Three Congruent Circles by Reflection III: What is this about?
A Mathematical Droodle
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Explanation
Copyright © 1996-2008 Alexander Bogomolny
The applet attempts to illustrate a theorem by Quang Tuan Bui
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Given ΔABC and a point P with the pedal ΔPaPbPc, as usual. Let Ba and Ca be reflections of B in Pc and C in Pb, respectively. (Oa) is the circumcircle of ΔABaCa. The points Ab, Cb, Ac, Bc, and the circles (Ob) and (Oc) are defined similarly. Then the three circles (Oa), (Ob) and (Oc) are congruent.
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Note that the theorem generalizes two previous results where the vertices of a triangle have been reflected in the altitudes and its angle bisectors. The former is obvious. The latter stems from an observation that, if I denotes the incenter of ΔABC with pedal points Ia, Ib and Ic, the reflection in IA of the reflection in IC of B is the reflection of B in IIc.
Proof of the theorem
We proceed very much in the spirit of the other two statements. Let O and O(P) be the circumcenter of ΔABC and the circle through P with center O. Introduce further
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| A' | midpoint of AP |
| B' | midpoint of BP |
| C' | midpoint of CP |
| Ap | reflection of Pa in midpoint Ma of BC |
| Bp | reflection of Pb in midpoint Mb of AC |
| Cp | reflection of Pc in midpoint Mc of AB |
| X | reflection of P in OMa |
| Y | reflection of P in OMb |
| Z | reflection of P in OMc |
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Naturally, X, Y, and Z all lie on O(P), so that, for example, PPaApX is a rectangle with one midline through O. We may also conclude that (Oa) is the reflection of O(P) in A' and similarly for (Ob) and (Oc) from which it follows that the latter three are indeed congruent and have the radius of OP.
Copyright © 1996-2008 Alexander Bogomolny
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