X OM - III - Task 3

Given is a pyramid with a square base $ABCD$ and apex $S$. Find the shortest path that starts and ends at point $S$ and passes through all the vertices of the base.
Let $ a $ denote the length of the side of square $ ABCD $; the lengths of the lateral edges of the pyramid will be denoted by the letters $ k $, $ l $, $ m $, $ n $ in such a way that $ k \leq l \leq m \leq n $.
Each path starting and ending at point $ S $ and passing through all the vertices of the square can be symbolically written as

where $ W_1, W_2, W_3, W_4 $ represent the vertices $ A $, $ B $, $ C $, $ D $ taken in some order. The part of the path from $ S $ to $ W_1 $ cannot be shorter than the edge $ SW_1 $, and the part of the path from $ W_4 $ to $ S $ cannot be shorter than the edge $ SW_4 $. These two parts of the path together cannot be shorter than the sum of the two shortest lateral edges, i.e., than $ k + l $. The part of the path from $ W_1 $ to $ W_2 $ cannot be shorter than the side of the square, i.e., than $ a $; the same applies to the path from $ W_2 $ to $ W_3 $ and from $ W_3 $ to $ W_4 $. The entire path (1) cannot, therefore, be shorter than

If we can find a path whose length is exactly $ k + l + 3a $, it will be the shortest path. To this end, we will determine the two shortest lateral edges of the pyramid.
Draw 4 axes of symmetry of the square $ ABCD $ (Fig. 31). They divide the plane of the square into 8 angular regions $ \textrm{I}-\textrm{VIII} $. The projection $ O $ of the vertex $ S $ of the pyramid onto the plane of the square lies inside or on the boundary of one of these regions, say region $ \textrm{I} $. Based on the properties of the perpendicular bisector of a segment, the following inequalities then hold

Thus, $ OA $ and $ OB $ are the two shortest of the segments $ OA $, $ OB $, $ OC $, $ OD $. In this case, $ SA $ and $ SB $ are the two shortest lateral edges of the pyramid (the shorter projection corresponds to the shorter slant), namely $ SA = k $, $ SB = l $, from which it follows that the path $ SADCBS $, whose length $ SA + AD + DC + CB + BS = k + l + 3a $ is the sought shortest path.
When the point $ O $ lies on the ray $ MA $ but does not coincide with the center $ M $ of the square, there are two shortest paths $ SADCBS $ and $ SABCDS $. When the point $ O $ coincides with the point $ M $, there are four such paths. Each path can, of course, be traversed in two directions.