Suppose I draw multiple rectangles, each can have different length and breadth but their areas are same. Also, there is one constraint that each rectangle should follow, the center of each rectangle should be the origin in the x-y coordinate plane.

Next, I have to pick any one given rectangle out of the multiple drawn rectangles and find its overlapping area with the other rectangles. How should I prove that the overlapping area will be maximum if the picked rectangle is a square

  • $\begingroup$ Perhaps it will have something to do with this: some rectangles will be taller than they are wide and others wider than they are tall. Rectangles that are not squares will overlap more with one type or the other, but the square will overlap equally with both. $\endgroup$ – Franklin Pezzuti Dyer May 29 '17 at 17:32
  • $\begingroup$ If your collection of "drawn" rectangles is given, then it need not be the case that any of them is a square. $\endgroup$ – hardmath May 29 '17 at 17:32
  • $\begingroup$ @Frpzzd - yep, its intuitive that way but I'm not sure how can I prove it mathematically. $\endgroup$ – Ulkurz May 29 '17 at 17:39
  • $\begingroup$ @hardmath - lets assume we can draw rectangles with equal length and breadth. :) $\endgroup$ – Ulkurz May 29 '17 at 17:39
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    $\begingroup$ I am wondering if your collection of rectangles is not a finite collection but the set of all possible rectangles centered at the origin with some fixed area. The Question doesn't suggest that interpretation, and indeed your drawing "multiple rectangles" suggests a finite number are in the collection. I'm also not clear on the objective to maximize "the overlapping area". While the overlapping area between two rectangles is unambiguous, I'd be at loss to define the area of overlap between one rectangle and many rectangles. Finally, are rotated rectangles allowed? $\endgroup$ – hardmath May 29 '17 at 17:48

Your claim is false under natural interpretations of it, hence you will not be able to prove it (at least until you clarify it).

Here is how I interpret it:

Given $n$ origin-centered rectangles $R_1, R_2, \ldots, R_n$, all of the same area, and given any index $i \leq n$, define $$f(i) = \sum_{j \neq i}Area(R_i \cap R_j).$$

The goal is to find an $i$ which maximizes $f$. Such an $i$ clearly exists since it is the maximum of a finite set. You claim that it is maximized at an index $i$ with $R_i$ a square. This is clearly false in general.

First of all, there is nothing in the assumptions that guarantees that any of the rectangles will be a square. If none of the rectangles are squares then obviously it can't be maximized at a square. Furthermore, even if you add the constraint that at least one of the rectangles is a square, there is no reason to think that f will be maximized at such a square.

As a counter-example: if $R_1$ is the axis-aligned square of area 1, $R_2$ is the $10 \times (1/10)$ axis-aligned rectangle with long side parallel to the $x$-axis, and $R_3$ is the $20 \times (1/20)$ axis-aligned rectangle, with the long side again parallel to the $x$-axis, then it is easy to see that $f(1) = \frac{1}{10} + \frac{1}{20} = \frac{3}{20}$, but $f(2) = \frac{1}{10} + \frac{1}{2} = \frac{6}{10}$. Thus $f(2) > f(1)$, so $f$ isn't maximized at the square.

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    $\begingroup$ +1 An even easier counterexample is a collection of several identical (or nearly identical) long thin rectangles and one square. $\endgroup$ – Ethan Bolker May 30 '17 at 19:28

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