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This is follow-on from Minimum area of Inscribed Square

If I have square S with perimeter 40, i.e. each side 10, and I have inscribed square T, what is the Maximum area of T?

How do I even go about solving this?

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  • $\begingroup$ $S$ is inscribed in itself, so $100$. If $T=S$ is unacceptable, we can get close, but there is no max. $\endgroup$ – André Nicolas Jan 23 '16 at 19:48
  • $\begingroup$ @AndréNicolas Based on k-jiang answer, the maximum area would be approaching 100 $\endgroup$ – Rhonda Jan 23 '16 at 19:59
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    $\begingroup$ As the comment above says, if we consider $S$ to be a square inscribed in itself, the answer is exactly $100$. If we do not allow that, there is no max, though the supremum is $100$. $\endgroup$ – André Nicolas Jan 23 '16 at 20:17
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If a square is inscribed in this larger square, all of its vertices must lie on the vertices of the larger square. This problem is trivial - just make the squares the same size by placing the vertices of the inscribed square atop the vertices of the original square. The answer is $10^{2} = \boxed{100}.$

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