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Problem 1234

AIME late

Geometry Difficulty 5.2 Find the answer HMMT February

Let $ABC$ be a triangle with circumcenter $O$ such that $AC=7$ . Suppose that the circumcircle of $AOC$ is tangent to $BC$ at $C$ and intersects the line $AB$ at $A$ and $F$ . Let $FO$ intersect $BC$ at $E$ . Compute $BE$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

$E B=\frac{7}{2} \quad O$ is the circumcenter of $\triangle ABC \Longrightarrow AO=CO \Longrightarrow \angle OCA=\angle OAC$ . Because $AC$ is an inscribed arc of circumcircle $\triangle AOC, \angle OCA=\angle OFA$ . Furthermore $BC$ is tangent to circumcircle $\triangle AOC$ , so $\angle OAC=\angle OCB$ . However, again using the fact that $O$ is the circumcenter of $\triangle ABC, \angle OCB=\angle OBC$ . We now have that $CO$ bisects $\angle ACB$ , so it follows that triangle $CA=CB$ . Also, by AA similarity we have $EOB \sim EBF$ . Thus, $EB^{2}=EO \cdot EF=EC^{2}$ by the similarity and power of a point, so $EB=BC / 2=AC / 2=7 / 2$ .

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.