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How so I find the area of the largest regular hexagon that can be inscribed in a unit square? I am using this http://www.drking.org.uk/hexagons/misc/deriv4.html to figure it out but not sure if it is correct or not.

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  • $\begingroup$ Looks like that page refers to other cases already treated, for some of the steps. $\endgroup$ – coffeemath Dec 16 '16 at 16:36
  • $\begingroup$ What are you asking? How to compute the area of the hexagon in the image? Or how to deduce that the hexagon in the picture really is the largest possible? $\endgroup$ – Karolis Juodelė Dec 16 '16 at 16:37
  • $\begingroup$ @KarolisJuodelė How do I deduce that it is really the largest? If possible, can you tell me what would be the area of the largest hexagon. $\endgroup$ – ImVikash_0_0 Dec 16 '16 at 16:40
  • $\begingroup$ If $ABCD$ is a square, $ABBCDD$ is a degenerate hexagon with the same area as $ABCD$. $\endgroup$ – Jack D'Aurizio Dec 16 '16 at 16:40
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    $\begingroup$ Maybe you are looking for the largest inscribed regular hexagon? $\endgroup$ – Jack D'Aurizio Dec 16 '16 at 16:41
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Consider the regular hexagon $H$ with vertices $(\pm1,0)$, $\left(\pm{1\over2},\pm{\sqrt{3}\over2}\right)$ and area $A(H)={3\sqrt{3}\over2}$. We have to determine the smallest circumscribed square. Due to symmetry it is sufficient to consider supporting lines tilted by an angle $\phi\in\bigl[0,{\pi\over6}\bigr]$ counterclockwise with respect to the vertical, and supporting lines orthogonal to these. The almost vertical lines hit $H$ at $\pm(1,0)$, and their orthogonals hit $H$ at $\pm\left(-{1\over2},-{\sqrt{3}\over2}\right)$. The side-length $s$ of the resulting square $Q$ is then given by $$s=2\max\left\{\cos\phi, \cos\left({\pi\over6}-\phi\right)\right\}\qquad\left(0\leq\phi\leq{\pi\over6}\right)\ ,$$ and is minimal when $\phi={\pi\over12}$. It follows that $${A(H)\over A(Q)}\leq {3\sqrt{3}\over2}\cdot{1\over4\cos^2{\pi\over12}}=3\sqrt{3}-{9\over2}\doteq 0.6962\ .$$

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