2. Let $S$ be the midpoint of the base $AB$ of a given isosceles triangle $ABC$. Suppose that the incircles of triangles $ACS$ and $BCS$ touch the line $AB$ at points that divide the base $AB$ into three equal parts. Calculate the ratio $|AB|:|CS|$.


2. Due to the symmetry with respect to the line $C S$, both inscribed circles touch the height $C S$ at the same point, which we denote as $D$. The points of tangency of these circles with the segments $A S, B S$, $A C, B C$ are denoted in order as $E, F, G, H$ (Fig. 1). To express all necessary lengths, we introduce the notation $x=|S D|$ and $y=|C D|$.

![](https://cdn.mathpix.com/cropped/2024_04_17_3b21ca17568ff452038cg-3.jpg?height=637&width=765&top_left_y=1001&top_left_x=654)

For the tangents from point $S$ to the given circle, the following equalities hold:

$$
|S D|=|S E|=|S F|=x \quad \text { and } \quad|C D|=|C G|=|C H|=y \text {. }
$$

The segment $E F$ therefore has a length of $2 x$, which is also the length of the segments $A E$ and $B F$, and thus the lengths of the segments $A G$ and $B H$ (again due to the symmetry of the tangents). From this, the following equalities immediately follow:

$$
|A B|=6 x, \quad|A C|=|B C|=2 x+y \quad \text { and } \quad|C S|=x+y .
$$

To find the relationship between the lengths $x$ and $y$, we apply the Pythagorean theorem to the right triangle $A C S$ (with the leg $A S$ of length $3 x$):

$$
(2 x+y)^{2}=(3 x)^{2}+(x+y)^{2} .
$$

By expanding and further simplifying, we get (since $x$ and $y$ are positive values):

$$
\begin{aligned}
4 x^{2}+4 x y+y^{2} & =9 x^{2}+x^{2}+2 x y+y^{2}, \\
2 x y & =6 x^{2}, \\
y & =3 x .
\end{aligned}
$$

The desired ratio is therefore

$$
|A B|:|C S|=6 x:(x+y)=6 x: 4 x=3: 2 .
$$

Note that the entire solution process can be written in a similar way using the standard notation $c=|A B|$ and $v=|C S|$. Since according to the problem, $|A E|=\frac{1}{3} c$, and thus $|S E|=\frac{1}{6} c$, from the equality $|S D|=|S E|$ it follows that $|C D|=|C S|-|S D|=v-\frac{1}{6} c$, hence

$$
|A C|=|A G|+|C G|=|A E|+|C D|=\frac{1}{3} c+\left(v-\frac{1}{6} c\right)=v+\frac{1}{6} c,
$$

so from the Pythagorean theorem for triangle $A C S$

$$
\left(v+\frac{1}{6} c\right)^{2}=\left(\frac{1}{2} c\right)^{2}+v^{2}
$$

we get $3 v=2 c$ or $c: v=3: 2$.

For a complete solution, award 6 points. For expressing the necessary lengths using two parameters $(x, y$ or $c, v)$, award 3 points, add 2 points for effectively using the Pythagorean theorem, and 1 point for determining the desired ratio.