Circles with radii $1$, $2$, and $3$ are mutually externally tangent. What is the area of the triangle determined by the points of tangency? $\textbf{(A)}\ \frac{3}{5} \qquad$ $ \textbf{(B)}\ \frac{4}{5} \qquad$ $ \textbf{(C)}\ 1 \qquad$ $ \textbf{(D)}\ \frac{6}{5} \qquad$ $ \textbf{(E)}\ \frac{4}{3}$
Solution
Let $A,B,C$ be the centers of the circles with radii $1,2,3$. Then $AB=3,AC=4,BC=5$ --- a right triangle with the right angle at $A$.
Place $A=(0,0), B=(3,0), C=(0,4)$.
The tangency point on $AB$ is at distance $1$ from $A$: $P=(1,0)$. The tangency point on $AC$ is at distance $1$ from $A$: $Q=(0,1)$.
The tangency point on $BC$ is at distance $2$ from $B$, along the unit vector $\frac15(-3,4)$ from $B$ to $C$: $R=(3,0)+2\cdot\frac15(-3,4)=\left(\frac95,\frac85\right)$.
By the shoelace theorem, the area of $\triangle PQR$ is $$\frac12\left|1\left(1-\frac85\right)+0\left(\frac85-0\right)+\frac95(0-1)\right|=\frac12\left|-\frac35-\frac95\right|=\frac12\cdot\frac{12}5=\boxed{\frac65}.$$
$\textbf{(A)}\ \frac{3}{5} \qquad$ $ \textbf{(B)}\ \frac{4}{5} \qquad$ $ \textbf{(C)}\ 1 \qquad$ $ \textbf{(D)}\ \frac{6}{5} \qquad$ $ \textbf{(E)}\ \frac{4}{3}$ 
Let $A,B,C$ be the centers of the circles with radii $1,2,3$. Then $AB=3,AC=4,BC=5$ --- a right triangle with the right angle at $A$.