Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Prove it Silk Road Mathematics Competition

In a triangle $ABC$ with incenter $I$ , let $P$ be the intersection point of the bisector of the angle $A$ with the circumcircle other than $A$ , $D$ the point of tangency of the incircle to the side $BC$ , and $Q$ the intersection point of $PD$ with the circumcircle other than $P$ . Show that $PI = QI$ if $PD$ is equal to the inradius.

Solution

In the triangle $BIA$ , $\angle BIP = \angle IBA + \angle IAB = \angle B/2 + \angle A/2$ . On the other hand, $\angle PBI = \angle PBC + \angle CBI = \angle A/2 + \angle B/2$ , as $\angle PBC = \angle PAC = \angle A/2$ . Hence, in the triangle $BPI$ , $BP = IP$ .

Since $\angle QPB = \angle BPD$ and $\angle BQP = \angle BAP = \angle A/2 = \angle PAC = \angle PBC = \angle PBD$ , the triangles $QPB$ and $BPD$ are similar. In particular, $\frac{PD}{PB} = \frac{BP}{QP}$ .

Combining this with the fact $BP = IP$ , gives
$\frac{PD}{PI} = \frac{PI}{PQ}.$
Therefore, the triangles $DPI$ and $IPQ$ are similar, and $\frac{QI}{PI} = \frac{ID}{PD} = 1$ .

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