MathDB

2

Part of 2011 All-Russian Olympiad

Problems(6)

Acute triangle

Source: All-Russian 2011

5/17/2011
Given is an acute angled triangle ABCABC. A circle going through BB and the triangle's circumcenter, OO, intersects BCBC and BABA at points PP and QQ respectively. Prove that the intersection of the heights of the triangle POQPOQ lies on line ACAC.
geometrycircumcirclegeometric transformationgeometry proposed
Notebooks

Source: All-Russian 2011

5/17/2011
In the notebooks of Peter and Nick, two numbers are written. Initially, these two numbers are 1 and 2 for Peter and 3 and 4 for Nick. Once a minute, Peter writes a quadratic trinomial f(x)f(x), the roots of which are the two numbers in his notebook, while Nick writes a quadratic trinomial g(x)g(x) the roots of which are the numbers in his notebook. If the equation f(x)=g(x)f(x)=g(x) has two distinct roots, one of the two boys replaces the numbers in his notebook by those two roots. Otherwise, nothing happens. If Peter once made one of his numbers 5, what did the other one of his numbers become?
quadraticsinvariantVietainductionalgebrapolynomialarithmetic sequence
Nine quadratics

Source: All-Russian 2011

5/17/2011
Nine quadratics, x2+a1x+b1,x2+a2x+b2,...,x2+a9x+b9x^2+a_1x+b_1, x^2+a_2x+b_2,...,x^2+a_9x+b_9 are written on the board. The sequences a1,a2,...,a9a_1, a_2,...,a_9 and b1,b2,...,b9b_1, b_2,...,b_9 are arithmetic. The sum of all nine quadratics has at least one real root. What is the the greatest possible number of original quadratics that can have no real roots?
quadraticsalgebra proposedalgebraQuadratic
Prove angle is right

Source: All-Russian 2011

5/17/2011
Given is an acute triangle ABCABC. Its heights BB1BB_1 and CC1CC_1 are extended past points B1B_1 and C1C_1. On these extensions, points PP and QQ are chosen, such that angle PAQPAQ is right. Let AFAF be a height of triangle APQAPQ. Prove that angle BFCBFC is a right angle.
geometrygeometry proposed
Parallelogram

Source: All-Russian 2011

5/17/2011
On side BCBC of parallelogram ABCDABCD (AA is acute) lies point TT so that triangle ATDATD is an acute triangle. Let O1O_1, O2O_2, and O3O_3 be the circumcenters of triangles ABTABT, DATDAT, and CDTCDT respectively. Prove that the orthocenter of triangle O1O2O3O_1O_2O_3 lies on line ADAD.
geometryparallelogramcircumcirclegeometric transformationreflectionsymmetrygeometry proposed
ARO 2011 11-6

Source:

5/6/2011
There are more than n2n^2 stones on the table. Peter and Vasya play a game, Peter starts. Each turn, a player can take any prime number less than nn stones, or any multiple of nn stones, or 11 stone. Prove that Peter always can take the last stone (regardless of Vasya's strategy).
S Berlov
combinatorics proposedcombinatoricsgame