AMC12 2025 A

AMC12 2025 A · Q24

AMC12 2025 A · Q24. It mainly tests Circle theorems, Trigonometry (basic).

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 $a, b, c$ are integers. What is $a+b+c?$

半径为 $r$ 的圆被 $12$ 个半径为 $1$ 的圆包围，这些圆与中心圆外切，并依次相切，如图所示。然后 $r$ 可以写成 $\sqrt a + \sqrt b + c$，其中 $a, b, c$ 是整数。求 $a+b+c$？

(A) 3 3

(B) 5 5

(D) 9 9

(E) 11 11

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let the center of the large circle be $O$ and the centers of the $12$ circles be $A_1, A_2, A_3, \dots, A_{12}$. Triangle $OA_1A_2$ has side lengths $r+1, r+1, 2$, with the angle opposite $2$ being $360/12 = 30$. Note that the line connecting $A_1$ and $A_2$ go through their common point of tangency, by definition, which causes $A_1A_2$ to have a length of $2$. Drawing the angle bisector of the $30$ degree angle, we split $OA_1A_2$ into two congruent right triangles, each with hypotenuse $r+1$ and side opposite the $15$ degree angle $1$. From here, note that $\sin{15} = \frac{\sqrt{6}-\sqrt{2}}{4}$, which be derived using the trigonometric identity $\sin{(A-B)} = \sin{A} \cos{B} - \sin{B} \cos{A}$, with $A=45$ and $B=30$. In our right triangle, $\sin{15} = \frac{1}{r+1} = \frac{\sqrt{6}-\sqrt{2}}{4}$. Let $x=r+1$. Solving for $x$, we get $x = \frac{4}{\sqrt{6}-\sqrt{2}}$. Rationalizing, we get that $x = \sqrt{6}+\sqrt{2}$. Remember $x = r+1 = \sqrt{6}+\sqrt{2}$, so $r = \sqrt{6}+\sqrt{2} - 1$. Therefore, our answer is $6+2-1 = \boxed{7}.$

设大圆中心为 $O$，$12$ 个小圆中心为 $A_1, A_2, A_3, \dots, A_{12}$。三角形 $OA_1A_2$ 边长为 $r+1, r+1, 2$，对边 $2$ 的角度为 $360/12 = 30^\circ$。注意 $A_1$ 和 $A_2$ 连线通过它们的切点，故 $A_1A_2=2$。画 $30^\circ$ 角的角平分线，将 $OA_1A_2$ 分成两个全等直角三角形，每个斜边 $r+1$，$15^\circ$ 角对边 $1$。注意到 $\sin{15} = \frac{\sqrt{6}-\sqrt{2}}{4}$，可由三角恒等式 $\sin{(A-B)} = \sin{A} \cos{B} - \sin{B} \cos{A}$ 推导，$A=45^\circ, B=30^\circ$。在直角三角形中，$\sin{15} = \frac{1}{r+1} = \frac{\sqrt{6}-\sqrt{2}}{4}$。设 $x=r+1$，解得 $x = \frac{4}{\sqrt{6}-\sqrt{2}}$。通分得 $x = \sqrt{6}+\sqrt{2}$。记住 $x = r+1 = \sqrt{6}+\sqrt{2}$，故 $r = \sqrt{6}+\sqrt{2} - 1$。因此，答案为 $6+2+(-1) = \boxed{7}$。

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