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Problem 1799

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.4 Prove it IMO Team Selection Test · United States

Let $ABC$ be an acute triangle. Circle $\omega_1$ , with diameter $AC$ , intersects side $BC$ at $F$ (other than $C$ ). Circle $\omega_2$ , with diameter $BC$ , intersects side $AC$ at $E$ (other than $C$ ). Ray $AF$ intersects $\omega_2$ at $K$ and $M$ with $AK < AM$ . Ray $BE$ intersects $\omega_1$ at $L$ and $N$ with $BL < BN$ . Prove that lines $AB$ , $ML$ , $NK$ are concurrent.

(This problem was suggested by Steve Dinh.)

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let $D$ be the foot of the perpendicular from $C$ to $AB$ and $H$ be the orthocenter of $\triangle ABC$ . Note first that $\omega_1$ and $\omega_2$ both intersect $AB$ at $D$ . By Power of a Point, $LH \cdot HN = CH \cdot HD = KH \cdot HM$ , implying that $KLMN$ is a cyclic quadrilateral. Noting that $AC$ and $BC$ are perpendicular bisectors of the diagonals of $KLMN$ , we conclude that the center of its circumcircle is $C$ . Observe now that since $\angle ANC = \angle ALC = 90^\circ$ , we have that $AN$ and $AL$ are tangent to the circumcircle of $KLMN$ . Thus, $H$ is on the polar of $A$ . Similarly, $H$ is on the polar of $B$ , and so by Brocard's theorem, $KN$ and $LM$ meet on the polar of $H$ , which is $AB$ .

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.