# Problem 7. Option 1.

Given a triangle with sides 6, 8, and 10. Find the length of the shortest segment connecting points on the sides of the triangle and dividing it into two equal areas.
Solution. 1) Note that the given triangle is a right triangle: $6^{2}+8^{2}=10^{2}$.

2) Suppose first (and we will justify this later) that the ends of the desired segment $D E=t$ lie on the larger leg $A C=8$ and the hypotenuse $A B=10$ (see figure). Let $A D=x$, $A E=y, \angle B A C=\alpha$. Then

$S_{A D E}=\frac{1}{2} x y \sin \alpha=\frac{3}{10} x y$.

3) By the problem's condition $S_{A D E}=\frac{1}{2} S_{A B C}$,

$\frac{3}{10} x y=\frac{1}{2} \cdot \frac{1}{2} \cdot 6 \cdot 8$,

$x y=40$.
4) $\triangle A D E: t^{2}=x^{2}+y^{2}-2 x y \cos \alpha=$

$$
\begin{aligned}
& =x^{2}+y^{2}-2 \cdot 40 \cdot \frac{8}{10}= \\
& =x^{2}+y^{2}-64= \\
& =x^{2}+\frac{1600}{x^{2}}-64
\end{aligned}
$$

By the Cauchy-Schwarz inequality $t^{2} \geq 2 \sqrt{x^{2} \cdot \frac{1600}{x^{2}}}-64=16$.

The minimum value $t_{\min }=4$ is achieved when $x=y=\sqrt{40}=2 \sqrt{10}$.

5) In an isosceles triangle with base $t$ and vertex angle $\alpha$, the area is

$$
S=\frac{1}{2} t \cdot \frac{1}{2} t \operatorname{ctg} \frac{\alpha}{2}=\frac{1}{4} t^{2} \operatorname{ctg} \frac{\alpha}{2} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1cdabbf1f496496d2444g-05.jpg?height=594&width=511&top_left_y=1162&top_left_x=1115)

From this, $t=2 \sqrt{S \operatorname{tg} \frac{\alpha}{2}}$.

It is clear that for a fixed area $S$, as $\alpha$ increases, the base $t$ will also increase. Therefore, the found value $t_{\min }$, corresponding to the segment opposite the smallest of the three angles of the triangle, is indeed the desired one.

Answer: 4.