"Consider triangle ABC. Point D is on AC, E is on BC and F is on AB. Given that triangle DEF is equilateral and that segments CD, BE and AF are equal in length, prove that triangle ABC must also be equilateral."

Here's one way of looking at this problem. Consider this graphic

If two circles of unequal radius share two points, then the angle subtended by those points and the center of the smaller circle is always greater than the angle subtended by same and any point on the larger circle. Now consider any equilateral triangle, and draw in three equal circles centered at its vertices. Then, given a point on any circle, draw a line through it and the center of the "next" circle, and extend it until it intersects it at a new point. Given a range of points on the first circle, the range of new points on the next will always be less. Repeat process from "next" circle to the next, with the result that the range of new points converges to a single point at all three circles. Since there will always exist an equilateral triangle that meets the conditions, there can be no other kind of triangle in which this process will converge to, as it would imply more than one limiting case.

Maybe later I'll post a graphic of this process with a fuller explanation.

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TopNewestHere's one way of looking at this problem. Consider this graphic

If two circles of unequal radius share two points, then the angle subtended by those points and the center of the smaller circle is always greater than the angle subtended by same and any point on the larger circle. Now consider any equilateral triangle, and draw in three equal circles centered at its vertices. Then, given a point on any circle, draw a line through it and the center of the "next" circle, and extend it until it intersects it at a new point. Given a range of points on the first circle, the range of new points on the next will always be less. Repeat process from "next" circle to the next, with the result that the range of new points converges to a single point at all three circles. Since there will always exist an equilateral triangle that meets the conditions, there can be no other kind of triangle in which this process will converge to, as it would imply more than one limiting case.

Maybe later I'll post a graphic of this process with a fuller explanation.

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Note: You will need the points to lie on the segments, instead of the (extended) line. Otherwise, this statement is not true.

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