Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Find the answer

8.5. In an isosceles triangle $ABC$ , the angle $A$ at the base is $75^\circ$ . The bisector of angle $A$ intersects side $BC$ at point $K$ . Find the distance from point $K$ to the base $AC$ , if $BK=10$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Answer: 5.

Solution. In triangle $ABC$ , $\angle B = (180^0 - 75^0 - 75^0) = 30^0$ . Drop perpendiculars from point $K$ : $KH$ to side $AB$ , and $KN$ to side $AC$ . In the right triangle $HBK$ , the hypotenuse $BK$ is 10. Angle $B$ is 30 degrees, and the side opposite to it is half the hypotenuse, so $HK = 5$ . Triangles $AHK$ and $ANK$ are equal by hypotenuse and acute angle, so $KN = KH = 5$ .

## Criteria.

1 point. A perpendicular is dropped from point $K$ to side $AB$ .

2 points. The length of the perpendicular dropped from point $K$ to side $AB$ is found. Comment. Points for criteria are summed.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.