2015 USAMO Problems

Day 1

Problem 1

Given a sequence of real numbers, a move consists of choosing two terms and replacing each with their arithmetic mean. Show that there exists a sequence of 2015 distinct real numbers such that after one initial move is applied to the sequence -- no matter what move -- there is always a way to continue with a finite sequence of moves so as to obtain in the end a constant sequence.

Solution

Problem 2

Solve in integers the equation $x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3.$

Solution

Problem 3

Quadrilateral $APBQ$ is inscribed in circle $\omega$ with $\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$ . Let $X$ be a variable point on segment $\overline{PQ}$ . Line $AX$ meets $\omega$ again at $S$ (other than $A$ ). Point $T$ lies on arc $AQB$ of $\omega$ such that $\overline{XT}$ is perpendicular to $\overline{AX}$ . Let $M$ denote the midpoint of chord $\overline{ST}$ . As $X$ varies on segment $\overline{PQ}$ , show that $M$ moves along a circle.

Solution

Day 2

Problem 4

Find all functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$ such that $f(x)+f(t)=f(y)+f(z)$ for all rational numbers $x<y<z<t$ that form an arithmetic progression. ( $\mathbb{Q}$ is the set of all rational numbers.)

Solution

Problem 5

Let $ABCD$ be a cyclic quadrilateral. Prove that there exists a point $X$ on segment $\overline{BD}$ such that $\angle BAC=\angle XAD$ and $\angle BCA=\angle XCD$ if and only if there exists a point $Y$ on segment $\overline{AC}$ such that $\angle CBD=\angle YBA$ and $\angle CDB=\angle YDA$ .

Solution

Problem 6

Steve is piling $m\geq 1$ indistinguishable stones on the squares of an $n\times n$ grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions $(i, k), (i, l), (j, k), (j, l)$ for some $1\leq i, j, k, l\leq n$ , such that $i<j$ and $k<l$ . A stone move consists of either removing one stone from each of $(i, k)$ and $(j, l)$ and moving them to $(i, l)$ and $(j, k)$ respectively,j or removing one stone from each of $(i, l)$ and $(j, k)$ and moving them to $(i, k)$ and $(j, l)$ respectively.