Question

图中二十个全等圆排成三行，并被一个长方形围住。这些圆彼此相切，并且如图所示与长方形的边相切。长方形较长边与较短边的比可以写成 $\frac{1}{2}\left(\sqrt{p} - q\right)$ ，其中 $p$ 和 $q$ 是正整数。求 $p + q$ 。

The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\frac{1}{2}\left(\sqrt{p} - q\right),$ where $p$ and $q$ are positive integers. Find $p + q.$

答案

答案：154

知识点：相切圆等边三角形分母有理化

难度评级：2020

解答：

设公共半径为 $r$ 。较长边容纳一行七个圆，所以长为 $14r$ 。相邻行中三个两两相切圆的圆心构成边长 $2r$ 的等边三角形，其高为 $r\sqrt{3}$ 。两段相邻行圆心间距合计贡献 $2r\sqrt{3}$ ，所以较短边为 $r + 2r\sqrt{3} + r = 2r + 2r\sqrt{3}$ 。

比值为所以 $p = 147$ ， $q = 7$ ， $p + q = 154$ 。 $\begin{aligned} \frac{14r}{2r\left(1 + \sqrt{3}\right)} &= \frac{7}{1 + \sqrt{3}} \\ &= \frac{7\left(\sqrt{3} - 1\right)}{2} \\ &= \frac{1}{2}\left(\sqrt{147} - 7\right), \end{aligned}$

Let $r$ be the common radius. The longer side holds a row of seven circles, so it equals $14r.$ The centers of three mutually tangent circles in adjacent rows form an equilateral triangle with side $2r,$ whose height is $r\sqrt{3},$ so the two gaps between rows of centers contribute $2r\sqrt{3},$ and the shorter side is $r + 2r\sqrt{3} + r = 2r + 2r\sqrt{3}.$

The ratio is $\begin{aligned} \frac{14r}{2r\left(1 + \sqrt{3}\right)} &= \frac{7}{1 + \sqrt{3}} \\ &= \frac{7\left(\sqrt{3} - 1\right)}{2} \\ &= \frac{1}{2}\left(\sqrt{147} - 7\right), \end{aligned}$ so $p = 147,$ $q = 7,$ and $p + q = 154.$

2002 AIME I 第 2 题

答案：154

难度评级：2020

解答：

其他年份的第 2 题