ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD = 2AC. If the poi

Q: Wait, why do we have a $\pm$ sign in the rotation formula?

That is a fantastic observation! In the complex plane, a $90^\circ$ rotation can be either counter-clockwise (multiplication by $i$) or clockwise (multiplication by $-i$). Since the problem asks for the position of $A$ and doesn't specify the orientation of the rhombus relative to the axes, both directions are valid, leading to two possible positions for $A$.

Q: Why can't we just use coordinate geometry with slopes?

You absolutely can! You could find the slope of $BD$, determine the perpendicular slope for $AC$, and use the distance formula. However, that approach is algebraically heavy and prone to calculation errors. Using complex numbers allows us to treat rotation and scaling as a single, elegant algebraic operation, which is a powerful tool in your JEE arsenal.

Q: Is $M$ definitely the midpoint of $AC$?

Yes, it is! A fundamental property of any rhombus is that its diagonals bisect each other at $90^\circ$. This means the intersection point $M$ is the midpoint of both $AC$ and $BD$. This geometric fact is the key that unlocks the entire problem.

Question

$ABCD$ is a rhombus. Its diagonals $AC$ and $BD$ intersect at the point $M$ and satisfy $BD = 2AC$. If the points $D$ and $M$ represent the complex numbers $1 + i$ and $2 - i$ respectively, then $A$ represents the complex number $\dots$ or $\dots$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Complex Plane

Imagine you are standing on the complex plane. You have two points, $D = 1 + i$ and $M = 2 - i$. These are the anchors of a rhombus, $ABCD$.

The intersection of the diagonals, $M$, is our pivot point. In the world of JEE geometry, a rhombus is a structure of perfect symmetry where the diagonals $AC$ and $BD$ bisect each other at exactly $90^{\circ}$. This perpendicularity is our golden ticket.

### Unlocking the Ratio

We are given the condition $BD = 2AC$. Because $M$ is the midpoint of both diagonals, we know that $MD = \frac{1}{2}BD$ and $MA = \frac{1}{2}AC$.

If we substitute these into our given condition, we get $2MD = 2(2MA)$, which simplifies beautifully to:

$$MD = 2MA$$

This tells us that the segment $MA$ is exactly half the length of $MD$. We have now established both the angular relationship ($90^{\circ}$) and the scaling factor ($1/2$).

### The Power of Rotation

This is where complex numbers shine. To move from $M$ to $A$, we are essentially taking the vector $\vec{MD}$, scaling it by $1/2$, and rotating it by $90^{\circ}$.

In the complex plane, a $90^{\circ}$ rotation is simply multiplication by $i$. Since we don't know if $A$ is to the left or right of the diagonal $BD$, we must account for both $\pm i$.

The vector $\vec{MD}$ is calculated as:

$$z_D - z_M = (1 + i) - (2 - i) = -1 + 2i$$

Now, we apply our rotation theorem:

$$z_A - z_M = \pm \frac{1}{2} i (z_D - z_M)$$

### The Final Calculation

Let's perform the arithmetic with care. We have $\vec{MA} = \pm \frac{1}{2} i (-1 + 2i)$.

For the positive case ($+i$):

$$\vec{MA} = \frac{1}{2} (-i + 2i^2) = \frac{1}{2} (-2 - i) = -1 - \frac{1}{2}i$$

Adding this to $z_M = 2 - i$, we get:

$$z_A = (2 - i) + (-1 - \frac{1}{2}i) = 1 - \frac{3}{2}i$$

For the negative case ($-i$):

$$\vec{MA} = -\frac{1}{2} (-i + 2i^2) = \frac{1}{2} (2 + i) = 1 + \frac{1}{2}i$$

Adding this to $z_M = 2 - i$, we get:

$$z_A = (2 - i) + (1 + \frac{1}{2}i) = 3 - \frac{1}{2}i$$

### Conclusion

By leveraging the geometric properties of the rhombus and the algebraic elegance of complex numbers, we have found the two possible positions for $A$:

**$z_A = 3 - \frac{i}{2}$ or $z_A = 1 - \frac{3}{2}i$**

This problem is a perfect reminder that in JEE, the most complex-looking geometry problems often collapse into simple, elegant rotations when viewed through the lens of complex numbers. Keep practicing this, and you will start seeing these rotations everywhere!

Visualized Solution

Plotting $M$ and $D$

Rhombus Geometry

Relating the Half-Diagonals

Calculating Vector $M D$

The Rotation Concept

Calculating the Rotated Vectors

Finding Coordinates of $A$

Final Rhombus Structure

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Geometry of the Complex Plane

Unlocking the Ratio

The Power of Rotation

The Final Calculation

Conclusion

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