2007 AMC 10A Problem 15

Question

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?

32

22 + 12\sqrt{2}

16 + 16\sqrt{3}

48

36 + 16\sqrt{2}

Connect the center of the large circle to the centers of two adjacent small circles

That triangle is isosceles right with legs

2 + 1 = 3,

so its hypotenuse is

3\sqrt2

Consider the isosceles right triangle joining the center of the radius-

2

circle to the centers of two adjacent small circles. Its legs have length

2 + 1 = 3,

so its hypotenuse is

3\sqrt2.

The side of the square exceeds this hypotenuse by

2

(one radius on each end), so

s = 2 + 3\sqrt2.

The area is

\begin{aligned} (2 + 3\sqrt2)^2 &= 4 + 12\sqrt2 + 18 \\ &= 22 + 12\sqrt2. \end{aligned}

Thus, the correct answer is B.

Problem 15 in other years