Unofficial archive — problems, solutions & results © IMC, reproduced with permission.

Studolymp / IMC / 2012

IMC 2012
contestants 316 · problems 10 (5+5) · scale 0–10 · per-problem yes

Problems

Day 1

P1

For every positive integer nn, let p(n)p(n) denote the number of ways to express nn as a sum of positive integers. For instance, p(4)=5p(4) = 5 because 4=3+1=2+2=2+1+1=1+1+1+1.4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1. Also define p(0)=1p(0) = 1.

Prove that p(n)p(n1)p(n) - p(n-1) is the number of ways to express nn as a sum of integers each of which is strictly greater than 1.

(Proposed by Fedor Duzhin, Nanyang Technological University)

P2

Let nn be a fixed positive integer. Determine the smallest possible rank of an n×nn \times n matrix that has zeros along the main diagonal and strictly positive real numbers off the main diagonal.

(Proposed by Ilya Bogdanov and Grigoriy Chelnokov, MIPT, Moscow)

P3

Given an integer n>1n > 1, let SnS_n be the group of permutations of the numbers 1,2,,n1, 2, \dots, n. Two players, A and B, play the following game. Taking turns, they select elements (one element at a time) from the group SnS_n. It is forbidden to select an element that has already been selected. The game ends when the selected elements generate the whole group SnS_n. The player who made the last move loses the game. The first move is made by A. Which player has a winning strategy?

(Proposed by Fedor Petrov, St. Petersburg State University)

P4

Let f:RRf : \mathbb{R} \to \mathbb{R} be a continuously differentiable function that satisfies f(t)>f(f(t))f'(t) > f(f(t)) for all tRt \in \mathbb{R}. Prove that f(f(f(t)))0f(f(f(t))) \le 0 for all t0t \ge 0.

(Proposed by Tomáš Bárta, Charles University, Prague)

P5

Let aa be a rational number and let nn be a positive integer. Prove that the polynomial X2n(X+a)2n+1X^{2^n} (X + a)^{2^n} + 1 is irreducible in the ring Q[X]\mathbb{Q}[X] of polynomials with rational coefficients.

(Proposed by Vincent Jugé, École Polytechnique, Paris)

Day 2

P6

Consider a polynomial f(x)=x2012+a2011x2011++a1x+a0.f(x) = x^{2012} + a_{2011} x^{2011} + \dots + a_1 x + a_0. Albert Einstein and Homer Simpson are playing the following game. In turn, they choose one of the coefficients a0,,a2011a_0, \dots, a_{2011} and assign a real value to it. Albert has the first move. Once a value is assigned to a coefficient, it cannot be changed any more. The game ends after all the coefficients have been assigned values.

Homer's goal is to make f(x)f(x) divisible by a fixed polynomial m(x)m(x) and Albert's goal is to prevent this.

(a) Which of the players has a winning strategy if m(x)=x2012m(x) = x - 2012?

(b) Which of the players has a winning strategy if m(x)=x2+1m(x) = x^2 + 1?

(Proposed by Fedor Duzhin, Nanyang Technological University)

P7

Define the sequence a0,a1,a_0, a_1, \dots inductively by a0=1a_0 = 1, a1=12a_1 = \frac{1}{2} and an+1=nan21+(n+1)anfor n1.a_{n+1} = \frac{n a_n^2}{1 + (n+1) a_n} \quad \text{for } n \ge 1. Show that the series k=0ak+1ak\sum\limits_{k=0}^{\infty} \dfrac{a_{k+1}}{a_k} converges and determine its value.

(Proposed by Christophe Debry, KU Leuven, Belgium)

P8

Is the set of positive integers nn such that n!+1n! + 1 divides (2012n)!(2012n)! finite or infinite?

(Proposed by Fedor Petrov, St. Petersburg State University)

P9

Let n2n \ge 2 be an integer. Find all real numbers aa such that there exist real numbers x1x_1, …, xnx_n satisfying x1(1x2)=x2(1x3)==xn1(1xn)=xn(1x1)=a.(1)\tag{1} x_1 (1 - x_2) = x_2 (1 - x_3) = \dots = x_{n-1} (1 - x_n) = x_n (1 - x_1) = a.

(Proposed by Walther Janous and Gerhard Kirchner, Innsbruck)

P10

Let c1c \ge 1 be a real number. Let GG be an abelian group and let AGA \subset G be a finite set satisfying A+AcA|A + A| \le c |A|, where X+Y:={x+yxX, yY}X + Y := \{ x + y \mid x \in X,\ y \in Y \} and Z|Z| denotes the cardinality of ZZ. Prove that A+A++Ak timesckA|\underbrace{A + A + \dots + A}_{k\ \text{times}}| \le c^k |A| for every positive integer kk. (Plünnecke's inequality)

(Proposed by Przemyslaw Mazur, Jagiellonian University)