# Pompeiu's theorem

The proof is quick. Consider a rotation of 60° about the point C. Assume A maps to B, and P maps to P '. Then we have $\scriptstyle PC\ =\ P'C$, and $\scriptstyle\angle PCP'\ =\ 60^{\circ}$. Hence triangle PCP ' is equilateral and $\scriptstyle PP'\ =\ PC$. It is obvious that $\scriptstyle PA\ =\ P'B$. Thus, triangle PBP ' has sides equal to PA, PB, and PC and the proof by construction is complete.