## Task A-2.3.

Let $A B C$ be a triangle such that $3|B C|=|A B|+|C A|$. Let $T$ be a point on the side $\overline{A C}$ such that $4|A T|=|A C|$, and let $K$ and $L$ be points on the sides $\overline{A B}$ and $\overline{C A}$, respectively, such that $K L \| B C$ and the line $K L$ is tangent to the incircle of triangle $A B C$.

In what ratio does the segment $\overline{B T}$ divide the segment $\overline{K L}$?
## Solution.

Let $v$ be the length of the altitude of triangle $ABC$ from vertex $A$, and $r$ be the radius of the inscribed circle. The area of triangle $ABC$ can then be expressed in two ways:

$$
r \cdot \frac{|AB|+|BC|+|CA|}{2}=P=\frac{v \cdot|BC|}{2}
$$

If we use the condition $3|BC|=|AB|+|CA|$ from the problem in the above equality, we get $4r=v$.

The distance between the lines $KL$ and $BC$ is $2r$, so we conclude that the altitude from vertex $A$ in triangle $AKL$ has a length of $v-2r=4r-2r=2r$.

![](https://cdn.mathpix.com/cropped/2024_05_30_b1b3186b66c6729c9326g-11.jpg?height=754&width=562&top_left_y=1505&top_left_x=750)

Since the lines $KL$ and $BC$ are parallel, triangles $AKL$ and $ABC$ have the same angles, so they are similar. The ratio of the lengths of the altitudes from vertex $A$ in these triangles is $4r:2r=2:1$, so $|KL|=\frac{1}{2}|BC|$, i.e., $\overline{KL}$ is the midsegment of triangle $ABC$. Therefore, $K$ and $L$ are the midpoints of segments $\overline{AB}$ and $\overline{AC}$, respectively.

Since $|AT|=\frac{1}{4}|AC|=\frac{1}{2}|AL|$, we conclude that $T$ is the midpoint of segment $\overline{AL}$. This means that $\overline{BT}$ and $\overline{KL}$ are medians of triangle $ABL$, so segment $\overline{BT}$ divides segment $\overline{KL}$ in the ratio $1:2$.