Let ABC be an equilateral triangle with orthocenter at the origin...

Let ABC be an equilateral triangle with orthocenter at the origin O(0,0) and side BC on the line $x + 2\sqrt{2}y = 4$.
Since $m_{BC} \cdot m_{AD} = -1$:
\left(-\frac{1}{2\sqrt{2}}\right)\left(\frac{\beta}{\alpha}\right) = -1 \Rightarrow \beta = 2\sqrt{2}\alpha$...(1)<br>Distance from O to BC:$OD = \frac{|-4|}{\sqrt{1+8}} = \frac{4}{3}$<br>For equilateral triangle:$AO = \frac{8}{3}$, so$AD = \frac{4}{3} + \frac{8}{3} = 4$<br>$\frac{|\alpha + 2\sqrt{2}\beta - 4|}{3} = 4 \Rightarrow \alpha = \frac{16}{9}$or$-\frac{8}{9}$<br>Since A(\alpha, \beta)$ and O(0,0) lie on the same side of BC (origin gives $0+0-4 < 0$), we need $\alpha + 2\sqrt{2}\beta - 4 < 0$.
$(\alpha, \beta) = \left(-\frac{8}{9}, -\frac{16\sqrt{2}}{9}\right)$
$\lfloor |\alpha + \sqrt{2}\beta| \rfloor = \left\lfloor \left|\frac{-8-32}{9}\right| \right\rfloor = \left\lfloor \frac{40}{9} \right\rfloor = 4$