If are the vertices of an equilateral triangle, whose centroid is , then is equal to [2025]

Question

If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2}$ is equal to [2025]

Smart Achievers · Accepted Answer

(1)

Centroid of triangle is $z_{0}$ , then $z_{1} + z_{2} + z_{3} = 3 z_{0}$

$\Rightarrow {(z_{1} + z_{2} + z_{3})}^{2} = 9 z_{0}^{2}$

$\Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2}) = 9 z_{0}^{2}$ [ $∵$ For equilateral $△, z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} = z_{1}^{2} + z_{2}^{2} + z_{3}^{2}$ ]

$\Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}$

Now, $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2} = {(z_{1} - z_{0})}^{2} + {(z_{2} - z_{0})}^{2} + {(z_{3} - z_{0})}^{2}$

$= z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 3 z_{0}^{2} - 2 (z_{1} + z_{2} + z_{3}) z_{0}$

$= 6 z_{0}^{2} - 6 z_{0}^{2} = 0$ .

Solution

Q.
If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2}$ is equal to [2025]

1 0

2 i

3 – i

4 1

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Solution

Q. If z1,z2,z3∈C are the vertices of an equilateral triangle, whose centroid is z0, then ∑k=13(zk–z0)2 is equal to [2025]

1 0

2 i

3 – i

4 1

Ans. (1) Centroid of triangle is z0, then z1+z2+z3=3z0 ⇒ (z1+z2+z3)2=9z02 ⇒ z12+z22+z32+2(z12+z22+z32)=9z02 [∵ For equilateral △, z1z2+z2z3+z3z1=z12+z22+z32] ⇒ z12+z22+z32=3z02 Now, ∑k=13(zk–z0)2=(z1–z0)2+(z2–z0)2+(z3–z0)2 =z12+z22+z32+3z02–2(z1+z2+z3)z0 =6z02–6z02=0.

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Q.
If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $\sum_{k = 1}^{3} {(z_{k} - z_{0})}^{2}$ is equal to [2025]