1. Prove the following inequality for positive real numbers x, y, z:
(xy + yz + zx) Solution: After clearing denominators, the given inequality be-comes
X sym
4x5y− x4y2− 3x3y3+ x4yz− 2x3y2z + x2y2z2≥ 0,
where the symmetric sum runs over all six permutations of x, y, z. (In particular, this means the coefficient of x3y3 in the final expression is -6, and that of x2y2z2 is 6.)
Recall Schur’s inequality:
x(x− y)(x − z) + y(y − z)(y − x) + z(z − x)(z − y) ≥ 0.
Multiplying by 2xyz and collecting symmetric terms, we get X
by two applications of AM-GM; combining the last two displayed inequalities gives the desired result.
2. Prove that for every pair m, k of natural numbers, m has a unique representation in the form
Solution: We first show uniqueness. Suppose m is represented by two sequences ak, . . . , atand bk, . . . , bt. Find the first position in which they differ; without loss of generality, assume this position is k and that ak > bk. Then
To show existence, apply the greedy algorithm: find the largest ak
such that akk ≤ m, and apply the same algorithm with m and k replaced by m− akk and k − 1. We need only make sure that the sequence obtained is indeed decreasing, but this follows because by assumption, m < akm+1, and so m − akk < ka−1k.
3. In triangle ABC, we have ∠A = 60◦. Let O, H, I, I0 be the circum-center, orthocircum-center, incircum-center, and excenter opposite A, respectively, of ABC. Let B0 and C0be points on the segments AC and AB such that AB = AB0 and AC = AC0. Prove that:
(a) The eight points B, C, H, O, I, I0, B0, C0 are concyclic.
(b) If OH intersects AB and AC at E and F , respectively, the perimeter of triangle AEF equals AB + AC.
(c) OH =|AB − AC|.
Solution:
(a) The circle through B, C, H consists of all points P such that
∠BP C = ∠BHC = 180◦− ∠CAB = 120◦ (as directed an-gles mod 180◦). Thus O lies on this circle, as does I because
∠BIC = 90◦+12∠A = 30◦. Note that the circle with diam-eter II0 passes through B and C (since internal and external angle bisectors are perpendicular). Hence I0also lies on the cir-cle, whose center lies on the internal angle bisector of A. This means reflecting B and C across this bisector gives two more points B0, C0 on the circle.
(b) Let R be the circumradius of triangle ABC. The reflection across AI maps B and C to B0 and C0, and preserves I. By
(a), the circle BCHO is then preserved, and hence H maps to O. In other words, AHO is isosceles with AH = AO = R and
∠HAO = |β − γ|, writing β for ∠B and γ for ∠C.
In particular, the altitude of AHO has length R cos β− γ and so the equilateral triangle AEF has perimeter
√3R cos(β−γ) = 2R sin(β+γ) cos(β−γ) = 2R(sin β+sin γ) = AB+AC.
(c) We use a, b, c to denote the lengths of BC, CA, AB. By a stan-dard computation using vectors, we find OH2= 9R2−(a2+b2+ c2), but since a = 2R sin 60◦, we have OH2= 2a2− b2− c2. By the Law of Cosines, a2= b2+ c2− bc, so OH2= b2+ c2− 2bc = (b− c)2, and so OH =|b − c|.
4. Let ABC be a scalene triangle. The medians from A, B, C meet the circumcircle again at L, M, N , respectively. If LM = LN , prove that 2BC2= AB2+ AC2.
Solution: Let G be the centroid of triangle ABC; then trian-gles N LG and AGL are similar, so LN/AC = LG/CG. Similarly LM/AB = GL/BG. Thus if LM = LN , then AB/AC = BG/CG.
Using Stewart’s theorem to compute the lengths of the medians, we have
AB2
AC2 = 2AB2+ 2BC2− AC2 2AC2+ 2BC2− AB2
which reduces to (AC2− AB2)(2BC2− AB2− AC2) = 0. Since the triangle is scalene, we conclude 2BC2= AB2+ AC2.
5. The top and bottom edges of a chessboard are identified together, as are the left and right edges, yielding a torus. Find the maximum number of knights which can be placed so that no two attack each other.
Solution: The maximum is 32 knights; if the chessboard is al-ternately colored black and white in the usual fashion, an optimal arrangement puts a knight on each black square. To see that this cannot be improved, suppose that k knights are placed. Each knight attacks 8 squares, but no unoccupied square can be attacked by more than 8 knights. Therefore 8k≤ 8(64 − k), whence k ≤ 32.
6. Find all nonnegative real numbers a1≤ a2≤ . . . ≤ an satisfying
Solution: Adding or removing zeroes has no effect, so we may assume the ai are positive. By Cauchy-Schwarz,
(a1+· · · + an)(a31+· · · + a3n)≥ (a21+· · · + a2n).
Since 96· 216 = 1442, we have equality, so the sequences a1,· · · , an
and a31,· · · , a3n are proportional, so that a1 =· · · = an = a. Now na = 96, na2= 144 so that a = 3/2, n = 32.
7. Points D and E lie on sides AB and AC of triangle ABC such that DE||BC. Let P be an arbitrary point inside ABC. The lines P B and P C intersect DE at F and G, respectively. If O1 is the circumcenter of P DG and O2 is the circumcenter of P F E, show that AP ⊥ O1O2.
Solution: (Note: angles are directed modulo π.) Let M be the second intersection of AB with the circumcircle of DP G, and let N be the second intersection of N with the circumcircle of EP F . Now ∠DM P = ∠DGP by cyclicity, and ∠DGP = ∠BCP by par-allelism, so ∠DM P = ∠BCP and the points B, C, P, M are con-cyclic. Analogously, B, C, P, N are concon-cyclic. Therefore the points B, C, M, N are concyclic, so ∠DM N = ∠BCN . Again by parallels,
∠BCN = ∠DEN , so the points D, E, M, N are concyclic.
We now apply the radical axis theorem to the circumcircles of DGP , EP F , and DEM N to conclude that DM ∩ EN = A lies on the radical axis of the circles P DG and P EF , so AP ⊥ O1O2as desired.
8. Let P (x) be a polynomial with rational coefficients such that P−1(Q)⊆ Q. Show that P is linear.
Solution: By a suitable variable substitution and constant factor, we may assume P (x) is monic and has integer coefficients; let P (0) = c0. If p is a sufficiently large prime, the equation P (x) = p + c0 has a single real root, which by assumption is rational and which we
may also assume is positive (since P has positive leading coefficient).
However, by the rational root theorem, the only rational roots of P (x)− p − c0 can be±1 and ±p. Since the root must be positive and cannot be 1 for large p, we have P (p)− p − c0= 0 for infinitely many p, so P (x) = x + c0is linear.
9. For S ={x1, x2, . . . , xn} a set of n real numbers, all at least 1, we count the number of reals of the form
n
X
i=1
ixi, i∈ {0, 1}
lying in an open interval I of length 1. Find the maximum value of this count over all I and S.
Solution: The maximum is n
bn/2c, achieved by taking xi = 1 + i/(n + 1). To see that this cannot be improved, note that for any permutation σ of{1, . . . , n}, at most one of the sets {σ(1), . . . , σ(i)}
for i = 1, . . . , n has sum lying in I. Thus if T is the set of subsets whose sum lies in I, we have
X
t∈T
t!(n− t)! ≤ n! ⇔X
t∈T
n t
−1
≤ 1.
In particular, we have|T | ≤ bn/2cn .