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Cyclic HexagonBui Quang Tuan has stated and proved a curious property of cyclic hexagons. He also observed that his proof extends to 2n-gons with n > 2. Given a cyclic hexagon A0B2A1B0A2B1, drop the perpendiculars AiKi on Bi-1Bi+1 and BiLi on Ai-1Ai+1, where indices are counted cyclically modulo 3. Denote for simplicity ki = AiKi and li = BiLi. Then
The proof is based on a lemma and is illustrated by the applet below:
ProofStill assuming the indices are considered cyclically modulo 3, let Pi be the midpoint of AiBi+1 and Qi the midpoint of BiAi+1. The six circles with centers at Pi and Qi and the sides of the hexagon as diameters form a chain of circles with common chords AiKi and BiLi. We denote them C(Pi) and C(Qi), respectively, R(Pi) and R(Qi) being their radii.
With every vertex of the hexagon we also associate the image of the given circle under homothety with center at that vertex and coefficient 1/2. All these circles have radius R/2, where R is the radius of the given circle, and the center at the midpoint of the segments joining the vertices with the center of the given circle. By the lemma below,
It them follows that
where ∏ stands for the product of the following terms with indices from the allowed range, Now, each of the radii in (3) is exactly half of one of the sides; so that if the sides are denoted ai, i = 0, ..., 5, and setting
RemarkNote that the hexagon does not need to be convex which implies that that the six points can be located on the circumcircle in any order. In other words, the statement is actually about two triangles with the same circumcircle and the distances from the vertices of one to the sides of the other and vice versa. As I mentioned at the outset Bui Quang Tuan generalized the problem from a hexagon to a cyclic 2n-gon with LemmaLet two circles C(O1, R1), C(O2, R2) meet in points A and B so that AB is the common chord of the two circles. Let R be the radius of the circle through their centers and A. Then
Proof of LemmaFor any triangle with side lengths a, b, c, area S and circumradius R we have
Apply that to ΔO1AO2 with sides R1, R2, and m, where m is the distance between the centers of the circles,
where, 2h = AB. It follows that
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