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

(Solution 1)
Line 8: Line 8:
 
===Solution 1===
 
===Solution 1===
  
----
 
  
 
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
 
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

Revision as of 07:39, 24 July 2017

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?

2007 AMC 10A -15 for wiki.png

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

Solution 1

The diagonal has length $\sqrt{2}+1+2+2+1+\sqrt{2}=6+2\sqrt{2}$. Therefore the sides have length $2+3\sqrt{2}$, and the area is

\[A=(2+3\sqrt{2})^2=4+6\sqrt{2}+6\sqrt{2}+18=22+12\sqrt{2}  \Rightarrow \text{(B)}\]

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 $3, 3, 3\sqrt{2}$. 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 \[A = (2+3\sqrt{2})^2 = 22 + 12\sqrt{2} \Rightarrow \text{(B)}\]

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

Invalid username
Login to AoPS