Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it Ukraine

A point $D$ is chosen on the side $BC$ of an acute triangle $ABC$ (and is distinct from the vertices). Points $P$ and $Q$ are defined as the centers of the circumcircles of the triangles $ABD$ and $ACD$ respectively. Prove that, as soon as the triangle $ABC$ is fixed, there exists a point in the plane, which is different from $A$ and belongs to the circumcircles of all the possible triangles $APQ$ (generated by the different positions of the point $D$ ).

Solution

We claim that this point is the circumcenter $O$ of triangle $ABC$ .

If $AD$ is an altitude, the statement is obvious. We may assume that triangle $ADC$ is acute-angled and triangle $ADB$ is obtuse-angled. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$ respectively. Since $\angle ADC > \angle ABC$ , as can be easily shown, $\angle NAQ < \angle NAO$ , and the point $O$ lies inside the angle $QAM$ . Then $\angle AQN = \angle ADC$ , $\angle APM = 180^\circ - \angle ADB = \angle ADC$ . It follows that the quadruple of points $A$ , $P$ , $O$ , $Q$ is concyclic.

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