JEE Advanced 1981

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: 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 $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}$ .

Progressive Hints

Hint 1

Hint 1: Visualize the triangle. In an equilateral triangle, the circumcenter, centroid, orthocenter, and incenter all collapse into a single, beautiful point. Start by identifying the relationship between the vertices

z_{1}, z_{2}, z_{3}

and the circumcenter

z_{0}

Hint 2

Hint 2: Recall the fundamental condition for an equilateral triangle in the complex plane:

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

. This is your bridge between the geometry of the triangle and the algebra of the vertices.

Hint 3

Hint 3: Begin your algebraic journey by expressing the centroid

z_{0}

as the average of the vertices:

z_{0} = \frac{z _{1} + z _{2} + z _{3}}{3}

. Now, square both sides of this equation and expand the trinomial

(z_{1} + z_{2} + z_{3})^{2}

. Watch how the cross-terms align with the identity from Hint 2.

Visualized Solution

Vertices of the Equilateral Triangle

Let $z_{1}, z_{2}, z_{3}$ be the vertices of an equilateral triangle in the complex plane.

The Circumcentre $z_{0}$

Let $z_{0}$ be the circumcentre.
For an equilateral triangle, the circumcentre and centroid coincide.

The Equilateral Identity

The fundamental identity for an equilateral triangle is:
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}$

Centroid Formula

Using the centroid formula, we can express $z_{0}$ as:
$z_{0} = \frac{z _{1} + z _{2} + z _{3}}{3}$

Rearranging the Centroid Equation

Multiplying both sides by $3$ to remove the fraction:
$z_{1} + z_{2} + z_{3} = 3 z_{0}$

Squaring Both Sides

Squaring both sides of the equation:
$(z_{1} + z_{2} + z_{3})^{2} = (3 z_{0})^{2}$

Expanding the Left Hand Side

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

Applying the Equilateral Condition

Substitute the equilateral identity $z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} = z_{1}^{2} + z_{2}^{2} + z_{3}^{2}$ :

Simplifying the Expression

Replacing the mixed products with pure squares:
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2}) = 9 z_{0}^{2}$

Combining Like Terms

Combining the like terms on the left side:
$3 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2}) = 9 z_{0}^{2}$

Final Conclusion

Dividing both sides by $3$ yields the final result:
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}$

00:00 / 00:00

The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

Imagine a perfect, equilateral triangle resting on the Argand plane. Its vertices,

z_{1}

z_{2}

, and

z_{3}

, are complex numbers representing positions in the complex plane.

Every triangle has a circumcenter, which we denote as

z_{0}

. Because the triangle is equilateral, its circumcenter is identical to its centroid.

The centroid is the arithmetic mean of the vertices, defined by the following relation:

z_{0} = \frac{z _{1} + z _{2} + z _{3}}{3}

By multiplying both sides by

3

, we establish the linear relationship:

z_{1} + z_{2} + z_{3} = 3 z_{0}

The Equilateral Identity

To proceed, we invoke the fundamental identity for equilateral triangles in the complex plane. This identity captures the essence of the triangle's rotational symmetry:

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

This identity serves as the bridge between the linear sum of the vertices and their quadratic interactions. It is the "magic key" for solving problems involving equilateral geometry.

The Grand Unification

We return to our centroid equation,

z_{1} + z_{2} + z_{3} = 3 z_{0}

, and square both sides to introduce the quadratic terms:

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

Expanding the left side using the trinomial expansion formula

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)

, we obtain:

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

We now substitute the pairwise product sum with the equilateral identity

z_{1}^{2} + z_{2}^{2} + z_{3}^{2}

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

Final Calculation

Combining the terms on the left side, we simplify the expression to:

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

Dividing both sides by

3

, we arrive at the final, elegant relationship:

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

We have successfully demonstrated that for an equilateral triangle with centroid

z_{0}

, the sum of the squares of the vertices is equal to three times the square of the centroid.

Common Traps & FAQs

Wait, why is the circumcenter the same as the centroid? Is this always true?

That is a fantastic question! In a general triangle, the circumcenter, centroid, and orthocenter are distinct points that form the Euler line. However, an equilateral triangle possesses perfect rotational symmetry. Every median is also an altitude and an angle bisector. Because the median passes through the centroid and the altitude passes through the orthocenter, they must coincide. Thus, for an equilateral triangle,

z_{0}

is both the centroid and the circumcenter.

How do I derive

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

if I forget it in the exam?

Do not panic! You can derive it using rotation. If you rotate

z_{1}

about

z_{2}

6 0^{\circ}

(or

\frac{π}{3}

radians), you get

z_{3}

. Using the rotation formula

z_{3} - z_{2} = (z_{1} - z_{2}) e^{iπ /3}

, you can manipulate this into the identity. It is a powerful exercise in complex geometry that reinforces your understanding of how complex numbers represent rotations!

Why can't we just use energy conservation or some other physics principle here?

While physics intuition is wonderful, this is a problem of pure complex geometry. We are dealing with the algebraic structure of the complex plane. The 'energy' here is the symmetry of the triangle itself. We are using the algebraic properties of complex numbers to prove a geometric truth. Stick to the complex algebra—it is the most elegant path to the solution.

Animated Solution for Mathematics - Complex Numbers: 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

Vertices of the Equilateral Triangle

The Circumcentre $z_{0}$

The Equilateral Identity

Centroid Formula

Rearranging the Centroid Equation

Squaring Both Sides

Expanding the Left Hand Side

Applying the Equilateral Condition

Simplifying the Expression

Combining Like Terms

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Equilateral Identity

The Grand Unification

Final Calculation

Similar Questions

Prove that the complex numbers $z_{1}, z_{2}$ and the origin form an equilateral triangle only if $z_{1}^{2} + z_{2}^{2} - z_{1} z_{2} = 0$ .

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

Show that the area of the triangle on the Argand diagram formed by the complex numbers $z, i z$ and $z + i z$ is $\frac{1}{2} ∣ z ∣^{2}$ .

The complex numbers $z_{1}, z_{2}$ and $z_{3}$ satisfying $\frac{z _{1} - z _{3}}{z _{2} - z _{3}} = \frac{1 - i 3}{2}$ are the vertices of a triangle which is

Suppose $Z_{1}, Z_{2}, Z_{3}$ are the vertices of an equilateral triangle inscribed in the circle $∣ Z ∣ = 2$ . If $Z_{1} = 1 + i 3$ then $Z_{2} = \dots, Z_{3} = \dots$

Let $Z_{1}$ and $Z_{2}$ be two roots of the equation $Z^{2} + a Z + b = 0$ , $Z$ being complex. Further, assume that the origin, $Z_{1}$ and $Z_{2}$ form an equilateral triangle. Then

If $a, b, c$ , are the numbers between 0 and 1 such that the points $z_{1} = a + i, z_{2} = 1 + bi$ and $z_{3} = 0$ form an equilateral triangle, then $a = \dots$ and $b = \dots$

Let a complex number be $w = 1 - 3 i$ . Let another complex number $z$ be such that $∣ z w ∣ = 1$ and $ar g (z) - ar g (w) = \frac{π}{2}$ . Then the area of the triangle with vertices origin, $z$ and $w$ is equal to:

If one the vertices of the square circumscribing the circle $∣ z - 1∣ = 2$ is $2 + 3 i$ . Find the other vertices of the square.

Let $z_{1} = 10 + 6 i$ and $z_{2} = 4 + 6 i$ . If $z$ is any complex number such that the argument of $\frac{z - z _{1}}{z - z _{2}}$ is $\frac{π}{4}$ , then prove that $∣ z - 7 - 9 i ∣ = 32$ .

Visualized Solution

Vertices of the Equilateral Triangle

The Circumcentre .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z0​

The Equilateral Identity

Centroid Formula

Rearranging the Centroid Equation

Squaring Both Sides

Expanding the Left Hand Side

Applying the Equilateral Condition

Simplifying the Expression

Combining Like Terms

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Equilateral Identity

The Grand Unification

Final Calculation

Similar Questions

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The Circumcentre $z_{0}$

Prove that the complex numbers $z_{1}, z_{2}$ and the origin form an equilateral triangle only if $z_{1}^{2} + z_{2}^{2} - z_{1} z_{2} = 0$ .

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

Show that the area of the triangle on the Argand diagram formed by the complex numbers $z, i z$ and $z + i z$ is $\frac{1}{2} ∣ z ∣^{2}$ .

The complex numbers $z_{1}, z_{2}$ and $z_{3}$ satisfying $\frac{z _{1} - z _{3}}{z _{2} - z _{3}} = \frac{1 - i 3}{2}$ are the vertices of a triangle which is

Suppose $Z_{1}, Z_{2}, Z_{3}$ are the vertices of an equilateral triangle inscribed in the circle $∣ Z ∣ = 2$ . If $Z_{1} = 1 + i 3$ then $Z_{2} = \dots, Z_{3} = \dots$

Let $Z_{1}$ and $Z_{2}$ be two roots of the equation $Z^{2} + a Z + b = 0$ , $Z$ being complex. Further, assume that the origin, $Z_{1}$ and $Z_{2}$ form an equilateral triangle. Then

If $a, b, c$ , are the numbers between 0 and 1 such that the points $z_{1} = a + i, z_{2} = 1 + bi$ and $z_{3} = 0$ form an equilateral triangle, then $a = \dots$ and $b = \dots$

If one the vertices of the square circumscribing the circle $∣ z - 1∣ = 2$ is $2 + 3 i$ . Find the other vertices of the square.

Let $z_{1} = 10 + 6 i$ and $z_{2} = 4 + 6 i$ . If $z$ is any complex number such that the argument of $\frac{z - z _{1}}{z - z _{2}}$ is $\frac{π}{4}$ , then prove that $∣ z - 7 - 9 i ∣ = 32$ .