Oblique Prisms

$\dfrac{x}{2} = r \sin(\dfrac{\pi}{4n}) \implies r = \dfrac{x}{2 \sin(\dfrac{\pi}{4n})} \implies h^* = \dfrac{x}{2} \cot(\dfrac{\pi}{4n}) \implies$ the area of the $4n$ -gon $A = nx^2\cot(\dfrac{\pi}{4n})$ .

The volume $V = nx^2\cot(\dfrac{\pi}{4n}) H = 1 \implies H = \dfrac{\tan(\dfrac{\pi}{4n})}{nx^2}$

The lateral surface area $S = 4nxs = 1 \implies s = \dfrac{1}{4nx} \implies Y(r) = y^2(r) = s^2 - H^2 = \dfrac{1}{n^2}(\dfrac{1}{16 x^2} - \dfrac{\tan^2(\dfrac{\pi}{4n})}{x^4}) \implies$

$\dfrac{dY}{dr} = \dfrac{32\tan^2(\dfrac{\pi}{4n}) - x^2}{8 n^2 x^5} = 0 \implies$ $x = 4\sqrt{2}\tan(\dfrac{\pi}{4n}) \implies r = 2\sqrt{2}\sec(\dfrac{\pi}{4n}),$ $s = \dfrac{\cot(\dfrac{\pi}{4n})}{16\sqrt{2} n}$ , $H = \dfrac{\cot(\dfrac{\pi}{4n})}{32 n}$ and $y = \sqrt{s^2 - H^2} = \dfrac{\cot(\dfrac{\pi}{4n})}{32 n} = H$ .

Note: $\dfrac{d^2Y}{dx^2}|_{x = 4\sqrt{2}\tan(\dfrac{\pi}{4n})} =$ $(\dfrac{1}{8n^2(4\sqrt{2}\tan(\dfrac{\pi}{4n}))^4})(3 - 5\cot(\dfrac{\pi}{4n})) < 0,$ since $\cot(\dfrac{\pi}{4n}) \geq 1$ and is increasing for increasing $n \implies$ we have a max at $x =4\sqrt{2}\tan(\dfrac{\pi}{4n})$

Let $m\angle{UWV} = \lambda \implies \cos(\lambda) = \dfrac{y}{s} = \dfrac{1}{\sqrt{2}}$ and $\theta = 180 - \lambda \implies \cos(\theta) = \cos(180 - \lambda) = -\cos(\lambda) = \dfrac{-1}{\sqrt{2}} \implies \boxed{\theta = 135^{\circ}}$

$r = 2\sqrt{2}\sec(\dfrac{\pi}{4n})$ and $s = \dfrac{\cot(\dfrac{\pi}{4n})}{16\sqrt{2} n} \implies$

$d^2(n) = s^2 + 4r^2 + 4rs\cos(\lambda) = \dfrac{\cos^2(\dfrac{\pi}{4n})}{2^9(n\sin(\dfrac{\pi}{4n}))^2}$ + $\dfrac{2^5}{\cos^2(\dfrac{\pi}{4n})}$ + $\dfrac{1}{2\sqrt{2}(n\sin(\dfrac{\pi}{4n}))} \implies$

$d_{n} = \sqrt{\dfrac{\cos^2(\dfrac{\pi}{4n})}{2^9(n\sin(\dfrac{\pi}{4n}))^2} + \dfrac{2^5}{\cos^2(\dfrac{\pi}{4n})} + \dfrac{1}{2\sqrt{2}(n\sin(\dfrac{\pi}{4n}))}}$

Using the inequality $\cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \dfrac{\pi}{4}\cos(\dfrac{\pi}{4n}) < n\sin(\dfrac{\pi}{4n}) < \dfrac{\pi}{4} \implies$

$\lim_{n \rightarrow \infty} d_{n} = \sqrt{\dfrac{1}{32\pi^2} + 32 + \dfrac{\sqrt{2}}{\pi}} =$ $\dfrac{\sqrt{32^2\pi^2 + 32\sqrt{2}\pi + 1}}{4\sqrt{2}\pi} \approx \boxed{5.69678}$

For $n = 1$ $d_{1} = \dfrac{1}{16}\sqrt{\dfrac{2^{15} + 2^8 + 1}{2}} \approx \boxed{8.03131}$

$\therefore \lfloor{d + \theta + d_{1}}\rfloor = \boxed{148}$ .

$$

Checking using oblique cylinder:

In the above Oblique Cylinder $m\angle{QPT} = \theta$ , $TPQU$ is a parallelogram and $\triangle{PRT}$ is a right triangle.

$V = \pi r^2 H = 1 \implies H = \dfrac{1}{\pi r^2}$ and $A = 2\pi rs = 1 \implies s = \dfrac{1}{2\pi r}$

$\implies Y(r) = y^2(r) = s^2(r) - H^2(r) = \dfrac{1}{4\pi^2 r^2 } - \dfrac{1}{\pi^2 r^4} \implies \dfrac{dY}{dr} = \dfrac{1}{\pi^2}(\dfrac{8 - r^2}{2r^5})$ $r > = 0 \implies r = 2\sqrt{2} \implies$ $s = \dfrac{1}{4\sqrt{2}\pi}$ and $H = \dfrac{1}{8\pi}$ and $y = \sqrt{s^2 - H^2} = \dfrac{1}{8\pi} = H$ .

$\dfrac{d^2Y}{dr^2}|_{r = 2\sqrt{2}} = \dfrac{-1}{(8\pi)^2} < 0 \implies$ max at $r = 2\sqrt{2}$ .

Let $m\angle{PTR} = \lambda \implies \cos(\lambda) = \dfrac{y(r)}{s(r)} = \dfrac{1}{\sqrt{2}}$ and $\theta = 180 - \lambda \implies \cos(\theta) = cos(180 - \theta) = -cos(\lambda) = -\dfrac{1}{\sqrt{2}} \implies \boxed{\theta = 135^\circ}$

$\implies d^2 = s^2 + 4r^2 + 4rs \cos(\lambda) = \dfrac{1}{32\pi^2} + 32 + \dfrac{\sqrt{2}}{\pi} =$ $\dfrac{32^2\pi^2 + 32\sqrt{2}\pi + 1}{32\pi^2} \implies$ $d = \dfrac{\sqrt{32^2\pi^2 + 32\sqrt{2}\pi + 1}}{4\sqrt{2}\pi} \approx \boxed{5.69678}$

Note: Using the above $x = 4\sqrt{2}\tan(\dfrac{\pi}{4n}) \implies r = 2\sqrt{2}\sec(\dfrac{\pi}{4n}),$ $s = \dfrac{\cot(\dfrac{\pi}{4n})}{16\sqrt{2} n}$ , and $y = \dfrac{\cot(\dfrac{\pi}{4n})}{32 n} = H$ we obtain:

$r = \lim_{n \rightarrow \infty} r_{n} = 2\sqrt{2}$ , $y = \lim_{n \rightarrow \infty} \dfrac{\cos(\dfrac{\pi}{4n})}{32(n\sin(\dfrac{\pi}{4n}))} = \dfrac{1}{8\pi} = H$ , and $s = \lim_{n \rightarrow \infty} \dfrac{\cos(\dfrac{\pi}{4n})}{16\sqrt{2}(n\sin(\dfrac{\pi}{4n}))} =\dfrac{1}{4\sqrt{2}\pi}$