## Task A-3.3.

Let $A^{\prime}, B^{\prime}, C^{\prime}$ be the points where the angle bisectors of triangle $A B C$ intersect the opposite sides $\overline{B C}, \overline{C A}, \overline{A B}$ respectively, and let $S$ be the center of the inscribed circle of triangle $A B C$.

If $|A S|:\left|S A^{\prime}\right|=3: 2,|B S|:\left|S B^{\prime}\right|=4: 3$ and if $|A B|=12$, determine the lengths of the other sides of the triangle.
## Solution.

Let $a, b$ and $c$ be the lengths of the sides $\overline{B C}, \overline{A C}$ and $\overline{A B}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_30_77c5fa69d8ea891e3e07g-19.jpg?height=520&width=811&top_left_y=754&top_left_x=634)

Using the Angle Bisector Theorem for triangle $A B A^{\prime}$, we get

$|A B|:\left|A^{\prime} B\right|=|A S|:\left|A^{\prime} S\right|=3: 2$, which implies $\left|A^{\prime} B\right|=\frac{2 c}{3}$.

Similarly, for triangle $A B B^{\prime}$, we get $|A B|:\left|A B^{\prime}\right|=|B S|:\left|B^{\prime} S\right|=4: 3$,

which implies $\left|A B^{\prime}\right|=\frac{3 c}{4}$.

For triangle $A C A^{\prime}$, we get $|A C|:\left|A^{\prime} C\right|=|A S|:\left|A^{\prime} S\right|=3: 2$,

which implies $b:\left(a-\frac{2 c}{3}\right)=3: 2$, hence $3 a-2 c=2 b$.

Similarly, for triangle $B C B^{\prime}$, we have $|B C|:\left|B^{\prime} C\right|=|B S|:\left|B^{\prime} S\right|=4: 3$, which implies $a:\left(b-\frac{3 c}{4}\right)=4: 3$, or $4 b-3 c=3 a$.

Considering $c=12$, solving the system

$$
\begin{aligned}
& 3 a-2 c=2 b \\
& 4 b-3 c=3 a
\end{aligned}
$$

we get $a=28, b=30$.

(2 points)