ABSTRACT: I made a simple geometric argument to demonstrate that ever circle with a positive radius can be realized or drawn as a cross-sectional slice of a hollow infinite right circular cone.I used simple Trigonometric , Geometric properties of cone and Rational Number Set for positivity of radius.

INTRODUCTION:

Motivation

I just discovered this idea while thinking about circles and hollow cones and thought if a hollow cone contains every circle. Surprisingly it does.

Well though the result may seem literally obvious , there are really very few theorems that deal with cones and infinite geometric surfaces in this manner.

I am writing this fron the perspective of an undergraduate exploring ideas who has very little experience in this field and open to feedback or correction.

Geometric Notations and Definitions:

We define a hollow infinite cone as a cone that has no solid volume and extends infinitely along its slant height. In this setup, we consider the slant height of the cone to be a variable denoted by x, and we assume it extends indefinitely. The cone has an apex angle of 2θ (two theta), which means the angle between the two sides of the cone at the tip is 2 times theta. Therefore, the half-angle at the apex is θ. For any given slant height x, the radius of the circular cross-section at that point is given by the equation:
r(x) = x.sin(θ)
Here:

r(x) is the radius of the circle at slant height x,
x is the slant height from the apex to the point on the cone,
θ is the fixed half-angle of the cone,
sin(θ) is the sine of that angle.

Since we're dealing with an idealized infinite shape, it's important to note that certain aspects may not be formally defined in conventional finite geometry. To address this, we will initially use a subset of rational numbers greater than zero to construct our argument in a more grounded way before generalizing the result.

Theorem and Proof:

Theorem Statement: For every positive real number r, there exists a point on the surface of a hollow infinite cone where the radius of the circular cross-section is exactly r.

Proof: We are working with a hollow cone that extends infinitely along its slant height. Let the slant height be represented by x, and the fixed half-angle of the cone be θ, where 0 < θ < 180° (excluding the endpoints to avoid degeneracy).
Let:

x = slant height (variable)
θ = fixed half-angle of the cone (0° < θ < 180°)
r(x) = radius at slant height x

Define the radius function:
r(x) = x .sin(θ)
Step 1: Rational radii (r ∈ Q, r > 0):

Choose any rational radius r > 0
Solve for slant height:
x = r / sin(θ)
Since sin(θ) is a constant > 0, x > 0
So, every positive rational radius has a valid location on the cone

Step 2: Extend to real radii (r ∈ R, r > 0):

Rational numbers are dense in real numbers
Function r(x) = x.sin(θ) is continuous
⇒ All positive real radii are also achieved on the cone

Since r is any positive real number and sin(θ) is also positive, this equation always yields a valid, positive slant height x. Therefore, for every positive radius r, there exists a corresponding point on the slant height of the cone where a circle of that radius lies. Thus, the surface of the cone contains a circle of every possible positive radius, as claimed.

Conclusion:

We have shown that a hollow infinite cone, with a fixed half-angle, can contain a circle of every positive radius as a cross-section along its slant height. By using a basic geometric formula and starting with rational numbers, we extended the idea to cover all positive real numbers. The result is both simple and intuitive, showing how an infinite geometric shape can represent an entire continuous set of circular sizes.

Every Positive Radius Circle Exists on a Hollow Infinite Cone

Conclusion: