Q30 / 75 · JEE Main 2026 (28 January Shift 1) · Mathematics
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
- A2
- B5
- C3
- Correct answerD4
Solution
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)
Distance from O to BC: $OD = \frac{|-4|}{\sqrt{1+8}} = \frac{4}{3}$
For equilateral triangle: $AO = \frac{8}{3}$, so $AD = \frac{4}{3} + \frac{8}{3} = 4$
$\frac{|\alpha + 2\sqrt{2}\beta - 4|}{3} = 4 \Rightarrow \alpha = \frac{16}{9}$ or $-\frac{8}{9}$
Since $A(\alpha, \beta)$ and $O(0,0)$ lie on the same side of BC (origin gives $0+0-4 < 0$), we need $\alpha + 2\sqrt{2}\beta - 4 < 0$.
$(\alpha, \beta) = \left(-\frac{8}{9}, -\frac{16\sqrt{2}}{9}\right)$
$\lfloor |\alpha + \sqrt{2}\beta| \rfloor = \left\lfloor \left|\frac{-8-32}{9}\right| \right\rfloor = \left\lfloor \frac{40}{9} \right\rfloor = 4$
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)
Distance from O to BC: $OD = \frac{|-4|}{\sqrt{1+8}} = \frac{4}{3}$
For equilateral triangle: $AO = \frac{8}{3}$, so $AD = \frac{4}{3} + \frac{8}{3} = 4$
$\frac{|\alpha + 2\sqrt{2}\beta - 4|}{3} = 4 \Rightarrow \alpha = \frac{16}{9}$ or $-\frac{8}{9}$
Since $A(\alpha, \beta)$ and $O(0,0)$ lie on the same side of BC (origin gives $0+0-4 < 0$), we need $\alpha + 2\sqrt{2}\beta - 4 < 0$.
$(\alpha, \beta) = \left(-\frac{8}{9}, -\frac{16\sqrt{2}}{9}\right)$
$\lfloor |\alpha + \sqrt{2}\beta| \rfloor = \left\lfloor \left|\frac{-8-32}{9}\right| \right\rfloor = \left\lfloor \frac{40}{9} \right\rfloor = 4$
Hence, the correct answer is option: 'D'.