Let $M$ and $N$ be the midpoints of the edges $AC$ and $A1B1$ of a regular triangular prism $ABC A1B1C1$ with bases $ABC$ and $A1B1C1$ ($AA1 \parallel BB1 \parallel CC1$), all edges of which are equal to 2; $M1$ - the orthogonal projection of point $M$ onto the plane $A1B1C1$. Then $M1$ is the midpoint of $A1C1$, and $M1N$ is the midline of triangle $A1B1C1$. From the right triangle $MM1N$, we find that

$$
MN = \sqrt{MM1^2 + NM1^2} = \sqrt{4 + 1} = \sqrt{5}
$$

It is clear that the distances between the midpoints of any other non-parallel sides of the base are also equal to $\sqrt{5}$.

## Answer

$\sqrt{5}$
3 [Pythagorean Theorem in Space]

Find the distance between the midpoints of the non-parallel sides of different bases of a regular triangular prism, all edges of which are equal to 2.

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