Pentagon inside the Square

Let one of the vertices $P$ of pentagon be at the distance of $a$ from one vertex $A$ of the square as shown in the figure. Now the two sides of pentagon from $P$ is at angle $\theta$ and $72\degree - \theta$ as shown. Let the side length of pentagon be $s$ .

From both the lower triangles we get

$s\text{ cos }\theta = 1 - a\hspace{50pt}$ and $\hspace{50pt} s\text{ cos}(72 - \theta) = a$

Combining both the equation,

$s \text{ cos }\theta + s\text{ cos}(72 - \theta) = 1\newline \Rightarrow s = \Large\frac{1}{\text{ cos }\theta\, + \text{ cos}(72 - \theta)} = \frac{1}{2\text{ cos }36\text{ cos}(36 - \theta)}$

Two sides $PQ$ and $PT$ are drawn. Now to construct the next two sides $QR$ and $TS$ within the boundary $BC$ and $AD$ respectively, two red marked angles in the figure should be greater than or equal to $108\degree$ .

$90 + \theta \geq 108\hspace{50pt}$ and $\hspace{50pt} 162 - \theta \geq 108$

$\Rightarrow 18\leq \theta \leq 54$

Now conditions are made to ensure that $QR$ and $TS$ lies within the boundary $CD$ .

$BQ + QL \leq BC \hspace{50pt}$ and $\hspace{50pt}AT + TM \leq AD\newline \Rightarrow s\text{ sin }\theta + s\text{ cos}(\theta - 18) \leq 1 \hspace{50pt}$ and $\hspace{50pt}s\text{ sin}(72 - \theta) + s\text{ cos}(54 - \theta)\leq 1$

We see that on replacing $\theta$ by $72 - \theta$ in one inequation we will get the other. This implies that on solving one inequation we can the solution of other inequation by just replacing $\theta$ by $72 - \theta$ . Now solving for 1st inequation by putting the value of $s$ in terms of $\theta$ .

$\Large\frac{\text{cos}(\theta - 18)\, + \text{ sin }\theta}{2\text{ cos }36\text{ cos}(36 - \theta)} = \frac{\text{cos}(\theta - 18)\, + \text{ cos}(90-\theta)}{2\text{ cos }36\text{ cos}(36 - \theta)} = \frac{2\text{ cos }36\,\text{ cos}(54-\theta)}{2\text{ cos }36\text{ cos}(36 - \theta)}$ $\,\leq 1$

$\Rightarrow \Large \frac{\text{cos}(54-\theta)}{\text{cos}(36-\theta)}$ $\,\leq 1 \newline\Rightarrow \text{cos}(54 - \theta) \leq \text{cos}(36 - \theta)\hspace{20pt}[\leq$ sign does not change because $\text{cos}(36-\theta)$ is positive for $18\leq\theta\leq54 ]\newline \Rightarrow \text{cos}(36-\theta) - \text{cos}(54 - \theta) \geq 0\newline \Rightarrow 2\,\text{sin}(45 - \theta)\,\text{sin }18 \geq 0\newline \Rightarrow \text{sin}(45 - \theta)\geq 0\newline \Rightarrow \theta \leq 45\degree$

Now for solving the other equation, $\theta$ is replaced by $72 - \theta$ in the final answer.

$72\degree - \theta \leq 45\degree\Rightarrow \theta \geq 27\degree$

Hence the final valid range of $\theta$ for constructing the regular pentagon inside the square is $27\degree\leq \theta\leq 45\degree$

Now, in order to maximize $s$ we have to minimize $\text{cos}(36-\theta)$ over this range of $\theta$ .

$27\degree\leq \theta\leq 45\degree\newline \Rightarrow -9\degree \leq 36\degree-\theta \leq 9\degree\newline\Rightarrow \text{cos }9\degree \leq \text{ cos}(36\degree-\theta)\leq1$

The minimum value of $\text{ cos}(36\degree-\theta)$ is $\text{cos }9\degree$ which occurs either at $\theta = 27\degree$ or $\theta = 45\degree$

So $s_{max} = \Large \frac{1}{2\text{ cos }36\degree\text{ cos }9\degree}$ $\approx 0.6257379$

The length of each side of the regular pentagon is $\dfrac{1}{2\cos 36\degree}=\dfrac{\sqrt 5-1}{2}\approx \boxed {0.618}$ .