2006 AMC 10A Problem 16

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

Circles inscribed in an angle are all similar from the vertex's point of view: their centers lie on the bisector and their radii scale with the distance to the vertex. That ratio, combined with the fixed distance between the centers, pins down every length. Drawing the radius to the tangent point is the standard move that turns tangency into a right triangle.

Coordinates: put $F$ at the origin and $A = (0, 8)$. Side $AC$ is tangent to the circle centered $(0, 2)$ with radius $2$; for the line through $A$ with slope $-m$, distance from $(0,2)$ equals $\frac{6}{\sqrt{1 + m^2}} = 2$, so $m = 2\sqrt 2$ and $C = (8/m, 0) = (2\sqrt 2, 0)$. Same area $16\sqrt 2$.

• Using the center-to-center distance $3$ as the apex-to-small-center distance, or forgetting to add the bottom radius $2$ below $O_2$ when finding the height.
• Treating the big circle as the incircle and applying $\text{Area} = rs$ with the wrong $r$ or without knowing the sides.