3
Part of 2002 Tuymaada Olympiad
Problems(3)
Circle + triangle = hexagon ==> two parallel lines.
Source: Tuymaada 2002, day 1, problem 3. - Author : S. Berlov.
5/3/2007
A circle having common centre with the circumcircle of triangle meets the sides of the triangle at six points forming convex hexagon ( and lie on , and lie on , and lie on ).
If is parallel to the bisector of angle , prove that is parallel to the bisector of angle .Proposed by S. Berlov
geometrycircumcircleincentergeometry proposed
trinomial with integer coefficients, and powers of two as natural values
Source: Tuymaada Junior 2002 p3
5/11/2019
Is there a quadratic trinomial with integer coefficients, such that all values which are natural to be natural powers of two?
algebrapolynomialInteger Polynomialtrinomialquadratic trinomialpower of 2
Acute triangle, circumcircle, isoceles triangle, altitudes.
Source: Tuymaada 2002, day 2, problem 3. - Author : D. Shiryaev.
5/3/2007
The points and on the circumcircle of an acute triangle are such that . Let be the common point of the altitudes of triangle .
It is known that .
Prove that lies on the segment .Proposed by D. Shiryaev
geometrycircumcircletrigonometrypower of a pointradical axisgeometry proposed