Star Pyramids 2

Case 1: $n$ is an odd positive integer $\geq 5$ .

To find $A_{star}:$

$m\angle{AOP} = \dfrac{\pi}{n}$ and by Inscribed Angle Theorem $m\angle{EPF} = \dfrac{1}{2}m\angle{EOF} = \dfrac{1}{2}\dfrac{2\pi}{n} = \dfrac{\pi}{n} \implies m\angle{APO} = \dfrac{\pi}{2n}$

$\dfrac{AC}{OC} = \tan(\dfrac{\pi}{n}) \implies OC = \dfrac{AC}{\tan(\dfrac{\pi}{n})}$ and $PC = \dfrac{AC}{\tan(\dfrac{\pi}{2n})} = r - OC = r - \dfrac{AC}{\tan(\dfrac{\pi}{n})} \implies$

$AC = \dfrac{\tan(\dfrac{\pi}{2n})\tan(\dfrac{\pi}{n})}{\tan(\dfrac{\pi}{n}) + \tan(\dfrac{\pi}{2n})}r$

$\implies A_{\triangle{AOP}} = \dfrac{1}{2}(\dfrac{\tan(\dfrac{\pi}{2n})\tan(\dfrac{\pi}{n})}{\tan(\dfrac{\pi}{n}) + \tan(\dfrac{\pi}{2n})})r^2$

$\tan(\dfrac{\pi}{n}) = \dfrac{2\tan(\dfrac{\pi}{2n})}{1 - \tan^2(\dfrac{\pi}{2n})} \implies$ $A_{\triangle{AOP}} = \dfrac{\tan(\dfrac{\pi}{2n})}{3 - \tan^2(\dfrac{\pi}{2n})}r^2$

$\implies A_{star} = \dfrac{2n\tan(\dfrac{\pi}{2n})}{3 - \tan^2(\dfrac{\pi}{2n})}r^2$

Let $f = \dfrac{2n\tan(\dfrac{\pi}{2n})}{3 - \tan^2(\dfrac{\pi}{2n})}$ .

It will be shown that $\lambda$ (slant angle) is independent $f$ .

Using the above diagram:

$EG^2 = 2r^2(1 - \cos(\dfrac{4\pi}{n}) = 4\sin^2(\dfrac{2\pi}{n})r^2 \implies EG = 2\sin(\dfrac{2\pi}{n})r$ $\implies OR = \sqrt{r^2 - r^2\sin^2(\dfrac{2\pi}{n})} = r\cos(\dfrac{2\pi}{n})$ and $RQ = \sqrt{h^2 + r^2\cos^2(\dfrac{2\pi}{n})}$

$\implies$ The lateral surface area $A = n\sin(\dfrac{2\pi}{n})r\sqrt{h^2 + r^2\cos^2(\dfrac{2\pi}{n})}$

The volume $V = \dfrac{1}{3}fr^2h = k \implies k = \dfrac{3k}{fr^2} \implies$ $A(r) = n\sin(\dfrac{2\pi}{n})r\sqrt{9k^2 + f^2\cos^2(\dfrac{2\pi}{n})r^6} \implies$ $\dfrac{dA}{dr} = (n\sin(\dfrac{2\pi}{n}))\dfrac{2f^2\cos^2(\dfrac{2\pi}{n})r^6 - 9k^2}{r^2\sqrt{9k^2 + f^2\cos^2(\dfrac{2\pi}{n})r^6}} \implies r = (\dfrac{3k}{\sqrt{2}f\cos(\dfrac{2\pi}{n})})^{\frac{1}{3}}$ $\implies h = \dfrac{3k}{f}(\dfrac{\sqrt{2} f\cos(\dfrac{2\pi}{n})}{3k})^{\frac{2}{3}}$

Let $\lambda$ be the measure of the angle between the slant height and the base.

$\implies \tan(\lambda) = \dfrac{h}{OR} = \sqrt{2} \implies \lambda \approx \boxed{54.7356}$ .

Case 2: $n$ is an even positive integer $\geq 6$ .

$m\angle{AOB} = \dfrac{\pi}{n}$ and by Inscribed Angle Theorem $m\angle{QBT} = \dfrac{1}{2}m\angle{QOT} = \dfrac{1}{2}\dfrac{4\pi}{n} = \dfrac{2\pi}{n} \implies m\angle{ABO} = \dfrac{\pi}{n}$ $\implies \triangle{AOB}$ is an isosceles triangle $\implies BC = OC = \dfrac{r}{2} \implies$ $\tan(\dfrac{\pi}{n}) = \dfrac{2h^{*}}{r} \implies h^{*} = \dfrac{r\tan(\dfrac{\pi}{n})}{2} \implies$ $A_{\triangle{AOB}} = \dfrac{1}{4}\tan(\dfrac{\pi}{n})r^2 \implies A_{star} = \dfrac{n}{2}\tan(\dfrac{\pi}{n})r^2$

$QT^2 = 2r^2(1 - \cos(\dfrac{4\pi}{n})) = 4\sin^2(\dfrac{2\pi}{n})r^2 \implies QT = 2\sin(\dfrac{2\pi}{n})r \implies a = \cos(\dfrac{2\pi}{n})r$

$\implies PM = \sqrt{h^2 + \cos^2(\dfrac{2\pi}{n})r^2} \implies$

The surface area $A = n\sin(\dfrac{2\pi}{n})r\sqrt{h^2 + \cos^2(\dfrac{2\pi}{n})r^2}$

The volume $V = \dfrac{n}{6}\tan(\dfrac{\pi}{n})r^2h = k \implies h = \dfrac{6k}{n\tan(\dfrac{\pi}{n})r^2}$

Let $f^{*} = n\tan(\dfrac{\pi}{n})\cos(\dfrac{2\pi}{n})$

$\implies A(r) = 2\cos^2(\dfrac{\pi}{n})$ $\dfrac{\sqrt{36k^2 + {f^{*}}^{2} r^6}}{r} \implies$

$\dfrac{dA}{dr} =\dfrac{4\cos^2(\dfrac{\pi}{n})({f^{*}}^{2} r^6 - 18k^2)}{r^2\sqrt{36k^2 + {f^{*}}^2r^6}} = 0$

$\implies r = (\dfrac{3\sqrt{2}k}{f^{*}})^{\frac{1}{3}} \implies h = \dfrac{6k\cos(\dfrac{2\pi}{n})}{f^{*}}(\dfrac{f^{*}}{3\sqrt{2}k})^{\frac{2}{3}}$

Let $\theta$ be the measure of the angle between the slant height and the base.

$\tan(\theta) = \dfrac{h}{a} = \sqrt{2} \implies \lambda = \theta = \boxed{54.7356}$