Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the l…

Question

Let $ABC$ be an equilateral triangle with orthocenter at the origin and the side $BC$ on the line $x+2\sqrt2y=4$. If the co-ordinates of the vertex $A$ are $(\alpha,\beta)$, then the greatest integer less than or equal to $|\alpha+\sqrt2\beta|$ is:

Accepted Answer

Since the orthocentre coincides with the centroid $G(0,0)$ for an equilateral triangle, the distance from $G$ to $BC$ is
$$r = \frac{|0+0-4|}{\sqrt{1+8}} = \frac43$$

The median (= altitude) from $A$ passes through $G$, and $AG = 2r = \dfrac83$ (since $G$ divides the altitude $A$--(foot on $BC$) in ratio $2:1$).

$BC$ has slope $-\dfrac1{2\sqrt2}$, so the altitude through $A$ (perpendicular to $BC$, through the origin) has slope $2\sqrt2$: points on it are $(k, 2\sqrt2 k)$.

$$AG = \sqrt{k^2+8k^2} = \frac83 \implies 3|k| = \frac83 \implies |k| = \frac89$$

So $\alpha = k,\ \beta = 2\sqrt2 k$, and (taking $A$ on the opposite side of $G$ from $BC$):
$$|\alpha+\sqrt2\beta| = |k+4k| = 5|k| = 5\cdot\frac89 = \frac{40}{9}\approx4.44$$

$$\lfloor 4.44\rfloor = 4$$

Solutions · 1