Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.2 National Olympiad, round 2 Prove it Croatia

Circles $k_1$ and $k_2$ meet at points $M$ and $N$ . The line $l$ meets the circle $k_1$ in points $A$ and $C$ , and the circle $k_2$ in points $B$ and $D$ so that the points $A, B, C$ and $D$ are on the line $l$ in that order. Let $X$ be a point on the line $MN$ such that the point $M$ is between points $X$ and $N$ . Let $P$ be the intersection of lines $AX$ and $BM$ , and $Q$ the intersection of lines $DX$ and $CM$ .
If $K$ is the midpoint of the segment $\overline{AD}$ , and $L$ the midpoint of the segment $\overline{BC}$ , prove that the lines $XK$ and $ML$ meet on the line $PQ$ .
(Matija Bucić)

Solution

Let $Y$ be the second intersection of the circle $k_1$ with the line $AX$ and let $Z$ be the second intersection of the circle $k_2$ with the line $DX$ .
Since lines $AY$ , $DZ$ and $MN$ pass through the point $X$ , by the converse of the radical centre theorem, quadrilateral $AYZD$ is cyclic. Hence $\angle YZX = \angle XAD$ .
Since $AYMC$ is a cyclic quadrilateral, it holds $\angle YMQ = \angle YAC = \angle XAD$ , so $\angle YMQ = \angle YZX = \angle YZQ$ and it follows that the quadrilateral $QYMZ$ is cyclic. Analogously, quadrilateral $PYMZ$ is cyclic, so points $P, Q, Z, Y$ and $M$ lie on the same circle. From this it follows that $\angle YPQ = \angle YZQ = \angle XAD$ , so $PQ \parallel AD$ .
Hence the quadrilaterals $ADPQ$ and $BCPQ$ are trapezia, and since the midpoints of bases of a trapezium and the intersection of its diagonals lie on the same line, we conclude the lines $KX$ and $LM$ pass through the midpoint of the segment $\overline{PQ}$ . This finishes the proof.

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