Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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?

[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

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 triangles has area $\frac{1}8$. It can be observed 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 area 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 one rectangle 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,

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

$\frac{\sqrt{2}+1}{2}$ is greater than the length of the square, so it is impossible for that to be the brush width. Thus $n=\frac{\sqrt{2}-1}{2}$. 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_19&oldid=155016"