67. The perimeter of triangle $ABC$ is $2p$. What is the greatest value that the length of a segment tangent to the inscribed circle of $ABC$ and parallel to $BC$ can have? For which triangle (triangles) is this value achieved?
67. Let the inscribed circle $s$ of the triangle touch its

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-182.jpg?height=291&width=409&top_left_y=434&top_left_x=616)

Fig. 154.

sides $B C, A C$, and $A B$ of lengths $a, b$, and $c$ at points $D, E$, and $F$, and the line $K L$ (where $K$ and $L$ are points on sides $A B$ and $A C$) at point $M$ (Fig. 154). From the equality of the lengths of tangents drawn from a point to a circle, it follows that

$$
K M=K F, L M=L E \text { and } B F=B D, C E=C D \text {; }
$$

thus, the perimeter $2 p^{\prime}$ of triangle $A K L$ is

$$
\begin{aligned}
2 p^{\prime} & =A K+A L+K L=A K+A L+(K M+L M)=A K+A L+K F+L E= \\
& =(A K+K F)+(A L+L E)=A F+A E=(A B-B F)+(A C-C E)= \\
& =(A B-B D)+(A C-C D)=A B+A C-(B D+C D)= \\
& =c+b-a=a+b+c-2 a=2 p-2 a=2(p-a) .
\end{aligned}
$$

Since the semiperimeters $p$ and $p^{\prime}$ of similar triangles $A B C$ and $A K L$ are in the same ratio as their bases $B C=a$ and $K L=m$ (new notation!), we have

$$
\frac{p}{p-a}=\frac{a}{m},
$$

and therefore,

$$
m=\frac{a(p-a)}{p}=\frac{a p-a^{2}}{p}=\frac{\frac{p^{2}}{4}-\left(\frac{p}{2}-a\right)^{2}}{p}
$$

From the last formula, it follows that the maximum possible value of the length of segment $K L$ is achieved for all triangles $A B C$ with base $B C=a=\frac{p}{2}$ (i.e., such that $b+c=3 a$); this value is $\frac{p^{2}}{4}: p=\frac{p}{4}$.