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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

Let $ABCD$ be a quadrilateral inscribed in a circle with center $O$ . Let $P$ denote the intersection of $AC$ and $BD$ . Let $M$ and $N$ denote the midpoints of $AD$ and $BC$ . If $AP=1$ , $BP=3$ , $DP=\sqrt{3}$ , and $AC$ is perpendicular to $BD$ , find the area of triangle $MON$ .

Solution

Solution:

Answer: $\frac{3}{4}$

We first prove that $ONPM$ is a parallelogram. Note that $APD$ and $BPC$ are both $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. Let $M'$ denote the intersection of $MP$ and $BC$ . Since $\angle BPM' = \angle MPD = 30^{\circ}$ , we have $MP \perp BC$ . Since $ON$ is the perpendicular bisector of $BC$ , we have $MP \parallel NO$ . Similarly, we have $MO \parallel NP$ . Thus $ONPM$ is a parallelogram.

It follows that the area of triangle $MON$ is equal to the area of triangle $MPN$ , which is equal to $\frac{1}{2} \cdot 1 \cdot 3 \cdot \sin \angle MPN = \frac{1}{2} \cdot 1 \cdot 3 \cdot \sin 150^{\circ} = \frac{3}{4}$ .

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