Question

If z1, z2, z3 are the vertices of the equilateral triangle and the z0 be its orthocentre, such that z12+z22+z32 = Kz02, then K equals

• A. 6
• B. 2
• C. 9
• D. 3

Answer 1

The Correct Option is D

Let's first understand the relationship between the vertices of an equilateral triangle and its orthocenter. For any equilateral triangle, the centroid, circumcenter, incenter, and orthocenter all coincide. In other words, for an equilateral triangle, the orthocenter is also its centroid.
Given that \(( z_1, z_2, )\) and \(( z_3 )\) are the vertices of the equilateral triangle, the centroid (which is also the orthocenter \(( z_0 )\) is given by:
\([ z_0 = \frac{z_1 + z_2 + z_3}{3} ]\)Squaring both sides: \([ z_0^2 = \frac{z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1)}{9} ]\) 
Given: \([ z_1^2 + z_2^2 + z_3^2 = Kz_0^2 ]\) Substituting for \(( z_0^2 )\) from the above equation: \([ z_1^2 + z_2^2 + z_3^2 = K\) c dot \(\frac{z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1)}{9} ]\)
Simplifying: \([ 9 = K(3 + 2[z_1z_2 + z_2z_3 + z_3z_1]/[z_1^2 + z_2^2 + z_3^2]) ]\) 
Because  \(( z_1, z_2, )\)and \(( z_3 )\) are vertices of an equilateral triangle, the product of any two vertices will have the same magnitude. 
Therefore, \(( z_1z_2 + z_2z_3 + z_3z_1 )\) will have a certain magnitude (let's call it\(( M )),\) and this magnitude is divided by the sum of the squares of the vertices, which is again a fixed magnitude. 
Thus, \(( K )\) will be a constant value. Comparing the equation to the given expression \(( z_1^2 + z_2^2 + z_3^2 = Kz_0^2 ),\) we see that: \([ K = 3 ]\).
So, the correct answer is: D