Trying to solve a problem with a colleague of mine, we prove a theorem that I'm sure someone else must have had to come across but couldn't find anything about it.
We needed a way to tell how far away from equilateral was any triangle. A measure for any triangle of its non-equilateral-ness.
One idea (that later turned out not to be the best) was this:
Let ABC be any triangle (the one we want to know how far away from equilateralness is).
Let's take any of its sides, say AB.
Let's take the point C' in which ABC' is equilateral, and which lies to the same side of AB as C.
Let's measure the distance CC'.
Let's take points B' and A' the same way we took C'.
What we found is that the distances CC', BB', and AA', are equal. No matter which side of the triangle we start with, the result is the same.
Is it known to be found earlier?
Edit
For the record, the proof goes this way:
We have the original triangle ABC and three new points A', B', C' such that ABC' is equilateral and so on.
Let's see triangles ABC' and AB'C. Both are equilateral and share the same point A.
Let's see the geometric transformation that brings C' into C, B into B' and A into itself. Given that the distances C'A and BA are the same, the distance traveled by C' to go into C must be the same as the distance traveled by B to go into B'. Ergo, distances C'C and BB' are the same.
By the same way we prove that the distance A'A is also the same.
The answer is on the comments.
According to Calvin Lin it's a "well known" olympiad problem.
A more elegant proof that mine involves 60 degrees rotations. IE, BCC' and BA'A are the same triangle rotated 60 degrees so AA' and CC'must be the same length.
This finding is related to Napoleons's theorem, more specifically, the inner Napoleon triangle.
Thanks a lot.