If I have a parallelogram $P$ symmetric around the origin, and a vector $v$, such that $(P+v)\cap (P-v)$ is not empty, is there a simple way to obtain the parallelogram $Q\subset (P+v) \cup (P-v)$, symmetric around the origin, with the biggest area?


It seems that if $v$ is very small it is better to take the sides of $Q$ parallel to the sides of $P$, and if $v$ is large, it might be better for one of the sides of $Q$ to be parallel to $v$. I'm interested in the generalization to parallelotopes.


Thanks!



Biggest parallelogram inside the union of two translated parallelograms
We may assume that the two parallelograms are squares (e.g. applying a suitable linear transformation, or, what is the same, choosing an Euclidean structure in the plane, that induce that measure, and for which the parallelograms are squares). At least, this simplifies the notation and reduces the number of data.


The initial problem may be restated as: what is the maximum area parallelogram among those included in the unit square $[0,1]\times [0,1]$, and disjoint from the rectangles $(0,a)\times(0,b)$ and $(1-a,1)\times(1-b,1)$? By the original assumptions, here $0<a\le b<1/2$.


The existence being clear, simple arguments by elementary variations tells us that any maximizing parallelogram 


* has all its vertices on the boundary of the unit squares,
* has the points $(a,b)$ and $(1-a, 1-b)$ in its boundary,


as in Joseph O'Rourke's picture (for the former fact, consider variations of the parallelograms that keep fixed one edge and move the opposite one parallel-wise. For the latter, consider e.g. variations that rotate the parallelogram keeping its vertices on the boundary of the unit square).


The above consideration reduce the search of the maximizer to a one-parameter family corresponding to a $90$ degrees rotation of the parallelogram, and subdivided into three subintervals (corresponding to the distribution of the vertices of the parallelogram in the edges of the square). In the middle sub-interval (parallelogram in slant position, of whom $(x,0)$ is a vertex, for $x$ in the interval $a< a/(1-b)\le x\le1$), the area is a convex function of the form $Ax+B/(x-a)+C$, and we conclude that the maximizing parallelogram


* has an edge included in one edge of the unit square,


which reduces the choice between the two parallelograms with vertices in $(0,1)$ and $(1,0)$; since $a\le b$, the maximizer is the one with horizontal base.


