Find the limit of the area bounded by the diagonals?

Let n be a positive integer and consider a regular (2n+1)-gon with sidelength 1.

Now, draw every diagonal the polygon has. Due to symmetry, this gives us a picture that is somewhat pleasing to the eye. For instance, see
http://en.wikipedia.org/wiki/File:4-simplex_graph.png
http://en.wikipedia.org/wiki/File:6-simplex_graph.png
http://en.wikipedia.org/wiki/File:8-simplex_graph.png
http://en.wikipedia.org/wiki/File:10-simplex_graph.png

Note that at the center of the original (2n+1)-gon sits another (2n+1)-gon bounded by the drawn diagonals.

Now the questions:
(1) Find the area of the smaller (2n+1)-gon as a function of n.
(2) As n goes to infinity, does this expression converge? If so, to what value?

2010-06-25T14:12:03Z

I should note that due to naming conventions, then 4-simplex can be drawn as a 5-gon. Don't let this confuse you though.

2010-06-26T15:18:00Z

Scythian: First off, thanks for answering. I was wondering whether it was harder than I had originally thought.

Something bothers me about your answer though. I want the sidelength of the outer polygons to be fixed at length 1. I believe this will make the limit nonzero.

In fact, I believe I already know the limit:
pi / 16
but I don't know the answer to the first part of my question, so I wasn't able to check it.

2010-06-26T15:29:34Z

Quick argument for my
pi / 16
comment (to show how I skipped the first part):

Note that the diagonals involved are the two diagonals coming from consecutive vertices going to the "opposite" vertex (in the sense of opposite to the side between the two consecutive vertices).

Now, as n goes to infinity (with the sidelength of the outside polygon held constant), these two diagonals bound the smaller polygon in smaller and smaller sides until they only bound the polygon in a point. By symmetry and regularity, the smaller polygon is a circle.

Now, what is the diameter of this circle? Well, the midpoints of these two diagonals should approach the point where the diagonals are tangent to the limit circle. The distance between these two points is always half the sidelength, or 1/2, so this would be the diameter of the circle.

This gives the previously stated pi / 16 as the limiting area.

I waved my hands a few more times than I am comfortable with doing this, though...

Scythian19502010-06-26T10:16:51Z

Favorite Answer

I'm not sure why this one is not getting any answers. This is a straightforward geometry problem, the ratio of the areas of the outer and inner polygons is just the square of the ratio of their apothems, or:

(Sin(a) / Sin(b))²

where we let the radius of the circle circumscribing the outer polygon = 1, and a is the base angle of the right triangle formed by the apothem of the outer polygon, and b the same with the apothem of the inner polygon. A little bit of geometry, with simplification, gets you:

(1 - Cos(π/n)) / 2(Cos(π/n))²

where n is the number of sides of the polygon, presumed to be odd. For n = 5, we have the ratio:

(3 - √5) / (3 + √5)

As n -> infinity, this ratio becomes 0.

Edit: Okay, I see now that the side length is fixed at 1. This is now a more interesting problem. Consider that n -> infinity, so that the polygon is nearly a circle, both outer and inner. But the diameter of the inner circle should be just 1/2, using a simple geometric ratio (draw the 2 diagonals from one vertex to the opposite side, which is fixed at 1). This means that the limiting value for this area should be π/16. Hmm, let me see if that agrees with the other equations. Give me a few.

Yes, it does work out nicely. The area of the n-polygon with side lengths of 1 is:

(n/4) Cot(π/n)

When this is multiplied by the expression given above for the ratio, and taking the limit n -> infinity, we do end up with π/16. A thumbs up for this nice problem.

120

Apratim R2010-06-29T02:57:16Z

Would the following argument do away with the need for handwaving?

Let the two diagonals mentioned above touch the incircle of the inner regular n-gon at A and B. Then as n goes to infinity, AB approaches the diameter of the incircle in length (more accurately, AB is fixed in length at 1/2, and the diameter approaches it, their ratio being cos (pi/n)). As you've noted, AB = 1/2, so the incircle approaches pi/16 in area.

(At the same time, as n goes to infinity, the inner regular n-gon also approaches its incircle in area, i.e. the ratio of their areas approaches 1.)

40

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