# Cherchez le quadrilatère cyclique IIWhat Is This About? A Mathematical Droodle

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Explanation

Copyright © 1996-2018 Alexander Bogomolny

The applet provides an illustration to a generalization of the following problem from an outstanding collection by T. Andreescu and R. Gelca:

Consider a semicircle of center O and diameter AB. A line intersects AB at M and the semicircle at C and D in such a way that BM < AM and CM < DM. The circumcircles of triangles AOD and BOC intersect a second time at K. Show that KM and KO are perpendicular.

(This is one of the problems from the 1996 Balkan Mathematical Olympiad.)

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The generalization is twofold:

• First, AB need not be a diameter of the given circle. Suffice it for the quadrilateral ABCD to be cyclic.

• Second, the inequalities BM < AM and CM < DM can be done away with, although the proof then may have to be split into several cases.

The outline of the proof is as follows:

• First we show that quadrilaterals ABLK and CDKL are cyclic. By this means we introduce two circles meeting at K and L and thus having KL as their radical axis. Adjoining the given circle gives a set of three circles with AB, CD, and KL pairwise radical axes implying M is a radical center of the three circles. Thus L lies on KM.

• Due to the first step, we need now prove that KO is perpendicular to KL (instead of KM.)

First step: ABLK and CDKL are cyclic quadrilaterals. Indeed,

 ∠AKB = ∠AKO + ∠BKO = ∠ADO + ∠BCO = (180° - ∠AOD)/2 + (180° - ∠BOC)/2 = 180° - ∠AOD/2 - ∠BOC/2 = 180° - arcAD/2 - arcBC/2 = 180° - ∠ABD - ∠BAC = 180° - ∠ABL - ∠BAL = ∠ALB,

where (and also later on) the arcs are those of the given circle. This says that quadrilateral ABLK is cyclic. Using the cyclicity of ABLK we proceed to establish the cyclicity of CDKL:

 ∠DKL = 360° - ∠AKD - ∠AKL = 360° - ∠AOD - (180° - ∠ABL) = 180° - ∠AOD + ∠ABL = 180° - ∠AOD + ∠ABD = 180° - arcAD + arcAD/2 = 180° - arcAD/2 = 180° - ∠ACD = 180° - ∠LCD.

Wherefrom we obtain the identity

∠DKL + ∠DCL = 180°,

so that quadrilateral CDKL is also cyclic. Thus L does lie on KM.

Second step: To prove the orthogonality of KM and KO note that

∠AKL + ∠ABL = 180°

implying that orthogonality of KL and KO is equivalent to

 (1) ∠AKO + ∠ABL = 90°.

But quadrilateral ADKO is cyclic: ∠AKO = ∠ADO. Now from (1):

 ∠AKO + ∠ABL = ∠ADO + ∠ABD = (180° - ∠AOD)/2 + ∠ABD = 90° - arcAD/2 + arcAD/2 = 90°,

as required. KO ⊥ KL.

### References

1. T. Andreescu, R. Gelca, Mathematical Olympiad Challenges, Birkhäuser, 2004, 5th printing, 1.3.12 (p. 13)

### Radical Axis and Radical Center

Copyright © 1996-2018 Alexander Bogomolny

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