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Problem 1825

National Olympiad, first round

Geometry Difficulty 6.3 Prove it Ukrainian National Mathematical Olympiad · Ukraine

On sides $AB$ , $BC$ , $AC$ of triangle $ABC$ with $\angle BAC = 120°$ there are points $M$ , $K$ , $N$ respectively so, that $\triangle MKN$ is equilateral, and $AM = 2,017$ , $AN = 2,018$ . Baron Munchausen assures, that $\triangle MKN$ has the smallest perimeter of all equilateral triangles, that have exactly one vertex on each of sides of $\triangle ABC$ . Is baron right?

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

According to the condition, all the angles of $\triangle MKN$ equal to $60°$ (Fig. 39). As $\angle BAC + \angle MKN = 180°$ , quadrilateral $AMKN$ is cyclic, so
$\angle KAC = \angle KMN = 60° = \angle MNK = \angle BAK.$
Let's consider points $M_1$ and $N_1$ – projections of point $K$ on sides $AB$ and $AC$ respectively. So $\triangle AKM_1 = \triangle AKN_1$ , so $AM_1 = AN_1$ . Apart from that, $\angle M_1KN_1 = 60°$ and $KM_1 = KN_1$ . So, $\triangle KM_1N_1$ is also equilateral. Case $N = N_1$ , $M = M_1$ is impossible, as otherwise we would have

$AM = AN$ . So $KM_1 < KM$ , so perimeter of $\triangle KM_1N_1$ is strictly less than perimeter of $\triangle KMN$ . So, baron Munchausen is not right.

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