8. In coordinate space, find the length of the shortest path between the points $(0 ; 1 ; 2)$ and $(22 ; 4 ; 2)$ along the surface of a rectangular parallelepiped bounded by the planes $x=22$, $y=5$, $z=4$ and the three coordinate planes.
Answer: $\sqrt{657}$.

Solution. Due to the symmetry of the configuration relative to the plane $\pi(x ; y ; 2)$ (passing through points $K$ and $L$ parallel to the base $A B C D$), we can assume that the shortest path does not go above the plane $\pi$ and thus does not have any common points with the upper base $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. Next, if the path does not pass through the interior points of the base $A B C D$, then its length is no less than the length of its projection onto the base plane, which is half the perimeter of $A B C D$, equal to $(22+5)=27$. This value is achieved on the horizontal path $K L$, lying in the plane $\pi$. It remains to consider the cases where the path passes through the face $A B C D$. Since the shortest path should not enter the same face twice, there are three possible unfoldings:

![](https://cdn.mathpix.com/cropped/2024_05_06_f47d17e1867b5208014dg-6.jpg?height=285&width=1151&top_left_y=840&top_left_x=404)

In this case, we have $K L=\sqrt{(2+22+2)^{2}+(5-1-1)^{2}}=\sqrt{26^{2}+3^{2}}=\sqrt{685}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f47d17e1867b5208014dg-6.jpg?height=425&width=1202&top_left_y=1255&top_left_x=381)

In this case, we have $K L=\sqrt{(2+22+1)^{2}+(4+2)^{2}}=\sqrt{25^{2}+6^{2}}=\sqrt{661}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f47d17e1867b5208014dg-6.jpg?height=559&width=1222&top_left_y=1805&top_left_x=363)

In this case, we have $K L=\sqrt{(1+22+1)^{2}+(2+5+2)^{2}}=\sqrt{24^{2}+9^{2}}=\sqrt{657}$.

Since $27>\sqrt{685}>\sqrt{661}>\sqrt{657}$, we get the answer $\sqrt{657}$.