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Ukrainian National Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Ukraine

Let the triangle ABC be such that $2AC = AB$ and $\angle A = 2\angle B$ . Let AL be its bisector and let M be the midpoint of AB. It turns out that $CL = ML$ . Show that $\angle B = 30^\circ$ .

(Danylo Khilko)

Fig. 4

Solution

Since AL is a bisector, then $\angle CAL = \angle LAB = \angle CBA$ (Fig. 4). Then $\triangle ALB$ is isosceles, so LM is its altitude and a median. Thus $\angle LMA = 90^\circ$ . Consider the triangles AML and ALC. Let C' be the projection of L on AC. Then right triangles AML and AC'L are equal by hypothenuse and the angle. Thus, $LC' = LM = LC$ . Therefore, $C = C'$ , since there exists only one projection.

Thus $\triangle ABC$ is a right triangle for which $2AC = AB$ , hence $\angle ABC = 30^\circ$ .

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