The triangle $ABC$ is inscribed in the circle. Construct a point $P$ such that the points of intersection of lines $AP, BP$ and $CP$ with this circle are the vertices of an equilateral triangle.
(A. Zaslavsky)
Construct a circle passing through $A$ and $C$ such that for any point $P$ of the circle which lies on the same side of the plane as $B$ with respect to $AC$ $\angle APC=60^{\circ}+\beta$. Consider an analogous circle passing through $A$ and $B$. Let $P$ be the point of their intersection. Then $\angle APC=60^{\circ}+\beta$; $\angle BPC=60^{\circ}+\gamma$; $\angle APB=60^{\circ}+\gamma$. It's not hard to see that this point suffices the conditions