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Romanian Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Romania

Triangle $ABC$ has $\angle ABC = 60^\circ$ . Points $M$ and $D$ are placed on the sides $(AC)$ and $(AB)$ respectively, so that $\angle BCA = 2\angle MBC$ and $BD = MC$ . Find the measure of the angle $\angle DMB$ .

Solution

Take $G$ so that $CG = CM$ , $C \in (BG)$ and denote $x = \angle MBC$ . Since $\angle MCB$ is exterior to the isosceles triangle $MCG$ , $\angle MGC = x$ . This means that triangle $MBG$ is isosceles and similar to triangle $MCG$ . Thus, $\frac{BG}{MG} = \frac{MB}{MC} = \frac{MG}{BD}$ .

Denote $H$ the common point of the line $AB$ and the perpendicular bisector of the segment $[BG]$ . Then triangle $HBG$ is isosceles and has an angle of $60^\circ$ , so it is equilateral. This yields
$\frac{BG}{MG} = \frac{GH}{MB} = \frac{MG}{BD}.$

Since $\angle MGH = \angle MBD = 60^\circ - x$ , triangles $MGH$ and $DBM$ are similar, therefore we get the answer $\angle DMB = \angle MHG = 30^\circ$ .

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