Competition problem — Geometry — Olympiad Maths Prep

Library / /21 of 55

Geometry Difficulty 5.6 AIME, harder Prove it Ukraine

Let $ABC$ be a triangle. Let $C_1$ , $A_1$ and $B_1$ be the points of $AB$ , $BC$ and $AC$ , respectively. Let $K$ be the projection of $B_1$ on the line $A_1C_1$ . Let the points $M$ and $N$ lie on the rays $B_1A$ and $B_1C$ respectively, so that $\angle B_1A_1C_1 = 2\angle KNB_1$ and $\angle B_1C_1A_1 = 2\angle KMB_1$ . Prove that the length of the segment $MN$ is not greater than the perimeter of the triangle $\Delta A_1B_1C_1$ .

Solution

Let $M_1$ be the point of the ray $A_1C_1$ such that $M_1C_1 = B_1C_1$ , $N_1$ be the point of the ray $C_1A_1$ such that $N_1A_1 = B_1A_1$ (Fig. 39). Then
$\angle B_1M_1K = \frac{1}{2} \angle A_1C_1B_1 = \angle KMB_1$ ,
whence $KM_1MB_1$ is inscribed and $M$ is the projection of $M_1$ on the line $AC$ .
Analogously, $N$ is the projection of $N_1$ . Then, the segment $MN$ is the projection $M_1N_1$ on the line $AC$ , hence
$MN \leq M_1N_1 = B_1C_1 + A_1C_1 + B_1A_1,$
which proves the problem statement.

Fig. 39

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.