### Hubert Shutrick

Given three equilateral triangles A_{1}B_{1}C, A_{2}B_{2}C and A_{3}B_{3}C with a common vertex, the midpoints M_{1} of B_{2}A_{3}, M_{2} of B_{3}A_{1} and M_{3} of B_{1}A_{2} are also vertices of an equilateral triangle.

This theorem is also proved at another site.

It is sufficient to prove that, if we have three triangles for which the theorem is true and we replace A_{1}B_{1}C by another equilateral triangle A'_{1}B'_{1}C, then the new midpoints M'_{2} and M'_{3} with M_{1} are vertices of an equilateral triangle. It is sufficient because we can start with three congruent triangles equally spaced, where it is obviously true, and replace them, one by one, to get the required configuration.

Since M_{1}M_{2}M_{3} is equilateral, M_{1}M'_{2}M'_{3} will also be if rotating M_{1}M_{2}M'_{2} through \frac{\pi}{3} about M_{1} takes it to M_{1}M_{3}M'_{3}. The rotation certainly takes M_{2} to M_{3} so the theorem will be proved if M_{2}M'_{2} goes to M_{3}M'_{3}. The segment M_{2}M'_{2} is half the length of A_{1}A'_{1}and parallel with it and similarly with M_{3}M'_{3} and B_{1}B'_{1}. The triangle B_{1}B'_{1}C is obtained by rotating A_{1}A'_{1}C through \frac{\pi}{3} about C which gives directly that the lengths of the two segments are equal. Moreover, M_{2}M'_{2} turns through the same angle, \frac{\pi}{3}, as A_{1}A'_{1} did and so M'_{2} goes to M'_{3} as required.