Library / /638 of 1394

, 2025

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:
Let $\triangle ABC$ be an equilateral triangle with side length $6$ . Let $P$ be a point inside triangle $\triangle ABC$ such that $\angle BPC = 120^\circ$ . The circle with diameter $\overline{AP}$ meets the circumcircle of $\triangle ABC$ again at $X \neq A$ . Given that $AX = 5$ , compute $XP$ .

Solution

Solution:

Let $A'$ be the antipode of $A$ . As $\angle AXA' = 90^\circ$ , we have $X$ , $P$ , and $A'$ are collinear. As $\angle BPC = 120^\circ$ and $\angle BA'C = 180^\circ - \angle BAC = 120^\circ$ , it follows that $P$ lies on the circle with center $A'$ passing through $B$ and $C$ , so $A'P = \frac{BC}{\sqrt{3}} = 2\sqrt{3}$ and $AA' = 2A'B = 4\sqrt{3}$ . By the Pythagorean theorem, $XA' = \sqrt{(4\sqrt{3})^2 - 5^2} = \sqrt{23}$ , so the answer is $XA' - PA' = \left[\sqrt{23} - 2\sqrt{3}\right]$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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