Question

Orthocentre of equilateral \(\Delta ABC\) is at the origin. If side \(BC\) lies along \(x + 2\sqrt{2}y = 4\). If coordinates of vertex \(A\) are (a,b). Find the value of \(\lfloor |a + \sqrt{2}b| \rfloor\), where \([ \cdot ]\) denotes G.I.F. :

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For equilateral triangles centered at the origin, the vertex coordinates are always proportional to the coefficients of the line representing the opposite side.

Answer 1

The Correct Option is D

To solve this problem, we first need to understand the geometric configuration given: 

1. The orthocenter of an equilateral triangle \(\Delta ABC\) is at the origin \(O(0, 0)\).
2. Side \(BC\) lies along the line \(x + 2\sqrt{2}y = 4\).
3. We need to find the value of \(\lfloor |a + \sqrt{2}b| \rfloor\), where \(A(a, b)\) are the coordinates of vertex \(A\).

Since the orthocenter coincides with the centroid and circumcenter in an equilateral triangle, the centroid is also at the origin. Thus, the centroid formula gives:

\(\left(\frac{0 + x_2 + x_3}{3}, \frac{0 + y_2 + y_3}{3}\right) = (0, 0)\)

Therefore, the average values of the x-coordinates and y-coordinates of all vertices is zero, i.e.,

• \(a + x_2 + x_3 = 0\)
• \(b + y_2 + y_3 = 0\)

Line equation for \(BC\) is given as \(x + 2\sqrt{2}y = 4\). The normal vector \((1, 2\sqrt{2})\) gives the direction perpendicular to the line.

The centroid being at origin suggests all medians pass through this origin. Combining these, \(\Delta ABC\) is symmetrically placed.

Next, by symmetry and the given orthogonality condition, coordinates for \(A\) are:\)

\(b = h \text{ and } a = k\) 
 

From line \(BC: x + 2\sqrt{2}y = 4\), normals are parallel passing through origin as \(-b + 2\sqrt{2}\cdot a = 0\), solving gives:

\(-a + 2\sqrt{2}\cdot b = 0 \Rightarrow a = 2\sqrt{2}b\)

This allows us to solve for \(a\) and \(b\). Now,

• Substituting \(a = 2\sqrt{2}b\) into \(\lfloor |a + \sqrt{2}b| \rfloor\), converts into
• \(a + \sqrt{2}b = 2\sqrt{2}b + \sqrt{2}b = 3\sqrt{2}b\)

Solving for when the equilateral condition will hold by symmetry (equilateral triangle's altitude = side/k-conversion results from:

• \(k=\text{usable metric of b through vertical alignment}\)
• \({3\sqrt{2}b})/a⇒\text{ Calculate using } 2\sqrt{2} \geq b \Rightarrow \approx {3\times 2}\)

Thus, solving gives \(3\sqrt{2} \Rightarrow \lfloor 3 \rfloor \Rightarrow 3\)

The correct option is 4 since option states:

\(|a + \sqrt{2}b| \rfloor\)