Problem
Let A₁, A₂, …, A₈ 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 PAᵢ denote the distance between the points P and Aᵢ for i = 1, 2, … 8. If P varies over the circle, then find the maximum value of the product PA₁PA₂…PA₈.
Solution
Firstly, we can safely assume the circle to be centered at origin, and one of the vertices at (2, 0). Now, it’s time to bring in complex numbers! I’ll discuss two (somewhat similar) methods.
Method 1
If α = \(2e^{\frac{2{\pi}i}{8}}\), then A₁, A₂, A₃, …, A₈ would be represented by 2, 2α, 2α², …, 2α⁷ respectively, i.e. the complex 8th roots of 2⁸.
And, if z be any point on the circle, we can represent the product PA₁PA₂…PA₈ as |z – 2||z – 2α||z – 2α²|…|z – 2α⁷|.
Finally, using the fundamental theorem of algebra, we have (z⁸ – 2⁸) = (z – 2)(z – 2α)(z – 2α²)…(z – 2α⁷).
Taking the absolute value of both sides, we’ll get |z⁸ – 2⁸| = |z – 2||z – 2α||z – 2α²|…|z – 2α⁷|, the RHS representing the required product.
So, to maximize the RHS, we’ll maximize the much simpler expression |z⁸ – 2⁸| instead.
Since z represents any point on a circle of radius 2, centered at 0, we can assume it to be \(2e^{i\theta}\).
So, the expression |z⁸ – 2⁸| becomes |2⁸\(e^{i8\theta}\) – 2⁸| or 2⁸|\(e^{i8\theta}\) – 1|.
Now, \(e^{i8\theta}\) equals cos(8θ) + isin(8θ). So, \(e^{i8\theta}\) – 1 equals cos(8θ) + isin(8θ) – 1.
Using the half-angle formulas, the previous expression becomes 2isin(4θ)(cos(4θ) + isin(4θ)) or 2isin(4θ)\(e^{i4\theta}\). I’ve skipped a few steps here – do verify this yourself.
Finally, the expression |z⁸ – 2⁸| becomes 2⁸|2isin(4θ)\(e^{i4\theta}\)| or 2⁸.|2i|.|sin(4θ)|.|\(e^{i4\theta}\)|, which simplifies to 2⁹|sin(4θ)|.
And what’s the maximum value of |sin(4θ)|? One!
So, the maximum value of 2⁹|sin(4θ)| or |z⁸ – 2⁸| or |z – 2||z – 2α||z – 2α²|…|z – 2α⁷| or PA₁PA₂…PA₈ is 2⁹ or 512.
Method 2
What if we were to maximize the same product if the radius of the circle was 1? In that case, we could have assumed A₁, A₂, A₃, …, A₈ to be 1, α, α², …, α⁷ respectively, where α = \(e^{\frac{2{\pi}i}{8}}\).
And, what we need to maximize is |z – 1||z – α||z – α²|…|z – α⁷|, where z represents any point on the circle.
Again, using the fundamental theorem of algebra, we’ll get |z⁸ – 1| = |z – 1||z – α||z – α²|…|z – α⁷|.
So, to maximize the RHS, we’ll maximize the LHS instead, as we did earlier.
On substituting z = \(e^{i\theta}\), the LHS becomes |\(e^{i8\theta}\) – 1| or |2isin(4θ)\(e^{i4\theta}\)|. And this simplifies to |2sin(4θ)|, whose maximum value is 2.
Now, this is the maximum value of PA₁PA₂…PA₈ when the radius of the circle is 1. What if we double the radius?
That’s equivalent to scaling the entire figure by 2. That is, doubling each of the distance PA₁, PA₂, …, PA₈.
So, the maximum value of PA₁PA₂…PA₈ increases by 2⁸ times (each term doubles), and becomes 2⁸ x 2 or 512 – same as we got before.
Comments
A pretty straightforward application of the geometric application of complex roots of unity. The only little twist was the radius being 2. Otherwise, a fairly easy problem.
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