Oblique Cylinder 3

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

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

For $r$ to minimize the distance $y$ $\dfrac{d^2Y}{dr^2}|_{r = 2\sqrt{2} b} > 0 \implies \dfrac{d^2Y}{dr^2}|_{r = 2\sqrt{2} b} = \dfrac{1}{128\pi^2}(3 - 5b) > 0 \implies \boxed{b < \dfrac{3}{5}} \implies \boxed{r < \dfrac{6\sqrt{2}}{5} = \alpha \approx 1.697056}$ .

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

$\implies d^2 = s^2 + 4r^2 + 4rs \cos(\lambda) = \dfrac{1}{32\pi^2} + 32 b^2 + \dfrac{\sqrt{2} b}{\pi} < \dfrac{1}{32\pi^2} + 32 (\dfrac{9}{25}) + \dfrac{\sqrt{2}}{\pi}(\dfrac{3}{5}) =$

$\dfrac{(96\pi)^2 + 480\sqrt{2}\pi + 25}{2(20\pi)^2} \implies \boxed{d < \dfrac{\sqrt{(96\pi)^2 + 480\sqrt{2}\pi + 25}}{20\sqrt{2}\pi} = \beta \approx 3.43413}$

$\therefore \lfloor{\alpha}\rfloor$ + $\lfloor{\beta}\rfloor + \theta = \boxed{139}$