[ Orthogonal pyramid edges, angle between them] [ Law of Cosines $\quad$]

In a regular triangular pyramid SKLM with vertex S, the median MP is drawn in triangle SLM. It is known that $K L=1$ and $S K=3$. Through the midpoint $N$ of edge SM, a line $N E$ is drawn parallel to $K L$. Through point $K$, a line is drawn intersecting lines MP and NE at points A and B, respectively. Find AB.
The straight line NE passes through point N and is parallel to the straight line KL. Therefore, this line lies in the plane KLN. Point K and the point of intersection of the lines MP and LN belong simultaneously to the planes KNL and KPM. Therefore, they intersect along a line $l$, passing through these two points. This line intersects the line $NE$, lying in the plane KNL, and the line MP. Since through a given point, one can draw no more than one line intersecting two given skew lines not passing through this point, the line mentioned in the problem condition is the line $l$. Then $A$ is the point of intersection of this line with $PM$, and $B$ is the point of intersection with the line $NE$. From the isosceles triangle KSL, using the formula for the median, we find that

$$
K P=\frac{1}{2} \sqrt{2 S K^{2}+2 K L^{2}-S L^{2}}=\frac{1}{2} \sqrt{2 \cdot 9+2 \cdot 1-9}=\frac{\sqrt{11}}{2}
$$

Similarly, $L N=\frac{\sqrt{11}}{2}$. From the isosceles triangle KNL, we find that

$$
\cos \angle K L N=\frac{\frac{1}{2} K L}{L N}=\frac{\frac{1}{\sqrt{11}}}{2}=\frac{1}{\sqrt{11}} .
$$

Since $A$ is the point of intersection of the medians of triangle LSM,

$$
A L=\frac{2}{3} L N=\frac{2}{3} \cdot \frac{\sqrt{11}}{2}=\frac{\sqrt{11}}{3}
$$

By the cosine theorem from triangle AKL, we find that

$$
\begin{gathered}
A K=\sqrt{K L^{2}+A L^{2}-2 K L \cdot A L \cos \angle K L N}= \\
=\sqrt{1+\frac{11}{9}-2 \cdot 1 \cdot \frac{\sqrt{11}}{3} \cdot \frac{1}{\sqrt{11}}}=\frac{\sqrt{14}}{3}
\end{gathered}
$$

Finally, from the similarity of triangles BAN and KAL, we find that

$$
A B=\frac{A K}{A L} \cdot A L=\frac{1}{2} A K=\frac{\sqrt{14}}{6}
$$

## Answer

$\sqrt{14}$