• 0

Central Area

Question

Posted · Report post

inspired by one of Bonanova's past problems:

What proportion of the area of a regular n-gon is closer to its center than to any edge?

0

Share this post


Link to post
Share on other sites

9 answers to this question

  • 0

Posted · Report post

Since I inspired this, ^_^ I guess I should solve it.

The locus of points equidistant from the center and any side of a regular n-gon has no straight portions.
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

is the area of the n-gon that is closer to the center than to any side.

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).

That line intersects the parabola at xn = 2(-m + sqrt(m2+1)), yn = xn2/4

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


n fraction

------+------------

3 0.1851851852
4 0.2189514165
5 0.2314757303
6 0.237604307
8 0.2432751885
10 0.2457666208
12 0.2470862357
20 0.2489644765
48 0.24982129
100000 0.2499999961

0

Share this post


Link to post
Share on other sites
  • 0

Posted · Report post

that area is bounded by a similar n-gon at 1/2 linear scale, thus 1/4 of the area.

0

Share this post


Link to post
Share on other sites
  • 0

Posted (edited) · Report post

that area is bounded by a similar n-gon at 1/2 linear scale, thus 1/4 of the area.

Are you sure? For some reason I am finding my area to be rounder in shape.

Edited by BMAD
0

Share this post


Link to post
Share on other sites
  • 0

Posted (edited) · Report post

.

Edited by BMAD
0

Share this post


Link to post
Share on other sites
  • 0

Posted · Report post

You're right. I drew a ray from the center and took the midpoint.

But that doesn't fit the question. I'll give it more thought.

0

Share this post


Link to post
Share on other sites
  • 0

Posted · Report post

I second Bonanova's first response.

0

Share this post


Link to post
Share on other sites
  • 0

Posted · Report post

hmm well it does appear to be approaching 1/4 so maybe that is the limit.

0

Share this post


Link to post
Share on other sites
  • 0

Posted · Report post

awesome problem and solution :thumbsup:

0

Share this post


Link to post
Share on other sites
  • 0

Posted · Report post

hmm well it does appear to be approaching 1/4 so maybe that is the limit.

Yes of course for the circle the locus of equidistant points is a circle with half the radius and thus 1/4 the area.

The mistake of the first guess was to apply that reasoning to a polygon.

m00li, agree it is an interesting problem, much deeper than first appearance.

0

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!


Register a new account

Sign in

Already have an account? Sign in here.


Sign In Now

  • Recently Browsing   0 members

    No registered users viewing this page.