Three concentric circles have radii $3,4,5$. An equilateral triangle with side length $s$ has one vertex on each circle, and the common center of the circles lies inside the triangle. What is $s^2$?
Let $O$ be the common center and $A,B,C$ the vertices with $OA=3$, $OB=4$, $OC=5$.
Rotate by $60^\circ$ about $A$ taking $B$ to $C$, and let $O'$ be the image of $O$. Then $\triangle AOO'$ is equilateral, so $OO'=AO'=3$, and $O'C=OB=4$. In $\triangle OO'C$ the sides are $3,4,5$, so $\angle OO'C=90^\circ$.
Since $\angle AO'O=60^\circ$, the angle $\angle AO'C$ equals $60^\circ+90^\circ=150^\circ$ or $90^\circ-60^\circ=30^\circ$. By the law of cosines in $\triangle AO'C$ (with $AO'=3$, $O'C=4$), $s^2=AC^2=9+16-24\cos\angle AO'C$, which equals $25+12\sqrt3$ for $150^\circ$ and $25-12\sqrt3$ for $30^\circ$.
Because $O$ is inside the triangle, $OC=5$ is less than the side $s$ (no point of a triangle is farther from a vertex than the farthest vertex), so $s^2>25$. Hence $s^2=\boxed{25+12\sqrt3}$.
