Suppose equilateral triangles ABE, ACF and ADG share a common vertex A as shown. Then the midpointss of the segments EC, FD and GB form the vertices of an equilateral triangle.
The special case where points BECFDG lie on a circe centred at A was proved by Alon Amit (April 5 2020) in response to a question posed on Quora:
A hexagon is inscribed in a unit circle such that three alternate sides are of unit length with the other three arbitrary. Connect the midpoints of the three arbitrary length sides. Can you prove those points form an equilateral triangle?
Amit's clever proof, using complex numbers, does not in fact depend on the points lying on the unit circle and holds as long as we have the configuration in the figure above, and this more general case is even more surprising than the original version.
A Lovely Geometry Theorem
Suppose equilateral triangles ABE, ACF and ADG share a common vertex A as shown. Then the midpointss of the segments EC, FD and GB form the vertices of an equilateral triangle.
The special case where points BECFDG lie on a circe centred at A was proved by Alon Amit (April 5 2020) in response to a question posed on Quora:
A hexagon is inscribed in a unit circle such that three alternate sides are of unit length with the other three arbitrary. Connect the midpoints of the three arbitrary length sides. Can you prove those points form an equilateral triangle?
Amit's clever proof, using complex numbers, does not in fact depend on the points lying on the unit circle and holds as long as we have the configuration in the figure above, and this more general case is even more surprising than the original version.