Question

如图，一把画刷沿正方形的两条对角线扫过，形成对称的涂色区域。正方形面积的一半被涂色。正方形边长与画刷宽度之比是多少？

A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width?

$2\sqrt{2} + 1$

$3\sqrt{2}$

$2\sqrt{2} + 2$

$3\sqrt{2} + 1$

$3\sqrt{2} + 2$

答案：C

知识点：面积分割特殊直角三角形分母有理化

难度评级：1820

解答：

设正方形边长为 $s$ ，画刷宽度为 $w$ ，一个未涂色等腰直角三角形的直角边为 $x$ 。每个三角形面积为 $\tfrac18 s^2$ ，所以 $\tfrac12 x^2 = \tfrac18 s^2$ ，得到 $x = \tfrac{s}{2}$ 。

直角边加上画刷宽度等于半条对角线： $x + w = \tfrac{\sqrt2}{2} s$ 。因此 $w = \tfrac{\sqrt2}{2} s - \tfrac{s}{2}$ 。

所以 $\dfrac{s}{w} = \dfrac{2}{\sqrt2 - 1} = 2\sqrt2 + 2.$

所以正确答案是 C。

Let $s$ be the side, $w$ the brush width, and $x$ the leg of one unpainted isosceles right triangle. Each triangle has area $\tfrac18 s^2,$ so $\tfrac12 x^2 = \tfrac18 s^2$ and $x = \tfrac{s}{2}.$

The leg plus the brush width is half the diagonal: $x + w = \tfrac{\sqrt2}{2} s.$ Thus $w = \tfrac{\sqrt2}{2} s - \tfrac{s}{2}.$

Therefore $\dfrac{s}{w} = \dfrac{2}{\sqrt2 - 1} = 2\sqrt2 + 2.$

Thus, the correct answer is C.

2007 AMC 10A 第 19 题

答案：C

难度评级：1820

解答：

其他年份的第 19 题