## Task A-3.6.

Circles $k_{1}, k_{2}$, and $k_{3}$ with centers $S_{1}, S_{2}, S_{3}$ and radii of lengths $r_{1}=1, r_{2}=2$, $r_{3}=3$, respectively, touch each other externally such that $A$ is the point of tangency of circles $k_{1}$ and $k_{2}$, $B$ is the point of tangency of circles $k_{2}$ and $k_{3}$, and $C$ is the point of tangency of circles $k_{3}$ and $k_{1}$. Determine the area of triangle $A B C$.
## Second Solution.

From the first solution, we conclude that $\left|S_{1} S_{2}\right|=3,\left|S_{2} S_{3}\right|=5$ and $\left|S_{3} S_{1}\right|=4$. Let $\alpha$ be the measure of the angle $\varangle S_{1} S_{2} S_{3}$. As in the previous solution, the area $P$ of triangle $S_{1} S_{2} S_{3}$ is 6 (we can calculate it using Heron's formula or by noticing and using the fact that the triangle is right-angled).

By applying the formula for the area of a triangle using the sine of the angle, we have

$$
\frac{P\left(A B S_{2}\right)}{P}=\frac{\frac{1}{2} \cdot 2 \cdot 2 \sin \alpha}{\frac{1}{2} \cdot 3 \cdot 5 \sin \alpha}=\frac{4}{15}
$$

from which it follows that $P\left(A C S_{1}\right)=\frac{4}{15} P=\frac{8}{5}$.

Similarly, we obtain $P\left(C A S_{1}\right)=\frac{1}{2}$ and $P\left(B C S_{3}\right)=\frac{27}{10}$.

Now we can finally determine the area of triangle $A B C$

$$
P(A B C)=P\left(S_{1} S_{2} S_{3}\right)-P\left(C A S_{1}\right)-P\left(A B S_{2}\right)-P\left(B C S_{3}\right)=\frac{6}{5}
$$