I was wondering if it was possible to calculate the coordinates of every vertex in a regular pentagon using only its side length and its center?
The pentagon's center will not be fixed at the origin, and thus could be any $(x,y)$.
Please let me know if you can help!
You would need at least one more piece of information to be able to solve your problem because you do not know the orientation of the pentagon. An equilateral pentagon is often illustrated with a vertex at the top and an edge at the base but without knowledge of the orientation, the coordinates can not be discerned. I will show you the general method that you can use though.
Firstly, we know that the distance to each vertex will be the same. If we say that the length of a side is a, then from a vertex (called A) to halfway along the edge ($\frac {1}{2} a$) then we can draw a perpendicular line from this point to the centroid (called C). The angle A is 108 degrees since a pentagon is a 5-sided polygon.
Now we have a right angle triangle wherein we can determine that $cos(54) = \frac {\frac {1}{2}a}{CA}$ and hence, $CA = \frac {a}{2*cos(54)} $
There are 5 vertices so the angle between each line to a vertex about C is 72 degrees. We can say C = ($x_c$, $x_c$). With $\theta = 0$ in the positive direction of the x-axis, you can use sin and cos functions to determine the x and y coordinates of each vertex relative to the centroid. Then simply add the x and y values of the centroid to account for the position of the centroid. For a vertex V: $V = (x_c + CA*sin(\theta), y_c + CA*cos(\theta))$ = $(x_c + \frac {a*sin(\theta)}{2*cos(54)}),(y_c + \frac {a*cos(\theta)}{2*cos(54)})$