Question

在直角三角形 $ABC$ 中，直角在 $C$ ，且 $CA = 30$ $CB = 16$ 。将它的两条直角边 $\overline{CA}$ 和 $\overline{CB}$ 分别越过 $A$ 和 $B$ 。延长。点 $O_1$ 和 $O_2$ 位于三角形外部，并且是两个半径相等的圆的圆心。以 $O_1$ 为圆心的圆与斜边以及边 $CA$ 的延长线相切，以 $O_2$ 为圆心的圆与斜边以及边 $CB$ 的延长线相切，且两个圆外切。任一圆的半径长度可表示为 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 为互素正整数。求 $p + q$ 。

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 $p + q.$

答案

答案：737

知识点：相切圆直角三角形三角恒等式

难度评级：2920

解答：

斜边长为 $AB = \sqrt{30^2 + 16^2} = 34$ 。设 $T_1$ 和 $T_2$ 分别为两个圆与 $AB$ 的切点。两个圆心都在距直线 $AB$ 为 $r$ 、远离三角形的一侧，所以 $\overline{O_1 O_2}$ 平行于 $AB$ ，并且 $T_1 T_2 = O_1 O_2 = 2r$ ，因为两个圆外切。因此 $AB = AT_1 + 2r + T_2 B$ 。

圆 $O_1$ 内切于 $A$ 处由射线 $AB$ 和 $\overline{CA}$ 越过 $A$ 的延长线形成的角，该角大小为 $180^\circ - \angle A$ 。因此从 $A$ 出发的切线长为 $AT_1 = r\big/\tan\left(90^\circ - \tfrac{A}{2}\right)$ $= r \tan\frac{A}{2}$ 。由 $\sin A = \frac{16}{34}$ 和 $\cos A = \frac{30}{34}$ ，半角公式给出 $\tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} = \frac{16}{64} = \frac{1}{4}$ ，同理 $\tan\frac{B}{2} = \frac{30}{34 + 16} = \frac{3}{5}$ 。

所以 $34 = \frac{r}{4} + 2r + \frac{3r}{5} = \frac{57r}{20}$ ，得 $r = \frac{680}{57}$ 。因为 $680 = 2^3 \cdot 5 \cdot 17$ 与 $57 = 3 \cdot 19$ 没有公因数，所以 $p + q = 680 + 57 = 737$ 。

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 第 9 题

答案：737

难度评级：2920

解答：

其他年份的第 9 题