Oblique Square Pyramid

I provided two solutions for the square base.

Solution using coordinate geometry:

For $BD: y = a - x$ and for $AQ: y = \dfrac{1}{a}x \implies x = \dfrac{a^2}{a + 1} \implies$

$A_{\triangle{ABP}} = \dfrac{a^3}{2(a + 1)} = \dfrac{a}{2}$ $\implies a(a^2 - a - 1) = 0 \:\ a > 0 \implies$

$\boxed{a = \dfrac{1 + \sqrt{5}}{2} = \phi}$ and $P:(1,\phi - 1)$ .

Solution using similar triangles:

Since vertical angles are congruent and $AB \parallel CD \implies$ alternate interior angles are congruent $\implies \triangle{ABP} \sim \triangle{DPQ} \implies$

$\dfrac{a}{1} = \dfrac{x}{a - x} \implies x = \dfrac{a^2}{a + 1} \implies A_{\triangle{ABP}} = \dfrac{a^3}{2(a + 1)} = \dfrac{a}{2}$

$\implies a(a^2 - a - 1) = 0 \:\ a > 0 \implies \boxed{a = \dfrac{1 + \sqrt{5}}{2} = \phi}$ and $P:(1,\phi - 1)$ .

(Ignore the $1$ in the diagram above. I have no idea how it got there.)

Using the above information and the diagram above $\implies \overline{PP_{1}} = \overline{PP_{2}} = \phi - 1$ and $\overline{PP_{3}} = \overline{PP_{4}} = 1$

$\implies$ the slant heights $s_{1} = s_{2} = \sqrt{\phi^2 + (\phi - 1)^2} =$ $\sqrt{\dfrac{(1 + \sqrt{5})^2}{4} + \dfrac{(\sqrt{5} - 1)^2}{4}} = \sqrt{3}$

and

$s_{3} = s_{4} = \sqrt{1 + \phi^2} = \sqrt{1 + \dfrac{(1 + \sqrt{5})^2}{4}} = \sqrt{\dfrac{10 + 2\sqrt{5}}{4}} =$ $\sqrt{\dfrac{5 + \sqrt{5}}{2}} = \sqrt{\sqrt{5}\phi}$

$\implies$ the lateral surface area $S = \phi(\sqrt{\sqrt{5}\phi} + \sqrt{3})$

and the height $PR = \overline{AB} = \phi \implies \dfrac{S}{\overline{PR}} = \sqrt{\sqrt{5}\phi} + \sqrt{3} = \sqrt{\sqrt{\alpha}\phi} + \sqrt{\beta}$

$\implies \alpha + \beta = \boxed{8}$ .