Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

Let $O$ be a circumcenter of a right triangle. The circle with smaller radius and center at point $O$ is tangent to the greater cathetus and the height of the triangle from the right angle.
Find the acute angles of the right triangle and the relation between the radii of the circumcircle and the other circle.
(Bogdan Rublyov)

Solution

The acute angles are $30^\circ$ and $60^\circ$ ; the ratio of the radii is $2:1$ .

Let the right triangle be $CMD$ (see fig. 16).
Let $ON$ be the radius of the smaller circle that is tangent to the cathetus.
Thus $\triangle ODN = \triangle OMN$ , and $\triangle AMO = \triangle OMN$ , therefore, $2AM = MN + ND = MD$ , hence $\angle ADM = 30^\circ$ .
Therefore, the acute angles equal $30^\circ$ and $60^\circ$ .
From $\triangle OND$ it is clear that $OD = R$ (hypotenuse), $ON = r$ (cathetus), that opposes angle $30^\circ$ , thus $\frac{R}{r} = 2$ .

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