JEE Main 2026 — Mathematics Coordinate Geometry

Let ABC be an equilateral triangle with orthocenter at the origin O(0,0) and side BC on the line $x + 2\sqrt{2}y = 4$x+22​y=4.

Since $m_{BC} \cdot m_{AD} = -1$mBC​⋅mAD​=−1:

$\left(-\frac{1}{2\sqrt{2}}\right)\left(\frac{\beta}{\alpha}\right) = -1 \Rightarrow \beta = 2\sqrt{2}\alpha$(−22​1​)(αβ​)=−1⇒β=22​α ...(1)

Distance from O to BC: $OD = \frac{|-4|}{\sqrt{1+8}} = \frac{4}{3}$OD=1+8​∣−4∣​=34​

For equilateral triangle: $AO = \frac{8}{3}$AO=38​, so $AD = \frac{4}{3} + \frac{8}{3} = 4$AD=34​+38​=4

$\frac{|\alpha + 2\sqrt{2}\beta - 4|}{3} = 4 \Rightarrow \alpha = \frac{16}{9}$3∣α+22​β−4∣​=4⇒α=916​ or $-\frac{8}{9}$−98​

Since $A(\alpha, \beta)$A(α,β) and $O(0,0)$O(0,0) lie on the same side of BC (origin gives $0+0-4 < 0$0+0−4<0), we need $\alpha + 2\sqrt{2}\beta - 4 < 0$α+22​β−4<0.

$$(\alpha, \beta) = \left(-\frac{8}{9}, -\frac{16\sqrt{2}}{9}\right)$$(α,β)=(−98​,−9162​​)

$$\lfloor |\alpha + \sqrt{2}\beta| \rfloor = \left\lfloor \left|\frac{-8-32}{9}\right| \right\rfloor = \left\lfloor \frac{40}{9} \right\rfloor = 4$$⌊∣α+2​β∣⌋=⌊​9−8−32​​⌋=⌊940​⌋=4