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

Difference between revisions of "2007 AMC 10A Problems/Problem 19"

(→‎Solution)
(→‎Solution)
Line 12: Line 12:
 
<math>\frac{a^2}2 = \frac{1}8 \rightarrow a= \frac{1}2</math>
 
<math>\frac{a^2}2 = \frac{1}8 \rightarrow a= \frac{1}2</math>
 
</center>
 
</center>
−
The diagonal of the triangle is <math>\frac{\sqrt{2}}2</math>. The corners of the painted areas are also isosceles right triangles with side length <math>\frac{1-\frac{\sqrt{2}}2}2 = \frac{1}2-\frac{\sqrt2}4</math>. Its diagonal is equal to the width of the paint, and is <math>\frac{\sqrt{2}}2-\frac{1}2</math>. The answer we are looking for is thus <math>\frac{1}{\frac{\sqrt{2}}2-\frac{1}2}</math>. Multiply the numerator and the denominator by <math>\frac{\sqrt{2}}2+\frac{1}2</math> to simplify, and you get <math>\frac{\frac{\sqrt{2}}{2}+\frac{1}{2}}{\frac{2}{4}-\frac{1}{4}}</math> or <math>4({\frac{\sqrt{2}}{2}+\frac{1}{2}}})</math> which is <math>2\sqrt{2}+2 \rightarrow C</math>
+
The diagonal of the triangle is <math>\frac{\sqrt{2}}2</math>. The corners of the painted areas are also isosceles right triangles with side length <math>\frac{1-\frac{\sqrt{2}}2}2 = \frac{1}2-\frac{\sqrt2}4</math>. Its diagonal is equal to the width of the paint, and is <math>\frac{\sqrt{2}}2-\frac{1}2</math>. The answer we are looking for is thus <math>\frac{1}{\frac{\sqrt{2}}2-\frac{1}2}</math>. Multiply the numerator and the denominator by <math>\frac{\sqrt{2}}2+\frac{1}2</math> to simplify, and you get <math>\frac{\frac{\sqrt{2}}{2}+\frac{1}{2}}{\frac{2}{4}-\frac{1}{4}}</math> or <math>4({\frac{\sqrt{2}}{2}+\frac{1}{2}}})</math> which is <math>2\sqrt{2}+2 \rightarrow C</math>.
  
 
==See also==
 
==See also==

Revision as of 04:40, 5 July 2008

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?

2007 AMC 10A Problems-Problem 19 Picture.png

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

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 simplify, and you 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 $2\sqrt{2}+2 \rightarrow C$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_19&oldid=26889"