Difference between revisions of "2007 AMC 10A Problems/Problem 15"
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<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 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> | ||
Revision as of 19:07, 1 May 2021
Contents
Problem
Four circles of radius 
are each tangent to two sides of a square and externally tangent to a circle of radius 
, 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]](http://latex.artofproblemsolving.com/2/3/7/23759fb5d0c074107da579ab5ce358f0aa30b5a0.png)
Solution 1
Draw a square connecting the centers of the four small circles of radius . This square has a diagonal of length 
, as it includes the diameter of the big circle of radius and two radii of the small circles of radius . Therefore, the side length of this square is ![\[\frac{6}{\sqrt{2}} = 3\sqrt{2}.\]](http://latex.artofproblemsolving.com/9/a/e/9ae6a28635f7bc2f1d620576f68cea4d8a917152.png)
The radius of the large square has a side length units larger than the one found by connecting the midpoints, so its side length is ![\[2 + 3\sqrt{2}.\]](http://latex.artofproblemsolving.com/2/d/8/2d8d2199d03dc28d51bf35318bfa8e414d50d207.png)
The area of this square is 
Solution 2
We draw the long diagonal of the square. This diagonal yields 
. We know that the side length 
in terms of the diagonal 
is 
, so our side length is 
. However, we are trying to look for the area, so squaring yields 
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. 
