πr2; 6r2+5r−6= 0 factors to(3r−2)(3r +3) = 0. Only the positive
root makes sense;r =23
.
4. Let x be TNYWR. If a solid has9x congruent faces and18x edges of length12x, what is the greatestdistance between two vertices on the solid?
Solution: The solid has 6 congruent faces and 12 edges of length 8; this is a cube of side length 8.The greatest distance between two vertices of a cube is the distance between opposite vertices, which
is√
3s= 8√
3 .
5. Let n be TNYWR. A (very) unfair coin has a probabilitypq of coming up heads, wherepq is in lowest
terms. If2pq= n2 what is the greatest possible probability that of two throws, at least one will comeup heads?
Solution: We are given that 2pq = (8√
3)2 =⇒ pq = 96. As we are given thatp andq have nocommon factors, the only possible pairs are(1,96),(3,32),(32,3),(96,1). But sincep
q is a probability,p≤ q, so only the first two make sense. Then, we want to find the greatest possible probability thatout of two throws at least one is heads. That’s the same as the probability that in two throws, not both
are tails, which is 1−(
q−pq
)2. This is larger ifq− p is smaller, so(3,32) is the right pair. So we want
1− 292
322 = 1024−8411024 =
1831024
.
39