[ Rectangular parallelepipeds ] [ Unfolding helps solve the task ]

In a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$, it is known that $A B = A A 1 = 12$ and $A D = 30$. Point $M$ is located in the face $A B B 1 A 1$ at a distance of 1 from the midpoint of $A B$ and at equal distances from vertices $A$ and $B$. Point $N$ lies in the face $D C C 1 D 1$ and is symmetric to point $M$ with respect to the center of the parallelepiped. Find the length of the shortest path along the surface of the parallelepiped between points $M$ and $N$.
Suppose the path crosses edges A1B1 and C1D1 (or edges $A B$ and CD). In this case (Fig.2), the length of the shortest path is $11+30+1=42$. Suppose the path sequentially crosses edges BB1, B1C1, and C1D1. Consider the part of such an unfolding of the parallelepiped that contains rectangles A1B1C1D1 and BB1C1C with a common side B1C1, square CDD1C1, which has a common side D1C1 with rectangle A1B1C1D1, and square AA1B1B, which has a common side BB1 with rectangle BB1C1C. By the Pythagorean theorem, we find that in this case the shortest path is $\sqrt{37^{2}+17^{2}}=\sqrt{1658}$. If the path sequentially crosses edges $A B, B C, B 1 C 1$ and $C 1 D 1$, then consider the part of such an unfolding of the parallelepiped (Fig.3) that contains rectangles $A B C D$ and ВСС1В1 with a common side $B C$, rectangles A1B1C1D1 and BB1C1C with a common side B1C1, square CDD1C1, which has a common side C1D1 with rectangle A1B1C1D1, and square ABB1A1, which has a common side AB with rectangle $A B C D$. By the Pythagorean theorem, we find that in this case the shortest path is $\sqrt{24^{2}+32^{2}}=40$. Thus, the shortest of the considered paths is 40. Other possible paths are obviously longer.

## Answer

40.00