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Brazilian Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Brazil

Given a circle and its center $O$ , a point $A$ inside the circle and a distance $h$ , construct a triangle $BAC$ with $\angle A = 90^\circ$ , $B$ and $C$ on the circle and the altitude from $A$ with length $h$ .

Solution

Let $H$ on $BC$ such that $AH = h$ . Since $BAC$ is a right angle, $BH \cdot CH = h^2$ .

But the power of $H$ with respect to the given circle is $HB \cdot HC = R^2 - OH^2$ .

So $OH = \sqrt{R^2 - h^2}$ is determined and $H$ is the intersection of the circle with center $O$ and radius $\sqrt{R^2 - h^2}$ and the circle with center $A$ and radius $h$ . So $H$ is determined. To determine $B$ and $C$ , it is enough to trace a perpendicular to $AH$ passing through $H$ .

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