Let the equilateral triangle be denoted as ABC. The orthocenter is at the origin O(0, 0). In an equilateral triangle, the orthocenter coincides with the centroid and the circumcenter.

1. Calculate Apothem (Distance to Side BC)

The side BC lies on the line $L: 2x + \sqrt{2}y = 4$. The distance from the origin (orthocenter/circumcenter) to the side BC is the apothem, denoted by $r$. The formula for the distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Here, $(x_0, y_0) = (0, 0)$, and the line is $2x + \sqrt{2}y - 4 = 0$. So, $A=2$, $B=\sqrt{2}$, $C=-4$.

$r = \frac{|2(0) + \sqrt{2}(0) - 4|}{\sqrt{2^2 + (\sqrt{2})^2}} = \frac{|-4|}{\sqrt{4 + 2}} = \frac{4}{\sqrt{6}}$

2. Calculate Circumradius

For an equilateral triangle, the circumradius $R$ (distance from the center to a vertex) is twice the apothem $r$.

$R = 2r = 2 \times \frac{4}{\sqrt{6}} = \frac{8}{\sqrt{6}}$

The vertex A has coordinates $(\alpha, \beta)$. Since the origin is the circumcenter, the distance OA is $R$. Thus, $\alpha^2 + \beta^2 = R^2$.

$ \alpha^2 + \beta^2 = \left(\frac{8}{\sqrt{6}}\right)^2 = \frac{64}{6} = \frac{32}{3} $

3. Determine Coordinates of Vertex A

The altitude from vertex A to the side BC passes through the orthocenter (origin). This altitude line is perpendicular to the line BC.

The slope of line BC ($2x + \sqrt{2}y = 4$) is $m_{BC} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.

The slope of the altitude line is $m_A = -\frac{1}{m_{BC}} = -\frac{1}{-\sqrt{2}} = \frac{1}{\sqrt{2}}$.

The equation of the altitude line passing through the origin (0, 0) is $y = m_A x \implies y = \frac{1}{\sqrt{2}}x$. This is incorrect. The direction vector of the altitude should be parallel to the normal vector of the line BC.

The normal vector to the line $2x + \sqrt{2}y - 4 = 0$ is $\vec{n} = (2, \sqrt{2})$. The altitude line passes through the origin and has direction $\vec{n}$.

So, the coordinates of vertex A $(\alpha, \beta)$ can be represented as $A = t\vec{n} = t(2, \sqrt{2}) = (2t, \sqrt{2}t)$ for some scalar $t$.

Therefore, $\alpha = 2t$ and $\beta = \sqrt{2}t$.

Substitute these into the distance equation $\alpha^2 + \beta^2 = \frac{32}{3}$:

$ (2t)^2 + (\sqrt{2}t)^2 = \frac{32}{3} $

$ 4t^2 + 2t^2 = \frac{32}{3} $

$ 6t^2 = \frac{32}{3} $

$ t^2 = \frac{32}{18} = \frac{16}{9} $

$ t = \pm \frac{4}{3} $

4. Evaluate the Expression

We need to find the greatest integer less than or equal to $|\alpha + \sqrt{2}\beta|$. Substitute $\alpha = 2t$ and $\beta = \sqrt{2}t$ into the expression:

$ |\alpha + \sqrt{2}\beta| = |(2t) + \sqrt{2}(\sqrt{2}t)| = |2t + 2t| = |4t| $

Using the possible values for $t$ ($t = \pm \frac{4}{3}$):

$ |4t| = 4 \times \left|\pm \frac{4}{3}\right| = 4 \times \frac{4}{3} = \frac{16}{3} $

5. Find the Greatest Integer

The value is $\frac{16}{3} \approx 5.333...$. The greatest integer less than or equal to this value is required.

$ \left\lfloor \frac{16}{3} \right\rfloor = 5 $