olymmath

IOM 2018

International Olympiad of Metropolises 2018.

Other problems

Problem 2

A convex quadrilateral ABCDABCD is circumscribed about a circle ω\omega . Let PQPQ be the diameter of ω\omega perpendicular to ACAC . Suppose lines BPBP and DQDQ intersect at point XX , and lines BQBQ and DPDP intersect at point YY . Show that the points XX and YY lie on the line ACAC .

Géza Kós

Problem 3

Let kk be a positive integer such that p=8k+5p = 8k + 5 is a prime number. The integers r1,r2,…,r2k+1r_1, r_2, \dots, r_{2k+1} are chosen so that the numbers 0,r14,r24,…,r2k+140, r_1^4, r_2^4, \dots, r_{2k+1}^4 give pairwise different remainders modulo pp . Prove that the product ∏1⩽i<j⩽2k+1(ri4+rj4)\prod_{1 \leqslant i < j \leqslant 2k+1} \big(r_i^4 + r_j^4\big) is congruent to (−1)k(k+1)/2(-1)^{k(k+1)/2} modulo pp .

(Two integers are congruent modulo pp if pp divides their difference.)

Fedor Petrov

Problem 4

Let 1=d0<d1<⋯<dm=4k1 = d_0 < d_1 < \dots < d_m = 4k be all positive divisors of 4k4k , where kk is a positive integer. Prove that there exists i∈{1,…,m}i \in \{1, \dots, m\} such that di−di−1=2d_i - d_{i-1} = 2 .

Ivan Mitrofanov

Problem 5

Ann and Max play a game on a 100×100100 \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?

Lev Shabanov

Problem 6

The incircle of a triangle ABCABC touches the sides BCBC and ACAC at points DD and EE , respectively. Suppose PP is the point on the shorter arc DEDE of the incircle such that ∠APE=∠DPB\angle APE = \angle DPB . The segments APAP and BPBP meet the segment DEDE at points KK and LL , respectively. Prove that 2KL=DE2KL = DE .

Dušan Djukić