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2007 AMC 10A Problems/Problem 19

Revision as of 09:45, 24 July 2017 by Asp211 (talk | contribs) (→‎Solution 1)

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? [asy] unitsize(3 cm); real w = (sqrt(2) - 1)/2; filldraw((0,0)--(w*sqrt(2)/2,0)--(0.5,0.5 - w*sqrt(2)/2)--(1 - w*sqrt(2)/2,0)--(1,0)--(1,w*sqrt(2)/2)--(0.5 + w*sqrt(2)/2,0.5)--(1,1 - w*sqrt(2)/2)--(1,1)--(1 - w*sqrt(2)/2,1)--(0.5,0.5 + w*sqrt(2)/2)--(w*sqrt(2)/2,1)--(0,1)--(0,1 - w*sqrt(2)/2)--(0.5 - w*sqrt(2)/2,0.5)--(0,w*sqrt(2)/2)--cycle,gray(0.7)); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); [/asy] $\text{(A)}\ 2\sqrt {2} + 1 \qquad \text{(B)}\ 3\sqrt {2}\qquad \text{(C)}\ 2\sqrt {2} + 2 \qquad \text{(D)}\ 3\sqrt {2} + 1 \qquad \text{(E)}\ 3\sqrt {2} + 2$

Solution 1

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 hypotenuse 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 hypotenuse 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 simplify, and you get $\frac{\frac{\sqrt{2}}{2}+\frac{1}{2}}{\frac{2}{4}-\frac{1}{4}}$ or $4\left(\frac{\sqrt{2}}{2}+\frac{1}{2}\right)$ which is $\boxed{2\sqrt{2}+2} \rightarrow C$.

Solution 2

Again, have the length of the square equal to $1$ and let the width of each individual stripe be $n$. Note that you can split each stripe into two rectangles and two isosceles right triangles at the corners. Then the area of each stripe is $n(\sqrt{2}-\frac{n}{2})=\sqrt{2}n-\frac{n^2}{2}$. The area covered by the two total stripes is twice the area of one stripe, minus the area in the intersection of the stripes, which is a square with side length $n$. This area is equal to $\frac{1}{2}$ So:

$2\sqrt2n-2n^2=\frac{1}{2}\rightarrow-2n^2+2\sqrt2n-\frac{1}{2}=0$.

By the quadratic formula,

$n=\frac{-2\sqrt{2}\pm\sqrt{(2\sqrt{2})^2-4(-2)(-\frac{1}{2})}}{2(-2)}\rightarrow n=\frac{-2\sqrt{2}\pm 2}{-4}\rightarrow n=\frac{\sqrt{2}\pm 1}{2}$

It's easy to tell that $\frac{\sqrt{2}+1}{2}$ is too large, so $n=\frac{\sqrt{2}-1}{2}$. We want to find $\frac{1}{n}$, and $\frac{1}{n}=\frac{2}{\sqrt{2}-1}$. Multiply the numerator and the denominator by $\sqrt{2}+1$,

$\frac{1}{n}=\frac{2(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=\boxed{2\sqrt{2}+2}\rightarrow C$

See also

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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