Wedged in equilateral triangles 2.

Let $m$ be a side of square $ABCD$ .

Let $A_{e}$ be the area of equilateral $\triangle{DFE}$ $\implies 4A_{e} = 4(\dfrac{1}{2})(a)(\dfrac{\sqrt{3}}{2}a) = \sqrt{3}a^2$

Let $A_{I}$ be the area of $\triangle{GFD}$

The height $h_{\triangle{GFD}} = (m - a)\sin(60^{\circ}) = \dfrac{\sqrt{3}}{2}(m - a) \implies$

$4A_{I} = 4(\dfrac{1}{2})(m - a)^2(\dfrac{\sqrt{3}}{2}) = \sqrt{3}(m - a)^2$

Using the law of cosines on $\triangle{GFD}$ with included $\angle{FGD} \implies$

$a^2 = 2(m - a)^2 (\dfrac{3}{2}) = 3(m - a)^2 = 3a^2 -6ma + 3m^2 \implies 2a^2 - 6ma + 3m^2 = 0$

$\implies a = \dfrac{3 \pm \sqrt{3}}{2}m$

$a = \dfrac{3 + \sqrt{3}}{2}m \implies m - a = -(1 + \sqrt{3})m < 0$

$a = \dfrac{3 - \sqrt{3}}{2}m \implies m - a = \dfrac{\sqrt{3} - 1}{2}m$

$\therefore$ we drop $a = \dfrac{3 + \sqrt{3}}{2}m$ and choose $a = \dfrac{3 - \sqrt{3}}{2}m$

$\implies 4A_{e} = \dfrac{\sqrt{3}}{4}(3 - \sqrt{3})^2 m^2 = \dfrac{3\sqrt{3}}{2}(2 - \sqrt{3}) m^2$

and

$4A_{I} = \sqrt{3}(m - a)^2 = \dfrac{\sqrt{3}}{4}(\sqrt{3} - 1)^2 m^2 = \dfrac{\sqrt{3}}{2}(2 - \sqrt{3})m^2$

Let $A_{s}$ be the area of the red square

$\implies A_{s} = m^2 - (4A_{e} + 4A_{I}) = (1 - (2\sqrt{3}(2 - \sqrt{3})))m^2 =$ $(1 - (4\sqrt{3} - 6))m^2$

$= (7 - 4\sqrt{3})m^2$

Let $y$ be a side of the red square $\implies y= \sqrt{7 - 4\sqrt{3}}m \implies \overline{OQ} = \dfrac{\sqrt{2}}{2}\sqrt{7 - 4\sqrt{3}}m = \sqrt{\dfrac{7 - 4\sqrt{3}}{2}}m \implies$

$V_{p} = \dfrac{1}{3}A_{s}\overline{OQ} = \dfrac{(7 - 4\sqrt{3})^{\dfrac{3}{2}}}{3\sqrt{2}}m^3 \implies$

$\dfrac{V_{P}}{V_{C}} = \dfrac{(7 - 4\sqrt{3})^{\dfrac{3}{2}}}{3\sqrt{2}} = \dfrac{(\alpha - \beta\sqrt{\lambda})^{\dfrac{\lambda}{\gamma}}}{\lambda\sqrt{\gamma}} \implies$

$\alpha + \beta + \lambda + \gamma = \boxed{16}$ .

For simplicity let $ABCD$ be a unit square. Then, the volume of the cube is ${{V}_{C}}=1$ .

Now, with the labelling of the figure where $a$ is the side length of the equilateral triangles and $s$ is the side length of the inner red square, we focus on $\triangle FEB$ :
$\angle FEB= 120{}^\circ$ and $\angle FBE= 30{}^\circ$ , hence $\angle EFB= 30{}^\circ$ , i..e. $\triangle FEB$ is isosceles, with $EB=EF=1-a$ .

By sine rule, $\dfrac{FB}{\sin E}=\dfrac{FE}{\sin B}\Rightarrow \dfrac{a}{\sin 120{}^\circ }=\dfrac{1-a}{\sin 30{}^\circ }\Rightarrow \dfrac{1}{2}a=\dfrac{\sqrt{3}}{2}\left( 1-a \right)\Rightarrow a=\dfrac{3-\sqrt{3}}{2}$ Furthermore, $\begin{aligned} & s=GF=GE-FE=a-\left( 1-a \right)=2a-1=2\dfrac{3-\sqrt{3}}{2}-1 \\ & \Rightarrow s=2-\sqrt{3} \\ \end{aligned}$ For the height $h$ of the pyramid, $h=OQ=\dfrac{GQ}{2}=\dfrac{s\sqrt{2}}{2}=\dfrac{s}{\sqrt{2}}$ Thus, the volume of the pyramid is ${{V}_{P}}=\dfrac{1}{3}{{s}^{2}}h=\dfrac{1}{3}\cdot \dfrac{{{s}^{3}}}{\sqrt{2}}=\dfrac{{{\left( 2-\sqrt{3} \right)}^{3}}}{3\sqrt{2}}=\dfrac{{{\left[ {{\left( 2-\sqrt{3} \right)}^{2}} \right]}^{\dfrac{3}{2}}}}{3\sqrt{2}}=\dfrac{{{\left( 7-4\sqrt{3} \right)}^{\dfrac{3}{2}}}}{3\sqrt{2}}$ $\Rightarrow \dfrac{{{V}_{P}}}{{{V}_{C}}}=\dfrac{{{V}_{P}}}{1}=\dfrac{{{\left( 7-4\sqrt{3} \right)}^{\dfrac{3}{2}}}}{3\sqrt{2}}$ Hence, $\alpha =7$ , $\beta =4$ , $\lambda =3$ , $\gamma =2$ and $\alpha+\beta+\lambda+\gamma =7+4+3+2=\boxed{16}$ .