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 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

(1) k0k1k2 = l0l1l2.

The proof is based on a lemma and is illustrated by the applet below:

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

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Proof

Still 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.

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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) ki = 2·R(Pi)R(Qi+1) / R and
li = 2·R(Qi)R(Pi+1) / R.

It them follows that

(3) ∏ki = 8·∏R(Pi)·∏R(Qi) / R3 = ∏li,

where ∏ stands for the product of the following terms with indices from the allowed range, i = 0, 1, 2 in this case.

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 D = 2R, the diameter of the circumcircle of the hexagon, (3) becomes

(4) ∏ki = ∏ai · D-3 = ∏li.

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 n ≥ 3. The above remark extends to that circumstance as well: we may talk of two individual n-gons that share a circumcircle. (1)-(4) have a natural interpretation in the general case. However, as Bui Quang Tuan has observed, there are special cases of interest. These are discussed on a separate page.

Lemma

Let 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

(5) AB = R1·R2 / 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 ΔO1AO2 with sides R1, R2, and m, where m is the distance between the centers of the circles, m = O1O2:

(6) R1·R2·m = 4RS = 4R(hm/2),

where, 2h = AB. It follows that

  AB = 2h = R1·R2 / R.

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Copyright © 1996-2017 Alexander Bogomolny

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