2004 IMO Shortlist Problems/G8

Problem

A cyclic quadrilateral $ABCD$ is given. The lines $AD$ and $BC$ intersect at $E$ , with $C$ between $B$ and $E$ ; the diagonals $AC$ and $BD$ intersect at $F$ . Let $M$ be the midpoint of the side $CD$ , and let $N \neq M$ be a point on the circumcircle of $\triangle ABM$ such that $\frac{AN}{BN} = \frac{AM}{BM}$ . Prove that $E, F, N$ are collinear.

Solution

Let $P = CD \cap EF$ . Let $Q = AB \cap CD$ . Let $R = AB \cap EF$ . Let $(ABM)$ denote the circumcircle of $\triangle ABM$ . Let $N' = EF \cap (ABM)$ . Note that $N' = PR \cap (ABM)$

Claim: $P$ is on $(ABM)$ . Proof: $(C, D; P, Q) = -1$ as complete quadrilaterals induce harmonic bundles. $QP \cdot QM = QC \cdot QD$ by Lemma 9.17 on Euclidean Geometry in Maths Olympiad. By power of a point theorem, $QP \cdot QM = QA \cdot QB$ and this is equivalent to our original claim.

$P$ is on $(ABM)$ by Lemma 9.17 on Euclidean Geometry in Maths Olympiad. $(A, B; R, Q) = -1$ as complete quadrilaterals induce harmonic bundles. By a projection through $P$ from $AQ$ onto $(ABM)$ , $(A, B; N', M) = -1$ . Since $\frac{AN}{BN} = \frac{AM}{BM}$ , $M$ and $N$ are on the intersections of $(ABM)$ and an Appollonian circle centered on AB, so N and M are on the opposite sides of AB. Therefore, $(A, B; N, M) = -\frac{AN}{BN}\div\frac{AM}{BM} = -1$ . By uniqueness of harmonic conjugate, $N = N'$

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2004 IMO Shortlist Problems/G8

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Solution