Here is a desmos graph that visualizes what I am about to say
- Okay, let's say we have a polygon with $s$ sides and $a = \frac{360°}{s}$.
- All of those polygon's angles are equal and all of it's sides are also equal.
- The polygon is formed using the sine and cosine methods (look at the desmos graph for refrence).
My hypothesis is that if a $\sin(x \times a)$ wave is run through the shape and $s$ is any even number, or $5$ or $7$ then the sine will cut the the polygon into two exact halves (which all the times I observed the halves will not be cut at the polygon's points)
TL;DR
What I need help with is proving that if you run a $\sin(x \times a)$ wave through a polygon with $s$ sides, and $a = \frac{360°}{s}$, and $s = \text{even}$ or $s = 5$ or $s = 7$ then the wave will cut the polygon exactly in half. Ergo all the postive $s$ values that won't cut it in half are $3, 9, 11, 13, 15, 17, 19, 21, ...$
Can someone help me prove this? Seems like a fun question to tackle.
Side note:
If you want to complicate it even more let's have not only $y = \sin(x \times a)$ but also $x = \sin(y \times a)$, have fun!
Suppose $A$ is an area symmetric about the $y$-axis and $x$-axis, and $f$ is an odd function. We then have that $$\begin{aligned} |(x,y)\in A : y >f(x)| &= |(-x,-y)\in A : y <f(x)| \\ &= |(x,-y)\in A : y <f(x)| \\ &= |(x,y)\in A : y <f(x)| \end{aligned}$$ where the first line comes from the fact that $f$ is odd, the second line comes from the fact that $(-x,-y)\in A$ if and only if $(x,-y)\in A$ (symmetric about $y$-axis), and the third line comes from $(x,-y)\in A$ if and only if $(x,y)\in A$ (symmetric about $x$-axis). Clearly if $s$ is even then the polygon is symmetric about both the axes and moreover $\sin(ax)$ is also odd for any $a$. Proving the cases for the odd numbers you mentioned could be done tediously with some calculus.