Two circles with centers $A$ and $B$ and radii 2 and 1, respectively, touch each other. Point $C$ lies on a line that is tangent to each of the circles and is at a distance of $\frac{3 \sqrt{3}}{2 \sqrt{2}}$ from the midpoint of segment $A B$. Find the area $S$ of triangle $A B C$, given that $S>2$.
Consider the following cases.

1) The circles touch each other internally.
2) The circles touch each other externally at point $K$, and point $C$ lies on their common tangent passing through point $K$.
3) The circles touch each other externally at point $K$, and point $C$ lies on their common tangent, not passing through point $K$. In this case, there are two possible positions of point $C$ relative to the midpoint of the segment of the common external tangent, enclosed between the points of tangency.

## Solution

Let $M$ be the midpoint of segment $A B$. Suppose the circles touch each other internally at point $K$. Then $C K$ is the height of triangle $A B C$. Therefore,

$$
C K=\sqrt{C M^{2}-K M^{2}}=\sqrt{\frac{27}{8}-\frac{9}{4}}=\frac{3 \sqrt{2}}{4}
$$

$$
S=\frac{1}{2} A B \cdot C K=\frac{1}{2} \cdot \frac{3 \sqrt{2}}{4}=\frac{3}{8} \cdot \sqrt{2}2
$$

Suppose the circles touch each other externally at point $K$, and point $C$ lies on the line tangent to the circles with centers $A$ and $B$ at different points $P$ and $Q$ respectively. Then

$$
P Q=\sqrt{(A P+B Q)^{2}-(A P-B Q)^{2}}=\sqrt{9-1}=2 \sqrt{2}
$$

If $\alpha$ is the angle between lines $A B$ and $P Q$, then

$$
\sin \alpha=\frac{1}{3}, \cos \alpha=\frac{2 \sqrt{2}}{3}
$$

If $F$ is the projection of point $M$ onto this line, then $M F$ is the midline of the right trapezoid $A P Q B$ with bases $A P=2$ and $B Q=1$, so

$$
M F=\frac{1}{2}(A P+B Q)=\frac{3}{2}, C F=\sqrt{C M^{2}-M F^{2}}=\sqrt{\frac{27}{8}-\frac{9}{4}}=\frac{3 \sqrt{2}}{4}
$$

If point $C$ lies between $F$ and $Q$, then

$$
C Q=Q F-C F=\frac{1}{2} P Q-C F=\sqrt{2}-\frac{3 \sqrt{2}}{4}=\frac{\sqrt{2}}{4}
$$

If point $C$ lies between $F$ and $P$, then

$$
C Q=Q F+C F=\frac{1}{2} P Q+C F=\sqrt{2}+\frac{3 \sqrt{2}}{4}=\frac{7 \sqrt{2}}{4}
$$

Draw a line through point $Q$ parallel to $A B$. Let $N$ be the point of intersection of this line with the height $C D$ of triangle $A B C$. Then

$$
C D=C N+D N=C Q \sin \alpha+B Q \cos \alpha
$$

If point $C$ lies between $F$ and $Q$, then

$$
\begin{gathered}
C D=C Q \sin \alpha+B Q \cos \alpha=\frac{\sqrt{2}}{12}+\frac{2 \sqrt{2}}{3}=\frac{3 \sqrt{2}}{4} \\
S=\frac{1}{2} A B \cdot C D=\frac{9 \sqrt{2}}{8}2
\end{gathered}
$$

## Answer

$\frac{15 \sqrt{2}}{8}$.