Problem 6. Answer: $|L|_{\text {min }}=\frac{31}{2}=15.5$.

## Variant 0

The base $A B C$ of the pyramid $A B C D$ lies on the plane $S$. Point $M$ is located on the extension of the line $A B$ beyond point $B$ such that $M B: A B=\lambda$. Point $N$ is the midpoint of the edge $D C$. The curve $L$, connecting points $M$ and $N$, lies either on the plane $S$ or on the lateral surface of the pyramid. Find the smallest possible length of the curve $L$, if all edges of the pyramid, including the base edges, have length $a$.
Answer: $|L|_{\min }=\frac{a}{2} \sqrt{4 \lambda^{2}+6 \lambda+3}$.

Solution. Let the curve $L$ be the desired one. It cannot fail to intersect the line $A B$ at some point $E$. Then the part of the curve from point $M$ to point $E$ is a straight line segment. Point $F$ is the midpoint of edge $A C$. The geometric locus of points equidistant from $N$ and $F$ is the median plane $P$, perpendicular to $N F$ and passing through the midpoint of segment $N F$. Plane $P$ coincides with plane $B C G$, where $G$ is the midpoint of edge $A D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ae4736b169c02f1b6df7g-4.jpg?height=654&width=737&top_left_y=187&top_left_x=594)

Point $T$ is the midpoint of segment $N F$ and median $C G$. Line $B C$ lies on plane $P$, so $|L| \geq M E + E N = M E + E F \geq M F$. Equality is achieved if point $E$ coincides with point $H$ - the intersection of lines $M F$ and $B C, \min |L| = M F$. The desired curve is the broken line $M H N$.

Calculations:

$$
\begin{aligned}
& M F^{2}=a^{2}(1+\lambda)^{2}+\frac{a^{2}}{4}-\frac{1}{2} a^{2}(1+\lambda)=\frac{a^{2}}{4}\left(4(1+\lambda)^{2}-2(1+\lambda)+1\right)=\frac{a^{2}}{4}\left(4 \lambda^{2}+6 \lambda+3\right) \rightarrow \\
& \rightarrow M F=\frac{a}{2} \sqrt{4 \lambda^{2}+6 \lambda+3}
\end{aligned}
$$

In Variant 1

$$
\lambda=2, a=\sqrt{31} \rightarrow|L|_{\min }=\left.\frac{a}{2} \sqrt{4 \lambda^{2}+6 \lambda+3}\right|_{\lambda=2, a=\sqrt{31}}=\frac{31}{2}=15.5
$$

## Variant 2