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colinear and fixed point

Source: Problem 3 VMO 2013 day 1

January 11, 2013
geometrygeometric transformationreflectionperpendicular bisectorgeometry proposed

Problem Statement

Let ABCABC be a triangle such that ABCABC isn't a isosceles triangle. (I)(I) is incircle of triangle touches BC,CA,ABBC,CA,AB at D,E,FD,E,F respectively. The line through EE perpendicular to BIBI cuts (I)(I) again at KK. The line through FF perpendicular to CICI cuts (I)(I) again at LL.JJ is midpoint of KLKL. a) Prove that D,I,JD,I,J collinear. b) B,CB,C are fixed points,AA is moved point such that ABAC=k\frac{AB}{AC}=k with kk is constant.IE,IFIE,IF cut (I)(I) again at M,NM,N respectively.MNMN cuts IB,ICIB,IC at P,QP,Q respectively. Prove that bisector perpendicular of PQPQ through a fixed point.
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