Track / Stage 6 / 73 of 400 #1073 of 1964

Problem 1073

National olympiad, first round

Geometry Difficulty 6.1 Prove it

Let's draw a right-angled triangle and construct a semicircle outward over each side, treating each side as the diameter. Then, find the intersection point of the extension of the altitude from the opposite vertex with the semicircle. Prove that any two such intersection points are equidistant from the common endpoint of the semicircles passing through them.

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

Let the altitudes of the acute triangle $ABC$ be $AA_1$ , $BB_1$ , and $CC_1$ , and let the intersection points of these altitudes with the semicircles drawn over the sides $BC$ , $CA$ , and $AB$ be $P$ , $Q$ , and $R$ . It is sufficient to prove the theorem for one vertex.

By Thales' theorem, $\triangle AQC$ and $\triangle ARB$ are right triangles. In a right triangle, the leg is the geometric mean of the hypotenuse and the projection of the leg on the hypotenuse, so:

$A Q^{2}=A C \cdot A B_{1} \quad \text { and } \quad A R^{2}=A B \cdot A C_{1} \text {. }$

However, $A C \cdot A B_{1}=A B \cdot A C_{1}$ , because both expressions represent the power of point $A$ with respect to the circle with diameter $CB$ of $\triangle ABC$ . (Points $B_1$ and $C_1$ lie on the circumference of this circle by Thales' theorem.)

Based on this: $A Q^{2}=A R^{2}$ and thus $A Q=A R$ , which was to be proven.

The theorem can be proven similarly for each vertex.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.