## Task A-2.2.

Inside the triangle $A B C$ are points $S$ and $T$. The distances from point $S$ to the lines $A B, B C$, and $C A$ are 10, 7, and 4, respectively. The distances from point $T$ to these lines are 4, 10, and 16, respectively.

Determine the radius of the inscribed circle of triangle $A B C$.
## First Solution.

Let $a=|B C|, b=|C A|, c=|A B|$. Let $P$ be the area, $s=\frac{1}{2}(a+b+c)$ the semiperimeter, and $r$ the radius of the inscribed circle of triangle $A B C$.

Divide triangle $A B C$ into three smaller triangles by connecting point $S$ to the vertices $A, B$, and $C$.

![](https://cdn.mathpix.com/cropped/2024_05_30_9bdde249022d274b7f19g-08.jpg?height=417&width=1196&top_left_y=2170&top_left_x=430)

By summing the areas of triangles $A B S, B C S$, and $C A S$, we obtain the area of triangle $A B C$, so we have

$$
2 P=10 c+7 a+4 b
$$

Similarly, considering point $T$, we get

$$
2 P=4 c+10 a+16 b
$$

If we multiply the first equation by 2 and add it to the second, we get

$$
6 P=24(a+b+c)=48 s, \quad \text { i.e. } \quad P=8 s
$$

Since $P=r s$, it follows that $r=8$.

## Second Solution.

As in the first solution, we obtain

$$
2 P=10 c+7 a+4 b \quad \text { and } \quad 2 P=4 c+10 a+16 b
$$

From $10 c+7 a+4 b=4 c+10 a+16 b$, we express $a=2 c-4 b$.

Substituting into the first equation, we get $2 P=24 c-24 b$.

On the other hand, we have $2 P=r(a+b+c)=r(3 c-3 b)$.

From this, we conclude that $r(3 c-3 b)=24 c-24 b$, i.e., $r=8$.