Of all parallelograms of a given area find the one with the shortest possible longer diagonal.
Let one of the longer sides be the base and be called L, and the height h.
If the parallelogram is not already a rectangle, slide the top edge in the direction of making it a rectangle. The longer side will get shorter all the way, at which point (when a rectange is formed) the "longer" side's length will match the "shorter" side, as the shorter side was getting larger through the process. If the sliding were to continue past making a rectangle, what had been the shorter side would become the longer side, and be longer than when in the rectangular state.
Now, if the parallelogram, now a rectangle, is not a square (L strictly greater than h), increase h while decreasing L, while maintaining L*h. The diagonals will decrease all the way to the square state, when L = h.
The shortest possible diagonal is that of a square: sqrt(L*h) aka sqrt(A), where A is the area.
If one insists that the parallelogram not be a rectangle, then there is no one shortest possible. Just approach the square configuration as a limit.
Posted by Charlie
on 2021-04-28 08:58:52