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I am trying to understand the proof of Copernicus' Theorem.

Consider two circles of radii R and R/2 with the smaller one rolling inside the bigger circle without slipping. Copernicus' Theorem states a surprising result that a point on the circumference of the small circle traces a straight line segment - a diameter of the big circle, to be precise.

If found the following useful resource with a proof.

I was able to understand everything of the proof besides the first argument:

Assume point M on the small circle has previously occupied the position of point N on the large circle. Since there is no slipping, arcs PM (on the small circle) and PN (on the large circle) have exactly the same length.

enter image description here

I can't conclude that the arcs of PM and PN (marked as red) should have the same lenght. Is there a proof for that?

enter image description here

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  • $\begingroup$ John Baez has a few blog posts on curves traced by rolling circles, and in particular the entry here describes the arrangment you're interested in. (Plus animations!) $\endgroup$ – Semiclassical May 8 '17 at 16:38
  • $\begingroup$ @Semiclassical thanks for your link! I just read it and he states The arc PN of the big circle has the same length as the arc PM of the small circle, since they are both the distance rolled between times t and t. Where t is the time of my bottom left picture and t. the time of the bottem right picture. But I still do not understand that argument. $\endgroup$ – Adam May 8 '17 at 16:49
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    $\begingroup$ Isn't that the definition of "no slipping"? Equality of distance? If not, what is the definition of "no slipping"? $\endgroup$ – Brian Tung May 8 '17 at 16:57
  • $\begingroup$ @BrianTung I only found the definition of no slipping here: real-world-physics-problems.com/rolling-without-slipping.html which says that there is no tangential l movement at the contact point (for P or N). This means the circle only moves through rotation. I can't see how one can conclude from this fact that the both arcs lengts from PM and PN are equal. $\endgroup$ – Adam May 8 '17 at 19:20
  • $\begingroup$ Do you mean that you can't see how to demonstrate that, or that you don't think it's true? $\endgroup$ – Brian Tung May 8 '17 at 19:29

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