Task 5. Given is a triangle $A B C$ with the property that $|A B|+|A C|=3|B C|$. Let $T$ be the point on line segment $A C$ such that $|A C|=4|A T|$. Let $K$ and $L$ be points on the interior of line segments $A B$ and $A C$ respectively, such that first $K L \| B C$ and second $K L$ is tangent to the inscribed circle of $\triangle A B C$. Let $S$ be the intersection of $B T$ and $K L$. Determine the ratio $\frac{|S L|}{|K L|}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_d99998af97a21a8e955bg-7.jpg?height=919&width=1350&top_left_y=787&top_left_x=255)
Solution I. The quadrilateral $K B C L$ touches the incircle of $\triangle A B C$ with all its sides. At each vertex, there are two equal tangent segments. Two opposite sides of the quadrilateral consist precisely of the four different tangent segments, so $|K B|+|L C|=|K L|+|B C|$.

Since $K L \| B C$, it follows that $\triangle A K L \sim \triangle A B C$. Let $t$ be the scaling factor of this similarity, so $|A K|=t|A B|,|A L|=t|A C|$, and $|K L|=t|B C|$. Then $|B K| = |A B| - |A K| = |A B| - t|A B| = (1-t)|A B|$. Similarly, $|C L| = (1-t)|A C|$. Now we have

$t|B C| + |B C| = |K L| + |B C| = |K B| + |L C| = (1-t)|A B| + (1-t)|A C| = 3(1-t)|B C|$,

where the last equality holds due to the given condition in the problem. Dividing by $|B C|$ gives $1 + t = 3 - 3t$, or $4t = 2$, thus $t = \frac{1}{2}$. We conclude that $K$ is the midpoint of $A B$ and $L$ is the midpoint of $A C$.

We already knew that $\triangle A K L \sim \triangle A B C$, which implies $\frac{|K L|}{|B C|} = \frac{|A L|}{|A C|}$. Since $K L \| B C$, it also follows that $\triangle T S L \sim \triangle T B C$, so $\frac{|S L|}{|B C|} = \frac{|T L|}{|T C|}$. Combining these ratios, we get

$$
\frac{|S L|}{|K L|} = \frac{|S L|}{|B C|} \cdot \frac{|B C|}{|K L|} = \frac{|T L|}{|T C|} \cdot \frac{|A C|}{|A L|}
$$

We know that $\frac{|A C|}{|A L|} = 2$, since $L$ is the midpoint of $A C$. This also means that $T$ is the midpoint of $A L$, as $4|A T| = |A C| = 2|A L|$. Thus, $\frac{|T L|}{|T C|} = \frac{\frac{1}{4}|A C|}{\frac{3}{4}|A C|} = \frac{1}{3}$. We conclude that $\frac{|S L|}{|K L|} = \frac{1}{3} \cdot 2 = \frac{2}{3}$.