Is it possible to construct a circle from line segments?

  • Thread starter Thread starter flatmaster
  • Start date Start date
  • Tags Tags
    Circle Inscribed
flatmaster
Messages
497
Reaction score
2
For the next critical mass ride, I'm going to try to mount a gong inside the front triangle of my bicycle. Naturally, I want to maximize the radius of the gong and learn some geometry. I spent a couple minutes trying to derive it myself. The number of terms to take care of became huge, so I googled it.

http://www.efunda.com/math/areas/CircleInscribeTriangleGen.cfm

Looking at the complxity of the final answer, I suppose I did myself a favor by not solving it the entire way for myself. At one point, I had 36 terms and nothing was canceling.

My question is, for higher dimentions,(a sphere inscribed in an arbitrary pyramid with triangluar base), does this get mathematically interesting? What about polygons with arbitrary numbers of sides and angles?
 
Physics news on Phys.org
For polygons with a higher number of sides, an interesting question is whether you can inscribe a circle. For example, for a quadrilateral you can only inscribe a circle if the four angle bisectors meet (assuming I remember the condition correctly :) )
 
Office_Shredder said:
For polygons with a higher number of sides, an interesting question is whether you can inscribe a circle. For example, for a quadrilateral you can only inscribe a circle if the four angle bisectors meet (assuming I remember the condition correctly :) )

I see no reason why you can't inscribe a circle within a regular polygon of n sides (no matter how large n is). However, I believe the inscribed circle is not the limit circle of the regular polygon as n approaches infinity. The limit circle is the circle circumscribing the polygon. Why?
 
Last edited:
SW VandeCarr said:
I see no reason why you can't inscribe a circle within a regular polygon of n sides (no matter how large n is). However, I believe the inscribed circle is not the limit circle of the regular polygon as n approaches infinity. The limit circle is the circle circumscribing the polygon. Why?

Considering the OP is about arbitrary triangles, I assumed he was referring to arbitrary polygons too.

I'm not sure what you mean with the inscribed and circumscribed circle thing... I would have thought the difference in area between them goes to zero. I'll have to check
 
Office_Shredder said:
Considering the OP is about arbitrary triangles, I assumed he was referring to arbitrary polygons too.

I'm not sure what you mean with the inscribed and circumscribed circle thing... I would have thought the difference in area between them goes to zero. I'll have to check

There's another post in the Set Theory forum entitled "an impossible circle" (yyttr2, Jun 2) where aspects of this were discussed. The initial opinion was that you can't construct a circle from line segments. The discussion then went to whether a point is a limiting "length" of a line segment. I argued that a line segment cannot be reduced to a point analytically and still be a line segment. To take a limit, you need a (conceptual) tangent line segment to a point on a differentiable function (for derivatives that are not constants). A point doesn't have a slope.

The edges of the regular polygon are tangent line segments to the inscribed circle. Can they go to zero length and still be tangent "line segments"? No scalar field is necessarily defined for this problem.

EDIT: If I'm correct, the area of the regular polygon equals the area of the circumscribed circle at the limit and the inscribed circle is not defined at the limit.
 
Last edited:
Hello! There is a simple line in the textbook. If ##S## is a manifold, an injectively immersed submanifold ##M## of ##S## is embedded if and only if ##M## is locally closed in ##S##. Recall the definition. M is locally closed if for each point ##x\in M## there open ##U\subset S## such that ##M\cap U## is closed in ##U##. Embedding to injective immesion is simple. The opposite direction is hard. Suppose I have ##N## as source manifold and ##f:N\rightarrow S## is the injective...

Similar threads

Replies
2
Views
1K
Replies
2
Views
2K
Replies
1
Views
3K
Replies
10
Views
3K
Replies
39
Views
6K
Replies
7
Views
3K
Replies
9
Views
4K
Back
Top