2007 AMC 10A Problems/Problem 19

Problem

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?

$\text{(A)}\ 2\sqrt {2} + 1 \qquad \text{(B)}\ 3\sqrt {2}\qquad \text{(C)}\ 2\sqrt {2} + 2 \qquad \text{(D)}\ 3\sqrt {2} + 1 \qquad \text{(E)}\ 3\sqrt {2} + 2$

Solution

Without loss of generality, let the side length of the square be $1$ unit. The area of the painted area is $\frac{1}2$ of the area of the larger square, so the total unpainted area is also $\frac{1}{2}$ . Each of the $4$ unpainted triangle has area $\frac{1}8$ . It is easy to tell that these triangles are isosceles right triangles, so let $a$ be the side length of one of the smaller triangles:

$\frac{a^2}2 = \frac{1}8 \rightarrow a= \frac{1}2$

The diagonal of the triangle is $\frac{\sqrt{2}}2$ . The corners of the painted areas are also isosceles right triangles with side length $\frac{1-\frac{\sqrt{2}}2}2 = \frac{1}2-\frac{\sqrt2}4$ . Its diagonal is equal to the width of the paint, and is $\frac{\sqrt{2}}2-\frac{1}2$ . The answer we are looking for is thus $\frac{1}{\frac{\sqrt{2}}2-\frac{1}2}$ . Multiply the numerator and the denominator by $\frac{\sqrt{2}}2+\frac{1}2$ to get $\frac{\frac{\sqrt{2}}{2}+\frac{1}{2}}{\frac{2}{4}-\frac{1}{4}}$ or $4({\frac{\sqrt{2}}{2}+\frac{1}{2}}})$ (Error compiling LaTeX. Unknown error_msg) which is $2\sqrt{2}+2 \rightarrow C$

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2007 AMC 10A Problems/Problem 19

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