5. The midline of a trapezoid is 4. A line parallel to the bases of the trapezoid and dividing its area in half intersects the lateral sides at points $M$ and $N$. Find the smallest possible length of the segment $M N$.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_dcb518c8ba3f5cf57563g-4.jpg?height=360&width=757&top_left_y=1262&top_left_x=638)

Notations: $A D=a, B C=b, C K=H, N E=h$ - heights of triangles $N C P$ and $D N Q$, $M N=x$

$$
S_{\text {AMND }}=S_{\text {MBCN }} \rightarrow \frac{x+b}{2} H=\frac{a+x}{2} h \rightarrow \frac{H}{h}=\frac{a+x}{x+b}
$$

$\square N C P \quad \square D N Q \rightarrow \frac{H}{h}=\frac{x-b}{a-x} \rightarrow \frac{a+x}{x+b}=\frac{x-b}{a-x} \rightarrow a^{2}-x^{2}=x^{2}-b^{2} \rightarrow x=\sqrt{\frac{a^{2}+b^{2}}{2}}$.

Since $a+b=8$, then $x=\sqrt{\frac{a^{2}+(8-a)^{2}}{2}}=\sqrt{a^{2}-16 a+32}=\sqrt{(a-4)^{2}+16}$.

Answer: $M N_{\text {min }}=4$.

## Answers for other variants.

## Variant 2