# [Solutions] International Olympiad of Metropolises 2018

1. Solve the system of equations in real numbers $\begin{cases} (x - 1)(y - 1)(z - 1) = xyz - 1,\\ (x - 2)(y - 2)(z - 2) = xyz - 2.\\ \end{cases}$
2. A convex quadrilateral $ABCD$ is circumscribed about a circle $\omega$. Let $PQ$ be the diameter of $\omega$ perpendicular to $AC$. Suppose lines $BP$ and $DQ$ intersect at point $X$, and lines $BQ$ and $DP$ intersect at point $Y$. Show that the points $X$ and $Y$ lie on the line $AC$.
3. Let $k$ be a positive integer such that $p = 8k + 5$ is a prime number. The integers $r_1, r_2, \dots, r_{2k+1}$ are chosen so that the numbers $0, r_1^4, r_2^4, \dots, r_{2k+1}^4$ give pairwise different remainders modulo $p$. Prove that the product $\prod_{1 \leqslant i < j \leqslant 2k+1} \big(r_i^4 + r_j^4\big)$is congruent to $(-1)^{k(k+1)/2}$ modulo $p$. (Two integers are congruent modulo $p$ if $p$ divides their difference.)
4. Let $1 = d_0 < d_1 < \dots < d_m = 4k$ be all positive divisors of $4k$, where $k$ is a positive integer. Prove that there exists $i \in \{1, \dots, m\}$ such that $d_i - d_{i-1} = 2$.
5. Ann and Max play a game on a $100 \times 100$ board. First, Ann writes an integer from 1 to 10 000 in each square of the board so that each number is used exactly once. Then Max chooses a square in the leftmost column and places a token on this square. He makes a number of moves in order to reach the rightmost column. In each move the token is moved to a square adjacent by side or vertex. For each visited square (including the starting one) Max pays Ann the number of coins equal to the number written in that square. Max wants to pay as little as possible, whereas Ann wants to write the numbers in such a way to maximise the amount she will receive. How much money will Max pay Ann if both players follow their best strategies?
6. The incircle of a triangle $ABC$ touches the sides $BC$ and $AC$ at points $D$ and $E$, respectively. Suppose $P$ is the point on the shorter arc $DE$ of the incircle such that $\angle APE = \angle DPB$. The segments $AP$ and $BP$ meet the segment $DE$ at points $K$ and $L$, respectively. Prove that $2KL = DE$.
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