Cyclic Hexagon
Bui 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 A_{0}B_{2}A_{1}B_{0}A_{2}B_{1}, drop the perpendiculars A_{i}K_{i} on B_{i-1}B_{i+1} and B_{i}L_{i} on A_{i-1}A_{i+1}, where indices are counted cyclically modulo 3.
Denote for simplicity k_{i} = A_{i}K_{i} and l_{i} = B_{i}L_{i}. Then
(1) | k_{0}k_{1}k_{2} = l_{0}l_{1}l_{2}. |
The proof is based on a lemma and is illustrated by the applet below:
What if applet does not run? |
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Copyright © 1996-2018 Alexander BogomolnyProof
Still assuming the indices are considered cyclically modulo 3, let P_{i} be the midpoint of A_{i}B_{i+1} and Q_{i} the midpoint of B_{i}A_{i+1}. The six circles with centers at P_{i} and Q_{i} and the sides of the hexagon as diameters form a chain of circles with common chords A_{i}K_{i} and B_{i}L_{i}. We denote them C(P_{i}) and C(Q_{i}), respectively, R(P_{i}) and R(Q_{i}) being their radii.
What if applet does not run? |
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,
(2) |
k_{i} = 2·R(P_{i})R(Q_{i+1}) / R and l_{i} = 2·R(Q_{i})R(P_{i+1}) / R. |
It them follows that
(3) | ∏k_{i} = 8·∏R(P_{i})·∏R(Q_{i}) / R^{3} = ∏l_{i}, |
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 a_{i}, i = 0, ..., 5, and setting
(4) | ∏k_{i} = ∏a_{i} · D^{-3} = ∏l_{i}. |
Remark
Note 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
Lemma
Let two circles C(O_{1}, R_{1}), C(O_{2}, R_{2}) 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
(5) | AB = R_{1}·R_{2} / R. |
Proof of Lemma
For any triangle with side lengths a, b, c, area S and circumradius R we have
abc = 4RS. |
Apply that to ΔO_{1}AO_{2} with sides R_{1}, R_{2}, and m, where m is the distance between the centers of the circles,
(6) | R_{1}·R_{2}·m = 4RS = 4R(hm/2), |
where, 2h = AB. It follows that
AB = 2h = R_{1}·R_{2} / R. |
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Copyright © 1996-2018 Alexander Bogomolny