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Problem 877

AMC 12 late, AIME early

Geometry Difficulty 4.6 Prove it Berkeley Math Circle: Monthly Contest 5 · United States

Let $ABC$ be a triangle, $I$ the incenter, and $D$ the intersection of lines $AI$ and $BC$ . The perpendicular bisector of $AD$ meets $BI$ and $CI$ at $P$ and $Q$ . Show that $I$ is the orthocenter of triangle $PQD$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

It suffices to show that $CI \perp PD$ .

Note that since $AP = PD$ and $BI$ is bisector of $\angle ABD$ , point $P$ lies on the circumcircle of $\triangle ABD$ (on the midpoint of the arc). From this one can compute $\angle PDC = \angle IDC - \angle ADP = \angle IDC - \angle ABI$ and show it is $90^\circ - \frac{1}{2} \angle C$ , which is all you need.

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