Competition problem — Geometry — Maths Olympiad Prep

Library / /453 of 520

Geometry Difficulty 7.2 National olympiad, round 2 Prove it

Points $M, N, K$ lie on the sides $B C, C A, A B$ respectively, of a triangle $A B C$ , and are different from its vertices. The triangle $M N K$ is called beautiful if the triangles $M N K$ and $A B C$ are similar (with the vertices respectively in this order). Show that if in the triangle $A B C$ there are two beautiful triangles with a common vertex, then $\triangle A B C$ is right-angled.

Solution

(L. Ploscaru) Say $M N_{1} K_{1}$ and $M N_{2} K_{2}$ are such two beautiful triangles. Let $T=N_{1} K_{1} \cap N_{2} K_{2} ; T$ exists, and belongs to the interior of $\triangle A B C$ , since angles at $M$ are equal to $\angle A$ . Then $\angle N_{1} M N_{2}=\angle K_{1} M K_{2}$ . We then have $\angle M N_{1} T=\angle M N_{2} T=\angle B$ and $\angle M K_{1} T=\angle M K_{2} T=\angle C$ . It follows that $M N_{2} N_{1} T$ and $M K_{1} K_{2} T$ are cyclic quadrilaterals.

So $\angle K_{2} K_{1} T=\angle K_{2} M T$ and $\angle A N_{1} T=\pi-\angle N_{2} N_{1} T=\angle N_{2} M T$ , whence $\pi-\angle A=\angle A K_{1} N_{1}+\angle A N_{1} K_{1}=\angle K_{2} M N_{2}=\angle A$ , yielding $\pi-\angle A=\angle A$ , so $\angle A=\pi / 2$ . Moreover, notice it forces $M$ to be the midpoint $M_{0}$ of $B C$ ; conversely then, all triangles $M_{0} N K$ with $\angle K M_{0} N=\pi / 2$ are beautiful.

Comments. What is called, in "English mathematical parlance", "simple angle chasing". The converse is also worth noting.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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