Comparing 3

We just need the top pyramid of the bi-pyramid, then double the angles in the final result.

For area of $n - gon$ :

Let $BC = x$ be a side of the $n - gon$ , $AC = AB= r$ , $AD = h^*$ , and $\angle{BAD} = \dfrac{180}{n}$ .

$\dfrac{x}{2} = r \sin(\dfrac{180}{n}) \implies r = \dfrac{x}{2 \sin(\dfrac{180}{n})} \implies h^* = \dfrac{x}{2} \cot(\dfrac{180}{n}) \implies A_{\triangle{ABC}} = \dfrac{1}{4} \cot(\dfrac{180}{n}) x^2 \implies$

$A_{n - gon} = \dfrac{n}{4} \cot(\dfrac{180}{n}) x^2 \implies$ the Volume of the pyramid $V_{p} = \dfrac{n}{12} \cot(\dfrac{180}{n}) x^2 H$

The lateral surface area $P^{*} = \dfrac{nx}{2} * s,$ where $s$ is the slant height.

$V_{p} = K$ (constant) $\implies H = \dfrac{12 K}{n * cot(\dfrac{180}{n}) x^2} \implies$ $P^{*}(x) = \dfrac{1}{2 m(n) x} \sqrt{(j(n) * m(n))^2 x^6 + (24 K)^2}$ , where $m(n) = \cot(\dfrac{180}{n})$ and $j(n) = n * m(n)$

$\implies \dfrac{dP^{*}}{dx} = \dfrac{2 (j(n) * m(n))^2 x^6 - (24 K)^2}{2 m(n) \sqrt{(j(n) * m(n))^2 x^6 + (24 K)^2} * x^2} = 0$ $\implies x = (\dfrac{(24 K)^2}{2 (j(n)* m(n))^2})^{\dfrac{1}{6}}$

$\implies H = \dfrac{12 K}{(j(n)) *( \dfrac{(24 K)^2}{2 (j(n) * m(n))^2})^{\dfrac{1}{3}}}$

$\implies \tan(\dfrac{\theta}{2}) = \dfrac{H}{h^{*}} = \sqrt{2} \implies \theta = 2\arctan(\sqrt{2}) \approx 109.47122$ and $\tan(\dfrac{\gamma_{n}}{2}) = \dfrac{H}{r} = \sqrt{2} \cos(\dfrac{180}{n}) \implies \tan(\dfrac{\gamma}{2}) = \lim_{n \rightarrow \infty} \tan(\dfrac{\gamma_{n}}{2}) = \sqrt{2} * \lim_{n \rightarrow \infty} \cos(\pi/n) = \sqrt{2}$ $\implies \gamma = 2\arctan(\sqrt{2}) \approx 109.47122$ .

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Let $x(t) = at^2 + b,\:\ y(t) = ct^3 + d \implies \dfrac{dy}{dx}|(t = t_{1}) = \dfrac{3c}{2a}t_{1} \implies$ the tangent line to the curve at $(x(t_{1}),y(t_{1}))$ is: $y - (ct_{1}^3 + d) = \dfrac{3c}{2a}t_{1}(x - (at_{1}^2+ b))$

Let the line be normal to the curve at $(x(t_{2}),y(t_{2})) \implies c(t_{2} - t_{1})(t_{2}^2 + t_{1}t_{2} + t_{1}^2) = \dfrac{3c}{2}t_{1}(t_{2} - t_{1})(t_{2} + t_{1}) \implies \dfrac{c}{2}(t_{2} - t_{1})(2t_{2}^2 - t_{1}t_{2} - t_{1}^2) = 0$ $t_{1} \neq t_{2} \implies t_{2} = -\dfrac{t_{1}}{2}$

Since the tangent is also normal to the curve at $(x(t_{2}),y(t_{2})) \implies \dfrac{9c^2}{4a^2}t_{1}t_{2} = -1$ $\implies \dfrac{9c^2}{8a^2}t_{1}^2 = 1 \implies t_{1} = \pm\dfrac{2\sqrt{2}a}{3c} \implies$ the two slopes are $\pm\sqrt{2}$ .

$\tan(\dfrac{\lambda}{2}) = \sqrt{2} \implies \lambda = 2\arctan(\sqrt{2}) \approx \boxed{109.47122}$ .

$\implies \dfrac{\lambda \gamma}{\theta^2} = \dfrac{\theta^2}{\theta^2} = \boxed{1}$ .

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