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(D) In an equilateral triangle, the orthocenter coincides with the centroid. Let $O(0,0)$ be the orthocenter. The altitude $AD$ passes through $O$. Th

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$4$

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(D) In an equilateral triangle, the orthocenter coincides with the centroid. Let $O(0,0)$ be the orthocenter. The altitude $AD$ passes through $O$. The slope of $BC$ is $m_{BC} = -\frac{1}{2\sqrt{2}}$. Since $AD \perp BC$, the slope of $AD$ is $m_{AD} = 2\sqrt{2}$. The equation of $AD$ is $y = 2\sqrt{2}x$, so $\beta = 2\sqrt{2}\alpha$. The distance from $O(0,0)$ to $BC$ $(x+2\sqrt{2}y-4=0)$ is $OD = \frac{|0+0-4|}{\sqrt{1^2+(2\sqrt{2})^2}} = \frac{4}{\sqrt{9}} = \frac{4}{3}$. In an equilateral triangle, the centroid divides the altitude in the ratio $2:1$. Thus, $AO = 2OD = 2(\frac{4}{3}) = \frac{8}{3}$. The total length of the altitude $AD = AO + OD = \frac{8}{3} + \frac{4}{3} = 4$. The vertex $A$ lies on the line $y = 2\sqrt{2}x$ at a distance $4$ from the line $BC$. The coordinates of $A$ are $(\alpha, 2\sqrt{2}\alpha)$. The distance from $A$ to $x+2\sqrt{2}y-4=0$ is $\frac{|\alpha+2\sqrt{2}(2\sqrt{2}\alpha)-4|}{\sqrt{1+8}} = \frac{|9\alpha-4|}{3} = 4$. This gives $9\alpha-4 = 12$ or $9\alpha-4 = -12$. So $\alpha = \frac{16}{9}$ or $\alpha = -\frac{8}{9}$. Since $O(0,0)$ and $A$ must be on the same side of $BC$ (as $O$ is the centroid), we test the sign of the expression $x+2\sqrt{2}y-4$ at $(0,0)$, which is $-4$. For $A(\alpha, 2\sqrt{2}\alpha)$, the expression is $9\alpha-4$. Thus $9\alpha-4 We need to find the greatest integer less than or equal to $|\alpha+\sqrt{2}\beta| = |-\frac{8}{9} + \sqrt{2}(-\frac{16\sqrt{2}}{9})| = |-\frac{8}{9} - \frac{32}{9}| = |-\frac{40}{9}| = \frac{40}{9} \approx 4.44$. The greatest integer is $4$.

Let $ABC$ be an equilateral triangle with orthocenter at the origin and the side $BC$ on the line $x+2\sqrt{2}y=4$. If the coordinates of the vertex $A$ are $(\alpha, \beta)$, then the greatest integer less than or equal to $|\alpha+\sqrt{2}\beta|$ is

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