I wondered, is there a geometrical way to find the center of a pentagon or a hexagon? I'm not talking about equal sides, just polygons with 5 or 6 corners.
Like, with a triangle you can take the intersection of two medians to find the center. With a quadrilateral, the center is the intersection of the bimedians.
Is it possible to construct the center of pentagons and hexagons in a similar way?
Edit: Apparently is rather difficult, so I probably have to settle for a formula to calculate the centroid. I always learned that the $x$ and $y$ values of the centroid are just the mean values of the $x_i$ and $y_i$ values of the corners respectively, but Wikipedia says otherwise (Wiki):
$C_x = \dfrac{1}{6A} \displaystyle \sum_{i=0}^{n-1} (x_i+x_{i+1})(x_iy_{i+1}-x_{i+1}y_i)$
$C_y = \dfrac{1}{6A} \displaystyle \sum_{i=0}^{n-1} (y_i+y_{i+1})(x_iy_{i+1}-x_{i+1}y_i)$
Where $A = \dfrac{1}{2} \displaystyle \sum_{i=0}^{n-1} (x_iy_{i+1}-x_{i+1}y_i)$
I'm not entirely sure, but wouldn't those $(x_iy_{i+1}-x_{i+1}y_i)$ terms cancel out because you divide by the summation over the same interval? That would leave:
$C_x = \dfrac{1}{12} \displaystyle \sum_{i=0}^{n-1} (x_i+x_{i+1})$
which is rubbish, except for when your polygon has 6 corners -- and that's exactly the case on the source from Wikipedia, here.
Therefore I wonder, is my math correct and is this formula just a very elaborate way to calculate the centroid of a hexagon (and no other polygons), or is it just coincidence? If so, please explain the formula.