Question

设 $ABCD$ 是等腰梯形，其边长为 $AB = 6$ 、 $BC = 5 = DA$ 、 $CD = 4$ 。以 $A$ 和 $B$ 为圆心作半径为 $3$ 的圆，以 $C$ 和 $D$ 为圆心作半径为 $2$ 的圆。有一个位于梯形内部的圆与这四个圆都相切。它的半径为 $\frac{-k + m\sqrt{n}}{p}$ ，其中 $k$ 、 $m$ 、 $n$ 、 $p$ 是正整数， $n$ 不被任何质数的平方整除，且 $k$ 与 $p$ 互质。求 $k + m + n + p$ 。

Let $ABCD$ be an isosceles trapezoid, whose dimensions are $AB = 6,$ $BC = 5 = DA,$ and $CD = 4.$ Draw circles of radius $3$ centered at $A$ and $B,$ and circles of radius $2$ centered at $C$ and $D.$ A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k + m\sqrt{n}}{p},$ where $k,$ $m,$ $n,$ and $p$ are positive integers, $n$ is not divisible by the square of any prime, and $k$ and $p$ are relatively prime. Find $k + m + n + p.$

答案

答案：134

知识点：相切圆梯形勾股定理根式

难度评级：3060

解答：

从 $C$ 和 $D$ 作垂线可知，每条长为 $5$ 的腰对应的水平偏移为 $\frac{6 - 4}{2} = 1$ ，所以梯形高度为 $\sqrt{25 - 1} = \sqrt{24}$ 。由对称性，内圆圆心 $O$ 位于连接 $\overline{AB}$ 中点 $E$ 与 $\overline{CD}$ 中点 $F$ 的竖直对称轴上。若内圆半径为 $x$ ，外切条件给出 $OA = x + 3$ 和 $OC = x + 2$ 。又因为 $AE = 3$ 、 $CF = 2$ ，所以 $\begin{aligned} OE &= \sqrt{(x+3)^2 - 9} \\ &= \sqrt{x^2 + 6x}, \\ OF &= \sqrt{(x+2)^2 - 4} \\ &= \sqrt{x^2 + 4x}. \end{aligned}$

因为 $OE + OF = \sqrt{24}$ ，移项后平方得 $\sqrt{24(x^2 + 4x)} = 12 - x$ ，再次平方得到 $24x^2 + 96x = 144 - 24x + x^2$ ，即 $23x^2 + 120x - 144 = 0$ 。

正根为 $\begin{aligned} x &= \frac{-120 + \sqrt{14400 + 13248}}{46} \\ &= \frac{-120 + 96\sqrt{3}}{46} \\ &= \frac{-60 + 48\sqrt{3}}{23}, \end{aligned}$ 所以 $k + m + n + p$ $= 60 + 48 + 3 + 23$ $= 134$ 。

Dropping perpendiculars from $C$ and $D$ shows each leg of length $5$ spans a horizontal offset of $\frac{6 - 4}{2} = 1,$ so the height of the trapezoid is $\sqrt{25 - 1} = \sqrt{24}.$ By symmetry the inner circle's center $O$ lies on the vertical axis through the midpoints $E$ of $\overline{AB}$ and $F$ of $\overline{CD}.$ If its radius is $x,$ external tangency gives $OA = x + 3$ and $OC = x + 2,$ so with $AE = 3$ and $CF = 2,$ $\begin{aligned} OE &= \sqrt{(x+3)^2 - 9} \\ &= \sqrt{x^2 + 6x}, \\ OF &= \sqrt{(x+2)^2 - 4} \\ &= \sqrt{x^2 + 4x}. \end{aligned}$

Since $OE + OF = \sqrt{24},$ moving one radical across and squaring gives $\sqrt{24(x^2 + 4x)} = 12 - x,$ and squaring again yields $24x^2 + 96x = 144 - 24x + x^2,$ that is $23x^2 + 120x - 144 = 0.$

The positive root is $\begin{aligned} x &= \frac{-120 + \sqrt{14400 + 13248}}{46} \\ &= \frac{-120 + 96\sqrt{3}}{46} \\ &= \frac{-60 + 48\sqrt{3}}{23}, \end{aligned}$ so $k + m + n + p$ $= 60 + 48 + 3 + 23$ $= 134.$

2004 AIME II 第 12 题

答案：134

难度评级：3060

解答：

其他年份的第 12 题