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.
Problem 2
Solve in integers the equation
Problem 3
Quadrilateral

is inscribed in circle

with

and

. Let

be a variable point on segment

. Line

meets

again at

(other than

). Point

lies on arc

of

such that

is perpendicular to

. Let

denote the midpoint of chord

. As

varies on segment

, show that

moves along a circle.
Day 2
Problem 4
Find all functions

such that
![\[f(x)+f(t)=f(y)+f(z)\]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uWtawyFy78_xVBEBVsn-XoBaPN1sfqLRlc-s-CC0XCbiuka6iLJitZYbN8PJftGmYst-Nw02cxuAPI2o6H7cvDI58s4goJNyasT7ZBHseAoMjmNymSF5ntdJ77FAfCDoi98iX742SMuMhmb1b0k5Gas9zfSj9aDvtR1A=s0-d)
for all rational numbers

that form an arithmetic progression. (

is the set of all rational numbers.)
Problem 5
Let

be a cyclic quadrilateral. Prove that there exists a point

on segment

such that

and

if and only if there exists a point

on segment

such that

and

.
Problem 6
Steve is piling

indistinguishable stones on the squares of an

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

for some

, such that

and

. A stone move consists of either removing one stone from
each of

and

and moving them to

and

respectively,j or removing one stone from each of

and

and moving them to

and

respectively.
Two ways of piling the stones are equivalent if they can be obtained from one
another by a sequence of stone moves.
How many different non-equivalent ways can Steve pile the stones on the grid?
Aops
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