The perimeter of triangle $ABC$ is 8. An incircle is inscribed in the triangle, and a tangent is drawn to it, parallel to side $AB$. The segment of this tangent, enclosed between sides $AC$ and $CB$, is equal to 1. Find the side $AB$.

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The truncated triangle is similar to the given one with a coefficient equal to the ratio of their perimeters.

## Solution

Let the points of intersection of the tangent with sides $A C$ and $C B$ be $M$ and $N$, and the points of tangency of these sides with the inscribed circle be $P$ and $Q$. Then the semiperimeter of triangle $C M N$ is $C P = C Q = 4 - A B$. From the similarity of triangles $C M N$ and $C A B$, it follows that ${ }^{M N} / A B = 1 / 4(4 - A B)$, or ${ }^{1 /}{ }_{A B} = 1 - { }^{A B} / 4$. From this, $A B = 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-34.jpg?height=398&width=803&top_left_y=1556&top_left_x=633)

## Answer

2.

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