Library / China Hong Kong Mathematical Olympiad /79 of 89

China Hong Kong Mathematical Olympiad

Geometry Difficulty 6.9 National Olympiad Prove it Hong Kong

Let $\triangle ABC$ be an acute-angled triangle. Let $D$ be a point on the segment $BC$ , $I$ the incentre of $\triangle ABC$ . The circumcircle of $\triangle ABD$ meets $BI$ at $P$ and the circumcircle of $\triangle ACD$ meets $CI$ at $Q$ . If the area of $\triangle PID$ and the area of $\triangle QID$ are equal, prove that $PI \times QD = QI \times PD$ .

Solution

Since $\angle AQI = \angle AQC = \angle ADC = 180^\circ - \angle ADB = 180^\circ - \angle APB = 180^\circ - \angle API$ , the points $A, Q, I, P$ are concyclic. Let $DI$ meet $PQ$ at $M$ , and let $AM$ meet $(AQIP)$ again at $X$ . Since $[PID] = [QID]$ , we know that $MP = MQ$ . Also, note that $AP = PD$ and $AQ = QD$ since they are chords opposite to some equal angles in $(ABDP)$ and $(AQDC)$ . This shows $PQ$ is the perpendicular bisector of $AD$ . Therefore, we have
$\angle IMP = \angle DMP = \angle AMP = \angle XMQ.$
As $M$ is the midpoint of $PQ$ and $Q, X, I, P$ are concyclic, this implies $QXIP$ is an isosceles trapezoid. Thus, $\angle QAX = \angle IAP$ . Since $AM$ is the $A$ -median of $\triangle AQP$ , $AI$ is the $A$ -symmedian of $\triangle AQP$ . It follows that $PI \times AQ = QI \times AP$ .
Finally,
$PI \times QD = PI \times AQ = QI \times AP = QI \times PD.$

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.