Let A₁, A₂, A₃, ..., A₈ be the vertices of a regular octagon that lie on a circle of radius 2. Let P

Q: Why can't we just use coordinate geometry with $(x,y)$ coordinates? It seems more intuitive.

You absolutely could, but you would be walking into a trap. Using $(x,y)$ coordinates would force you to calculate the distance from $(x,y)$ to each vertex $(\cos\theta_k, \sin\theta_k)$, leading to a massive product of square roots and trigonometric terms. It would be an algebraic nightmare. Complex numbers are the 'superpower' here because they transform geometric distance into the modulus of a difference, allowing us to use the power of polynomial factorization.

Q: Wait, why is the product of distances equal to $|z^8 - 2^8 e^{i\alpha}|$?

This is the heart of the problem! Since $z_1, z_2, \dots, z_8$ are the roots of the polynomial $w^8 - 2^8 e^{i\alpha} = 0$, the Fundamental Theorem of Algebra tells us that we can factor this polynomial as $(w-z_1)(w-z_2)\dots(w-z_8)$. When we set $w=z$ (our point $P$), the product of the differences $(z-z_1)(z-z_2)\dots(z-z_8)$ becomes exactly $z^8 - 2^8 e^{i\alpha}$. Taking the modulus of both sides gives us the product of the distances.

Q: Why do we use the Triangle Inequality for the maximum? Is there a simpler way?

The Triangle Inequality is the most elegant way to find the bound. We have the expression $|z^8 - 2^8 e^{i\alpha}|$. This represents the distance between the complex number $z^8$ and the complex number $2^8 e^{i\alpha}$ in the complex plane. Since $|z|=2$, $z^8$ lies on a circle of radius $2^8$. The maximum distance between a point on a circle and another fixed point is achieved when they are diametrically opposite. The Triangle Inequality $|a - b| \le |a| + |b|$ naturally captures this geometric reality.

Question

Let $A_1, A_2, A_3, \dots, A_8$ be the vertices of a regular octagon that lie on a circle of radius 2. Let $P$ be a point on the circle and let $PA_i$ denote the distance between the points $P$ and $A_i$ for $i = 1,2,\dots,8$. If $P$ varies over the circle, then the maximum value of the product $PA_1 \cdot PA_2 \cdot \dots \cdot PA_8$, is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing before a regular octagon inscribed in a circle of radius $R=2$. You are asked to place a point $P$ anywhere on the circumference and calculate the product of the distances from $P$ to every single vertex $A_1, A_2, \dots, A_8$.

If you try to solve this using traditional Euclidean geometry—drawing lines, using the Law of Cosines, or attempting to multiply eight separate trigonometric expressions—you will quickly find yourself lost in a forest of identities.

This is the classic JEE trap: the problem is designed to look like a geometry question, but it is actually a test of your ability to recognize **algebraic structures**.

### The Complex Plane: Our Superpower

To solve this, we must shift our perspective. We leave the world of $(x,y)$ coordinates and enter the complex plane. Let the center of the circle be the origin $(0,0)$.

Now, our circle is defined by the simple equation $|z| = 2$. Any point $P$ on this circle is represented by a complex number $z$ such that $|z| = 2$.

Why is this so powerful? Because the vertices of a regular $n$-gon inscribed in a circle are not just random points; they are the roots of unity.

Specifically, if the octagon is rotated by an angle $\alpha$, the vertices $A_1, A_2, \dots, A_8$ are the roots of the equation:

$$w^8 = 2^8 e^{i\alpha}$$

This is the bridge between geometry and algebra. We have transformed a spatial arrangement into a polynomial equation: $w^8 - 2^8 e^{i\alpha} = 0$.

### The Polynomial Bridge

Now, let us look at the distance $PA_k$. In the complex plane, the distance between two points $z$ and $z_k$ is simply the modulus of their difference: $|z - z_k|$.

We want to find the product of these distances for all $k=1$ to $8$:

$$\prod_{k=1}^8 PA_k = \prod_{k=1}^8 |z - z_k|$$

Using the property that the product of moduli is the modulus of the product, we can write this as:

$$\left| \prod_{k=1}^8 (z - z_k) \right|$$

Here is the moment of clarity. Since $z_1, z_2, \dots, z_8$ are the roots of the polynomial $w^8 - 2^8 e^{i\alpha} = 0$, we know from the Fundamental Theorem of Algebra that the polynomial can be factored as:

$$w^8 - 2^8 e^{i\alpha} = (w - z_1)(w - z_2) \dots (w - z_8)$$

If we substitute $w = z$ into this identity, the entire product collapses into a single, beautiful expression:

$$\prod_{k=1}^8 (z - z_k) = z^8 - 2^8 e^{i\alpha}$$

### The Final Collapse

We have reduced the entire geometric problem to finding the value of $|z^8 - 2^8 e^{i\alpha}|$. We know that $P$ lies on the circle $|z|=2$, which implies $|z^8| = |z|^8 = 2^8 = 256$.

To find the value of $|z^8 - 2^8 e^{i\alpha}|$, we note that $z^8$ is a complex number with magnitude $256$, and $2^8 e^{i\alpha}$ is also a complex number with magnitude $256$.

Depending on the position of $P$ (the value of $z$), the product can vary. However, if we seek the maximum possible value, we invoke the Triangle Inequality, which states that for any complex numbers $a$ and $b$, $|a - b| \le |a| + |b|$.

Applying this to our expression:

$$\left| z^8 - 2^8 e^{i\alpha} \right| \le |z^8| + \left| 2^8 e^{i\alpha} \right|$$

Since $|z^8| = 256$ and $|2^8 e^{i\alpha}| = 2^8 \cdot |e^{i\alpha}| = 256 \cdot 1 = 256$, the maximum value is:

$$256 + 256 = 512$$

### Conclusion: The Elegance of JEE Mathematics

Look at what we have achieved. We started with a complex geometric configuration and, by choosing the right mathematical language—the language of complex numbers—we reduced it to a simple addition.

This is the essence of JEE Advanced preparation. It is not about brute force; it is about finding the elegant path.

When you see a problem involving regular polygons and distances, do not reach for your ruler; reach for your complex roots. You have the tools to solve anything. The **maximum product is 512**.

Enter Numerical Value:

Visualized Solution

The Geometric Setup

Shifting to Complex Plane

Vertices as Roots

Distance in Complex Plane

Product of Distances

Polynomial Factorization

The Simplified Expression

Triangle Inequality

Substituting the Modulus

Final Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Complex Plane

The Polynomial Bridge

The Final Collapse

Conclusion

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