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Geometry Difficulty 6.3 National olympiad Prove it Saudi Arabia

Let $ABCD$ be a quadrilateral inscribed in a circle ( $O$ ) and some point $I$ lies inside $ABCD$ . Consider the lines $d_1, d_2, d_3, d_4$ pass through the midpoints of segments $IA, IB, IC, ID$ respectively and perpendicular to lines $OA, OB, OC, OD$ . Line $d_1$ cuts $d_2$ at $P$ , line $d_2$ cuts $d_3$ at $Q$ , line $d_3$ cuts $d_4$ at $R$ and line $d_4$ cuts $d_1$ at $S$ . Suppose that the quadrilateral $PQRS$ is convex and points $I, O$ lie inside it. Prove that $PQRS$ circumscribes a circle.

Solution

Let $R$ be the radius of the circle ( $O$ ) and let $M, K$ be the midpoints of $IO, AI$ respectively. Then, according to the property of the midline in triangle $AIO$ , we have $MK = \frac{AO}{2}$ and $MK \parallel AO$ . Thus, we immediately have $MK \perp SP$ and $MK = \frac{R}{2}$ . Similarly for the other sides.

Therefore, the point $M$ is equidistant from the sides of the quadrilateral $PQRS$ and is also inside the quadrilateral (since $I, O$ lie inside it), so $M$ is the incenter of the quadrilateral $PQRS$ . $\square$

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