at 2008 mocks solutions

Mock Olympiad 1

Time: 4 hours

1. Find all polynomials f with non-negative real coefficients such that f(1) = 1 and

f(x)f(1

x) ≤ 1

for all x ∈ R, x > 0.

Solution: Let f by a polynomial with degree n that satisfies the condition given inthe problem. Let

f(x) =n∑

i=0

aixi

with ai ≥ 0 and an 6= 0. Since the coefficients of f are non-negative, by Cauchy-SchwarzInequality, we have

f(x)f(1

x) = (

n∑i=0

aixi)(

n∑i=0

aix−i) ≥ (a1 + a2 + · · ·+ an)2 = f(1)2 = 1.

Since f(x) ≤ 1 as stated in the problem, we conclude that equality holds in the givenequality. In order for this equality to hold, we need that

aixi

aix−i= x2i

to equal for all ai 6= 0 and for all x. This is impossible if there are at least two co-efficients in f which are strictly positive. Hence, f(x) = anx

n. Since f(1) = 1, thenan = 1. Therefore, f(x) = xn for any n ∈ Z≥0. This is easily verified as a solution. �

2. An acute-angled triangle ABC has circumcentre O. Let K, L, M be the intersectionpoints formed by the three circles passing through O whose centres are the midpointsof the sides AB, BA, CA. . Prove that O is the incircle of triangle KLM .

Solution: Let D, E, F be the midpoints of BC, CA, AB respectively. Let ω1, ω2, ω3

be the circles passing through O centred at the midpoints of D, E, F respectively. LetK = ω2 ∩ ω3, which is not O. Define similarly L = ω3 ∩ ω1 and M = ω1 ∩ ω2.Then DL = DM,EM = EK, FK = FL. Since EF ||BC and OD is perpendicularto BC, then OD ⊥ EF . Similarly, OE ⊥ FD and OF ⊥ DE. Therefore, O isthe orthocentre of ∆DEF . But EO = EK and FO = FK implies that EKFO isa rhombus. Hence, OK ⊥ EF . Hence, D, O, K are collinear. Similarly, E, O,L arecollinear and F, O, M are collinear. Since O is the orthocentre of ∆DEF , we have

1

∠FEK = ∠FEO = 90 − ∠EFD = ∠FDO = ∠FDK. This implies F, K, E, Dare concyclic. By symmetry, we conclude that D, E, F, K, L, M are concyclic. Hence,∠LKO = ∠LKD = ∠MKD = ∠MKO, implying O is on the internal angle bisectorof ∠K in ∆KLM . Similarly, O is on the internal angle bisector of ∠L and that of ∠Min ∆KLM . Hence, O is the incircle of ∆KLM . �

3. Let a, b, c be non-negative real numbers such that a + b + c = 3. Prove that

(a2 − ab + b2)(b2 − bc + c2)(c2 − ca + a2) ≤ 12.

Solution: Without loss of generality, suppose that c = min{a, b, c}. Then

(a2 − ab + b2)(b2 − bc + c2)(c2 − ca + a2) = (a2 − ab + b2)(b2 − c(b− c))(a2 − c(a− c))

≤ (a2 − ab + b2)a2b2 = ((a + b)2 − 3ab)a2b2 ≤ (9− 3ab)a2b2 = −3(ab)3 + 9(ab)2

= −3(ab− 2)2(ab + 1) + 12 ≤ 12.

Careful analysis of the inequalities show that equality holds if and only if (a, b, c) =(2, 1, 0) and its permutation solutions. �

4. Find all finite set of points S in the plane such that for every three distinct pointsA, B, C in S, there exists a fourth point D in S such that A, B, C,D are the verticesof a parallelogram.

Solution: I claim that S must consists of exactly four points, which are vertices ofthe parallelogram. Let A, B, C be three vertices of S that are vertices of the triangleof maximum area amongst all triangles whose vertices are in S. Draw a line l1 throughA parallel to BC, a line l2 through B parallel to CA and a line l3 through C passingthrough AB. Let A′ = l2 ∩ l3, B

′ = l3 ∩ l1, C′ = l1 ∩ l2. Since ∆ABC is the triangle of

maximum area amongst all triangles whose vertices are in S, then all points in S mustbe in the interior or the boundary of triangle A′B′C ′. Let D be a point in S that formsa parallelogram with A, B, C. Without loss of generality, suppose this point is D = A′.By performing the same operation on ABC to form A′B′C ′ as on BCD, we concludethat the vertices of S must be in the interior or the boundary of the parallelogramABCD. Suppose X ∈ S and X 6= A, B, C,D. Then X is not on a side of ABCD, sayAB. Consider the three points X, A,B and Y ∈ S such that X, A,B, Y are verticesof a parallelogram. If Y is opposite X in the parallelogram, then X, Y are in differenthalf planes formed by line AB, implying Y is external of ABCD, which is impossible.If Y is opposite A in the parallelogram, then A, Y are in different half planes formed byline BX. Then ∠CAX = ∠DBY as directed angles, implying Y is external of ABCD,which is impossible. Similarly, Y cannot be opposite B in the parallelogram. Hence,

2

X cannot exist. Hence, A, B, C,D are the only points in S. �

Source: USA Team Selection Test 2005

5. Let m, n with m ≥ n be non-negative integers with the same parity (i.e. both odd orboth even) such that m2−n2+1 divides n2−1. Prove that m2−n2+1 is a perfect square.

Solution: Without loss of generality, suppose m ≥ n. Let a, b ∈ Z≥0, a > b such thatm = a + b and n = a− b. This is possible since m, n are of the same parity. Therefore,m2 − n2 + 1 = 4ab + 1. Suppose m2 − n2 + 1 divides n2 − 1. Then m2 − n2 + 1 dividesm2 = (a + b)2. Hence, 4ab + 1 divides (a + b)2 (*). Suppose a, b are chosen minimallysuch that 4ab + 1 divides (a + b)2. Let

k =(a + b)2

4ab + 1.

Then we can rewrite this as a quadratic in a as

a2 + (2b− 4k)a + b2 − k = 0.

Consider the quadratic equation x2− (2b−4k)x+ b2−k = 0. One root of this equationis a. Let a′ be the other root, which is an integer. Since aa′ = b2 − k and a > b andk ≥ 0, then a′ < b < a. If k < b2, then a′ > 0 and (a′, b) satisfies (*), contradicting theminimality of (a, b). If k = b2, then k is a perfect square and we are done. If k > b2,then a′ ≤ −1. Then 0 ≥ (a′ + 1)(a + 1) = aa′ + (a + a′) + 1 = b2 − k − 2b + 4k + 1 =(b − 1)2 + 3k > 0, contradiction. Therefore, k = b2 is the only remaining case andtherefore k is a perfect square. �

Source: Irish Math Olympiad Round 2, 2005

3

Mock Olympiad 2

Time: 4 hours

1. Let n be an odd positive integer. There are n computers in a room with exactly onecable joining each pair of computers. Colours are assigned to the computers and thecables such that

(1) Every computer is assigned a distinct colour.(2) Two cables joined to a common computer are assigned distinct colours.(3) A computer and any cable joined to the computer are assigned distinct colours.

Prove that this colouring can be done using a total of n distinct colours.

Solution 1: Label the n computers C0, C1, · · · , Cn−1. Let {0, 1, · · · , n − 1} be the ncolours. Colour the cable joining computers Ci, Cj with colour (i+j) mod n and colourcomputer Ci with colour (i+ i) mod n. We claim this colouring works. Clearly, no twoincident cables are assigned the same colour and no computer is the same colour asany cable connected to it. Suppose Ci, Cj are assigned the same colour. Then 2i ≡ 2jmod n. Since n is odd, then gcd(2, n) = 1. This implies i ≡ j mod n implying i = j.Therefore, each computer are assigned different colours. �

Solution 2: (Hunter Spink) Place the n computers around the circle as vertices of aregular n-gon. Colour the computers {1, 2, · · · , n}. Let Ci be the computer coloured i.Consider a pair of computers Ci, Cj; since n is odd, there is exactly one other computerCk such that Ck is equidistant to Ci and Cj. Colour the cable joining Ci, Cj colourk. I claim that this colouring works. Clearly, no two computers are assigned the samecolour and no computer is assigned the same colour as any cable joining it. Supposethe cable joining Ci and Cj and the cable joining Ci and Cl are assigned the samecolour k. But there is exactly one other computer that is the same distance to Ck asCi. Hence j = l. This is a proper colouring. �

Source: Italy Team Selection Test 2007

2. Let f : R+ → R be a strictly increasing function 1 on the positive reals such that

f

(2xy

x + y

)=

f(x) + f(y)

2for all x, y ∈ R+ .

Show that f(x) < 0 for some value of x.

1To avoid confusion, a function is strictly increasing if f(x) < f(y) for x < y.

4

Solution: If x, y, z is a harmonic sequence (i.e. y = 2xzx+z

), then the given con-dition f implies that f(x), f(y), f(z) is an arithmetic sequence. Consider the har-monic sequence 1, 1/2, 1/3, 1/4, · · ·; since f is a strictly increasing sequence, thenf(1), f(1/2), f(1/3), f(1/4), · · · is a strictly decreasing arithmetic sequence, which musteventually become negative. �

Source: Brazilian Math Olympiad 2003

3. Let x, y, z be any positive real numbers such that x + y + z = 3. Prove that

x3

(y + 2z)2+

y3

(z + 2x)2+

z3

(x + 2y)2≥ 1

3.

as desired. �

Solution: Note that by AM −GM inequality, we have

x3

(y + 2z)2+

y + 2z

27+

y + 2z

27≥ 3x

9=

x

3.

Hence, ∑cyc

x3

(y + 2z)2≥

∑cyc

9x− 2y − 4z

[27 =

3(x + y + z)

27=

1

3.

4. Let ABC be an acute-angled triangle and let P be the point inside ∆ABC such that∠APB = ∠BPC = ∠CPA = 120o. Prove that the Euler lines of ∆APB, ∆BPC and∆CPA are concurrent. 2

Solution: Let A′ be external to ABC such that ∆A′BC is an equilateral triangle.Define B′, C ′ analogously. Since ∠BPC = 120o and ∠BA′C = 60o, then BPCA′ iscyclic. Therefore, ∠A′PB = ∠A′CB = 60o. Since ∠BPA = 120o, this implies A, P, A′

are collinear. Let D be the midpoint of BC, O be the circumcentre of PBC (whichis also the circumcentre of A′BC, since BPCA′ is cyclic), G be the centroid of ABCand G′ be the centroid of PBC. Note that AG/GD = 2, PG′/G′D = 2, A′O/OD = 2,since the circumcentre of A′BC is also the centroid of ABC since ABC is equilateral.Therefore, G, G′, O are collinear and the Euler line of ∆PBC passes through G. Simi-larly, the Euler lines of ∆PCA and that of ∆PAB passes through G. Hence, the threeEuler lines are concurrent. �

Source: Brazilian Math Olympiad 2005

2Recall that the Euler line of a triangle is the line that passes through the circumcentre, orthocentre andcentroid of the triangle.

5

5. Let n be a positive integer. Let S1, S2, · · · , Sn be subsets of {1, 2, · · · , n} such that forall m ∈ {1, 2, · · ·n}, the union of any m of these subsets contains at least m elements.Prove that there exists a permutation (a1, a2, · · · , an) of (1, 2, · · · , n) such that ai ∈ Si

for all i ∈ {1, 2, · · · , n}.

Solution: We will prove this by strong induction on n. For n = 1, S1 = {1} sinceS1 must contain at least one element. Now for a given n that this statement is truefor all smaller n. Suppose the union of every m of these subsets contains at leastm + 1 elements. Without loss of generality, suppose n ∈ Sn. Then by removing n,amongst S1, S2, · · · , Sn−1, the union of m of these subsets contains at least m elements.Then there exists a permutation (a1, · · · , an−1) of (1, 2, · · · , n− 1) such that ai ∈ Si for1 ≤ i ≤ n− 1 and n ∈ Sn to give the desired permutation.

Otherwise, there exists a fixed integer t, with 1 ≤ t < n such that the union of t ofthe subsets contains exactly t elements. Without loss of generality, suppose the unionof S1, S2, · · · , St is exactly {1, 2, · · · , t}. Let C be a collection of subsets chosen from{St+1, St+2, · · · , Sn. Without loss of generality, suppose the union of the sets in C con-tains {t + 1, t + 2, · · · , t + p} for some p ≥ 0. Then S1, S2, · · · , St and the subsets inC contains at least t + |C| elements, which implies p ≥ |C|. Since S1, · · · , St does notcontain any elements at least t + 1, then C contains {t + 1, t + 2, · · · , t + p}. Hence, wehave shown that the subsets St+1, St+2, · · · , Sn, using elements {t + 1, · · · , n} satisfiesthe condition given the in problem. Hence, by strong induction, there exists a permu-tation of (a1, · · · , at) of {1, 2, · · · , t} such that ai ∈ Si for 1 ≤ i ≤ t and a permutation(bt+1, bt+2, · · · , bn} of {t + 1, t + 2, · · · , n} such that bi ∈ Si for t + 1 ≤ i ≤ n. Hence,(a1, · · · , at, bt+1, · · · , bn} is the desired permutation. �

Source: In graph theory, this is known as Hall’s Theorem. There is a famousresult related to this for bipartite graph. Google it up to read more aboutit.

6

Mock Olympiad 3

Time: 4 hours

1. For all reals 0 ≤ x, y, z ≤ 1, find the maximum possible value of

x + y + z − xy − yz − zx.

and determine all solutions (x, y, z) for which this maximum is attained.

Solution: Note that x + y + z − xy − yz − zx = 1 − (1 − x)(1 − y)(1 − z) − xyz.Since 0 ≤ x, y, z ≤ 1, then the maximum possible value is 1. This is obtained when atleast one of x, y, z is 1 and at least one of x, y, z is 0. i.e. (x, y, z) = (1, 0, t) for some0 ≤ t ≤ 1 and its permutation solutions.

2. There are n > 2 islands where for some pairs of islands, there is a bridge joining thetwo islands. It is known that the islands are connected; i.e. One can go from anyone island to any other island via the bridges (possibly by going through intermediateislands). There are at least n bridges total. Prove that there is a bridge whose removalwill keep the islands connected.

Solution: This is a well-known result in graph theory; a graph on n vertices with nedges contains a cycle. Hence, removing any edge from this cycle, keeps the graphconnected. We will elaborate in this solution.

We say that a sequence of distinct islands I1, I2, · · · , It (with t ≥ 3) forms a cycle if Ii

is connected to Ii+1 for 1 ≤ i ≤ t− 1 and It is connected to I1. Note that if the islandscontains a cycle and the islands are connected, then removing any bridge in the cyclekeeps the islands connected. We will now show that a cycle always exist. Suppose not.

Label the n bridges B1, B2, · · · , Bn. A group of islands is said to be connected if anytwo islands in the group are connected. (Note that if A, B are connected and B, Care connected, then A, C are connected.) Let mi be the number of groups of islandsthat are connected if only bridges B1, B2, · · · , Bi exists and the other bridges do not.Then m0 = n since initially there are n isolated islands. Then m1 = n − 1, becausetwo islands became connected and the number of groups decreased by 1. Observe thatmi − 1 ≤ mi+1 ≤ mi. Suppose mi = mi+1. Then the number of groups of islands didnot decrease and hence, bridge Bi+1 connects two islands that are already connected.This implies that a cycle is formed. Contradiction. Therefore, mi+1 = mi − 1. Sub-sequently,. mn−1 = 1. Since mn > 0, then mn−1 = mn and a cycle must have beenformed. �

7

Source: Standard graph theory fact.

3. Find all functions f : R → R such that

(x + y)(f(x)− f(y)) = f(x2)− f(y2).

Solution: Note that if f(x) satisfies the given condition, then f(x) + c also satisfiesthe given condition for any constant c. Without loss of generality, suppose f(0) = 0.Then substituting y = 0 yields xf(x) = f(x2) (1)

Expanding the equation given in the problem yields

xf(x)− xf(y) + yf(x)− yf(y) = f(x2)− f(y2)

which by (1) simplies toyf(x) = xf(y)

for all x, y,∈ R. For all x ∈ R − {0}, f(x)/x is constant. Let f(x) = kx for someconstant k. Therefore, f(x) = kx. This is consistent with f(0) = 0. By an earlierremark, f(x) = kx + c is a general solution. This is easily verified. �

Source: I do not recall where I got this from.

4. For |x| < 1, evaluate the sum

∑a,b∈N,gcd(a,b)=1

xa+b

1− xa+b.

in terms of x (in closed form), where (a, b) is over all pairs of relatively prime positiveintegers.

Note that ∑a,b∈N,gcd(a,b)=1

xa+b

1− xa+b=

∞∑d=2

φ(d)xd

1− xd

where φ(d) is the number of positive integers less than or equal to d which is relativelyprime to d. We recall an important fact regarding the function φ, which is that for allpositive integers n, we have ∑

d|n

φ(d) = n.

8

Therefore,

∞∑d=2

φ(d)xd

1− xd=

∞∑d=2

φ(d)xd(1 + xd + x2d + · · ·) =∞∑

d=2

φ(d)(xd + x2d + x3d + · · ·+)

As a power series, consider the coefficient of n for any positive integer n ≥ 2; It appears∑d|n,d6=1

φ(d) = n− 1 times. Hence, this power series equals

∞∑n=2

(n− 1)xn =∞∑

n=2

∞∑j=n

xj =∞∑

n=2

xn

1− x=

x2

(1− x)2.

Hence, the sum is equal to x2

(1−x)2.

Source: University of Waterloo Big E Competition 1998.

5. Let s be the semiperimeter of a triangle ABC and ω be the excircle of ∆ABC touchingside BC. Let P, Q be the two points on line BC such that |AP | = |AQ| = s. Provethat ω is tangent to the circumcircle of ∆APQ.

Solution: Let X, Y be where ω touch the lines AB and AC. Recall that |AX| =|AY | = s. Invert every point in the diagram about A with radius s. Given a point T ,let T ′ the its image of the inversion. The excircle ω remains tangent to AB and AC.Since X = X ′ and Y = Y ′, then ω is fixed in the inversion map. The circumcircle of∆APQ is mapped to a straight line passing through P ′ = P and Q′ = Q, which is theline BC. The image of the circumcircle of ∆APQ is tangent to the image of ω throughthe inversion. Hence, these two circles are tangent. �

Source: Spanish Math Olympiad 2001?

9

Mock Olympiad 4

Time: 4 Hours

1. Let n be a positive integer. Prove that any 2n × 2n board with exactly one emptysquare, can be completely tiled using the L-shaped tiles consisting of three squares.

Solution: We will prove this by induction on n. For n = 1, this problem is clear.Suppose given a positive integer n > 1, that the statement holds for n− 1. Divide the2n × 2n board into four quadrants, each which has dimenstion 2n−1 × 2n−1. Considerwhich quadrant the empty square is in. Consider the middle four squares, where eachquadrant contains exactly one of the four squares. Place an L such that the squarescorresponding to quadrants not containing the empty square is tiled. Each of the fourquadrants, with dimensions 2n−1× 2n−1, now has one square covered. (Well, one of thequadrants contains the empty square. We can consider this one square covered.) Thenby induction, we can cover each of the 2n−1 × 2n−1 quadrants and therefore, the wholeboard. �

2. Find all polynomials P with real coefficients such that

1 + P (x) =1

2(P (x− 1) + P (x + 1))

for all real x.

Solution: Let Q(x) = P (x + 1)− P (x). Then the given condition is equivalent to

Q(x)−Q(x− 1) = 2.

Therefore, Q has degree 1 and is of the form Q(x) = 2x + d. Since Q(x) = P (x +1) − P (x), then P (x) has degree two and is a quadratic. Let P (x) = ax2 + bx + cfor some a 6= 0. Plugging this into Q(x) = P (x + 1) − P (x) yields a = 1. Hence,P (x) = x2 + bx + c. It is easy to verify that this is a solution. �

Source: Swedish Math Olympiad, unknown year.

3. Let ABC be a triangle and let D be a point in its interior. Construct a circle ω1 pass-ing through B and D and a circle ω2 passing through C and D such that the point ofintersection of ω1 and ω2 other than D lies on line AD. Denote by E and F the pointswhere ω1 and ω2 intersect side BC, respectively, and by X and Y the intersections oflines DF, AB and DE,AC respectively. Prove that XY ||BC.

Solution: Let ω1 intersect AB at M and ω2 intersect AC at N . By radical axistheorem, the quadrilateral MNCB is cyclic. Since MDEB is cyclic, then ∠ANM =

10

∠ABC = ∠MDY , meaning MDY N is cyclic. Similarly NDXM is cyclic. The corre-sponding circles are both circumcircles of MND, therefore MNXY is cyclic. Meaning∠MXY = ∠ANM = ∠MBC. Therefore, XY ||BC as desired. �


4. Let n be a positive integer and p be an odd prime such that n divides p − 1 and pdivides n3 − 1. Prove that 4p− 3 is a perfect square.

Solution: Solution: Note that n < p. Since p|(n3 − 1) = (n − 1)(n2 + n + 1), and pcannot divide n − 1, then p|n2 + n + 1. I claim that p = n2 + n + 1, which solves theproblem since 4p− 3 = 4(n2 + n + 1)− 3 = (2n + 1)2 which is a perfect square.

Let p− 1 = qn. If n ≤ √p, then p ≤ n2 + n + 1 < 2p, which means p = n2 + n + 1. If

n >√

p, then q ≤ √p. Therefore, p|n2 + n + 1 = (p−1)2

q2 + p−1q

+ 1. The denominator of

this expression is divisible by p. i.e. p|(p− 1)2 + q(p− 1) + q2 ⇒ p|q2 − q + 1 < p sinceq ≤ √

p, contradiction. (Clearly q2 − q + 1 6= 0). Therefore, p = n2 + n + 1. �

Source: Iran Math Olympiad 2005

5. Let S be a finite set of points in the plane such that no three points are collinear. Foreach convex polygon whose vertices are in S, let a(P ) be the number of vertices of Pand b(P ) be the number of points of S outside of P . Prove that for all real numbersx, the polynomial, ∑

P

xa(P )(1− x)b(P ) = 1.

where sum is over all convex polygons P with vertices in S. (Note: A line segment, apoint and the empty-set are considered convex polygons of 2, 1, 0 vertices respectively.)

Solution: For each dot, colour it red with probability x and colour it blue otherwise.Note that given a convex polygon P , the expression xa(P )(1− x)b(P ) is the probabilitythat the vertices of the polygon are all red and all the vertices outside the polygon isblue. The summation expression is the expected number of convex polygons with thisproperty. But note that there always exist exactly one polygon with this property. Ifevery dot is blue, we take the empty polygon. Otherwise, the polygon whose verticesare that of the convex hull of the red dots is the only polygon satisfying this property.Therefore, for 0 ≤ x ≤ 1, the summation in the question is equal to 1. But sincethis is a polynomial, and it agrees with f(x) = 1 for infinitely many points, then this

11

summation is equal to 1 for all x. �

Source: IMO Shortlist 2006

12

Mock Olympiad 5

Time: Take-Home

1. Let ABC be a triangle, M is the midpoint of BC, Q be the point on BC such thatthe line AQ is the reflection of the line AM across the angle bisector of ∠BAC. (Theline AQ is called a symmedian of ∆ABC.) Prove that AB2/AC2 = BQ/QC.

Solution: Note that ∠BAQ = ∠CAM . We compare areas of triangles whose base isalong the side BC. Let [XY Z] denote the area of triangle XY Z.

Then 2[BAQ]/[ABC] = [BAQ]/[MAC] = (AB·AQ)/(AM ·AC). Also, [ABC]/2[QAC] =[BAM ]/[QAC] = (AB · AM)/(AQ · AC).

Multiplying the equations give us

BQ/QC = [BAQ]/[QAC] = AB2/AC2 as desired.

2. Let n be a positive integer and a1, · · · , an be real numbers. Let

f(x) =n∑

k=1

akxk, g(x) =

n∑k=1

ak

2k − 1xk.

Suppose that g(1) = 0 and g(2n+1) = 0. Prove that there exists a real number m inthe range 1 ≤ m ≤ 2n such that f(m) = 0.

Solution: The observation required is that f(2l) = g(2l+1)−g(2l). Therefore, summingthe telescoping sums yields

n∑l=0

f(2l) =n∑

l=0

g(2l+1)− g(2l) = 0.

Hence, if any of f(2l) = 0, we are done. Otherwise, f takes on positive and negativevalues in the range [1, 2n]. Since polynomials are continuous, f must take on zero inthis range.


3. Let A, B, C be angles of a triangle. Prove that

sin3A

2+ sin

3B

2+ sin

3C

2≤ cos

A−B

2+ cos

B − C

2+ cos

C − A

2.

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Solution: Note that cos A−B2

= sin 2B+C2

. Therefore, it suffices to show that∑cyc

sin 3A2≤∑

cyc

sin 2A+B2

which is equivalent to

∑cyc

sin A cosA

2+

∑cyc

sinA

2cos A ≤

∑cyc

sin A cosB

2+

∑cyc

sinA

2cos B.(∗)

WLOG, suppose A ≤ B ≤ C. Then sin A ≤ sin B ≤ sin C and cos A ≥ cos B ≥ cos C.Then ∑

cyc

sin A cosA

2≤

∑cyc

sin A cosB

2

and ∑cyc

sinA

2cos A ≤

∑cyc

sinA

2cos B

follow from rearrangement inequality. Therefore, (*) holds as desired.


4. Find explicitly all pairs of positive integers (m, n) such that mn− 1 divides m2 + n2.

Solution: Let S = {(m, n)|mn − 1 divides m2 + n2}. Clearly, (m, n) ∈ S if and onlyif (n, m) ∈ S.

First, we shall handle the case where m = n. This yields m2 − 1 divides 2m2, whichmeans m2 − 1 divides 2. Thus m2 − 1 = 1 or 2, which is impossible. Henceforth, weshall assume that m 6= n.

Also, if (m, n) ∈ S. Let m2 + n2 = k(mn − 1). Then m2 − kn · m + (n2 + k) = 0.Then there exists another solution (m′, n) where m′ is the integer root of the quadraticequation

x2 − knx + (n2 + k) = 0.

Since mm′ = n2 + k > 0, then both m, m′ are positive.

Lemma 1: Suppose (m, n) ∈ S with 1 < m < n. Suppose (m′, n) ∈ S for m′ 6= m.Then m′ > n.

Proof of lemma 1: m, m′ < n is impossible since mm′ = n2 + k > n2.

14

Lemma 2: Suppose (m, n) ∈ S with 1 < n < m. Suppose (m′, n) ∈ S for m′ 6= m.Then m′ < n.

Proof of lemma 2: Suppose m′, m > n. Then by quadratic equation, we have that

kn−√

k2n2 − 4(n2 + k)

2> n.

which simplies to 2n2 > k(n2 − 1). Since n > 1, this means k = 1, 2 or 3.

If k = 1, then m2 + n2 = mn − 1 ⇒ (m − n)2 + mn = −1, which is impossible. Ifk = 2, then m2 + n2 = 2mn− 2 ⇒ (m− n)2 = −2, which is also impossible. If k = 3,then 2n2 > 3(n2− 1) meaning n2 ≤ 3 ⇒ n = 1, contradiction. Therefore, m, m′ cannotboth be greater than n. This proves Lemma 2.

Therefore, for any solution (m, n) ∈ S, we can create a descent of solutions until wereach a solution where m = 1 or n = 1. If m = 1, then n − 1|n2 + 1, which meansn − 1|2. Therefore, n = 2, 3. Therefore, all solutions with at least one of m, n = 1are (1, 2), (1, 3), (2, 1), (3, 1). In all cases, k = 5. Conversely, all solutions should canconstructed using an ascent starting from (1, 2), (1, 3).

Therefore, if (m, n) is a solution with m < n, then m′+m = 5n, which means (5n−m, n)is also a solution. Therefore, solutions can be formed by the sequence (ai, ai+1) wherea1 = 1, a2 = 2 with ai+2 = 5ai+1 − ai, and the sequence (bi, bi+1) where b1 = 1, b2 = 3and bi+2 = 5bi+1 − bi.

Bashing yields that

an =21−

√21

42(5 +

√21

2)n +

21 +√

21

42(5−

√21

2)n,

and

bn =21 +

√21

42(5 +

√21

2)n +

21−√

21

42(5−

√21

2)n,

All solutions are (an, an+1), (an+1, an), (bn+1, bn), (bn, bn+1) over all n ∈ N0.


5. Let n ≥ 3, k ≥ 2 be positive integers. There are n people in a group. On each of kdays, at least two people from the group went together to buy ice cream. It turns out

15

that each pair of persons went together for ice cream on exactly one of the k days.Prove that k ≥ n.

Solution: Let P1, P2, · · · , Pn be the n people and G1, G2, . . . , Gk be the k groups. Fori ∈ {1, 2, · · · , n} let xi be the number of groups person Pi belongs in. By doublecounting we have the identity

n∑i=1

xi =k∑

j=1

|Gj|.(1)

Let S = {(i, j)|Pi 6∈ Gj}.

Lemma: For all (i, j) ∈ S, xi ≥ |Gj|.

Proof: Suppose Pi belongs in t groups denoted by G. Then note that for everyone exceptPi, each of them belong to exactly one of the t groups. Therefore, a group not con-taining Pi cannot have more than two members belonging in G. This proves the lemma.

We will now use (1).

n∑i=1

xi =n∑

i=1

(k − xi)xi

k − xi

=n∑

i=1

1

k − xi

∑Gj 63Pi

xi =∑

(i,j)∈S

xi

k − xi

(2).

k∑j=1

|Gj| =k∑

j=1

(n− |Gj|)|Gj|

n− |Gj|=

k∑j=1

1

n− |Gj|∑

Pi 6∈Gj

|Gj| =∑

(i,j)∈S

|Gj|n− |Gj|

(3).

By (1), the expressions in (2) and (3) are equal. But by our lemma, in (2) we havexi

k−xi≥ |Gj |

k−|Gj | , which means

∑(i,j)∈S

|Gj|n− |Gj|

≥∑

(i,j)∈S

|Gj|k − |Gj|

.

which means that n ≤ k as desired.

Source: St. Petersburg 2001

16

Mock Olympiad 6

Time: 4 hours

1. Let n be a positive integer. Suppose there are n points in the plane with no threecollinear such that if we give labels P1, P2, · · · , Pn to the points in any way at all, thebroken line P1P2 · · ·Pn does not intersect itself. Find the maximum possible value ofn.

Solution: The answer is n = 4. Consider four points that does not form a convexquadrilateral. Any broken line on these four vertices does not intersect itself.

The observation is that if A, B, C,D are four vertices of a convex quadrilateral, thenpicking points in the order A, C,B,D will have AC intersecting BD. I claim that forany five points, there exists four points that are in convex position. This solves theproblem. Consider the convex hull of the five points. If the boundary contains fouror five points, we are done. If the boundary of the convex hull contains three pointsA, B, C, let P be a fourth point. Let AP, BP,CP intersect the sides of triangle ABCat XY Z respectively. WLOG, suppose the fifth point Q is in ∆APY . Then segmentsBQ and AP intersect. Then Q,P, B, A are convex. �

2. A positive integer n is said to be friendly if there exist non-empty subsets A1, A2, · · · , An

of {1, 2, · · · , n} such that

(I) i 6∈ Ai for all i ∈ {1, 2, · · · , n}(II) i ∈ Aj if and only if j 6∈ Ai for all distinct i, j ∈ {1, 2, · · · , n}(III) Ai ∩ Aj 6= ∅ for all i, j ∈ {1, 2, · · · , n}

Prove that n ≥ 7 if and only if n is friendly.

Solution: We first prove that n = 7 is friendly. We construct the following chartwhere each row represents a number from 1 to n and each column represents the setsA1, · · · , An. We mark (i, j) with a ∗ if i ∈ Aj and an X otherwise.

A1 A2 A3 A4 A5 A6 A7

1 X * * * X X X2 X X * X * * X3 X X X * * X *4 X * X X X * *5 * X X * X * X6 * X * X X X *7 * * X X * X X

17

Note that X appears in every (i, i) entry and the chart is anti-symmetrical. i.e. ∗appears in (i, j) if and only if X appears in (j, i).

b.) We first observe that |Ai| ≥ 3 for all collection of subsets satisfying (I), (II), (III).If |Ai| = 1, then if |Ai| = {k}, then since k 6∈ Ak then Ai ∩Ak = ∅ contradicting (III).If |Ai| = 2, then if Ai{k, l}, then Ak, Al must contain k or l. But by (I), we have l ∈ Ak

and k ∈ Al, contradicting (II).

Now, note that∑

Ai = n(n − 1)/2 by property (I) and (II). Combining this with|Ai| ≥ 3 gives us n(n− 1)/2 ≥ 3n which means n ≥ 7.

To prove the other direction, suppose n ≥ 7. Set A1, A2, · · · , A7 as in (a). For m ≥ 8,set Am = {1, 2, · · · , m − 1}. Clearly, (I) and (III) is satisfied. Note that (II) is alsosatisfied since m 6∈ Ai for i ≤ 7. For i > j ≥ m, j ∈ Ai and i 6∈ Aj. �

Source: Romania Math Competitions

3. Let a, b be positive integers such that for all positive integers n, an + n divides bn + n.Prove that a = b.

We first show that there exists a positive integer n not divisible by p − 1 such thatan + n is divisible by p.

If a is divisible by p, set n = p and we are done. Otherwise, set n ≡ 1 mod (p − 1)and n ≡ −a mod p. This is possible by Chinese Remainder Theorem. Then an ≡ amod p by Fermat’s Little Theorem. Then an+n mod p ≡ a+(−a) = 0 mod p as desired.

Let p be a prime larger than a, b. Choose n as in (a). Then an + n, bn + n is alsodivisible by p. But bn + n ≡ b− a mod p. Therefore, p divides b− a and both a, b areless than p. Therefore, b− a = 0 as desired. �


4. Find all functions f : Q → Q such that

f(x)f(y) + f(x + y) = f(xy) + f(x) + f(y).

for all x, y ∈ Q.

18

Let’s state some properties. Set y = xx−1

. (Note xy = x + y.) Then

f(x)f(x

x− 1) = f(x) + f(

x

x− 1).(1)

Set y = 1, gives usf(x + 1) = (2− f(1))f(x) + f(1).(2)

Set y = 0. Then f(x)f(0)+ f(x) = f(x)+2f(0) ⇒ f(0)(f(x)− 2) = 0. Then f(0) = 0or f(x) = 2 for all x ∈ Q. The latter can be verified as a solution. Setting x, y = 0fields f(0)2 = 2f(0). Therefore, f(0) = 0 or f(0) = 2.

If f(0) = 2, let k = f(1). Then by (2), k = f(1) = (2− k)2 + k. Meaning −k + 4 = kand k = 2. Therefore, f(x+1) = 2 ⇒ f(x) = 2 for all x ∈ Q, a solution already found.Henceforth suppose f(0) = 0.

Let k = f(1). Then f(2) = (2 − k)k + k = (3 − k)k. Putting x = 2 into (1) gives usf(2)2 = 2f(2), meaning f(2) = 0 or f(2) = 2. If f(2) = 0, then 0 = (2 − k)k + k ⇒(3− k)k = 0. If f(1) = 0, then f(x + 1) = 2f(x). Thus f(n) = 0 for all n ∈ N. Settingx = n ∈ N gives f(n/(n−1) = 0. Thus f(1/n) = 0 for all n ∈ N. For m/n ∈ Q and in-ducting on m, we have f(m/n+1/n) = −f(m/n)f(1/n)+f(m/n2)+f(m/n)+f(1/n) =0. Therefore, f(x) = 0 for all x ∈ Q.

If f(2) = 2, then (3 − k)k = 2. Then f(1) = 1 or f(1) = 2. If f(1) = 2, thenf(x + 1) = f(1) = 2 for all x meaning f(0) = 2, which we solved already. Henceforth,we suppose f(1) = 1. But by (2), we have f(x + 1) = f(x) + 1. Meaning f(n) = n forall integers n. Again by (1), we have f(n/(n−1) = n/(n−1) meaning f(1/n) = 1/n forall n. For m/n ∈ Q and inductively on m, we have f(m/n+1/n) = −f(m/n)f(1/n)+f(m/n2) + f(m/n) + f(1/n) = −m/n2 + m/n2 + m/n + 1/n = (m + 1)/n as desired.Therefore, f(x) = x for all x ∈ Q. �

Source: Indian Math Olympiad, unknown year.

Hence, the only solutions are f ≡ 0, f ≡ 2 and f(x) = x for all x ∈ Q. �

5. Let ABC be an acute-angled triangle with circumcentre O and orthocentre H. LetXA, YA be points on BC and OA respectively such that OXAHYA is a parallelogram.Define XB, YB, XC , YC similarly. Prove that

OXA

OYA

+OXB

OYB

+OXC

OYC

≥ 3.

Let ZA be where the altitude AD from A meet the circumcircle of ABC. Note thatHD = DZA. Since OZA = OA. Therefore, O, XA, ZA are collinear. Therefore,

19

OXA/OYA = YAH/OYA = YAA/OYA = AH/HZA. Let MA be the midpoint of BC,then AH/HZA = 2OMA/2BHcosC = 2R·cos A/4R·cos B cos C = cos A/(2 cos B cos C).It suffices to show that ∑

cyc

cos A

cos B cos C≥ 6.

By AM/GM, it suffices to show that cos A cos B cos C ≤ 18.

We use the fact that ln(cos(x)) is a convex function. Thus,∑cyc

ln(cos A) ≤ 3ln(cosA + B + C

3) = 3ln(1/2).

meaning cos A cos B cos C ≤ (1/2)3 = 1/8 as desired. �


20

Mock Olympiad 7

Time: 4.5 hours

1. Find all primes p such that the number of ordered pairs of integers (x, y) with 0 ≤x, y < p satisfying the following equation, is exactly p.

y2 ≡ x3 − x(mod p).

Solution: The answer is p = 2 and all primes p ≡ 3 mod 4.

For p = 2, the only solutions are (0, 0) and (1, 0).

If p ≡ 3 mod 4, then since −1 is not a square mod p, Then x3 − x is a square modp if and only if −x3 + x is not a square mod p. For x = 0, 1, p − 1 yields 3 solutions.And half of x = 2, 3, · · · , p − 2 yields solution for y, and exactly two solutions each.Therefore, this yields p− 3 solutions. This total p solutions.

If p ≡ 1 mod 4, then x = 0, 1, p − 1 yields 3 solutions. Here, −1 is a square mod p,then x yields a solution for y if and only if p−x yields a solution for y. They yield twosolutions each. Therefore, the number of solutions for x = 2, 3, · · · , p− 2 is divisible by4, thus the number of solution is 3 mod 4, which cannot be p. �

Source: Turkish Math Olympiad, unknown year.

2. Let x1, x2, · · · , xn be positive real numbers. Prove that

1

1 + x1

+1

1 + x1 + x2

+ · · ·+ 1

1 + x1 + x2 + · · ·+ xn

<

√1

x1

+1

x2

+ · · ·+ 1

xn

.

Solution: By Cauchy Schwarz inequality, and squaring both sides of the equation, weget

(1

x1

+ · · ·+ 1

xn

)(x1

(1 + x1)2+ · · ·+ xn

(1 + x1 + · · ·+ xn)2)

≥ (1

1 + x1

+1

1 + x1 + x2

+ · · ·+ 1

1 + x1 + · · ·+ xn

)2.

It suffices to show that

x1

(1 + x1)2+ · · ·+ xn

(1 + x1 + · · ·+ xn)2< 1.

21

This is true becausen∑

i=1

xi

(1 + x1 + · · ·+ xi)2<

n∑i=1

xi

(1 + x1 + · · ·+ xi−1)(1 + x1 + · · ·+ xi)=

n∑i=1

1

1 + x1 + · · ·+ xi−1

− 1

1 + x1 + · · ·+ xi

< 1.

by telescoping sums. As desired. �

Source: Romania Math Olympiad 2005, Grade 9

3. Let n be a positive integer. A company has n workers, n tasks and a boss. Each workeris assigned two real numbers for each task, called preference and ability. Each workeris to be assigned a unique task.

(i) Suppose worker A is assigned task U and worker B is assigned task V . If A has ahigher ability value at task V than B, and A has higher preference value for task Vthan for task U , then A is not happy.

(ii) If everyone can be reassigned tasks so that the ability of the worker of every taskgoes up, then the boss is not happy.

Prove that an assignment is possible where the workers and the boss are all happy.

Solution: Let W1, · · · , Wn be the workers and T1, · · · , Tn be the tasks. Let (pij, aij)be the preference number and ability number for Wi with task Tj respectively. We willassign tasks using the following procedure. In the procedure, certain pairs (i, j) will becrossed out.

Pick an unassigned task Tj and choose the worker Wi with the highest ability valuesuch that (i, j) is NOT crossed out.

(I) If Wi is not yet assigned a task, assign Wi to Tj.

(II) If Wi is already assigned a task Tk and pi,j > pi,k, then assign Wi to Tj, unassigntask Tk and cross out entry (i, k).

(III) Otherwise, have Wi keep task Tk and cross out (i, j).

22

Repeat these three steps until all tasks are assigned workers.

This procedure ends since at each step, since no one is asked to accept a task more thanonce. If (i, j) is crossed out, it means Wi and Tj was already assigned, and wouldn’thave the possibility to be re-assigned.

I claim that this procedure works. WLOG, suppose W1 is assigned T1 and W2 is as-signed T2. Suppose W1 is not happy. That means W1 has higher ability than W2 at T2

and W1 prefers T2 over T1. But this is impossible since tasks are first given to workerswith higher ability. And a worker’s task changes only if the new task has higher pref-erence. Therefore W1 must be happy.

To show that the boss is happy, for each task, consider the moment that the task islast assigned. It is chosen amongst the person with the highest ability for that task.(Let me try to write this better later) �

Source: Japan Math Olympiad, Final Round, 2005

23

Mock Olympiad 8

Time: 3 hours

1. Let n ≥ 4 be a positive integer. Let S be a set of points P1, P2, · · · , Pn in the planesuch that no three are collinear and no four are concyclic. Let ai be the number ofcircles that passes through three of the points in S that contains Pi in its interior. Letm(S) = a1 + a2 + · · ·+ an. Prove that there exists a function f(n), depending only onn, such that the following statement holds;

The points in S are the vertices of a convex n-gon if and only if f(n) = m(S).

Solution: First we solve this for n = 4. If S are the vertices of a convex quadri-lateral ABCD, since the quadrilateral ABCD is not cyclic, then a pair of oppositeangles has sum larger than 180o and the other pair has sum less than 180O. Suppose∠A+∠C > 180o. Then the circumcircle of ∆ABD will contain C and the circumcircleof ∆CBD will contain A. Conversely, the circumcircle of ABC will not contain Dand that of DBC will not contain A. Hence, m(S) = 2 for n = 4. This is because ifthe vertices of S does not form a convex quadrilateral, then the convex hull of S is atriangle and contains one interior point. Then m(S) = 1. This then holds for n = 4.

For n ≥ 4, I claim that f(n) = 2(

n4

)satisfies the given condition. If S are the vertices

of a convex n-gon, then every subset of four points form a convex quadrilateral. Eachsuch subset contributes 2 to the sum m(S) by the argument used for n = 4. And eachsuch circle-point pair is counted exactly once. Therefore, m(S) = 2

(n4

).

If S are not the vertices of a convex n-gon, I claim that there exist four points in Snot in convex position. Consider the convex hull of S, which does contain an interiorpoint P . Let A be any fixed vertex on the convex hull. Then P is in the interior oftriangle AMN where MN is an edge of the polygon formed by the convex hull. ThenA, M, N, P are not in convex position. Then when we compute m(S), a set of fourpoints contributes one to m(S) instead of 2, meaning m(S) < 2

(n4

).

Our claim is proved. �


2. Let ABC be an acute-angled scalene triangle with circumcentre O and an interior pointP such that ∠PAB = ∠PBC and ∠PAC = ∠PCB. Let Q be the point on line BCsuch that QA = QP . Prove that ∠AQP = 2∠OQC.

24

Solution: (By Jarno Sun) WLOG, suppose Q is closer to B than C. Let T be themidpoint of AP , X be on QT such that XB is perpendicular to BC and Y on QTsuch that Y C is perpendicular to BC.

Note the the circumcircle of APB is tangent to BC since ∠BAP = ∠PBC. Since thecentre of this circle is on the perpendicular bisector of AP , then its centre is X. Simi-larly, the centre of the circumcircle of APC is Y . Let E, F be the midpoints of AC, ABrespectively. Then O, E, Y are collinear and O,F, X are collinear. Furthermore,A, T, E, Y are concyclic and A, T, F, X are concyclic. Meaning ∠PAC = ∠XY O and∠PAB = ∠Y XO. Therefore, ∆OXY ∼ ∠PBC. Therefore, OY/PC = XY/BC =QY/QC. Therefore, ∆QOY ∼ QPC. Particularly, ∠OQY = ∠PQC. These an-gles are isogonal conjugates of angle ∠Y QC. Therefore, ∠TQP = ∠OQC, meaning2∠OQC = ∠AQP as desired. �


3. Find all triples of positive integers (a, m, n) such that am + 1 divides (a + 1)n.

Setting m = 1 or a = 1 certainly give a set of solutions. i.e. (1, m, n) and (a, 1, n) aresolutions, which is easily verified. If n = 1, then m = 1. We will now assume thata, m, n ≥ 2.

If a = 2, then 2m+1 divides 3n. This means 2m+1 = 3l for some non-negative integer l.Since m ≥ 2, then 3l ≡ 1 mod 4, meaning l is even. Therefore, 2m = (3l/2−1)(3l/2 +1),which is the product of two consecutive even numbers, which are both powers of 2.Meaning 3l/2 = 3, which implies l = 2 thus m = 3. We need n ≥ 2 for this to work.Therefore, (2, 3, n) with n ≥ 2 is also a solution.

Now suppose a ≥ 3. I claim that no other solutions exist. The first claim is that mis odd. First suppose am + 1 is a power of 2. i.e. am + 1 = 2k. Since a, m > 1, thenk ≥ 2, meaning am ≡ −1 mod 4, meaning m cannot be even. Therefore, m is odd.Otherwise, let p be an odd prime divisor of am + 1. Then p divides (a + 1)n meaningp divides a + 1. Then a ≡ −1 mod p. Since am ≡ −1 mod p, then m must be odd.

Let p be a (odd) prime divisor of m. Then ap +1 divides am +1 which divides (a+1)n.Then (ap−1 − ap−2 + · · · + a2 − a + 1 divides (a + 1)n−1. (Recall n ≥ 2). Let f(x) =xp−1−xp−2+· · ·−x+1. Then note f(−1) = p. Then we can write f(x) = (x+1)g(x)+pfor some g ∈ Z[x]. Any prime q that divides f(a) will divides (a+1), which then dividesp, meaning q = p. Therefore, f(a) = pk for some k ∈ Z. Then pk = (ap + 1)/(a + 1).Let a + 1 = pt. Then pk((pt − 1)p + 1)/pt = (terms with higher powers of p) + p,meaning k = 1. Therefore, f(a) = p. This means ap + 1 = p(a + 1). For a ≥ 3,

25

ap + 1 > ap > (a + 1)p. Therefore, no solutions can exist.

Therefore, the only solutions are (1, m, n), (a, 1, n) and (2, 3, n) with n ≥ 2. �


26

Mock Olympiad 9

Time: 3 hours

1. A country has n cities where some pairs of cities are joined by roads. It is known thatyou can go from any one city to any other city by using the roads (by possibly goingthrough intermediate cities). A path is defined to be a sequence of pairwise distinctcities such that successive cities in the sequence are joined by a road. Two paths, bothcontaining the maximum possible number of cities in a path, are chosen. Prove thatthe two paths have at least one city in common.

2. Let ω, l be a circle and a line respectively that do no intersect. Let AB be a diameterof ω such that AB is perpendicular to l and B is closer to l than A. Let C be a pointon ω which is not A or B. Let AC intersect l at D, let E be on ω such that DE istangent to ω and B, E are on the same side of line AC. Let F be the intersection ofEB and l and G be the intersection of AF and ω (which is not A itself). Let G′ be thereflection of G across AB. Prove that F, C,G′ are collinear.

3. Find all functions f : R → R such that

|f(x)− f(y)| ≤ (x− y)2

for all x, y ∈ R.

4. Let p ≥ 5 be prime. Let bxc denote the greatest integer less than or equal to x. Provethat

b2p/3c∑k=1

(p

k

).

is divisible by p2.

5. Let n ≥ 4 be a positive integer. Given a set S of n points in the plane such that nothree points are collinear and no four points are concyclic, let f(S) be the number of(unordered) pairs of points in S such that there exist a circle that contains these twopoints but no other points in S. Prove that f(S) ≤ 3n− 6.

27

Mock Olympiad 10

Time: 3 hours

1. Let a, b, c be positive real numbers in the range [12, 1]. Prove that

2 ≤ a + b

1 + c+

b + c

1 + a+

c + a

1 + b≤ 3.

2. Let n ≥ 3 be a positive integer. There is one bird at each vertex of a regular n-gon. Atthe sound of a whistle, all the birds fly away and they return one to each vertex, suchthat not all birds return to their former vertex. Find all positive integers n for whichthe following statement is true;

There always exists three birds such that the triangles formed from their former loca-tions and their new locations are either both acute-angled, both right-angled or bothobtused-angled triangles.

3. Let ABCDEF be a convex hexagon such that AB = BC, CD = DE,EF = FA.Prove that

BC

BE+

DE

DA+

FA

FC≥ 3

2.

Determine when equality holds.

4. Let p be a prime and m be a positive integer. Prove that there exists a positive integern such that the decimal representation of pn contains a string of m consecutive zeroes. 1

5. Let ABC be a triangle and L be a point on side BC. Extend rays AB and AC topoints M, N respectively such that ∠AMC = 2∠ALC and ∠ANB = 2∠ALB. Let Obe the circumcentre of ∆AMN . Prove that OL is perpendicular to BC.

1The infinitely many leading zeroes to the left of a number does not count!

28

Mock Olympiad 11

Time: 4.5 hours

1. Prove that there are infinitely many pairs of positive integers (a, b) such that gcd(a, b) =1, a divides b2 − 5 and b divides a2 − 5.

2. Each edge of a complete graph G on n vertices is oriented with an arrow and is colouredeither red or blue. A directed path is defined to be a finite sequence of distinct verticeswhere each vertex in the sequence has an arrow pointing towards the next vertex in thesequence. Prove that there is a vertex v in G such that there exists a monochromaticdirected path starting at v and ending at any other vertex. 1

3. Let ABC be a triangle with incircle γ. Let TA be the point on γ such that thecircumcircle γA of TABC is tangent to γ at TA. Let the common tangent at TA of γand γA intersect the line BC at PA. Define TB, TC and PB, PC similarly.

a.) Prove that PA, PB, PC are collinear.

b.) Prove that ATA, BTB, CTC are concurrent.

1A path is monochromatic if the edges in the path are of the same colour.

29

Mock Olympiad 12

Time: 4.5 hours

1. Let ABC be a triangle with ∠ABC = 30o. Consider the three closed discs with radius|AC|/3 centered at A, B and C respectively. Prove that there exists a unique equi-lateral triangle whose three vertices lie one each in each of the three discs. (Note: Aclosed disc is a circle that includes its perimeter and its interior.)

2. Let a, b, c be positive real numbers such that a + b + c = 3. Prove that

1

a2+

1

b2+

1

c2≥ a2 + b2 + c2.

3. A simple polygon is a polygon whose perimeter does not intersect itself, but is notnecessarily convex. Let n ≥ 4 be a positive integer. Prove that that a simple n-goncontains a diagonal which is strictly inside the polygon such that the diagonal dividesthe perimeter into two parts where both parts have at least n/3− 1 vertices.

30

at 2008 mocks solutions

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