# 2006 AMC 12B Problems/Problem 15

(Redirected from 2006 AMC 12B Problem 15)

## 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 $AOBCPD$? $[asy] // from amc10 problem series unitsize(0.4 cm); defaultpen(linewidth(0.7) + fontsize(11)); pair A, B, C, D; pair[] O; O = (6,0); O = (12,0); A = (32/6,8*sqrt(2)/6); B = (32/6,-8*sqrt(2)/6); C = 2*B; D = 2*A; draw(Circle(O,2)); draw(Circle(O,4)); draw((0.7*A)--(1.2*D)); draw((0.7*B)--(1.2*C)); draw(O--O); draw(A--O); draw(B--O); draw(C--O); draw(D--O); label("A", A, NW); label("B", B, SW); label("C", C, SW); label("D", D, NW); dot("O", O, SE); dot("P", O, SE); label("2", (A + O)/2, E); label("4", (D + O)/2, E);[/asy]$ $\textbf{(A) } 18\sqrt {3} \qquad \textbf{(B) } 24\sqrt {2} \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 24\sqrt {3} \qquad \textbf{(E) } 32\sqrt {2}$

## 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 $2\cdot12\sqrt{2} = \boxed{(B)24\sqrt{2}}$

## Solution 2 $ADPO$ 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 $\frac{1}{2}\cdot(4+2)\cdot4\sqrt{2} = 12\sqrt{2}$, and the total area is two trapezoids, or $\boxed{24\sqrt{2}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 