Polar Equation of a Hexagon

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Have you ever heard of a squircle? It’s kind of like a circle, but with more pronounced “edges.” When you plot one out, it looks like a beveled square (at least in the first quadrant). Here’s an example:

\sqrt[p]{x^p+y^p}=1 where p>2

When p=4:

    MSP10791gh815119g79g4c700003eb5a454hi49eic3

Isn’t that nice? It looks so happy!

But, in a wildly unrelated tangent, I was wondering, how would you represent a hexagon? More specifically, how would you represent a hexagon in the polar coordinate system?

Just for fun, I’m going to couch this in terms of a scientific investigation.

Question: What is the polar equation for a hexagon?
Hypothesis: It’ll involve an absolute-value sine function. That’s because the x-y graph should look something like this:

hypothesis

Variables: Independent = x, \theta; Dependent = y. (See what I did there? Okay…)

Investigation: I decided to begin by dealing with the equilateral triangle in the first quadrant. It looks like this:

diagram

I then solved for the value of x.

x=\frac{\sin60^\circ}{\sin(120^\circ-\theta)}=\frac{\frac{\sqrt3}{2}}{\sin120^\circ\cos\theta-\sin\theta\cos120^\circ}=\frac{\frac{\sqrt3}{2}}{\frac{\sqrt3}{2}\cos\theta+\frac{1}{2}\sin\theta}

This solution can be converted into a different form:

x=\frac{\sqrt3}{\sqrt3\cos\theta+\sin\theta}=\frac{\sqrt3}{2\sin(\theta+60^\circ)}=\frac{\sqrt3}{2}\csc\left(\theta+60^\circ\right)

Now, if you add a floor function, you can have it repeat for angles up to 360 degrees!

x=\frac{\sqrt3}{2}\csc(\theta+60^\circ-60^\circ\lfloor\frac{\theta}{60^\circ}\rfloor)=\boxed{\frac{\sqrt3}{2}\csc(\theta-60^\circ\lfloor\frac{\theta}{60^\circ}+1\rfloor)}

Conclusion: The hypothesis was incorrect – the graph is actually a concatenation of segments from a cosecant graph. Below is a graph of this equation in Wolfram Alpha!

equation

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2 thoughts on “Polar Equation of a Hexagon”

    • I’m working on it right now actually! I have an answer, but it’s really complex and I want to simplify it first. If I can’t find a way to simplify, I’ll go ahead and post the long equation.

      Reply

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