4 [ Described quadrilaterals $]$

Non-intersecting circles $S 1, S 2$ and $S 3$ are sequentially inscribed in an angle of $60^{\circ}$. Find the area of the quadrilateral with vertices at the points of intersection of the common internal tangents of circles $S 1$ and $S 2$ and circles $S 2$ and $S 3$, given that the radius of circle $S 2$ is $r$, and the difference in radii of circles $S 3$ and $S 1$ is $a$.
Let the common internal tangent of circles $S 1$ and $S 2$ intersect one side of the given angle with vertex $O$ at point $A$, the other side at point $D$, touch circle $S 1$ at point $Q$, and circle $S 1$ touches lines $O A$ and $O D$ at points $M$ and $N$, respectively. Then $A D = A Q + Q D = A M + D N$. Similarly, if the common internal tangent of circles $S 3$ and $S 2$ intersects side $O A$ of the given angle at point $B$, the other side at point $C$, touches circle $S 3$ at point $E$, and circle $S 3$ touches lines $O A$ and $O D$ at points $K$ and $L$, respectively, then $B C = B E + E C = B K + C L$. Therefore, if $p$ is the semiperimeter of quadrilateral $A B C D$, then

$$
\begin{gathered}
p = \frac{1}{2}(A B + B C + C D + A D) = \frac{1}{2}(A B + (B E + E C) + C D + (A Q + Q D)) = \\
= \frac{1}{2}(A B + (B K + C L) + C D + (A M + D N)) = \\
\left.= \frac{1}{2}(A B + B K + A M) + (D N + C D + C L)\right) = \frac{1}{2}(M K + N L) = \frac{1}{2} \cdot 2 N L = N L
\end{gathered}
$$

From the center $O 1$ of circle $S 1$ with radius $r 1$, drop a perpendicular $O 1 T$ to the radius $O 3 L$ of circle $S 3$ with radius $r 3$. From the right triangle $O 1 T O 3$, we find that

$$
O 1 T = O 3 T \operatorname{ctg} \angle O 3 O 1 T = (r 3 - r 1) \operatorname{ctg} 30^{\circ} = a \sqrt{3}.
$$

Therefore,

$$
S_{A B C D} = p r = N L \cdot r = O 1 T \cdot r = a \sqrt{3} \cdot r = a r \sqrt{3}
$$

## Answer

$\operatorname{ar} \sqrt{3}$.