IMC 2012

Consider a polynomial $$f(x)=x^{2012}+a_{2011}x^{2011}+\dots+a_1x+a_0.$$ Albert Einstein and Homer Simpson are playing the following game. In turn, they choose one of the coefficients $a_0,a_1,\dots,a_{2011}$ 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 $f(x)$ divisible by a fixed polynomial $m(x)$ and Albert's goal is to prevent this. (a) Which of the players has a winning strategy if $m(x)=x-2012$ ? (b) Which of the players has a winning strategy if $m(x)=x^2+1$ ?

Proposed by Fedor Duzhin, Nanyang Technological University.

Define the sequence $a_0,a_1,\dots$ inductively by $a_0=1$ , $a_1=\frac{1}{2}$ , and $$a_{n+1}=\dfrac{n a_n^2}{1+(n+1)a_n}, \forall n \ge 1.$$ Show that the series $\sum_{k=0}^\infty \dfrac{a_{k+1}}{a_k}$ converges and determine its value.

Proposed by Christophe Debry, KU Leuven, Belgium.

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

Proposed by Fedor Petrov, St. Petersburg State University.

Let $n \ge 2$ be an integer. Find all real numbers $a$ such that there exist real numbers $x_1,x_2,\dots,x_n$ satisfying $$x_1(1-x_2)=x_2(1-x_3)=\dots=x_n(1-x_1)=a.$$

Proposed by Walther Janous and Gerhard Kirchner, Innsbruck.

Let $c \ge 1$ be a real number. Let $G$ be an Abelian group and let $A \subset G$ be a finite set satisfying $|A+A| \le c|A|$ , where $X+Y:= \{x+y| x \in X, y \in Y\}$ and $|Z|$ denotes the cardinality of $Z$ . Prove that $$|\underbrace{A+A+\dots+A}_k| \le c^k |A|$$ for every positive integer $k$ .

Proposed by Przemyslaw Mazur, Jagiellonian University.