2006 AMC 12B Problems/Problem 15

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Problem

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 $\overline{AD}$ and $\overline{BC}$ are common external tangents to the circles. What is the area of hexagon $AOBCPD$ ?

$\textbf{(A) } 18\sqrt {3} \qquad \textbf{(B) } 24\sqrt {2} \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 24\sqrt {3} \qquad \t...$

Solution

Draw the altitude from $O$ onto $DP$ and call the point $H$ . Because $\angle OAD$ and $\angle ADP$ are right angles due to being tangent to the circles, and the altitude creates $\angle OHD$ as a right angle. $ADHO$ is a rectangle with $OH$ bisecting $DP$ . The length $OP$ is $4+2$ and $HP$ has a length of $2$ , so by pythagorean's, $OH$ is $\sqrt{32}$ .

$2 \cdot \sqrt{32} + \frac{1}{2}\cdot2\cdot \sqrt{32} = 3\sqrt{32} = 12\sqrt{2}$ , which is half the area of the hexagon, so the area of the entire hexagon is $2\cdot12\sqrt{2} = \boxed{(B)} \qquad24\sqrt{2}$

Solution 2

$ADOP$ and $OPBC$ are congruent right trapezoids with legs $2$ and $4$ and with $OP$ equal to $6$ . Draw an altitude from $O$ to either $DP$ or $CP$ , creating a rectangle with width $2$ and base $x$ , and a right triangle with one leg $2$ , the hypotenuse $6$ , and the other $x$ . Using the Pythagorean theorem, $x$ is equal to $4\sqrt{2}$ , and $x$ is also equal to the height of the trapezoid. The area of the trapezoid is thus $\frac{1}{2}\cdot(4+2)\cdot4\sqrt{2} = 12$ , and the total area is two trapezoids, or $\boxed{24\sqrt{2}}$ .

2006 AMC 12B (Problems • Resources)
Preceded by Problem 14	Followed by Problem 16
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2006 AMC 12B Problems/Problem 15

From AoPSWiki

Contents

Problem

Solution

Solution 2

See also