JEE Mathematics 2026 Question with Solution

Given: $ABC$ is an equilateral triangle, its orthocenter is at the origin, and side $BC$ lies on $x+2\sqrt{2}\,y=4$.

Find: The greatest integer less than or equal to $|\alpha+\sqrt{2}\beta|$ where $A=(\alpha,\beta)$.

In an equilateral triangle, the orthocenter, centroid, and circumcenter coincide. Therefore, the centroid is also at the origin.

Let

Since the centroid is at the origin,

The midpoint of $BC$ is

Because this midpoint lies on the line $x+2\sqrt{2}\,y=4$,

So,

that is,

Since the altitude from $A$ is perpendicular to $BC$, and the line $x+2\sqrt{2}\,y=4$ has normal vector $\left(1,2\sqrt{2}\right)$, the point $A$ lies on the line through the origin in the direction $\left(1,2\sqrt{2}\right)$. Hence,

So,

Substituting into $\alpha+2\sqrt{2}\beta=-8$,

Now,

Therefore,

Hence,

So the greatest integer less than or equal to this value is $4$.

The solution concludes with option D, but the working gives $\frac{40}{9}$, whose floor is $4$. Therefore, the defensible correct option from the given options is B.

Given: The centroid of the equilateral triangle is at the origin and side $BC$ lies on $x+2\sqrt{2}\,y=4$.

Find: $\lfloor |\alpha+\sqrt{2}\beta| \rfloor$.

For an equilateral triangle, the median from $A$ to side $BC$ is also the altitude. Since the centroid is at the origin, the centroid lies on this altitude. Therefore, the altitude from $A$ passes through the origin and is perpendicular to $BC$.

The line $BC$ is

so a perpendicular direction is its normal vector

Hence,

Also, the midpoint of $BC$ is the point on the altitude between the centroid and the base. From the centroid relation, that midpoint is

Substituting in the line equation gives

which simplifies to

Now put $\alpha=t$ and $\beta=2\sqrt{2}\,t$:

Thus,

So,

and therefore

Hence the correct option by calculation is B.