Question

一个半径为 $6$ 的圆与一个半径为 $24$ 的圆外切。求这两个圆的三条公切线围成的三角形区域的面积。

A circle with radius $6$ is externally tangent to a circle with radius $24.$ Find the area of the triangular region bounded by the three common tangent lines of these two circles.

答案

答案：192

知识点：相切圆切线相似

难度评级：2510

小提示：

两条外公切线交于中心连线上的一点 $P$ ，它到两个圆心的距离之比为 $24 : 6$

The two external tangents meet at a point $P$ on the line through the centers, whose distances to the centers are in ratio $24 : 6$

大提示：

第三条边是在切点处的公切线，垂直于中心连线。对 $P$ 处的半角使用 $\sin\theta = \frac{24}{40}$ 。

The third side is the common tangent at the point of tangency, perpendicular to the center line. Use $\sin\theta = \frac{24}{40}$ for the half-angle at $P.$

解答：

两个圆心 $O_1$ （半径 $24$ ）和 $O_2$ （半径 $6$ ）相距 $30$ 。两条外公切线交于 $O_1O_2$ 线上小圆外侧的一点 $P$ ，并满足 $\frac{PO_1}{PO_2} = \frac{24}{6} = 4$ 。结合 $PO_1 - PO_2 = 30$ ，得到 $PO_1 = 40$ 且 $PO_2 = 10$ 。每条外公切线与中心连线成角 $\theta$ ，其中 $\sin\theta = \frac{24}{40} = \frac{3}{5}$ ，所以 $\tan\theta = \frac{3}{4}$ 。

第三条公切线是在两圆切点 $T$ 处的切线，它在距 $O_1$ 为 $24$ 的位置垂直于 $O_1O_2$ 。三条切线围成的三角形以 $P$ 为顶点，底边在这条直线上，高为 $PT = 40 - 24 = 16$ ，半底长为 $16\tan\theta = 12$ 。

面积为 $\frac{1}{2} \cdot 24 \cdot 16 = 192$ 。

The centers $O_1$ (radius $24$ ) and $O_2$ (radius $6$ ) are $30$ apart. The two external tangents meet at a point $P$ on line $O_1O_2$ beyond the small circle, with $\frac{PO_1}{PO_2} = \frac{24}{6} = 4.$ Combined with $PO_1 - PO_2 = 30,$ this gives $PO_1 = 40$ and $PO_2 = 10.$ Each external tangent makes angle $\theta$ with the center line, where $\sin\theta = \frac{24}{40} = \frac{3}{5},$ so $\tan\theta = \frac{3}{4}.$

The third common tangent is the tangent at the point of tangency $T,$ which is perpendicular to $O_1O_2$ at distance $24$ from $O_1.$ The triangle bounded by the three tangents has apex $P$ and base on this line, with height $PT = 40 - 24 = 16$ and half-base $16\tan\theta = 12.$

Its area is $\frac{1}{2} \cdot 24 \cdot 16 = 192.$

2022 AIME II 第 7 题

答案：192

难度评级：2510

小提示：

大提示：

解答：

其他年份的第 7 题