If a regular polygon has a fixed edge length, can I know how many edges it has by knowing the length from corner to its center?

Solution 1:

Yes. The quantity you are referring to is the radius of the polygon, and it has the formula $$r=\frac{s}{2\sin\left(\frac{180}{n}\right)}$$ where $s$ is the side length of the polygon and $n$ is the number of sides. So given $r$ and $s$, you can simply solve the above equation for $n$.

Solution 2:

Yes, and here is the logic of it.

To show how, I've drawn a square, pentagon and hexagon.

In each case, I've drawn the same construction lines - a circle that touches all their corners (we can do this with any regular polygon). I've labelled the length of the "defined edge" as $s$ and the length from corner to centre as $r$. I'm going to call the number of sides, $n$.

I've also marked a grey line that bisects the edge, from the centre. Because of the way I've positioned the line, it is a perpendicular bisector of the edge (crosses it at a right angle) so each half is a right angle triangle. The right angle triangle has one side $r$, and one side $s\big/2$, and I've labelled the angle these make at the centre, $A$.

Basic trigonometry says that for the right angle triangles, $$\begin{align} \sin (A) &= \frac{\left[\dfrac s2\right]}r = \frac s{2r}\\ A &= sin^{-1} \left(\frac s{2r}\right) \end{align}$$

But we also know that each edge, "takes up" $2\times A$ degrees, and so $n$ sides will "take up" $2\times A\times n$ degrees. But all $n$ sides must take up 360 degrees, the number of degrees at the centre. So $2\cdot A\cdot n = 360$.

Now we can solve the problem

Since $$\begin{align} 2An &= 360\\ An &= 180\\ \sin^{-1}\left(\frac s{2r}\right)\cdot n &= 180\\ n &= \frac{180}{\sin^{-1}\left(\dfrac s{2r}\right)} \end{align}$$

Testing this with your square:

$s=1, r=\frac{\sqrt2}{2}$

$$\implies n = \frac{180}{\sin^{-1}\left(\dfrac{1}{2\cdot\dfrac{\sqrt2}{2}}\right)} = \frac{180}{45} = 4$$

So your example object was a square (4 sides).

Testing this with your hexagon:

$s=1, r=1$

$$\implies n = \frac{180}{\sin^{-1}\left(\dfrac{1}{2\cdot1}\right)} = \frac{180}{30} = 6$$

So your example object was a hexagon (6 sides).