Difference between revisions of "2007 AMC 10A Problems/Problem 15"

(Solution 2)
m (why is that other problem on the top)
(20 intermediate revisions by 4 users not shown)
Line 1: Line 1:
==Problem==
+
== Problem ==
 
Four circles of radius <math>1</math> are each tangent to two sides of a square and externally tangent to a circle of radius <math>2</math>, as shown. What is the area of the square?
 
Four circles of radius <math>1</math> are each tangent to two sides of a square and externally tangent to a circle of radius <math>2</math>, as shown. What is the area of the square?
 
+
<center>
[[Image:2007 AMC 10A -15 for wiki.png]]
+
<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>
 +
</center>
 
<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 1===
+
==Solution==
 
 
 
 
The diagonal has length <math>\sqrt{2}+1+2+2+1+\sqrt{2}=6+2\sqrt{2}</math>. Therefore the sides have length <math>2+3\sqrt{2}</math>, and the area is
 
 
 
<cmath>A=(2+3\sqrt{2})^2=4+6\sqrt{2}+6\sqrt{2}+18=22+12\sqrt{2}  \Rightarrow \text{(B)}</cmath>
 
 
 
=== Solution 2 ===  
 
  
 +
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>
  
 +
== Solution 2 ==
  
Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides <math>3, 3, 3\sqrt{2}</math>. The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then <cmath> A = (2+3\sqrt{2})^2 = 22 + 12\sqrt{2} \Rightarrow \text{(B)}</cmath>
+
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>
  
 
==See Also==
 
==See Also==

Revision as of 18:26, 12 December 2020

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

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 radius of the large square has a side length $2$ units larger than the one found by connecting the midpoints, so its side length is \[2 + 3\sqrt{2}.\] The area of this square is $(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}$ $(B).$

Solution 2

We draw the long 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 (ProblemsAnswer KeyResources)
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