Question

一个半径为 $r$ 的圆被 $12$ 个半径为 $1$ 的圆围住，这些圆与中央圆外切，并且相邻的圆彼此相切，如图所示。那么 $r$ 可以写成 $\sqrt{a} + \sqrt{b} + c$ ，其中 $a$ ， $b$ ，和 $c$ 是整数。 $a + b + c$ 是多少？

A circle of radius $r$ is surrounded by $12$ circles of radius $1,$ externally tangent to the central circle and sequentially tangent to each other, as shown. Then $r$ can be written as $\sqrt{a} + \sqrt{b} + c,$ where $a,$ $b,$ and $c$ are integers. What is $a + b + c?$

$3$

$5$

$7$

$9$

$11$

答案：C

知识点：相切圆正多边形三角学

难度评级：2410

解答：

$12$ 个外圆的圆心位于半径为 $r + 1$ 的圆上，形成一个正 $12$ 边形。相邻圆心距离为 $2$ （两个圆半径都为 $1$ ），它们之间的圆心角为 $30^\circ$ 。

因此 $2(r + 1)\sin 15^\circ = 2$ ，所以 $r + 1 = \dfrac{1}{\sin 15^\circ}$ 。由于 $\sin 15^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4}$ ， $r + 1 = \frac{4}{\sqrt{6} - \sqrt{2}} = \sqrt{6} + \sqrt{2}.$

于是 $r = \sqrt{6} + \sqrt{2} - 1$ ，所以 $a + b + c = 6 + 2 - 1 = 7$ 。

因此，正确答案是 C。

The centers of the $12$ outer circles lie on a circle of radius $r + 1,$ forming a regular $12$ -gon. Adjacent centers are $2$ apart (both circles have radius $1$ ), and the central angle between them is $30^\circ.$

Thus $2(r + 1)\sin 15^\circ = 2,$ so $r + 1 = \dfrac{1}{\sin 15^\circ}.$ Since $\sin 15^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4},$ $r + 1 = \frac{4}{\sqrt{6} - \sqrt{2}} = \sqrt{6} + \sqrt{2}.$

Then $r = \sqrt{6} + \sqrt{2} - 1,$ so $a + b + c = 6 + 2 - 1 = 7.$

Thus, the correct answer is C.

2025 AMC 12A 第 24 题

答案：C

难度评级：2410

解答：

其他年份的第 24 题