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On the table lies a rectangular piece of paper $ABCD$ of area $100$. Cosmo folds the rectangle once along a straight line, so that afterwards corner $C$ lies exactly on top of corner $A$. The folded piece of paper forms a pentagon.

Question: What is the largest possible area of this pentagon?

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There is no largest area, but the area can be made as close as wanted to 75.

  1. Let's imagine our rectangle. It has diagonal $AC$, centre $O$; let's say $AB$ is the longest side and $H$ is the middle of $AB$.
  2. The rectangle will be folded along a line $p$, which is perpendicular to $AC$ at point $O$. $p$ also crosses $AB$ at some point $M$.
  3. It is easy to note that when we fold the overlapped area will be of size $2*Size(AOM)$. The pentagon area is $Sp = 100-2*Size(AOM)$. In order to achieve biggest pentagon we need to make overlap as small as possible.
  4. But, $AOM$ always includes $AOH$ and $Size(AOH) = \frac{100}{8}$. Thereby $Sp = 100-2*Size(AOM) < 100-\frac{100}{4} = 75$.
  5. Finally, let's note that making $\frac{AB}{AD}$ ratio big enough we can make point $M$ as close as needed to point $H$ making $Size(AOM)$ as close as needed to $\frac{100}{8}$ and thereby $Sp$ to 75.
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  • $\begingroup$ Can someone make a diagram for this? I'm having trouble following. It seems obvious to me that a trivial max area is 100 and, based on your text, the largest area the pentagon could have would be arbitrarily close to 75 with that acting more as a limit than an actual (achievable) max, right? $\endgroup$ – Engineer Toast Nov 13 '15 at 13:47
  • $\begingroup$ @EngineerToast, yes, the largest area the pentagon could have would be arbitrarily close to 75 with that acting more as a limit than an actual (achievable) max $\endgroup$ – klm123 Nov 13 '15 at 13:55

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