MathDB

2

Part of 2014 All-Russian Olympiad

Problems(6)

Maximum number of perfect squares

Source: AllRussian-2014, Grade 9, day1, P2

5/3/2014
Sergei chooses two different natural numbers aa and bb. He writes four numbers in a notebook: aa, a+2a+2, bb and b+2b+2. He then writes all six pairwise products of the numbers of notebook on the blackboard. Let SS be the number of perfect squares on the blackboard. Find the maximum value of SS.
S. Berlov
number theorynumber theory proposed
Similar to P:11.2

Source: All Russian 2014 Grade 9 Day 2 P2

5/3/2014
Let ABCDABCD be a trapezoid with AB∥CDAB\parallel CD and Ω \Omega is a circle passing through A,B,C,DA,B,C,D. Let ω \omega be the circle passing through C,DC,D and intersecting with CA,CBCA,CB at A1A_1, B1B_1 respectively. A2A_2 and B2B_2 are the points symmetric to A1A_1 and B1B_1 respectively, with respect to the midpoints of CACA and CBCB. Prove that the points A,B,A2,B2A,B,A_2,B_2 are concyclic.
I. Bogdanov
geometrytrapezoidsymmetrygeometry proposed
Prove that f(x) lies in [0,1]

Source: All Russian 2014 Grade 10 Day 1 P2

5/3/2014
Given a function f ⁣:R→Rf\colon \mathbb{R}\rightarrow \mathbb{R} with f(x)2≤f(y)f(x)^2\le f(y) for all x,y∈Rx,y\in\mathbb{R} , x>yx>y, prove that f(x)∈[0,1]f(x)\in [0,1] for all x∈Rx\in \mathbb{R}.
functionalgebraRussiaFunctional inequality
BXMY is cyclic

Source: All Russian 2014 Grade 10 Day 2 P2

5/3/2014
Let MM be the midpoint of the side ACAC of △ABC \triangle ABC. Let P∈AMP\in AM and Q∈CMQ\in CM be such that PQ=AC2PQ=\frac{AC}{2}. Let (ABQ)(ABQ) intersect with BCBC at X≠BX\not= B and (BCP)(BCP) intersect with BABA at Y≠BY\not= B. Prove that the quadrilateral BXMYBXMY is cyclic.
F. Ivlev, F. Nilov
geometrypower of a pointgeometry proposed
Peter and Bob

Source: All Russian 2014 Grade 11 Day 1 P2

4/30/2014
Peter and Bob play a game on a n×nn\times n chessboard. At the beginning, all squares are white apart from one black corner square containing a rook. Players take turns to move the rook to a white square and recolour the square black. The player who can not move loses. Peter goes first. Who has a winning strategy?
combinatorics proposedcombinatorics
on a sphere

Source: All Russian 2014 Grade 11 Day 2 P2

4/30/2014
The sphere ω \omega passes through the vertex SS of the pyramid SABCSABC and intersects with the edges SA,SB,SCSA,SB,SC at A1,B1,C1A_1,B_1,C_1 other than SS. The sphere Ω \Omega is the circumsphere of the pyramid SABCSABC and intersects with ω \omega circumferential, lies on a plane which parallel to the plane (ABC)(ABC). Points A2,B2,C2A_2,B_2,C_2 are symmetry points of the points A1,B1,C1A_1,B_1,C_1 respect to midpoints of the edges SA,SB,SCSA,SB,SC respectively. Prove that the points AA, BB, CC, A2A_2, B2B_2, and C2C_2 lie on a sphere.
geometry3D geometryspherepyramidsymmetrycircumcircleparallelogram