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Iberoamerican Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Ibero-American Mathematical Olympiad

Problem:

The incircle of the triangle $ABC$ touches $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ respectively. $AD$ meets the circle again at $Q$ . Show that the line $EQ$ passes through the midpoint of $AF$ iff $AC = BC$ .

Solution

Solution:

$\angle AQ M = \angle EQ D$ (opposite angle) $= \angle EDC$ ( $CD$ tangent to circle $EQD$ ) $= \left(180^\circ - \angle C\right) / 2 = \angle A / 2 + \angle B / 2$ $(*)$ .

$MF^2 = MQ \cdot ME$ ( $MF$ tangent to circle $FQE$ ). So $AM = AF$ is equivalent to $AM^2 = MQ \cdot ME$ or $AM / MQ = ME / AM$ . But since triangles $AMQ$ and $EMA$ have a common angle $M$ , $AM / MQ = ME / AM$ iff they are similar, and hence iff $\angle AQ M = \angle A$ . Using $(*)$ , $AM = AF$ iff $\angle A = \angle B$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.