Question

设 $\omega_1$ 和 $\omega_2$ 分别表示圆 $x^2 + y^2 + 10x - 24y - 87 = 0$ 与 $x^2 + y^2 - 10x - 24y + 153 = 0$ ，令 $m$ 为正数 $a$ 的最小值，使得直线 $y = ax$ 包含某个圆的圆心，且该圆内切于 $\omega_1$ 、外切于 $\omega_2$ 。已知 $m^2 = \frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质正整数，求 $p + q$ 。

Let $\omega_1$ and $\omega_2$ denote the circles $x^2 + y^2 + 10x - 24y - 87 = 0$ and $x^2 + y^2 - 10x - 24y + 153 = 0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y = ax$ contains the center of a circle that is internally tangent to $\omega_1$ and externally tangent to $\omega_2.$ Given that $m^2 = \frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers, find $p + q.$

答案

答案：169

知识点：椭圆相切圆配方法二次方程

难度评级：3160

解答：

配方得 $\omega_1\colon (x+5)^2 + (y-12)^2 = 256$ 和 $\omega_2\colon (x-5)^2 + (y-12)^2 = 16$ ，圆心分别为 $F_1 = (-5, 12)$ 与 $F_2 = (5, 12)$ ，半径分别为 $16$ 和 $4$ 。若一个圆的圆心为 $P$ 、半径为 $r$ ，且它内切于 $\omega_1$ 、外切于 $\omega_2$ ，则 $PF_1 = 16 - r$ 且 $PF_2 = 4 + r$ ，所以 $PF_1 + PF_2 = 20$ 。

因此 $P$ 位于焦点为 $F_1$ 和 $F_2$ 、长轴为 $20$ 的椭圆上：半长轴为 $10$ ，焦距为 $5$ ，所以半短轴平方为 $100 - 25 = 75$ ，得到 $\frac{x^2}{100} + \frac{(y - 12)^2}{75} = 1,$ 即 $3x^2 + 4y^2 - 96y + 576 = 300$ 。代入 $y = ax$ ，得到 $(3 + 4a^2)x^2 - 96ax + 276 = 0$ 。

直线 $y = ax$ 含有这样的圆心，当且仅当这个二次方程有实根，即 $(96a)^2 - 4 \cdot 276\,(3 + 4a^2) \ge 0$ ，化简为 $4800a^2 \ge 3312$ ，所以 $a^2 \ge \frac{69}{100}$ 。最小的正 $a$ 满足 $m^2 = \frac{69}{100}$ ，因而 $p + q = 69 + 100 = 169$ 。

Completing the square gives $\omega_1\colon (x+5)^2 + (y-12)^2 = 256$ and $\omega_2\colon (x-5)^2 + (y-12)^2 = 16,$ with centers $F_1 = (-5, 12)$ and $F_2 = (5, 12)$ and radii $16$ and $4.$ If a circle with center $P$ and radius $r$ is internally tangent to $\omega_1$ and externally tangent to $\omega_2,$ then $PF_1 = 16 - r$ and $PF_2 = 4 + r,$ so $PF_1 + PF_2 = 20.$

Thus $P$ lies on the ellipse with foci $F_1$ and $F_2$ and major axis $20:$ the semimajor axis is $10,$ the center-to-focus distance is $5,$ so the semiminor axis squared is $100 - 25 = 75,$ giving $\frac{x^2}{100} + \frac{(y - 12)^2}{75} = 1,$ i.e. $3x^2 + 4y^2 - 96y + 576 = 300.$ Substituting $y = ax$ yields $(3 + 4a^2)x^2 - 96ax + 276 = 0.$

The line $y = ax$ contains such a center exactly when this quadratic has a real root, i.e. $(96a)^2 - 4 \cdot 276\,(3 + 4a^2) \ge 0,$ which simplifies to $4800a^2 \ge 3312,$ so $a^2 \ge \frac{69}{100}.$ The smallest positive such $a$ has $m^2 = \frac{69}{100},$ and $p + q = 69 + 100 = 169.$

2005 AIME II 第 15 题

答案：169

难度评级：3160

解答：

其他年份的第 15 题