What is the Radius?

For $\overline{AC}: y = x$ and $m_{\overline{ED}} = -2 \implies y = -2x + 2a \implies x = y = \dfrac{2a}{3}$

$\implies G(\dfrac{2a}{3},\dfrac{2a}{3}) \implies \overline{AG} = \dfrac{2\sqrt{2}}{3}a$ and $\overline{GD} = \dfrac{\sqrt{5}}{3}a$

$\implies A_{\triangle{AGD}} = \dfrac{1}{2}(a)(\dfrac{2a}{3}) = \dfrac{a^2}{3} =$ $\dfrac{ar}{2}(\dfrac{3 + 2\sqrt{2} + \sqrt{5}}{3}) \implies$

$2a^2 - (3 + 2\sqrt{2} + \sqrt{5})ar = 0 \implies a(2a - (3 + 2\sqrt{2} + \sqrt{5})r) = 0$ and

$a \neq 0 \implies r = \dfrac{2a}{3 + 2\sqrt{2} + \sqrt{5}} \implies \dfrac{r}{a} = \dfrac{2}{3 + 2\sqrt{2} + \sqrt{5}} = \dfrac{\alpha}{\beta + \alpha\sqrt{\alpha} + \sqrt{\lambda}}$

$\implies \alpha + \beta + \lambda = \boxed{10}$ .

We can find inradius of a triangle without finding the area and perimeter of the triangle. Check out this solution.

Let $a=1$ , $\angle CAD = \alpha$ , and $\angle EDA = \beta$ . Note that $AO$ bisects $\angle CAD$ and $OD$ bisects $\angle EDA$ . Then we have:

$\begin{aligned} AF + FD & = AD \\ OF \cdot \cot \angle OAF + OF \cot ODF & = AD \\ r \cot \frac \alpha 2 + r \cdot \cot \frac \beta 2 & = 1 & \small \blue{\text{Using }\tan \frac \theta 2 = \frac {1-\cos \theta}{\sin \theta} \text{ (see note).}} \\ \frac r{\sqrt 2 -1} + \frac {2r}{\sqrt 5 -1} & = 1 \\ r & = \frac 1{\frac 1{\sqrt 2 - 1} + \frac 2{\sqrt 5 -1}} & \small \blue{\text{Since }a = 1} \\ \implies \frac ra & = \frac 1{\frac {\sqrt 2 + 1}{2-1} + \frac {2(\sqrt 5 +1)}{5-1}} \\ & = \frac 1{\sqrt 2 + 1 + \frac {\sqrt 5 +1}2} \\ & = \frac 2{3+2\sqrt 2 + \sqrt 5} \end{aligned}$

Therefore the required answer is $3+2+5 = \boxed{10}$ .

---

Note:

• We note that $\tan \alpha = 1 \implies \sin \alpha = \cos \alpha = \dfrac 1{\sqrt 2} \implies \tan \dfrac \alpha 2 = \dfrac {1-\cos \alpha}{\sin \alpha} = \sqrt 2 - 1$ .
• Similarly, $\tan \beta = 2 \implies \sin \beta = \dfrac 2{\sqrt 5}$ , $\cos \alpha = \dfrac 1{\sqrt 5} \implies \tan \dfrac \beta 2 = \dfrac {1-\cos \beta}{\sin \beta} = \dfrac{\sqrt 5 - 1}2$ .