2006 AMC 10B Problem 24

1. Use the mirror line. The line $OP$ through both centers splits the figure into two matching halves. So the hexagon is two copies of the four-sided shape $OADP$. Find that area and double it.
2. Right angles. A radius drawn to the point where a tangent line touches the circle meets that line at a right angle. So $OA$ and $PD$ both meet $AD$ at right angles. That makes $OA$ and $PD$ parallel, and $OADP$ is a trapezoid with parallel sides $OA = 2$ and $PD = 4$. Its height is the length $AD$.
3. Find $AD$. From $O$, draw a line parallel to $AD$ until it hits $PD$ at a point $X$. Then $OADX$ is a rectangle, so $DX = OA = 2$ and $OX = AD$.
4. That leaves $PX = 4 - 2 = 2$. The circles touch each other from outside, so the distance between centers is $OP = 2 + 4 = 6$. Triangle $OXP$ has a right angle at $X$, so
$$OX = \sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt2.$$
So $AD = 4\sqrt2$.
5. Trapezoid area. Average of the parallel sides times the height:
$$\frac{2 + 4}{2} \cdot 4\sqrt2 = 3 \cdot 4\sqrt2 = 12\sqrt2.$$
6. Hexagon. Two copies: $2 \cdot 12\sqrt2 = 24\sqrt2$.

The answer is $\boxed{\textbf{(B)}\ 24\sqrt2}$.

Two facts do all the work. A radius meets a tangent line at a right angle, which turns the picture into right-angled trapezoids. And the length of a tangent line touching two circles comes from a right triangle whose longest side joins the centers and whose short side is the difference of the radii. The mirror line through the centers means only half the figure needs to be computed.

Extend $OP$ to the left until it meets both tangent lines at a point $Q$. Triangle $QAO$ and triangle $QDP$ have the same shape, and $PD$ is twice $OA$, so every length in the big one is twice the small one. Since $QP = QO + 6$ and $QP = 2 \cdot QO$, we get $QO = 6$ and $QP = 12$. Then $QA = \sqrt{36 - 4} = 4\sqrt2$ and $QD = 8\sqrt2$. The kite $QAOB$ is two right triangles of area $\frac12 \cdot 4\sqrt2 \cdot 2$, so $8\sqrt2$; the kite $QDPC$ is two of area $\frac12 \cdot 8\sqrt2 \cdot 4$, so $32\sqrt2$. The hexagon is the big kite minus the small one: $32\sqrt2 - 8\sqrt2 = 24\sqrt2$.

The line $OP$ is an axis of symmetry of the whole figure, so the hexagon is two congruent copies of quadrilateral $OADP$; find that area and double it.

Because $AD$ is tangent to both circles, $OA \perp AD$ and $PD \perp AD$. Thus $OA \parallel PD$ and $OADP$ is a right trapezoid with parallel sides $OA = 2$ and $PD = 4$ and height $AD$.

To get $AD$, drop a perpendicular from $O$ to $PD$, meeting it at $X$. Then $OAD X$ is a rectangle, so $DX = 2$ and $PX = 4 - 2 = 2$, while $OP = 2 + 4 = 6$ since the circles are externally tangent. In right triangle $OXP$,
$$OX = \sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt2 = AD.$$

Trapezoid area: $\frac{(2 + 4)}{2} \cdot 4\sqrt2 = 12\sqrt2$. The hexagon has area $2 \cdot 12\sqrt2 = 24\sqrt2$.

The answer is $\boxed{\textbf{(B)}\ 24\sqrt2}$.