Suppose $\triangle ABC$ is equilateral with side length $s$, with a unique point $P$ inside such that $AP=5, BP=4, CP=3$. What is $s$?

1. Rotate $\triangle CPA$ $60^\circ$ about $C$ to $\triangle CQB$; then $CP=CQ=3$, $\angle PCQ=60^\circ$ so $\triangle CPQ$ is equilateral with $PQ=3$, and $QB=PA=5$.
2. In $\triangle PQB$ (sides $PQ=3,QB=5,PB=4$): since $3^2+4^2=5^2$, $\angle QPB=90^\circ$.
3. $\angle BPC=\angle QPB+\angle QPC=90^\circ+60^\circ=150^\circ$, so $BC^2=PB^2+PC^2-2\cdot PB\cdot PC\cos150^\circ=16+9+12\sqrt3=25+12\sqrt3$.
4. $s=\boxed{\sqrt{25+12\sqrt3}}$.