Question

In right triangle $ABC$ with right angle $C,$ $CA = 30$ and $CB = 16.$ Its legs $\overline{CA}$ and $\overline{CB}$ are extended beyond $A$ and $B.$ Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center $O_1$ is tangent to the hypotenuse and to the extension of leg $CA,$ the circle with center $O_2$ is tangent to the hypotenuse and to the extension of leg $CB,$ and the circles are externally tangent to each other. The length of the radius of either circle can be expressed as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

Answer

Answer: 737

Concepts:tangent circles right triangle trigonometric identity

Difficulty rating: 2920

Solution:

The hypotenuse is $AB = \sqrt{30^2 + 16^2} = 34.$ Let $T_1$ and $T_2$ be the points where the circles touch $AB.$ Both centers lie at distance $r$ from line $AB$ on the side away from the triangle, so $\overline{O_1 O_2}$ is parallel to $AB$ and $T_1 T_2 = O_1 O_2 = 2r,$ since the circles are externally tangent. Thus $AB = AT_1 + 2r + T_2 B.$

Circle $O_1$ is inscribed in the angle at $A$ between ray $AB$ and the extension of $\overline{CA}$ beyond $A,$ which measures $180^\circ - \angle A.$ Its tangent length from $A$ is therefore $AT_1 = r\big/\tan\left(90^\circ - \tfrac{A}{2}\right)$ $= r \tan\frac{A}{2}.$ With $\sin A = \frac{16}{34}$ and $\cos A = \frac{30}{34},$ the half-angle formula gives $\tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} = \frac{16}{64} = \frac{1}{4},$ and similarly $\tan\frac{B}{2} = \frac{30}{34 + 16} = \frac{3}{5}.$

So $34 = \frac{r}{4} + 2r + \frac{3r}{5} = \frac{57r}{20},$ giving $r = \frac{680}{57}.$ Since $680 = 2^3 \cdot 5 \cdot 17$ and $57 = 3 \cdot 19$ share no common factor, $p + q = 680 + 57 = 737.$

2007 AIME I Problem 9

Answer: 737

Difficulty rating: 2920

Solution:

Problem 9 in Other Years