Maths Q5 – JEE-MAIN 2025

Home Maths Complex & QuadraticQuestion #90

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

Clue 1 of 3: Centroid Property

Understand how the centroid of a triangle is defined in terms of its vertices in the complex plane.

🥷

Ninja Strategy•Symmetry and Common Properties

Recognize that the geometric symmetry of an equilateral triangle implies the result must be real, eliminating complex options. Furthermore, sums of squares in such contexts often simplify to zero due to properties of complex roots of unity.

🧠 Full Solution Path

Step 1: Translate the triangle to the origin✦ Active

Let $w_{k} = z_{k} - z_{0}$ for $k = 1, 2, 3$ . These new vertices $w_{1}, w_{2}, w_{3}$ also form an equilateral triangle. The centroid of this new triangle is $w_{0} = \frac{w_{1} + w_{2} + w_{3}}{3}$ . Substituting the definition of $w_{k}$ and the centroid $z_{0} = \frac{z_{1} + z_{2} + z_{3}}{3}$ :

w_{0} = \frac{(z_{1} - z_{0}) + (z_{2} - z_{0}) + (z_{3} - z_{0})}{3} = \frac{(z_{1} + z_{2} + z_{3}) - 3 z_{0}}{3}

Since $z_{1} + z_{2} + z_{3} = 3 z_{0}$ , we have $w_{0} = \frac{3 z_{0} - 3 z_{0}}{3} = 0$ . Thus, the new triangle is centered at the origin, which implies $w_{1} + w_{2} + w_{3} = 0$ .

Step 2: Apply the property of equilateral triangles○ Expand

For any equilateral triangle with vertices $w_{1}, w_{2}, w_{3}$ , the following identity holds:

w_{1}^{2} + w_{2}^{2} + w_{3}^{2} = w_{1} w_{2} + w_{2} w_{3} + w_{3} w_{1}

💡 Teacher's Secret Hint

This property is crucial for equilateral triangles in the complex plane.

Step 3: Combine identities to find the sum○ Expand

We know the algebraic identity for the square of a sum:

(w_{1} + w_{2} + w_{3})^{2} = w_{1}^{2} + w_{2}^{2} + w_{3}^{2} + 2 (w_{1} w_{2} + w_{2} w_{3} + w_{3} w_{1})

From Step 1, we have $w_{1} + w_{2} + w_{3} = 0$ . From Step 2, we have $w_{1}^{2} + w_{2}^{2} + w_{3}^{2} = w_{1} w_{2} + w_{2} w_{3} + w_{3} w_{1}$ . Let $S = w_{1}^{2} + w_{2}^{2} + w_{3}^{2}$ . Then $w_{1} w_{2} + w_{2} w_{3} + w_{3} w_{1} = S$ . Substituting these into the algebraic identity:

0^{2} = S + 2 (S) ⟹ 0 = 3 S ⟹ S = 0

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

💡 Teacher's Secret Hint

Remember that $1 + ω + ω^{2} = 0$ is a fundamental property of cube roots of unity, which can also be used to derive this result by setting $w_{k} = R ω^{k - 1}$ .

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Maths Question 5 – JEE-MAIN 2025

Clue 1 of 3: Centroid Property

🧠 Full Solution Path