Question

圆 $\mathcal{C}_1$ 和 $\mathcal{C}_2$ 外切，并且它们都内切于圆 $\mathcal{C}_3$ 。 $\mathcal{C}_1$ 和 $\mathcal{C}_2$ 的半径分别为 $4$ 和 $10$ ，且三个圆的圆心共线。 $\mathcal{C}_3$ 的一条弦同时也是 $\mathcal{C}_1$ 和 $\mathcal{C}_2$ 的一条公外切线。已知这条弦的长度为 $\frac{m\sqrt{n}}{p}$ ，其中 $m$ ， $n$ ，和 $p$ 是正整数， $m$ 与 $p$ 互质，且 $n$ 不被任何质数的平方整除，求 $m + n + p$ 。

Circles $\mathcal{C}_1$ and $\mathcal{C}_2$ are externally tangent, and they are both internally tangent to circle $\mathcal{C}_3.$ The radii of $\mathcal{C}_1$ and $\mathcal{C}_2$ are $4$ and $10,$ respectively, and the centers of the three circles are all collinear. A chord of $\mathcal{C}_3$ is also a common external tangent of $\mathcal{C}_1$ and $\mathcal{C}_2.$ Given that the length of the chord is $\frac{m\sqrt{n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$

答案

答案：405

知识点：相切圆弦勾股定理

难度评级：2710

解答：

设 $P_1$ 、 $P_2$ 、 $P_3$ 为三个圆的圆心， $R$ 为 $\mathcal{C}_3$ 的半径。外切给出 $P_1P_2 = 4 + 10 = 14$ ，内切给出 $P_3P_1 = R - 4$ 和 $P_3P_2 = R - 10$ 。由于圆心共线， $(R - 4) + (R - 10) = 14$ ，所以 $R = 14$ ，且 $P_3$ 位于 $\overline{P_1P_2}$ 上，满足 $P_3P_1 = 10$ 和 $P_3P_2 = 4$ 。

向弦作垂线 $P_1X$ ， $P_2Y$ ， $P_3Z$ ，于是 $P_1X = 4$ ， $P_2Y = 10$ ，且 $Z$ 是弦的中点。沿着直线 $P_1P_2$ 移动时，到切线的距离线性变化，而 $P_3$ 位于从 $P_1$ 到 $P_2$ 的 $\frac{10}{14}$ 处，所以 $P_3Z = 4 + \frac{10}{14}(10 - 4) = \frac{58}{7}.$

半弦长为 $\sqrt{14^2 - \left(\frac{58}{7}\right)^2}$ $= \frac{\sqrt{9604 - 3364}}{7}$ $= \frac{4\sqrt{390}}{7}$ ，所以弦长为 $\frac{8\sqrt{390}}{7}$ 。由于 $390 = 2 \cdot 3 \cdot 5 \cdot 13$ 无平方因子且 $\gcd(8, 7) = 1$ ，答案为 $m + n + p = 8 + 390 + 7 = 405$ 。

Let $P_1,$ $P_2,$ $P_3$ be the centers of the circles and $R$ the radius of $\mathcal{C}_3.$ External tangency gives $P_1P_2 = 4 + 10 = 14,$ and internal tangency gives $P_3P_1 = R - 4$ and $P_3P_2 = R - 10.$ Since the centers are collinear, $(R - 4) + (R - 10) = 14,$ so $R = 14$ and $P_3$ lies on $\overline{P_1P_2}$ with $P_3P_1 = 10$ and $P_3P_2 = 4.$

Drop perpendiculars $P_1X,$ $P_2Y,$ $P_3Z$ to the chord, so $P_1X = 4,$ $P_2Y = 10,$ and $Z$ is the midpoint of the chord. The distance from a point moving along line $P_1P_2$ to the tangent line changes linearly, and $P_3$ is $\frac{10}{14}$ of the way from $P_1$ to $P_2,$ so $P_3Z = 4 + \frac{10}{14}(10 - 4) = \frac{58}{7}.$

The half-chord is $\sqrt{14^2 - \left(\frac{58}{7}\right)^2}$ $= \frac{\sqrt{9604 - 3364}}{7}$ $= \frac{4\sqrt{390}}{7},$ so the chord has length $\frac{8\sqrt{390}}{7}.$ Since $390 = 2 \cdot 3 \cdot 5 \cdot 13$ is squarefree and $\gcd(8, 7) = 1,$ the answer is $m + n + p = 8 + 390 + 7 = 405.$

2005 AIME II 第 8 题

答案：405

难度评级：2710

解答：

其他年份的第 8 题