Inscribed Circles

Let the center of the inscribed circle be $I$ and let $\theta = \angle ICF = \angle ICE$ . Then from $\triangle ICE$ , $IC = \sqrt{(r + 1)^2 + r^2} = \sqrt{2r^2 + 2r + 1}$ , $\cos \theta = \cfrac{r + 1}{\sqrt{2r^2 + 2r + 1}}$ , $\sin \theta = \cfrac{r}{\sqrt{2r^2 + 2r + 1}}$ , and $\cos 2\theta = \cos^2 \theta - \sin^2 \theta = \cfrac{(r + 1)^2 - r^2}{2r^2 + 2r + 1} = \cfrac{2r + 1}{2r^2 + 2r + 1}$ .

As tangents from the same point, $CF = CE = r + 1$ , $BD = BE = r + 2\sqrt{2} - 3$ , and $x = AF = AD$ . Therefore, the perimeter of $\triangle ABC$ is $P = 2(r + 1) + 2(r + 2\sqrt{2} - 3) + 2x$ , and the semiperimeter is $s = 2r + 2\sqrt{2} - 2 + x$ .

Since $A_{\triangle ABC} = 2r(2r + 2(\sqrt{2} - 1))$ and $A_{\triangle ABC} = rs = r(2r + 2\sqrt{2} - 2 + x)$ , these equations combine and solve to $x = 2r + 2\sqrt{2} - 2$ .

The sides of $\triangle ABC$ are then $AB = 2r + 2\sqrt{2} - 2 + r + 2\sqrt{2} - 3 = 3r + 4\sqrt{2} - 5$ , $AC = 2r + 2\sqrt{2} - 2 + r + 1 = 3r + 2\sqrt{2} - 1$ , and $BC = r + 2\sqrt{2} - 3 + r + 1 = 2r + 2\sqrt{2} - 2$ .

Then by the law of cosines on $\triangle ABC$ , $\cos 2\theta = \cfrac{(3r + 2\sqrt{2} - 1)^2 + (2r + 2\sqrt{2} - 2)^2 - (3r + 4\sqrt{2} - 5)^2}{2(3r + 2\sqrt{2} - 1)(2r + 2\sqrt{2} - 2)} = \cfrac{r - 2\sqrt{2} + 5}{3r + 2\sqrt{2} - 1}$ .

That means $\cos 2\theta = \cfrac{2r + 1}{2r^2 + 2r + 1} = \cfrac{r - 2\sqrt{2} + 5}{3r + 2\sqrt{2} - 1}$ , which rearranges to $(r - 1)(r + \sqrt{2} - 1)^2 = 0$ and solves to $r = \sqrt{2} - 1$ for $r > 0$ .

Therefore, $V_T = \frac{1}{3} \cdot A_{\triangle ABC} \cdot r = \frac{1}{3} \cdot 2r(2r + 2(\sqrt{2} - 1)) \cdot r = \cfrac{2^3}{3}(\sqrt{2} - 1)^3$ , so $\alpha = 2$ , $\beta = 3$ , and $\alpha + \beta = \boxed{5}$ .

I actually did the problem in general to obtain $r = \sqrt{2} - 1$ , so I will present the way I actually did it.

Let $a = 2\sqrt{2} - 3$ and $b = 1$ .

$A_{\triangle{ABC}} = \dfrac{1}{2}(2x + 4r + 2(a + b)) = (x + 2r + a + b)r$

Using Heron's formula with $s = x + 2r + a + b \implies$

$A_{\triangle{ABC}} = \sqrt{(x + 2r + a + b)(r + a)(r + b)x} \implies$

$(x + 2r + a + b)^2 r^2 = (x + 2r + a + b)(r + a)(r + b)x \implies$

$r^2x = (2r + a + b)r^2 = (r^2 + (a + b)r + ab)x \implies$

$((a + b)r + ab)x = (2r + a + b)r^2 \implies x = \dfrac{(2r + a + b)r^2}{(a + b)r + ab}$

$\implies A_{\triangle{ABC}} = (2r + a + b)(\dfrac{r^2}{(a + b)r + ab} + 1)r =$

$\dfrac{(2r + a + b)(r^2 + (a + b)r + ab)r}{(a + b)r + ab} = 2(2r + a + b)r \implies$

$r^2 + (a + b)r + ab = 2(a + b)r + 2ab \implies r^2 - (a + b)r - ab = 0$

Letting $a = 2\sqrt{2} - 3$ and $b = 1 \implies$ $r^2 - 2(\sqrt{2} - 1)r - (2\sqrt{2} - 3) = 0 \implies r = \sqrt{2} - 1 \implies$

$A_{\triangle{ABC}} = \dfrac{8(\sqrt{2} - 1)^4}{3 - 2\sqrt{2}} = 8(\sqrt{2} - 1)^2 \implies$

$\dfrac{8}{3}(\sqrt{2} - 1)^3 = \dfrac{2^3}{3}(\sqrt{2} - 1)^3 = \dfrac{\alpha^{\beta}}{\beta}(\sqrt{\alpha} - 1)^{\beta} \implies$ $\alpha + \beta = \boxed{5}$ .