2006 AMC 10A Problem 16

Let $O_1$ be the center of the small circle (radius $1$) and $O_2$ the center of the big circle (radius $2$). Both lie on the line of symmetry from $A$ straight down to $F$, the middle of $BC$.

1. Right angles. Draw a radius from each center to the point where its circle touches side $AC$. A radius meets a tangent line at a right angle, so we get two right triangles, both with a corner at $A$.
2. Same shape. The two right triangles share the angle at $A$, so they have the same shape. The side opposite $A$ is $1$ in the small one and $2$ in the big one, so the big triangle is exactly twice the small one. That means
$$AO_2 = 2 \times AO_1.$$
3. Distance between centers. The circles touch, so $O_1O_2 = 1 + 2 = 3$. So $AO_2 = AO_1 + 3$.
4. Solve. Putting steps 2 and 3 together, $2 \times AO_1 = AO_1 + 3$, so $AO_1 = 3$ and $AO_2 = 6$.
5. Height. From $O_2$ down to $BC$ is one big radius, $2$. So the height is $AF = 6 + 2 = 8$.
6. Half the base. In the small right triangle, the hypotenuse is $AO_1 = 3$ and one leg is $1$, so the other leg is $\sqrt{9 - 1} = 2\sqrt{2}$. Triangle $AFC$ has the same shape (right angle at $F$, same angle at $A$), with the side $AF = 8$ playing the role of the leg $2\sqrt2$. So
$$\frac{FC}{8} = \frac{1}{2\sqrt 2}, \qquad FC = \frac{8}{2\sqrt2} = 2\sqrt2.$$
7. Area. The base is $BC = 2 \times 2\sqrt2 = 4\sqrt2$, so the area is
$$\tfrac12 \times 4\sqrt2 \times 8 = 16\sqrt2.$$

The answer is $\boxed{\textbf{(D)}\ 16\sqrt{2}}$.

Circles that fit snugly into the same corner are scaled copies of each other as seen from the corner: double the radius, double the distance to the corner. Together with the known distance between the centers, that fixes every length. Drawing the radius to the touching point is the standard move that turns "tangent" into a right angle.

Let $O_1$ (radius $1$) and $O_2$ (radius $2$) be the centers; both lie on the axis of symmetry from $A$ down to $F$, the midpoint of $BC$. Draw the radii from $O_1$ and $O_2$ to their tangent points on $AC$; each is perpendicular to $AC$.

This creates two right triangles sharing the angle at $A$, with legs $1$ and $2$ opposite that angle. They are similar with ratio $1 : 2$, so $AO_2 = 2 \cdot AO_1$. The circles touch, so $O_1O_2 = 1 + 2 = 3$, i.e. $AO_2 = AO_1 + 3$. Together: $AO_1 = 3$ and $AO_2 = 6$.

The height of the triangle is then $AF = AO_2 + 2 = 8$.

In the small right triangle, hypotenuse $AO_1 = 3$ and leg $1$ give the other leg $\sqrt{9 - 1} = 2\sqrt{2}$, so $\tan(\angle FAC) = \dfrac{1}{2\sqrt 2}$. In right triangle $AFC$,
$$FC = AF \tan(\angle FAC) = \frac{8}{2\sqrt 2} = 2\sqrt 2, \qquad BC = 2 \cdot FC = 4\sqrt 2.$$
Area: $\tfrac12 \cdot BC \cdot AF = \tfrac12 \cdot 4\sqrt2 \cdot 8 = 16\sqrt 2$.

The answer is $\boxed{\textbf{(D)}\ 16\sqrt{2}}$.