Oblique Triangular Prism

$V = \dfrac{\sqrt{3}}{4} r^2 H = b^2 \implies H = \dfrac{4b^2}{\sqrt{3}r^2}$ $A = 3rs = b \implies s = \dfrac{b}{3r} \implies Y(r) = y^2(r) = \dfrac{b^2}{3}(\dfrac{1}{3 r^2} - \dfrac{16b^2}{r^4}) \implies$ $\dfrac{dY}{dr} = \dfrac{2b^2}{3}(\dfrac{24 b^2}{r^5} - \dfrac{1}{3r^3}) = \dfrac{2b^2}{9r^5}(72 b^2 - r^2) = 0 \implies r = 6\sqrt{2} b$ for $r > 0 \implies H = \dfrac{1}{18\sqrt{3}}$ and $s = \dfrac{1}{18\sqrt{2}} \implies y = \sqrt{s^2 - H^2} = \dfrac{1}{18\sqrt{6}}$

Let $\lambda = m\angle{DFE}$ .

$\cos(\lambda) = \dfrac{y}{s} = \dfrac{1}{\sqrt{3}}$ and $\cos(\theta) = \cos(180 - \lambda) = -\cos(\lambda) = \dfrac{-1}{\sqrt{3}}$

$\boxed{\cos(\theta) = \dfrac{-1}{\sqrt{3}} \implies \theta \approx 125.26438968^\circ}$

For $r$ to minimize $y$ $\dfrac{d^2Y}{dr^2}|_{r = 6\sqrt{2}b} > 0$ $\implies \dfrac{d^2Y}{dr^2}|_{r = 6\sqrt{2}b} = \dfrac{2b^2}{3}(\dfrac{72 - 120b^2}{72^3}) > 0$ $\implies \boxed{b < \sqrt{\dfrac{3}{5}}} \implies \boxed{r = 6\sqrt{2}b < 6\sqrt{\dfrac{6}{5}} = \alpha \approx 6.57267}$

$d^2 = s^2 + r^2 +2sr\cos(\lambda) = \dfrac{1}{2(18^2)} + 72b^2 + \dfrac{2b}{3\sqrt{3}} <$ $\dfrac{1}{2(18^2)} + 72(\dfrac{3}{5}) + \dfrac{2}{3\sqrt{5}} =$

$\dfrac{5\sqrt{5} + 139968\sqrt{5} + 2160}{10\sqrt{5}(18)^2} \implies$ $\boxed{d < \dfrac{\sqrt{5\sqrt{5} + 139968\sqrt{5} + 2160}}{18\sqrt{500}} = \beta \approx 43.4996856}$

$\implies \lfloor{\alpha}\rfloor$ + $\lfloor{\beta}\rfloor$ + $\lfloor{\theta}\rfloor = \boxed{174}$