# 1987 AIME Problems/Problem 9

## Problem

Triangle $ABC$ has right angle at $B$, and contains a point $P$ for which $PA = 10$, $PB = 6$, and $\angle APB = \angle BPC = \angle CPA$. Find $PC$.

$[asy] unitsize(0.2 cm); pair A, B, C, P; A = (0,14); B = (0,0); C = (21*sqrt(3),0); P = intersectionpoint(arc(B,6,0,180),arc(C,33,0,180)); draw(A--B--C--cycle); draw(A--P); draw(B--P); draw(C--P); label("A", A, NW); label("B", B, SW); label("C", C, SE); label("P", P, NE); [/asy]$

## Solution

Let $PC = x$. Since $\angle APB = \angle BPC = \angle CPA$, each of them is equal to $120^\circ$. By the Law of Cosines applied to triangles $\triangle APB$, $\triangle BPC$ and $\triangle CPA$ at their respective angles $P$, remembering that $\cos 120^\circ = -\frac12$, we have

$$AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x$$

Then by the Pythagorean Theorem, $AB^2 + BC^2 = CA^2$, so

$$x^2 + 10x + 100 = x^2 + 6x + 36 + 196$$

and

$$4x = 132 \Longrightarrow x = \boxed{033}.$$

### Note

This is the Fermat point of the triangle.