Task 7. II variant.

Given a triangle with sides 5, 12, and 13. Find the length of the shortest segment connecting points on the sides of the triangle and dividing it into two equal areas.
Solution. 1) By the converse of the Pythagorean theorem, the given triangle is a right triangle.

2) Consider triangle $ABC$, where $BC=5$, $AC=12$, $AB=13$ (see figure). Take points $D \in AC$, $E \in AB$. Let $AD=x$, $AE=y$, $DE=t$, $\angle BAC=\alpha$. Then

$S_{ADE}=\frac{1}{2} x y \sin \alpha=\frac{5}{26} x y$.
3) $S_{ADE}=\frac{1}{2} S_{ABC}$,

$\frac{5}{26} x y=\frac{1}{2} \cdot \frac{1}{2} \cdot 5 \cdot 12$

![](https://cdn.mathpix.com/cropped/2024_05_06_1cdabbf1f496496d2444g-06.jpg?height=614&width=414&top_left_y=447&top_left_x=1409)

$x y=78$.
4) $\triangle ADE: t^{2}=x^{2}+y^{2}-2 x y \cos \alpha=$

$$
\begin{aligned}
& =x^{2}+y^{2}-2 \cdot 78 \cdot \frac{12}{13}= \\
& =x^{2}+y^{2}-144= \\
& =x^{2}+\frac{78^{2}}{x^{2}}-144
\end{aligned}
$$

By the Cauchy-Schwarz inequality, $t^{2} \geq 2 \sqrt{x^{2} \cdot \frac{78^{2}}{x^{2}}}-144=12$. The minimum value $t_{\min }=2 \sqrt{3}$ is achieved when $x=y=\sqrt{78}$.

Answer: $2 \sqrt{3}$.