MathDB

4

Part of 2011 All-Russian Olympiad

Problems(6)

Dividing squares

Source: All-Russian 2011

5/17/2011
Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers?
number theoryrelatively primenumber theory proposed
ARO 2011 9-8

Source:

5/6/2011
There are some counters in some cells of 100×100100\times 100 board. Call a cell nice if there are an even number of counters in adjacent cells. Can exactly one cell be nice?
K. Knop
algebrapolynomialmodular arithmeticsymmetryrectanglecombinatorics proposed
ARO 2011 10-4

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4/28/2011
Perimeter of triangle ABCABC is 44. Point XX is marked at ray ABAB and point YY is marked at ray ACAC such that AX=AY=1AX=AY=1. Line segments BCBC and XYXY intersectat point MM. Prove that perimeter of one of triangles ABMABM or ACMACM is 22.
(V. Shmarov).
geometryRussiaAll Russian Math Olympiad
Aro 2011 10-8

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4/28/2011
A 2010×20102010\times 2010 board is divided into corner-shaped figures of three cells. Prove that it is possible to mark one cell in each figure such that each row and each column will have the same number of marked cells.
I. Bogdanov & O. Podlipsky
invariantgraph theorycombinatorics proposedcombinatorics
ARO 2011 11-4

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4/26/2011
Ten cars are moving at the road. There are some cities at the road. Each car is moving with some constant speed through cities and with some different constant speed outside the cities (different cars may move with different speed). There are 2011 points at the road. Cars don't overtake at the points. Prove that there are 2 points such that cars pass through these points in the same order.
S. Berlov
combinatorics proposedcombinatorics
ARO 2011 11-8

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5/3/2011
Let NN be the midpoint of arc ABCABC of the circumcircle of triangle ABCABC, let MM be the midpoint of ACAC and let I1,I2I_1, I_2 be the incentres of triangles ABMABM and CBMCBM. Prove that points I1,I2,B,NI_1, I_2, B, N lie on a circle.
M. Kungojin
geometrycircumcircletrigonometry