12. (POL 5) Given a nonequilateral triangle $A B C$, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles $A^{\prime} B^{\prime} C^{\prime}$ (the vertices listed counterclockwise) for which the triples of points $A, B^{\prime}, C^{\prime} ; A^{\prime}, B, C^{\prime}$; and $A^{\prime}, B^{\prime}, C$ are collinear.
12. Here all angles will be oriented and measured counterclockwise.

Note that $\angle C A^{\prime} B=\angle A B^{\prime} C=$ $\angle B C^{\prime} A=\pi / 3$. Let $a^{\prime}, b^{\prime}, c^{\prime}$ denote respectively the inner bisectors of angles $A^{\prime}, B^{\prime}, C^{\prime}$ in triangle $A^{\prime} B^{\prime} C^{r}$. The lines $a^{\prime}, b^{\prime}, c^{\prime}$ meet at the centroid $X$ of $A^{\prime} B^{\prime} C^{\prime}$, and $\angle\left(a^{\prime}, b^{\prime}\right)=$ $\angle\left(b^{\prime}, c^{\prime}\right)=\angle\left(c^{\prime}, a^{\prime}\right)=2 \pi / 3$. Now let $K, L, M$ be the points such that $K B=K C, L C=L A, M A=M B$, and $\angle B K C=\angle C L A=\angle A M B=$ $2 \pi / 3$, and let $C_{1}, C_{2}, C_{3}$ be the circles circumscribed about triangles $B K C, C L A$, and $A M B$ respectively. These circles are characterized by $C_{1}=\{Z \mid \angle B Z C=2 \pi / 3\}$, etc.; hence we deduce that they meet at a point $P$ such that $\angle B P C=\angle C P A=\angle A P B=2 \pi / 3$ (Torricelli's point). Points $A^{\prime}, B^{r}, C^{r}$ run over $C_{1} \backslash\{P\}, C_{2} \backslash\{P\}, C_{3} \backslash\{P\}$ respectively. As for $a^{\prime}, b^{\prime}, c$, we see that $K \in a^{r}, L \in U^{\prime}, M \in c^{\prime}$, and also that they can take all possible directions except $K P, L P, M P$ respectively (if $K=P, K P$ is assumed to be the corresponding tangent at $K$ ). Then, since $\angle K X L=$ $2 \pi / 3, X$ runs over the circle defined by $\{Z \mid \angle K Z L=2 \pi / 3\}$, without $P$. But analogously, $X$ runs over the circle $\{Z \mid \angle L Z M=2 \pi / 3\}$, from which we can conclude that these two circles are the same, both equal to the circumcircle of $K L M$, and consequently also that triangle $K L M$ is equilateral (which is, anyway, a well-known fact). Therefore, the locus of the points $X$ is the circumcircle of $K L M$ minus point $P$.