AMC10 2025 A

AMC10 2025 A · Q22

AMC10 2025 A · Q22. It mainly tests Circle theorems, Geometry misc.

A circle of radius $r$ is surrounded by three circles, whose radii are 1, 2, and 3, all externally tangent to the inner circle and externally tangent to each other, as shown in the diagram below. What is $r$?

一个半径为 $r$ 的圆被三个圆包围，这些圆的半径分别为 1、2 和 3，它们都与内部圆外切，并且彼此外切，如下图所示。 $r$ 是多少？

(A) \frac{1}{4} \frac{1}{4}

(B) \frac{6}{23} \frac{6}{23}

(D) \frac{5}{17} \frac{5}{17}

(E) \frac{3}{10} \frac{3}{10}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Descartes' Circle Formula (curvatures $k_i = \frac{1}{r_i}$) \[k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{k_1k_2 + k_2k_3 + k_3k_1}.\] For radii 1, 2, 3 we have \[k_1 = 1,\quad k_2 = \frac{1}{2},\quad k_3 = \frac{1}{3}.\] Compute the sum and the square-root term \[k_1+k_2+k_3 = \frac{11}{6},\qquad k_1k_2+k_2k_3+k_3k_1 = 1.\] Therefore \[k_4 = \frac{11}{6} \pm 2.\] Choose the plus sign for the small circle tangent externally to the three given circles \[k_4 = \frac{11}{6} + 2 = \frac{23}{6}, \qquad r_4 = \frac{1}{k_4} = \boxed{\textbf{(B) }\frac{6}{23}}.\]

笛卡尔圆公式（曲率 $k_i = \frac{1}{r_i}$） \[k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{k_1k_2 + k_2k_3 + k_3k_1}.\] 对于半径 1、2、3，我们有 \[k_1 = 1,\quad k_2 = \frac{1}{2},\quad k_3 = \frac{1}{3}.\] 计算和与平方根项 \[k_1+k_2+k_3 = \frac{11}{6},\qquad k_1k_2+k_2k_3+k_3k_1 = 1.\] 因此 \[k_4 = \frac{11}{6} \pm 2.\] 选择加号以得到与三个给定圆外切的小圆 \[k_4 = \frac{11}{6} + 2 = \frac{23}{6}, \qquad r_4 = \frac{1}{k_4} = \boxed{\textbf{(B) }\frac{6}{23}}.\]

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