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A rectangle is inscribed within another rectangle of dimensions 6 units by 8 units. The rectangle has been inscribed in it with its vertices on the sides of the rectangle in such a way that if the smaller rectangle is rotated slightly about the centre it still is confined within the boundary of the larger rectangle. How can I find the minimum possible perimeter of such a rectangle? I am interested in an intuitive proof by simple geometry and not by coordinate geometry.

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    $\begingroup$ You mean "maximum" instead of "minimum"? $\endgroup$ – QuIcKmAtHs Aug 10 '18 at 9:31
  • $\begingroup$ nope minimum perimeter only $\endgroup$ – saisanjeev Aug 10 '18 at 9:35
  • $\begingroup$ that way, we can have a infinitesimally small rectangle! $\endgroup$ – QuIcKmAtHs Aug 10 '18 at 9:37
  • $\begingroup$ By inscribed, do you mean that all four vertices are on the boundary of the given rectangle? $\endgroup$ – Jaap Scherphuis Aug 10 '18 at 9:37
  • $\begingroup$ @JaapScherphuis yes I edited the question accordingly $\endgroup$ – saisanjeev Aug 10 '18 at 9:38
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Intuitively, I think you are asking for the smallest circle that will fit into the 6x8 rectangle, and still touch all 4 sides. Your rectangle is then the corresponding touch points that actually form a rectangle. Again, intuitively (without calculation) this seems to be the minimum perimeter. I think the calculation might be provided by Jaap in a comment. enter image description here

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