Three Pentagons Cover a Fourth

I think I found a better solution, correct me if you find a mistake. How did you proof that your solution gives the largest $s$ ?

First $D$ , $H$ and $\varphi$ can be described easily with basic pentagon geometry:

$\begin{aligned} & \varphi = 108^\circ - 90^\circ = 18 ^\circ \\ & D = \frac{1}{2} ( 1 + \sqrt{5}) \\ & H = s \cdot \frac{1}{2} \sqrt{ 5 + 2 * \sqrt{5}} \end{aligned}$

Now I describe the base of the yellow triangle and the height of the green pentagon in a different way:

$\begin{aligned} & D = s + 2 \cdot h \tan{\varphi} \\ & H = h + \frac{\sqrt{3}}{2} \cdot D \end{aligned}$

This gives a system of two equations and two unknowns $h, s$ which solves to $s = 1.264313...$

Note: See Yannick Ruffiner's solution

For this lesser solution, refer to graphic at bottom for details

$s_1 = \dfrac{1}{2}(1+\sqrt{5})$
$s_2=\dfrac{1}{2}s\sqrt{4+\sqrt{5}}$
$s_3=\dfrac{1}{2} \sqrt{ 4-(5+\sqrt{5})s + (5+2\sqrt{5})s^2 }$
${ \theta }_{ 1 }=\dfrac { 3 }{ 5 } \pi -ArcSin\left( \dfrac { { s }_{ 1 } }{ { s }_{ 2 } } Sin\left( \dfrac { 2 }{ 5 } \pi \right) \right)$
${ \theta }_{ 2 }=ArcSin\left( \dfrac { s-1 }{ { s }_{ 3 } } Sin\left( \dfrac { 3 }{ 5 } \pi -ArcSin\left( \dfrac { s }{ 2{ s }_{ 2 } } Sin\left( \dfrac { 3 }{ 5 } \pi \right) \right) \right) \right)$
${ \theta }_{ 3 }=\dfrac { 2 }{ 5 } \pi -ArcSin\left( \dfrac { 1 }{ { s }_{ 3 } } Sin\left( \dfrac { 3 }{ 5 } \pi \right) \right)$

Find by numerical means $s$ such that

$\theta_1 + \theta_2 + \theta_3 = \dfrac{2}{5} \pi$

which works out to $s = 1.25219071389528224....$