(A) 5

(B) 4

(C) 2

(D) 3

Correct Answer: 4

Explanation

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

The side BC lies on the line

x + 2√2y − 4 = 0

The vertex A is opposite to side BC. In an equilateral triangle, the distance of the centroid from any side is one–third of the altitude.

Hence, the perpendicular distance of the origin from the line BC equals one–third of the altitude of the triangle.

Distance of origin from the line x + 2√2y − 4 = 0 is

|−4| / √(1² + (2√2)²) = 4 / √(1 + 8) = 4 / 3

Thus, the altitude of the equilateral triangle is

3 × (4 / 3) = 4

Vertex A lies on the perpendicular to BC passing through the origin. The direction ratios of the normal to BC are (1, 2√2).

Hence, the coordinates of A can be written as

(α, β) = t(1, 2√2)

Since the distance of A from the line BC equals the altitude 4:

|t(1) + 2√2 · t(2√2) − 4| / 3 = 4

Simplifying,

|9t − 4| = 12

This gives

t = 16/9 or t = −8/9

Now compute |α + √2β|:

α + √2β = t(1 + 4) = 5t

So the possible values are

|5 × 16/9| = 80/9 ≈ 8.88 |5 × (−8/9)| = 40/9 ≈ 4.44

The required greatest integer less than or equal to |α + √2β| is

4

Related JEE Main Questions