Tetrahedrons!

Let the center of the incircle be $I$ , let the top of the tetrahedron be $G$ , let $AB = c$ , $AC = b$ , and $BC = a$ , and let the semiperimeter of $\triangle ABC$ be $s$ .

Then $IG = ID = IE = IF = r$ , $GD = GE = GF = \sqrt{2}r$ , so $A_{\triangle BCG} = \frac{\sqrt{2}}{2}ar$ , $A_{\triangle ACG} = \frac{\sqrt{2}}{2}br$ , and $A_{\triangle ABG} = \frac{\sqrt{2}}{2}cr$ .

Since $A_{\triangle ABC} = rs$ , the surface area $S = \frac{\sqrt{2}}{2}ar + \frac{\sqrt{2}}{2}br + \frac{\sqrt{2}}{2}cr + rs = \frac{1}{2}(a + b + c)r\sqrt{2} + rs = rs\sqrt{2} + rs = rs(\sqrt{2} + 1)$ .

That means $\cfrac{S}{A_{\triangle ABC}} = \cfrac{rs(\sqrt{2} + 1)}{rs} = \sqrt{2} + 1$ , so $\alpha = 2$ , $\beta = 1$ , and $\alpha + \beta = 3$ .

$A_{\triangle{ABC}} = \dfrac{1}{2}(x + r + 10)r + \dfrac{1}{2}(x + r + 11) + \dfrac{1}{2}(2r + 21) = (x + 2r + 21)r$

and using Heron's formula with $s = x + 2r + 21 \implies$

$A_{\triangle{ABC}} = \sqrt{(x + 2r + 21)(x)(r + 10)(r + 11)} \implies$

$(x + 2r + 21)^2r^2 = (x + 2r + 21)(x)(r + 10)(r + 11) \implies$

$(x + 2r + 21)r^2 = (r + 10)(r + 11)x \implies r^2x + (2r + 21)r^2 = (r + 10)(r + 11)x$

$\implies (21r + 110)x = (2r + 21)r^2 \implies x = \dfrac{(2r + 21)r^2}{21r + 110} \implies$

$A_{\triangle{ABC}} = (2r + 21)(\dfrac{r^2}{21r + 110} + 1)r =\dfrac{(2r + 21)(r^2 + 21r + 110)r}{21r + 110} =$

$r(\dfrac{2r^2 + 1188}{21}) \implies 42r^3 + 1323r^2 + 13881r + 48510 = 42r^3 + 220r^2 + 24948r+ 130680$

$\implies 1103r^2 - 11067r - 82170 = 0$ dropping the negative root

$\implies r = \dfrac{11067 + 22023}{2206} = \dfrac{33090}{2206} = 15 \implies x = 27 \implies$

$A_{\triangle{ABC}} = (x + 2r + 21)r = (78)(15) = 1170$ .

Below I use the vector cross product to find the lateral surface area, but first I need to find $B(x^{*},y^{*})$ .

${x^{*}}^{2} + {y^{*}}^{2} = 2809$

$(x^{*} - 52)^2 + {y^{*}}^{2} = 2601$

$\implies$

${x^{*}}^{2} - 104x^{*} + 2704 + {y^{*}}^2 = 2601$

${x^{*}}^{2} + {y^{*}}^{2} = 2809$

$\implies 104x^{*} = 2912 \implies x^{*} = 28 \implies y^{*} = 45 \implies B(28,45,0)$

$\vec{AP} = 27\vec{i} + 15\vec{j} + 15\vec{k}$ and $\vec{AC} = 52\vec{i} + 0\vec{j} + 0\vec{k} \implies$

$\implies \vec{AP} X \vec{AC} = 0\vec{i} + 780\vec{j} - 780\vec{k} \implies |\vec{AP} X \vec{AC}| = 780\sqrt{2}$

$\implies A_{APC} = \dfrac{1}{2}|\vec{AP} X \vec{AC}| = \dfrac{780\sqrt{2}}{2}$

Similarly $\vec{AB} = 28\vec{i} + 45\vec{j} + 0\vec{k} \implies \vec{AP} X \vec{AB} = -675\vec{i} + 420\vec{j} + 795\vec{k}$

$\implies |\vec{AP} X \vec{AB}| = 795\sqrt{2} \implies A_{APB} = \dfrac{1}{2}|\vec{AP} X \vec{AB}| = \dfrac{795\sqrt{2}}{2}$

and

$\vec{CP} = -25\vec{i} + 15\vec{j} + 15\vec{k}$ and $\vec{CB} = -24\vec{i} + 45\vec{j} + 0\vec{k} \implies$

$\vec{CP} X \vec{CB} = -675\vec{i} - 360\vec{j} - 765\vec{k} \implies |\vec{CP} X \vec{CB}| = 765\sqrt{2} \implies$

$A_{CPB} = \dfrac{1}{2}|\vec{AP} X \vec{AB}| = \dfrac{765\sqrt{2}}{2}$

$\implies L.S. = 1170\sqrt{2} \implies S = 1170(\sqrt{2} + 1) \implies \dfrac{S}{A_{\triangle{ABC}}} = \sqrt{2} + 1 =$

$\sqrt{\alpha} + \beta \implies \alpha + \beta = \boxed{3}$ .