USAMO 2013

In triangle $ABC$ , points $P$ , $Q$ , $R$ lie on sides $BC$ , $CA$ , $AB$ respectively. Let $\omega_A$ , $\omega_B$ , $\omega_C$ denote the circumcircles of triangles $AQR$ , $BRP$ , $CPQ$ , respectively. Given the fact that segment $AP$ intersects $\omega_A$ , $\omega_B$ , $\omega_C$ again at $X$ , $Y$ , $Z$ , respectively, prove that $YX/XZ=BP/PC$ .

For a positive integer $n\geq 3$ plot $n$ equally spaced points around a circle. Label one of them $A$ , and place a marker at $A$ . One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2n$ distinct moves available; two from each point. Let $a_n$ count the number of ways to advance around the circle exactly twice, beginning and ending at $A$ , without repeating a move. Prove that $a_{n-1}+a_n=2^n$ for all $n\geq 4$ .

Let $n$ be a positive integer. There are $\tfrac{n(n+1)}{2}$ marks, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing $n$ marks. Initially, each mark has the black side up. An operation is to choose a line parallel to the sides of the triangle, and flipping all the marks on that line. A configuration is called admissible if it can be obtained from the initial configuration by performing a finite number of operations. For each admissible configuration $C$ , let $f(C)$ denote the smallest number of operations required to obtain $C$ from the initial configuration. Find the maximum value of $f(C)$ , where $C$ varies over all admissible configurations.

Find all real numbers $x,y,z\geq 1$ satisfying $$\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.$$

Given positive integers $m$ and $n$ , prove that there is a positive integer $c$ such that the numbers $cm$ and $cn$ have the same number of occurrences of each non-zero digit when written in base ten.

Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$ , then $$\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.$$