HMMT February, 2018

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Suppose $\triangle ABC$ has lengths $AB = 5$ , $BC = 8$ , and $CA = 7$ , and let $\omega$ be the circumcircle of $\triangle ABC$ . Let $X$ be the second intersection of the external angle bisector of $\angle B$ with $\omega$ , and let $Y$ be the foot of the perpendicular from $X$ to $BC$ . Find the length of $YC$ .

Solution

Solution:

Extend ray $\overrightarrow{AB}$ to a point $D$ . Since $BX$ is an angle bisector, we have $\angle XBC = \angle XBD = 180^{\circ} - \angle XBA = \angle XCA$ , so $XC = XA$ by the inscribed angle theorem. Now, construct a point $E$ on $BC$ so that $CE = AB$ . Since $\angle BAX \cong \angle BCX$ , we have $\triangle BAX \cong \triangle ECX$ by SAS congruence. Thus, $XB = XE$ , so $Y$ bisects segment $BE$ . Since $BE = BC - EC = 8 - 5 = 3$ , we have $YC = EC + YE = 5 + \frac{1}{2} \cdot 3 = \frac{13}{2}$ .

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