62. What is the greatest value that the length of the segment cut off by the lateral sides of the triangle on the tangent to the inscribed circle, drawn parallel to the base, can take if the perimeter of the triangle is $2 p$?
62. Let $x$ be the length of the desired segment, $b$ be the length of the base $AC$ of triangle $ABC$ (Fig. 44). The perimeter of triangle $BDE$

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-127.jpg?height=374&width=676&top_left_y=1123&top_left_x=256)

Fig. 44

is $2p - 2b$ (convince yourself of this using the property of tangents to a circle).

From the similarity of $\triangle BDE$ and $\triangle ABC$, we get

$$
x = \frac{1}{p} b(p - b) = \frac{1}{p} \left( \frac{p^2}{4} - \left( b - \frac{p}{2} \right)^2 \right).
$$

The maximum value of $\boldsymbol{x}$ is $p / 4$ and is achieved when $b = p / 2$.