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Problem 1689

National Olympiad, first round

Geometry Difficulty 6.3 Prove it Saudi Arabian Mathematical Competitions · Saudi Arabia

Let $ABC$ be a triangle whose incircle $(I)$ touches $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ , respectively. The line passing through $A$ and parallel to $BC$ cuts $DE$ , $DF$ at $M$ , $N$ , respectively. The circumcircle of triangle $DMN$ cuts $(I)$ again at $L$ .

1. Let $K$ be the intersection of $NE$ and $MF$ . Prove that $K$ is the orthocenter of the triangle $DMN$ .

2. Prove that $A$ , $K$ , $L$ are collinear.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

1) Because $MN \parallel BC$ so $\angle ANF = \angle FDB = \angle DFB = \angle AFN$ , this deduces $AN = AF$ . Similarly, $AM = AE = AF = AN$ .

Thus, $M$ , $N$ , $E$ , $F$ lie on the circle center $A$ . Let $MF$ cut $NE$ at $K$ , because $E$ , $F$ lie on circle diameter $MN$ so
$\angle KFD = \angle KED = 90^\circ,$
this means $DK$ is diameter of $(AEF)$ . So $K$ lies on $(I)$ . Easily seen, $K$ is orthocenter of triangle $DMN$ .

2) Let $P$ be the reflection of $K$ through $A$ then $KMPN$ is parallelogram, so $PM \parallel NK \perp MD$ and $PN \parallel KM \perp DN$ . Thus, $DP$ is diameter of $(DMN)$ . But $DK$ is diameter of $(I)$ so $\angle DLK = 90^\circ$ , we deduce $L$ lies on $AK$ . $\square$

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.