Since I inspired this, I guess I should solve it.
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is the area of the n-gon that is closer to the center than to any side. That line intersects the parabola at xn = 2(-m + sqrt(m2+1)), yn = xn2/4 ------+------------ 3 0.1851851852
It is rather a sequence of n parabolic sections centered on the apothems (perpendicular lines to the sides.)
Sketch an n-gon with center at (0,1) and bottom side on the line y = -1.
The origin lies at the midpoint of the apothem drawn to the bottom side.
The origin is thus one of the equidistant points.
The complete locus of points equidistant from the center and the bottom side is y = x2/4.
If we do this for each side, we join n parabolic segments, and the area of their combined interior
Lines drawn from the center to the n vertices define the points where the parabolic segments join.
From the center, draw a line downward at an angle pi/n from the vertical.
Call the slope of that line m where m = 1/tan(pi/n).
Look at the rectangle defined by x = 0, xn and y = 0, 1 - its area is just Ar = xn, comprising three parts
A triangular portion above the diagonal line with area At = 1/2 xn (1-yn)
A lower area under the parabola with area Ap = (1/12) xn3
The middle area comprises the points closer to the center than to any side.
The total area of interest is then 2n(Ar - At - Ap) = n xn (1 + (1/12) xn2)
Finally we find the total area of a regular n-gon with apothem = 2.
From standard geometry, side = 2 apothem tan(pi/n)
perimeter = n x side
Arean = 1/2 apothem perimeter = 4 n tan(pi/n)
Note that n tan(pi/n) goes to pi as n goes to infinity so Arean=inf = Areacircle = 4pi.
Because the apothem becomes the radius for a circle.
The fraction of area of a regular n-gon closer to its center than to any side is thus.
fn = xn(1+1/12 xn2) / 4 tan(pi/n) -> 1/4 as n -> infinity (circle).
Results for various values of n
is the area of the n-gon that is closer to the center than to any side.
That line intersects the parabola at xn = 2(-m + sqrt(m2+1)), yn = xn2/4