It's Golden!

Let $A$ be the origin $(0,0)$ of the $xy$ -plane with $AD$ and $AB$ be along the $x$ -axis and $y$ -axis respectively., and $\angle AGH = \theta$ . Then $J=(\sin \theta, \sin \theta + \cos \theta)$ and $C = (\sin \theta + \cos \theta + 1, 1)$ . The equation of the line $BC$ is given by:

$\begin{aligned} \frac {y-1}{x-\sin \theta - \cos \theta -1} & = \frac {\sin \theta + \cos \theta -1}{\sin \theta -\sin \theta - \cos \theta -1} & \small \blue{\text{Putting }x=0} \\ \implies y & = 1 + \frac {2\sin \theta \cos \theta}{\cos \theta +1} & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ & = 1 + \frac {2 \cdot \frac {2t}{1+t^2}\cdot \frac {1-t^2}{1+t^2}}{\frac {1-t^2}{1+t^2}+1} \\ & = 1 + \frac {2t(1-t^2)}{1+t^2} & \small \blue{\text{To find }\max(y)} \\ \frac {dy}{dt} & = \frac {2(t^4+4t^2-1)}{(1+t^2)^2} & \small \blue{\text{Putting }\frac {dy}{dt} =0} \\ t^4 + 4t^2 & = 1 \\ (t^2+2)^2 & = 5 & \small \blue{\text{Since }\theta < 90^\circ} \\ \implies t & = \sqrt{\sqrt 5 -2} & \small \blue{\text{Golden ratio }\varphi = \frac {1+\sqrt 5}2} \\ & = \sqrt{2\varphi - 3} = \frac 1{\varphi\sqrt \varphi} = \frac {\sqrt{\varphi-1}}\varphi & \small \blue{\implies \sqrt 5 = 2\varphi -1} \\ \implies \max (y) & = 1 + \frac {2t(1-t^2)}{1+t^2} \\ & = 1 + \frac {2\sqrt{\varphi-1}}\varphi \cdot \frac {\varphi^2- \varphi + 1}{\varphi^2 + \varphi - 1} \\ & = 1 + \frac {2\sqrt{\varphi-1}}\varphi \cdot \frac 2{2 \varphi} \\ & = 1 + \frac {2(\varphi-1)^\frac 32}\varphi \end{aligned}$

The required answer $\alpha + \beta + \gamma = 1 + 2 + 3 = \boxed 6$ .

Using the above diagram $\overline{IK} = (m + n)\sin(\theta) = \sin(\theta)$ and $\overline{JI} = \sin(\theta) + \cos(\theta) - 1$

and $\triangle{JIC} \sim \triangle{BPC} \implies \dfrac{\sin(\theta) + \cos(\theta) + 1}{y - 1} = \dfrac{\sin(\theta) + 1}{\sin(\theta) + \cos(\theta) - 1}$

$\implies y = 1 + \dfrac{\sin(2\theta)}{\sin(\theta) + 1} \implies$

$\dfrac{dy}{dx} = \dfrac{(2\cos(2\theta))(\sin(\theta) + 1) - \sin(2\theta)\cos(\theta)}{(\sin(\theta) + 1)^2} = 0 \implies$

$(2\cos^2(\theta) - 1)(\sin(\theta) + 1) = \sin(\theta)\cos^2(\theta) \implies \sin^3(\theta) + 2\sin^2(\theta) - 1 = 0$

Let $u = \sin(\theta) \implies u^3 + 2u^2 - 1 = 0 \implies (u + 1)(u^2 + u - 1) = 0$ and

$u \neq -1 \implies u = \dfrac{\sqrt{5} - 1}{2} = \phi - 1 = \sin(\theta) \:\$ dropping the negative root

$\implies \cos(\theta) = \sqrt{\phi - 1} \implies y(max) = 1 + \dfrac{2\sin(\theta)\cos(\theta)}{\sin(\theta) + 1} =$

$1 + \dfrac{2(\phi - 1)^{\frac{3}{2}}}{\phi} = \alpha + \dfrac{\beta(\phi - \alpha)^{\frac{\lambda}{\beta}}}{\phi}$ $\implies \alpha + \beta + \lambda = \boxed{6}$ .

You can check that a max does occur at $\theta = \arcsin(\phi - 1)$ .

I checked and $\dfrac{d^2y}{d\theta^2}|_{\theta = \arcsin(\phi - 1)} < 0$ .