AMC10 2007 A

AMC10 2007 A · Q19

AMC10 2007 A · Q19. It mainly tests Triangles (properties), Area & perimeter.

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?

画笔沿正方形的两条对角线扫过，产生如图所示的对称涂漆区域。正方形的一半面积被涂漆。求正方形边长与画笔宽度的比值？

(A) $2\sqrt{2} + 1$ $2\sqrt{2} + 1$

(B) $3\sqrt{2}$ $3\sqrt{2}$

(D) $3\sqrt{2} + 1$ $3\sqrt{2} + 1$

(E) $3\sqrt{2} + 2$ $3\sqrt{2} + 2$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $s$ be the side length of the square, let $w$ be the width of the brush, and let $x$ be the leg length of one of the congruent unpainted isosceles right triangles. Since the unpainted area is half the area of the square, the area of each unpainted triangle is $1/8$ of the area of the square. So $\frac{1}{2}x^2=\frac{1}{8}s^2$ and $x=\frac{1}{2}s$. The leg length $x$ plus the brush width $w$ is equal to half the diagonal of the square, so $x+w=(\sqrt{2}/2)s$. Thus $w=\frac{\sqrt{2}}{2}s-\frac{1}{2}s$ and $\frac{s}{w}=\frac{2}{\sqrt{2}-1}=2\sqrt{2}+2$.

答案（C）：设 $s$ 为正方形的边长，设 $w$ 为刷子的宽度，设 $x$ 为其中一个全等的未涂色等腰直角三角形的直角边长度。由于未涂色面积是正方形面积的一半，所以每个未涂色三角形的面积是正方形面积的 $1/8$。因此 $\frac{1}{2}x^2=\frac{1}{8}s^2$，且 $x=\frac{1}{2}s$。直角边长 $x$ 加上刷宽 $w$ 等于正方形对角线的一半，所以 $x+w=(\sqrt{2}/2)s$。因此 $w=\frac{\sqrt{2}}{2}s-\frac{1}{2}s$，并且 $\frac{s}{w}=\frac{2}{\sqrt{2}-1}=2\sqrt{2}+2$。

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