2006 AMC 10B Problem 24

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

Two standard moves carry the problem: radii to points of tangency are perpendicular to the tangent (turning the figure into right trapezoids), and the length of a common external tangent comes from the right triangle whose hypotenuse joins the centers and whose short leg is the difference of the radii. Symmetry across the line of centers means only half the figure needs computing.

Extend $OP$ past $O$ to meet the tangent lines at $Q$. Similar triangles $QAO \sim QDP$ with ratio $1 : 2$ give $QO = 6$, $QP = 12$, so $QA = \sqrt{36 - 4} = 4\sqrt2$ and $QD = 8\sqrt2$. Kite $QAOB$ has area $2 \cdot \frac{1}{2} \cdot 4\sqrt2 \cdot 2 = 8\sqrt2$ and kite $QDPC$ has area $32\sqrt2$; the hexagon is their difference, $24\sqrt2$.

• Using the center distance 6 as the tangent length AD, or assuming the tangent segment equals the sum of the radii, which gives 36.
• Treating $OADP$ as a rectangle or a parallelogram (parallel sides are $2$ and $4$, not equal), or forgetting to double the half-figure.