2007 AMC 10A Problem 15

Below is the professionally curated solution for Problem 15 of the 2007 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10A solutions, or check the answer key.

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Concepts:tangent circles special right triangle square (geometry)

Difficulty rating: 1540

15.

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?

$32$

$22 + 12\sqrt{2}$

$16 + 16\sqrt{3}$

$48$

$36 + 16\sqrt{2}$

Solution:

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 $(2 + 3\sqrt2)^2 = 4 + 12\sqrt2 + 18 = 22 + 12\sqrt2.$

Thus, the correct answer is B.

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