AMC12 2018 B

AMC12 2018 B · Q25

AMC12 2018 B · Q25. It mainly tests Angle chasing, Circle theorems.

Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius 4 and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$, respectively, so that $P_1P_2 = P_2P_3 = P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i = 1, 2, 3$, where $P_4 = P_1$. See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a} + \sqrt{b}$, where $a$ and $b$ are positive integers. What is $a + b$?

圆 $\omega_1$，$\omega_2$ 和 $\omega_3$ 各半径为 4，放置在平面内，使得每两个圆外部相切。点 $P_1$，$P_2$ 和 $P_3$ 分别位于 $\omega_1$，$\omega_2$ 和 $\omega_3$ 上，使得 $P_1P_2 = P_2P_3 = P_3P_1$，且直线 $P_iP_{i+1}$ 与 $\omega_i$ 相切，其中 $i = 1, 2, 3$，$P_4 = P_1$。见下图。$\triangle P_1P_2P_3$ 的面积可写成 $\sqrt{a} + \sqrt{b}$ 的形式，其中 $a$ 和 $b$ 是正整数。求 $a + b$？

(A) 546 546

(B) 548 548

(D) 552 552

(E) 554 554

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $O_i$ be the center of circle $\omega_i$ for $i=1,2,3$, and let $K$ be the intersection of lines $O_1P_1$ and $O_2P_2$. Because $\angle P_1P_2P_3=60^\circ$, it follows that $\triangle P_2KP_1$ is a $30-60-90$ triangle. Let $d=P_1K$; then $P_2K=2d$ and $P_1P_2=\sqrt{3}d$. The Law of Cosines in $\triangle O_1KO_2$ gives \[ 8^2=(d+4)^2+(2d-4)^2-2(d+4)(2d-4)\cos 60^\circ, \] which simplifies to $3d^2-12d-16=0$. The positive solution is $d=2+\frac{2}{3}\sqrt{21}$. Then $P_1P_2=\sqrt{3}d=2\sqrt{3}+2\sqrt{7}$, and the required area is \[ \frac{\sqrt{3}}{4}\cdot(2\sqrt{3}+2\sqrt{7})^2=10\sqrt{3}+6\sqrt{7}=\sqrt{300}+\sqrt{252}. \] The requested sum is $300+252=552$.

答案（D）：设 $O_i$ 为圆 $\omega_i$（$i=1,2,3$）的圆心，$K$ 为直线 $O_1P_1$ 与 $O_2P_2$ 的交点。因为 $\angle P_1P_2P_3=60^\circ$，可知 $\triangle P_2KP_1$ 是一个 $30-60-90$ 三角形。令 $d=P_1K$，则 $P_2K=2d$ 且 $P_1P_2=\sqrt{3}d$。在 $\triangle O_1KO_2$ 中应用余弦定理得 \[ 8^2=(d+4)^2+(2d-4)^2-2(d+4)(2d-4)\cos 60^\circ, \] 化简为 $3d^2-12d-16=0$。其正解为 $d=2+\frac{2}{3}\sqrt{21}$。于是 $P_1P_2=\sqrt{3}d=2\sqrt{3}+2\sqrt{7}$，所求面积为 \[ \frac{\sqrt{3}}{4}\cdot(2\sqrt{3}+2\sqrt{7})^2=10\sqrt{3}+6\sqrt{7}=\sqrt{300}+\sqrt{252}. \] 所求和为 $300+252=552$。

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