Inscribed Circle!

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

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

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

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

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

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

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

$2(2r + 21)r \implies r^2 + 21r + 110 = 42r + 220 \implies r^2 - 21r - 110 = 0$

dropping the negative root $\implies r = \dfrac{21 + \sqrt{881}}{2} = \dfrac{\alpha + \sqrt{\beta}}{\lambda}$

$\implies \alpha + \beta + \lambda = \boxed{904}$ .

I set a coordinate system with origin E an B (p, 0) and C (q, 0) on the x-axis. Calculated the tangents through B and C, and then the coordinates of A ((p+q)/(pq+R^2) * R^2; 2pqR/(pq+R^2)). Then I set the y-co-ordinate to 4R, p to -(R+10) and q to R+11.

Let $AD=AF = x$ . Then the semiperimeter of $\triangle ABC$ , $s = x + 2r + 21$ . Since the $r$ is the inradius of $\triangle ABC$ , then $A_\triangle = sr = (x+2r+21)r = 2r(2r+21) \implies x = 2r + 21$ , $AB = c = x + r + 11 = 3r+32$ , and $CA = b = x + r + 10 = 3r + 31$ .

Note that $A_\triangle = \dfrac {bc \sin A}2$ . Therefore,

$\begin{aligned} \frac {bc \sin A}2 & = 2r(2r+21) \\ (3r+31)(3r+32) \cdot \frac {2 \tan \frac A2}{1+\tan^2 \frac A2} & = 4r(2r+21) \\ (3r+31)(3r+32) \cdot \frac {2\left(\frac rx\right)}{1+\left(\frac rx\right)^2} & = 4r(2r+21) \\ (9r^2 + 189r + 992) \cdot \frac {rx}{x^2 + r^2} & = 2r(2r+21) \\ (9r^2 + 189r + 992) \cdot \frac {r(2r+21)}{(2r+21)^2 + r^2} & = 2r(2r+21) \\ \frac {9r^2 + 189r + 992}{5r^2 + 84r + 441} & = 2 \\ 9r^2 + 189r + 992 & = 10r^2 + 168r + 882 \\ r^2 - 21r - 110 & = 0 \\ \implies r & = \frac {21 + \sqrt{881}}2 \end{aligned}$

Therefore $\alpha + \beta + \gamma = 21 + 881 + 2 = \boxed{904}$ .

Let the center of the inscribed circle be $I$ and let $\theta = \angle ICF = \angle ICE$ . Then from $\triangle ICE$ , $IC = \sqrt{(r + 10)^2 + r^2} = \sqrt{2r^2 + 20r + 100}$ , $\cos \theta = \cfrac{r + 10}{\sqrt{2r^2 + 20r + 100}}$ , $\sin \theta = \cfrac{r}{\sqrt{2r^2 + 20r + 100}}$ , and $\cos 2\theta = \cos^2 \theta - \sin^2 \theta = \cfrac{(r + 10)^2 - r^2}{2r^2 + 20r + 100} = \cfrac{10(r + 5)}{r^2 + 10r + 50}$ .

As tangents from the same point, $CF = CE = r + 10$ , $BD = BE = r + 11$ , and $x = AF = AD$ . Therefore, the perimeter of $\triangle ABC$ is $P = 2(r + 10) + 2(r + 11) + 2x$ , and the semiperimeter is $s = 2r + 21 + x$ .

Since $A_{\triangle ABC} = 2r(2r + 21)$ and $A_{\triangle ABC} = rs = r(2r + 21 + x)$ , these equations combine and solve to $x = 2r + 21$ .

The sides of $\triangle ABC$ are then $AB = 2r + 21 + r + 11 = 3r + 32$ , $AC = 2r + 21 + r + 10 = 3r + 31$ , and $BC = r + 11 + r + 10 = 2r + 21$ .

Then by the law of cosines on $\triangle ABC$ , $\cos 2\theta = \cfrac{(3r + 31)^2 + (2r + 21)^2 - (3r + 32)^2}{2(3r + 31)(2r + 21)} = \cfrac{r + 9}{3r + 31}$ .

That means $\cos 2\theta = \cfrac{10(r + 5)}{r^2 + 10r + 50} = \cfrac{r + 9}{3r + 31}$ , which rearranges to $(r + 10)(r^2 - 21r - 110) = 0$ , and solves to $r = \cfrac{21 + \sqrt{881}}{2}$ for $r > 0$ .

Therefore, $\alpha = 21$ , $\beta = 881$ , $\lambda = 2$ , and $\alpha + \beta + \lambda = \boxed{904}$ .