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, 2016

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle with $AB = 13$ , $BC = 14$ , $CA = 15$ . Let $H$ be the orthocenter of $ABC$ . Find the distance between the circumcenters of triangles $AHB$ and $AHC$ .

Solution

Solution:

Let $H_{B}$ be the reflection of $H$ over $AC$ and let $H_{C}$ be the reflection of $H$ over $AB$ . The reflections of $H$ over $AB$ , $AC$ lie on the circumcircle of triangle $ABC$ . Since the circumcenters of triangles $AH_{C}B$ , $AH_{B}C$ are both $O$ , the circumcenters of $AHB$ , $AHC$ are reflections of $O$ over $AB$ , $AC$ respectively. Moreover, the lines from $O$ to the circumcenters in question are the perpendicular bisectors of $AB$ and $AC$ . Now we see that the distance between the two circumcenters is simply twice the length of the midline of triangle $ABC$ that is parallel to $BC$ , meaning the distance is $2\left(\frac{1}{2} BC\right) = 14$ .

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