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 15"

m (why is that other problem on the top)
m (→‎Solution 2)
 
(4 intermediate revisions by 2 users not shown)
Line 22: Line 22:
 
<math>\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}</math>
 
<math>\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}</math>
  
−
==Solution==
+
==Solution 1==
  
−
Draw a square connecting the centers of the four small circles of radius <math>1</math>. This square has a diagonal of length <math>6</math>, as it includes the diameter of the big circle of radius <math>2</math> and two radii of the small circles of radius <math>1</math>. Therefore, the side length of this square is <cmath>\frac{6}{\sqrt{2}} = 3\sqrt{2}.</cmath> The radius of the large square has a side length <math>2</math> units larger than the one found by connecting the midpoints, so its side length is <cmath>2 + 3\sqrt{2}.</cmath> The area of this square is <math>(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}</math> <math>(B).</math>
+
Draw a square connecting the centers of the four small circles of radius <math>1</math>. This square has a diagonal of length <math>6</math>, as it includes the diameter of the big circle of radius <math>2</math> and two radii of the small circles of radius <math>1</math>. Therefore, the side length of this square is <cmath>\frac{6}{\sqrt{2}} = 3\sqrt{2}.</cmath> The side length of the larger square is <math>2</math> units greater than the one found by connecting the midpoints, so its side length is <cmath>2 + 3\sqrt{2}.</cmath> The area of the larger square is <math>(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}</math> <math>(B).</math>
  
 
== Solution 2 ==
 
== Solution 2 ==
  
−
We draw the long diagonal of the square. This diagonal yields <math>2\sqrt{2}+1+1+2+2=2\sqrt{2}+6</math>. We know that the side length <math>s</math> in terms of the diagonal <math>d</math> is <math>s=\frac{d}{\sqrt{2}}</math>, so our side length is <math>\frac{2\sqrt{2}+6}{\sqrt{2}}</math>. However, we are trying to look for the area, so squaring <math>\frac{2\sqrt{2}+6}{\sqrt{2}}</math> yields <math>\frac{44+24\sqrt{2}}{2}=\boxed{\text{(B)}22+12\sqrt{2}}</math>
+
We draw the diagonal of the square. This diagonal yields <math>2\sqrt{2}+1+1+2+2=2\sqrt{2}+6</math>. We know that the side length <math>s</math> in terms of the diagonal <math>d</math> is <math>s=\frac{d}{\sqrt{2}}</math>, so our side length is <math>\frac{2\sqrt{2}+6}{\sqrt{2}}</math>. However, we are trying to look for the area, so squaring <math>\frac{2\sqrt{2}+6}{\sqrt{2}}</math> yields <math>\frac{44+24\sqrt{2}}{2}=\boxed{\text{(B)}22+12\sqrt{2}}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 09:29, 4 June 2021

Problem

Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2$, as shown. What is the area of the square?

[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); real h=3*sqrt(2)/2; pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h); pair X=O0+2*dir(30), Y=O2+dir(45); draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); draw(Circle(O0,2)); draw(Circle(O1,1)); draw(Circle(O2,1)); draw(Circle(O3,1)); draw(Circle(O4,1)); draw(O0--X); draw(O2--Y); label("$2$",midpoint(O0--X),NW); label("$1$",midpoint(O2--Y),SE); [/asy]

$\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}$

Solution 1

Draw a square connecting the centers of the four small circles of radius $1$. This square has a diagonal of length $6$, as it includes the diameter of the big circle of radius $2$ and two radii of the small circles of radius $1$. Therefore, the side length of this square is \[\frac{6}{\sqrt{2}} = 3\sqrt{2}.\] The side length of the larger square is $2$ units greater than the one found by connecting the midpoints, so its side length is \[2 + 3\sqrt{2}.\] The area of the larger square is $(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}$ $(B).$

Solution 2

We draw the diagonal of the square. This diagonal yields $2\sqrt{2}+1+1+2+2=2\sqrt{2}+6$. We know that the side length $s$ in terms of the diagonal $d$ is $s=\frac{d}{\sqrt{2}}$, so our side length is $\frac{2\sqrt{2}+6}{\sqrt{2}}$. However, we are trying to look for the area, so squaring $\frac{2\sqrt{2}+6}{\sqrt{2}}$ yields $\frac{44+24\sqrt{2}}{2}=\boxed{\text{(B)}22+12\sqrt{2}}$

See Also

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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_15&oldid=155096"