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 ( $\alpha,\beta$ ), then the greatest integer less than or equal to $|\alpha + \sqrt{2}\beta|$ is
Options
Solution
$${\because m_{BC} \cdot m_{AD} = - 1 }{\Rightarrow \left( - \frac{1}{2\sqrt{2}} \right)\left( \frac{\beta}{\alpha} \right) = - 1}$$
$$\begin{array}{r} \Rightarrow \beta = 2\sqrt{2}\alpha\#(1) \end{array}$$
$$\because OD = \left| \frac{- 4}{\sqrt{1 + 8}} \right| = \frac{4}{3} \Rightarrow AO = \frac{8}{3} $$So $AD = \frac{8}{3} + \frac{4}{3} = 4$
$\Rightarrow \frac{|\alpha + 2\sqrt{2}\beta - 4|}{3} = 4 \Rightarrow \alpha = \frac{16}{9}$ or $- \frac{8}{9}$
$$\begin{Bmatrix} \because A(\alpha,\beta)\&(0,0)\text{~}\text{lies on same}\text{~} \\ \text{~}\text{side of given line}\text{~} \end{Bmatrix} $$$\therefore(\alpha,\beta) = \left( \frac{16}{9},\frac{32\sqrt{2}}{9} \right);($ Rejected $)$
so $(\alpha,\beta) = \left( - \frac{8}{9},\frac{- 16\sqrt{2}}{9} \right)$
$$= \lbrack|\alpha + \sqrt{2}\beta|\rbrack = \left\lbrack \left| \frac{- 8 - 32}{9} \right| \right\rbrack = 4$$
Create a free account to view solution
View Solution Free