Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it

3. (Yugoslavia) Let $\triangle A_{1} B_{1} C_{1}$ be the orthic triangle of the non-equilateral acute triangle $A B C$ , and let $A_{2}, B_{2}, C_{2}$ be the points where the incircle of $\triangle A_{1} B_{1} C_{1}$ touches its sides. Prove that the Euler line of $\triangle A_{2} B_{2} C_{2}$ coincides with the Euler line of $\triangle A B C$ .

Note: The orthic triangle of a given triangle is the triangle formed by the feet of the altitudes of the given triangle. The Euler line of a triangle is the line passing through the orthocenter and circumcenter of the triangle, and in fact, it also passes through the incenter.

Solution

Prove that, as shown in the figure, $\triangle A_{1} B_{1} C_{1}$ is the pedal triangle of $\triangle A B C$ , and $A A_{1}$ is the altitude on side $B C$ . It is easy to prove that $A A_{1}$ is the angle bisector of $\angle B_{1} A_{1} C_{1}$ . Since points $B_{2}, C_{2}$ are two tangent points, then $A_{1} B_{2}=A_{1} C_{2}$ , which means $\triangle A_{1} B_{2} C_{2}$ is an isosceles triangle, and thus $A A_{1}$ perpendicularly bisects $B_{2} C_{2}$ . Therefore, the orthocenter of $\triangle A B C$ is the circumcenter of $\triangle A B_{2} C_{2}$ , and $\triangle A_{2}^{\prime} B_{2} C_{2}$ is parallel to $\triangle A B C$ respectively.

Since $\triangle B_{2} B_{2} C_{2}$ is homothetic to $\triangle A B C$ with a common point, the two corresponding lines can only coincide.

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