Library / /613 of 740

, 2015

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle that satisfies $AB = 13$ , $BC = 14$ , $AC = 15$ . Given a point $P$ in the plane, let $P_A$ , $P_B$ , $P_C$ be the reflections of $A$ , $B$ , $C$ across $P$ . Call $P$ good if the circumcircle of $P_A P_B P_C$ intersects the circumcircle of $ABC$ at exactly 1 point. The locus of good points $P$ encloses a region $\mathcal{S}$ . Find the area of $\mathcal{S}$ .

Solution

Solution:

By the properties of reflection, the circumradius of $P_A P_B P_C$ equals the circumradius of $ABC$ . Therefore, the circumcircle of $P_A P_B P_C$ must be externally tangent to the circumcircle of $ABC$ . Now it's easy to see that the midpoint of the 2 centers of $ABC$ and $P_A P_B P_C$ lies on the circumcircle of $ABC$ . So the locus of $P$ is simply the circumcircle of $ABC$ .

Since $[ABC] = \frac{abc}{4R}$ , we find the circumradius is $R = \frac{13 \cdot 14 \cdot 15}{84 \cdot 4} = \frac{65}{8}$ , so the enclosed region has area $\frac{4225}{64} \pi$ .

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