Circle, Isosceles Triangle and a Fixed Point
What is this about?
A Mathematical Droodle

22 October 2016, Created with GeoGebra

Explanation

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Copyright © 1996-2017 Alexander Bogomolny

The applet may suggest the following a problem from the 2006 Irish MO:

P and Q are points on the equal sides AB and AC respectively of an isosceles triangle ABC such that AP = CQ. Moreover, neither P nor Q is a vertex of ABC. Prove that the circumcircle of the triangle APQ passes through the circumcenter of the triangle ABC.

Let C(APQ) be the circumcircle of Δ. Let O be the second intersection of C(APQ) with the altitude from the apex A. Consider two triangles: BPO and AQO.

Circle, Isosceles Triangle and a Fixed Point

The idea is to prove that the triangles are congruent, from which it would follow that AO = BO. Since AO is the axis of summetry of the isosceles triangle ABC, BO = CO, implying that the three distances from O to the vertices of ΔABC are equal and making O the circumcenter.

For details see a wiki write-up. There are also five extra solutions.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2017 Alexander Bogomolny

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