8,9}

A circle is inscribed in a triangle with a perimeter of $2 p$. A tangent is drawn to this circle, parallel to one side of the triangle. Find the greatest possible length of the segment of this tangent that is contained within the triangle.

#
The distance from the vertex of a triangle to the nearest point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side.

## Solution

Let $P Q$ be the specified tangent segment to the circle inscribed in triangle $A B C$ with perimeter $2 p, P Q$ $\| A C$. Denote $P Q=x, A C=b$.

Triangles $B P Q$ and $B A C$ are similar with a ratio of ${ }^{x} /{ }_{b}$. If $p_{1}-$ is the semiperimeter of triangle $B P Q$, then $p_{1}=$ $p \cdot{ }^{x}{ }_{b}$. On the other hand, if $K$ is the point of tangency of the inscribed circle with side $A B$, then $p_{1}=B K=p-b$. Therefore, $p^{x / / b}=p-b$. Consequently, $x=\frac{1}{p} b(p-b) \leq \frac{1}{p}\left(\frac{b+(p-b)}{2}\right)^{2}=\frac{1}{p} \cdot \frac{p^{2}}{4}=\frac{p}{4}$, and equality is achieved when $b=p / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_17014d53b52b1543ee40g-17.jpg?height=406&width=644&top_left_y=1778&top_left_x=707)

## Answer