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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ Maybe you are looking for the largest inscribed regular hexagon? $\endgroup$ – Jack D'Aurizio Dec 16 '16 at 16:41

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 $${A(H)\over A(Q)}\leq {3\sqrt{3}\over2}\cdot{1\over4\cos^2{\pi\over12}}=3\sqrt{3}-{9\over2}\doteq 0.6962\ .$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.