UkraineMO

Geometry Difficulty 4.7 AIME Prove it Ukraine

In acute-angled triangle $ABC$ bisector $AL$ , height $BH$ and the perpendicular bisector of line $AB$ intersect at one point. Find the angle $BAC$ .

Answer: $\angle BAC = 60^\circ$ .

Solution

Let the angle $BAC$ be $2\alpha$ (fig. 15). Hence $\triangle APB$ is isosceles, thus $\angle PBA = \alpha$ . Since $\triangle AHB$ is right triangle, $3\alpha = 90^\circ$ , thus $\angle BAC = 2\alpha = 60^\circ$ .

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