Aspire Faculty ID #18091 · Topic: JEE Main 2026 (28 January Morning Shift) · Just now
JEE Main 2026 (28 January Morning Shift)

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

Solution

$ m_{BC} \cdot m_{AD} = -1 $

$ \Rightarrow \left( -\frac{1}{2\sqrt{2}} \right)\left( \frac{\beta}{\alpha} \right) = -1 $

$ \Rightarrow \beta = 2\sqrt{2}\alpha \quad ...(1)$

$ \therefore OD = \frac{|4|}{\sqrt{1 + 8}} = \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} \text{ or } -\frac{8}{9} $

$ \because A(\alpha,\beta) $ & $(0,0)$ lies on same side of given line

$ \therefore (\alpha,\beta) = \left( -\frac{8}{9}, -\frac{16\sqrt{2}}{9} \right) $

$ \Rightarrow [\alpha + \sqrt{2}\beta] = \left[ \frac{-8 - 32}{9} \right] = 4 $
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