Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it Czech-Polish-Slovak Mathematical Match

Let a convex quadrilateral $ABCD$ be inscribed in a circle with center $O$ and circumscribed to a circle with center $I$ , and let its diagonals $AC$ and $BD$ meet at a point $P$ . Prove that the points $O$ , $I$ and $P$ are collinear.

Solution

Assume that the lines $AI$ , $BI$ , $CI$ , $DI$ meet the circumcircle of the quadrilateral $ABCD$ at $E$ , $F$ , $G$ , $H$ , respectively. Since the lines $AI$ , $BI$ , $CI$ , $DI$ are the bisectors of the respective angles of the quadrilateral $ABCD$ , the lines $EG$ and $FH$ are the diameters of the circumcircle of $ABCD$ . Thus $EG$ and $FH$ meet at $O$ .
Denote by $X$ the point of intersection of $EB$ and $CH$ .
Using Pascal's theorem for the hexagon $ACHDBE$ we see that $P$ , $X$ and $I$ are collinear. Once again Pascal's theorem applied to the hexagon $GCHFBE$ yields that $O$ , $X$ and $I$ are collinear. Thus $O$ , $I$ and $P$ are collinear, as desired.

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