Prove the theorem on analytic geometry in the picture.

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I discovered this elegant theorem in my facebook feed. Does anyone have any idea how to prove?

Formulations of this theorem can be found in the answers and the comments. You are welcome to join in the discussion.

Current progress: The theorem is proven. There is a bound to curvature to be satisfied before the theorem can hold.

Some unresolved issues and some food for thought:

(1) Formalise the definition (construction) of the parallel curve in the case of concave curves. Thereafter, consider whether this theorem is true under this definition, where the closed smooth curve is formed by convex and concave curves linked alternatively.

(2) Reaffirm the upper bond of r proposed, i.e. r=ρ, where ρ is the radius of the curvature, to avoid self-intersection.

(3) What is the least identity of the curves in order for this theorem to be true?
This is similar to the first part of question (1).

(4) Can this proof be more generalised? For example, what if this theorem is extended into higher dimensions? (Is there any analogy in higher dimensions?)

Also, I would like to bring your attention towards some of the newly posted answers.

Meanwhile, any alternative approach or proof is encouraged. Please do not hesitate in providing any insights or comments in further progressing this question.


Example for irregular pentagon

Consider the irregular pentagon above. I have drawn it as well as it’s “extended version,” with lines connecting the two. Notice that the perimeter of the extended version is the same as the original, save for several arcs — arcs which fit together, when translated, to form a perfect circle! Thus, the difference in length between the two is the circumference of that circle, 2\pi r.

Any convex shape can be approximated as well as one wishes by convex polygons. Thus, by taking limits (you have to be careful here but it works out), it works for any convex shape.

I’ll get back to you on the concave case, but it should still work. EDIT: I’m not sure it does… EDIT EDIT: Ah, the problem with my supposed counterexample — the triomino — is that the concave vertex was more than r away from any point on its extended version. If you round out the triomino at that vertex, it works again. TL;DR, it works for concave shapes provided there are no “creases.”

Source : Link , Question Author : user122049 , Answer Author : Akiva Weinberger

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