Straight Lines
Equilateral Triangle Orthocentre at Origin — Floor of |α+√2β|
nta_pyq_2026_jan
Grade 11
Question:
Let $ABC$ be an equilateral triangle with orthocentre 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
Step-by-Step Solution
Key Concept: Slope of $BC=-1/(2\sqrt{2})$. Since $OA\perp BC$: slope of $OA=2\sqrt{2}\Rightarrow\beta=2\sqrt{2}\alpha$. Distance $O$ to $BC$: $OD=\tfrac{4}{\sqrt{1+8}}=\tfrac{4}{3}$. For equilateral: $OA=2OD=\tfrac{8}{3}$.
$\lfloor|\alpha+\sqrt{2}\beta|\rfloor=4$.
Correct Answer: 4