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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!

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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.

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  • $\begingroup$ Nice analysis, but the conclusion ("an edge included in one edge of the unit square") seems contradicted by the example in my drawing: the slanted parallelogram has larger area than the vertical or horizontal parallelograms. $\endgroup$
    – Joseph O'Rourke
    Commented Apr 18, 2015 at 11:46
  • $\begingroup$ Even larger than the area of the parallelogram \ \ with the lower horizontal edge moved to the right (therefore, not the green one // ) ? $\endgroup$
    – Pietro Majer
    Commented Apr 18, 2015 at 13:59
  • $\begingroup$ Mea culpa. I see your point. Those two parallelograms (slanted and lower right) have nearly the same area. $\endgroup$
    – Joseph O'Rourke
    Commented Apr 18, 2015 at 14:47
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    $\begingroup$ I guess the only thing that remains to be answered is if this argument generalizes to more dimensions, i.e. is it true that the maximizing parallelotope will share a normal with the original. If I was only concerned with the 2-dimensional question I would considered myself answered. Is the right thing to accept the answer and open a second question? $\endgroup$
    – funda
    Commented Apr 18, 2015 at 15:36
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    $\begingroup$ @funda: That does make sense to me: Accept and re-ask. $\endgroup$
    – Joseph O'Rourke
    Commented Apr 18, 2015 at 15:59
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Not an answer; just an illustration:

          Parallelograms
It is clear the maximum area is determined by four variables, e.g., $\{a, b, x, y \}$, but I don't see an easy way to find this maximum short of computing the pointwise maximum of the areas of the various combinatorially different candidate parallelograms.

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    $\begingroup$ Your illustration suggests to me a simple linear program which probably can be made simpler. Consider the intersection U of the two parallelograms, and look at the point of U closest to the letter P in your diagram; call this point R. A line drawn through R at various angles will form the side of an inscribed centrally symmetric parallelogram, and you can compare its area with one of the two answers with sides parallel to the original parallelograms: it will always be larger and will increase until it meets a corner. Gerhard "Stay Tuned For Part II" Paseman, 2015.04.17 $\endgroup$
    – Gerhard Paseman
    Commented Apr 17, 2015 at 16:25
  • $\begingroup$ It convinces me that a maximal solution will be a parallelogram with no sides parallel to the original, and that it is a matter of computing the area when the line through R induces such a parallelogram. This area should vary as a simple quadratic in the x-intercept of the line through R, whose maximum is easily determined. Gerhard "Still Working On The Details" Paseman, 2015.04.17 $\endgroup$
    – Gerhard Paseman
    Commented Apr 17, 2015 at 16:31
  • $\begingroup$ Nice observations. So, fixing $a$, $b$, and $v$, the potential parallelogram solutions form a one-parameter family. $\endgroup$
    – Joseph O'Rourke
    Commented Apr 17, 2015 at 17:12
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    $\begingroup$ Since linear maps preserves ratios of measures, it may be assumed wlog that the two parallelograms are squares. This simplifies the problem... $\endgroup$
    – Pietro Majer
    Commented Apr 17, 2015 at 17:16

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