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1998 APMO Problems/Problem 4

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Problem

(Răzvan Gelca) Let $ABC$ be a triangle and $D$ the foot of the altitude from $A$ . Let $E$ and $F$ be on a line through $D$ such that $AE$ is perpendicular to $BE$ , $AF$ is perpendicular to $CF$ , and $E$ and $F$ are different from $D$ . Let $M$ and $N$ be the midpoints of the line segments $BC$ and $EF$ , respectively. Prove that $AN$ is perpendicular to $NM$ .

Solution

We use directed angles mod $\pi$ .

Since $\angle ADB$ and $\angle AEB$ are both right angles, points $ABDE$ are concyclic. It follows that $\angle CBA \equiv \angle DBA \equiv \angle DEA \equiv \angle FEA .$ Similarly, the quadrilateral $ACDF$ is cyclic, so $\angle ACB \equiv \angle ACD \equiv \angle AFD \equiv \angle AFE.$ Thus $ACB$ and $AFE$ are similar triangles, so $AFEN$ and $ACBM$ are similar figures. It follows that $\angle AND \equiv \angle ANE \equiv \angle AMB \equiv \angle AMD,$ so points $AMND$ are concyclic. Since $ADM$ is a right angle, it then follows that $ANM$ is a right angle, as desired. $\blacksquare$

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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