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Saudi Arabian IMO Booklet

Geometry Difficulty 7.8 National Olympiad, round 2 Prove it Saudi Arabia

Let $BB_1$ and $CC_1$ be the altitudes of acute-angled triangle $ABC$ , and $A_0$ is the midpoint of $BC$ . Lines $A_0B_1$ and $A_0C_1$ meet the line passing through $A$ and parallel to $BC$ in points $P$ and $Q$ . Prove that the incenter of triangle $PA_0Q$ lies on the altitude of triangle $ABC$ .

Solution

Since triangles $BCB_1$ and $BCC_1$ are right-angled, their medians $B_1A_0$ , $B_1C_0$ are equal to the half of hypotenuse $B_1A_0 = A_0C = A_0B = C_1A_0$ .

Now
$\angle PB_1A = \angle CB_1A_0 = \angle B_1CA_0 = \angle PAC,$
thus $PA = PB_1$ . Similarly, $QA = QC_1$ . Then the incircle of triangle $A_0PQ$ touches its sides in points $A, B_1, C_1$ , which yields the assertion of the problem. $\square$

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