MathDB

3

Part of 2013 Turkey Team Selection Test

Problems(3)

AO/OD - BC/AT=4

Source: Turkey TST 2013 - Day 1 - P3

4/2/2013
Let OO be the circumcenter and II be the incenter of an acute triangle ABCABC with m(B^)≠m(C^)m(\widehat{B}) \neq m(\widehat{C}). Let DD, EE, FF be the midpoints of the sides [BC][BC], [CA][CA], [AB][AB], respectively. Let TT be the foot of perpendicular from II to [AB][AB]. Let PP be the circumcenter of the triangle DEFDEF and QQ be the midpoint of [OI][OI]. If AA, PP, QQ are collinear, prove that ∣AO∣∣OD∣−∣BC∣∣AT∣=4.\dfrac{|AO|}{|OD|}-\dfrac{|BC|}{|AT|}=4.
geometrycircumcircleincentergeometric transformationreflectionparallelograminradius
z(xz+yz+y)/(xy+y^2+z^2+1) <= K

Source: Turkey TST 2013 - Day 2 - P3

4/2/2013
For all real numbers x,y,zx,y,z such that −2≤x,y,z≤2-2\leq x,y,z \leq 2 and x2+y2+z2+xyz=4x^2+y^2+z^2+xyz = 4, determine the least real number KK satisfying z(xz+yz+y)xy+y2+z2+1≤K.\dfrac{z(xz+yz+y)}{xy+y^2+z^2+1} \leq K.
inequalitiestrigonometryinequalities proposed
Flights distributed to airway companies

Source: Turkey TST 2013 - Day 3 - P3

4/2/2013
Some cities of a country consisting of nn cities are connected by round trip flights so that there are at least kk flights from any city and any city is reachable from any city. Prove that for any such flight organization these flights can be distributed among n−kn-k air companies so that one can reach any city from any city by using of at most one flight of each air company.
inductiongraph theorycombinatorics proposedcombinatorics