2007 AMC 10A Problems/Problem 15

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?

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

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

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

Alternate Solution

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

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2007 AMC 10A Problems/Problem 15

Contents

Problem

Solution 1

Alternate Solution

See Also