CMO

Geometry Difficulty 5.2 AIME, harder Prove it Canada

Problem:

Let $ABC$ be a triangle with incenter $I$ . Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$ . Prove that $XI$ is perpendicular to $BC$ .

(The incenter is the point where the three angle bisectors meet.)

Solution

Solution:

Suppose the reflection of $AC$ across $BI$ intersects $BC$ at $E$ . Define $F$ similarly for the reflection of $AB$ across $CI$ . Also suppose $CI$ intersects $AB$ at $M$ and $BI$ intersects $AC$ at $N$ . Since $CA$ and $CF = BC$ are reflections across $CI$ , and so are $MA$ and $MF = XM$ , we have that $A$ and $F$ are reflections across $CI$ . Similarly $A$ and $E$ are reflections across $BI$ .

Thus $\angle XFC = \angle BAC = \angle XEB$ if $\angle BAC$ is acute (and $\angle XFC = \angle XEB = \pi - \angle BAC$ , when $\angle BAC$ is obtuse), so $XF = XE$ . Moreover we also find that $IF = IA = IE$ by the aforementioned reflection properties, so thus $XI$ is the perpendicular bisector of $EF$ and is hence perpendicular to $BC$ .

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