Question

Circles with centers $O$ and $P$ have radii $2$ and $4,$ respectively, and are externally tangent. Points $A$ and $B$ are on the circle centered at $O,$ and points $C$ and $D$ are on the circle centered at $P,$ such that $AD$ and $BC$ are common external tangents to the circles. What is the area of hexagon $AOBCPD?$

$18\sqrt{3}$

$24\sqrt{2}$

$36$

$24\sqrt{3}$

$32\sqrt{2}$

Solution:

The circles are externally tangent, so $OP = 2 + 4 = 6.$ In quadrilateral $AOPD,$ both $OA = 2$ and $PD = 4$ are perpendicular to the tangent line $AD,$ making it a right trapezoid.

Drawing the line through $O$ parallel to $AD$ creates a right triangle with hypotenuse $OP = 6$ and one leg $PD - OA = 2,$ so $AD = \sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt2.$

The trapezoid $AOPD$ has area $\frac{1}{2}(2 + 4)(4\sqrt2) = 12\sqrt2.$

By symmetry the hexagon $AOBCPD$ is made of two such trapezoids, so its area is $2 \cdot 12\sqrt2 = 24\sqrt2.$

Thus, the correct answer is B.

2006 AMC 12B Problem 15

Solution:

Problem 15 in Other Years