MathDB
MTB - CTM does not depend on choice of X

Source: ISL 2007, G2, AIMO 2008, TST 1, P3, Ukrainian TST 2008 Problem 1

June 3, 2008
geometrycircumcirclereflectioncyclic quadrilateralIMO Shortlistgeometry solvedmixtilinear incircle

Problem Statement

Denote by M M midpoint of side BC BC in an isosceles triangle △ABC \triangle ABC with AC=AB AC = AB. Take a point X X on a smaller arc \overarc{MA} of circumcircle of triangle △ABM \triangle ABM. Denote by T T point inside of angle BMA BMA such that ∠TMX=90 \angle TMX = 90 and TX=BX TX = BX.
Prove that ∠MTB−∠CTM \angle MTB - \angle CTM does not depend on choice of X X.
Author: Farzan Barekat, Canada
Back to ProblemsView on AoPS