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 $x+2 \sqrt{2} y=4$. If the co-ordinates of the vertex A are $(\alpha, \beta)$

Accepted Answer

$(O):(0,0)$

Side BC Line : $x+2\sqrt{2}y=4$.

Vertex A : $(\alpha ,\beta )$.

Find the distance from the Centroid to side BC

The side $BC$ lies on the line $x+2\sqrt{2}y-4=0$. The distance from the origin $G(0,0)$ to this line is the length of the "lower" part of the median ( $r$ ).

Using the point-to-line distance formula :

$r=\frac{|0+2\sqrt{2}(0)-4|}{\sqrt{{1}^{2}+(2\sqrt{2}{)}^{2}}}=\frac{|-4|}{\sqrt{1+8}}=\frac{4}{3}$

Find the coordinates of Vertex $A(\alpha ,\beta )$

The centroid $G$ divides the segment $AD$ (where $D$ is the foot of the altitude on $BC$ ) in the ratio $2:1$. Therefore, the distance from $A$ to $G$ is $2r$.

$AG=2	imes \frac{4}{3}=\frac{8}{3}$

The line $AG$ is perpendicular to side $BC$. The slope of $BC$ is $m=-\frac{1}{2\sqrt{2}}$, so the slope of the line $AG$ (the altitude) is the negative reciprocal:

${m}_{\perp }=2\sqrt{2}$

Since $A$ lies on the line passing through the origin with slope $2\sqrt{2}$, its coordinates can be written as:

$\alpha =k,\, \beta =2\sqrt{2}k$

Because $A$ and the side $BC$ are on opposite sides of the origin (the centroid), and the constant term in our line equation is negative, $A$ must satisfy the condition that its position yields an opposite sign. Substituting these into the distance formula $AG=\frac{8}{3}$ :

$\sqrt{{k}^{2}+(2\sqrt{2}k{)}^{2}}=\frac{8}{3}\implies \sqrt{9{k}^{2}}=\frac{8}{3}\implies 3|k|=\frac{8}{3}\implies |k|=\frac{8}{9}$

Calculate $|\alpha +\sqrt{2}\beta |$

Substitute $\alpha =k$ and $\beta =2\sqrt{2}k$ :

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

$=|k+4k|=|5k|$

$=5	imes \frac{8}{9}=\frac{40}{9}\approx 4.44$

greatest integer less than or equal to $|\alpha +\sqrt{2}\beta |=[4.44]=4$.

Solutions · 1