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For a triangle T = ABC we take the point X on the side (AB) such that AX/AB=4/5, the point Y on the segment (CX) such that CY = 2YX and, if possible, the point Z on the ray (CA such that \widehat{CXZ} = 180 - \widehat{ABC}. We denote by \Sigma the set of all triangles T for which
\widehat{XYZ} = 45. Prove that all triangles from \Sigma are similar and find the measure of their smallest angle.

Slični zadaci

Let triangle ABC be such that its circumradius is R = 1. Let r be the inradius of ABC and let p be the inradius of the orthic triangle A'B'C' of triangle ABC. Prove that p \leq 1 - \frac{1}{3 \cdot (1+r)^2}.
Given a triangle ABC, let D and E be points on the side BC such that \angle BAD = \angle CAE. If M and N are, respectively, the points of tangency of the incircles of the triangles ABD and ACE with the line BC, then show that
\frac{1}{MB}+\frac{1}{MD}= \frac{1}{NC}+\frac{1}{NE}.
Let ABC be a triangle and M be an interior point. Prove that

\min\{MA,MB,MC\}+MA+MB+MC<AB+AC+BC.
Let ABC be a triangle, \Omega its incircle and \Omega_{a}, \Omega_{b}, \Omega_{c} three circles orthogonal to \Omega passing through (B,C),(A,C) and (A,B) respectively. The circles \Omega_{a} and \Omega_{b} meet again in C'; in the same way we obtain the points B' and A'. Prove that the radius of the circumcircle of A'B'C' is half the radius of \Omega.
In the coordinate plane consider the set S of all points with integer coordinates. For a positive integer k, two distinct points a, B\in S will be called k-friends if there is a point C\in S such that the area of the triangle ABC is equal to k. A set T\subset S will be called k-clique if every two points in T are k-friends. Find the least positive integer k for which there exits a k-clique with more than 200 elements.

Proposed by Jorge Tipe, Peru
In an acute triangle ABC segments BE and CF are altitudes. Two circles passing through the point A anf F and tangent to the line BC at the points P and Q so that B lies between C and Q. Prove that lines PE and QF intersect on the circumcircle of triangle AEF.



Proposed by Davood Vakili, Iran