[Triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) $ABC$ has [right angle](https://artofproblemsolving.com/wiki/index.php/Right_angle) at $B$, and contains a [point](https://artofproblemsolving.com/wiki/index.php/Point) $P$ for which $PA = 10$, $PB = 6$, and $\angle APB = \angle BPC = \angle CPA$.  Find $PC$.
Let $PC = x$. Since $\angle APB = \angle BPC = \angle CPA$, each of them is equal to $120^\circ$.  By the [Law of Cosines](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/Fermat_point) of the triangle.