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Austrian Mathematical Olympiad

Geometry Difficulty 4.8 AIME Prove it Austria

Let $\mathcal{ABC}$ be an acute triangle with orthocenter $H$ . The circumcircle of the triangle $\mathcal{BHC}$ intersects $\mathcal{AC}$ a second time in point $P$ and $\mathcal{AB}$ a second time in point $Q$ .
Prove that $H$ is the circumcenter of the triangle $\mathcal{APQ}$ .

Solution

Figure 2: Problem 6

Let $H_a$ be the foot of the altitude on $BC$ . With the angle sum in triangle $AH_aC$ , we get
$\angle HAC = 90^\circ - \angle BCA.$
Let $H_b$ be the foot of the altitude on $AC$ . With the angle sum in triangle $CH_bB$ , we get
$\angle CBH = 90^\circ - \angle BCA.$
The inscribed angle theorem gives us
$\angle CPH = \angle CBH,$
therefore
$\angle CPH = \angle HAC.$
We conclude that the triangle $AHP$ is isosceles and we have $AH = PH$ . Analogously, we can prove that $AH = QH$ . Therefore, $H$ is the circumcenter of the triangle $APQ$ .

(Karl Czakler) ☐

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