Let the complex number z1, z2, z3 be...

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 1981Moderate

Let the complex number $z_{1}, z_{2}, z_{3}$ be the vertices of an equilateral triangle. Let $z_{0}$ be the circumcentre of the triangle. Then prove that $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}$ .

Visualized Solution (Hindi)

Visualizing the Equilateral Triangle

Let $z_{1}, z_{2}, z_{3}$ be the vertices of an equilateral triangle.
Let $z_{0}$ be the circumcentre (which is also the centroid for an equilateral triangle).

The Equilateral Condition

For an equilateral triangle, the following identity holds:
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}$

Defining the Circumcentre .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z_{0}

Since the triangle is equilateral, the circumcentre $z_{0}$ is the same as the centroid:
$z_{0} = \frac{z _{1} + z _{2} + z _{3}}{3}$

Squaring the Relation

Squaring both sides of the circumcentre equation:
$z_{0}^{2} = (\frac{z _{1} + z _{2} + z _{3}}{3})^{2}$
Which simplifies to:
$9 z_{0}^{2} = (z_{1} + z_{2} + z_{3})^{2}$

Expanding the Square

Expanding the right side using $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)$ :
$9 z_{0}^{2} = z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1})$

The Final Substitution

Substitute $z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} = z_{1}^{2} + z_{2}^{2} + z_{3}^{2}$ into the equation:
$9 z_{0}^{2} = (z_{1}^{2} + z_{2}^{2} + z_{3}^{2}) + 2 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2})$
$9 z_{0}^{2} = 3 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2})$

Conclusion and Takeaway

Dividing both sides by $3$ :
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}$
Key Takeaway: For an equilateral triangle, the sum of squares of vertices is thrice the square of the circumcentre.
Next Challenge: Can you prove this using the rotation theorem ( $e^{iπ /3}$ )?
Next Challenge: What happens if the circumcentre $z_{0}$ is at the origin ( $0$ )?
Next Challenge: Does a similar relation exist for a square?

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