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Austrian Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Austria

Let $ABC$ be an acute triangle. Let $H$ denote its orthocenter and $D$ , $E$ and $F$ the feet of its altitudes from $A$ , $B$ and $C$ , respectively. Let the common point of $DF$ and the altitude through $B$ be $P$ . The line perpendicular to $BC$ through $P$ intersects $AB$ in $Q$ . Furthermore, $EQ$ intersects the altitude through $A$ in $N$ .
Prove that $N$ is the mid-point of $AH$ .
(Karl Czakler)

Figure 3: Problem 13

Solution

See Figure 3. As usual, let $\beta = \angle ABC$ and $\gamma = \angle ACB$ . Since we know that

Figure 3: Problem 13

$\angle AFH = \angle AEH = 90^\circ$ holds, the quadrilateral $AFHE$ is cyclic, and because $DA$ is parallel to $PQ$ we obtain
$\angle FQP = \angle FAH = \angle FEH = \angle FEP.$
It follows that $QFPE$ is also cyclic. Since $\angle AFC = \angle ADC = 90^\circ$ , $AFDC$ is also cyclic, and we have $\angle QFP = \angle AFD = 180^\circ - \angle ACD = 180^\circ - \gamma$ . We therefore have $\angle QEP = \gamma$ . From this, we obtain $\angle EAN = 90^\circ - \gamma = \angle AEP - \angle QEP = \angle AEN$ , which shows us that triangle $ANE$ is isosceles. It therefore follows that $N$ is the circumcenter of the right triangle $AHE$ , and we therefore have $NA = NH$ , as claimed.

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