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.
[Math] Find the area of the largest hexagon that can be inscribed in a unit square
areageometryoptimization
Related Solutions
The vertices of any rectangle inscribed in an ellipse is given by $$(\pm a \cos(\theta), \pm b \sin(\theta))$$ The area of the rectangle is given by $$A(\theta) = 4ab \cos(\theta) \sin(\theta) = 2ab \sin(2 \theta)$$ Hence, the maximum is when $\sin(2 \theta) = 1$. Hence, the maximum area is when $2\theta = \dfrac{\pi}2$ i.e. $\theta = \dfrac{\pi}4$. The maximum area is $$A = 2ab$$
In the attached image, note that: $$\angle EBI = 180^\circ - \angle ABG - 60^\circ$$ Moreover, to minimize the square you want point $E$ to be on the square, so that $$2a\sin (\angle EBI)= x$$ Where $x$ is the square side length. The same logic works for point $D$, so that: $$2a\sin (\angle DAK)= x$$ Leading to: $$\angle EBI = \angle DAK \ \to \ \angle ABG = \angle GAB \ \ \blacksquare$$
Note that in the proof I did not assume a particular orientation of the hexagon, so that this also covers your second question - i.e. why can't the hexagon be parallel to the square.
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Best Answer
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 $${A(H)\over A(Q)}\leq {3\sqrt{3}\over2}\cdot{1\over4\cos^2{\pi\over12}}=3\sqrt{3}-{9\over2}\doteq 0.6962\ .$$