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Suppose that I have two distinct simple closed curves, $C_1$ & $C_2$, and each is inscribed in a convex polygon, D. By inscribed, I mean tangent to each side of D. In particular, I am most interested in the case when D is a triangle or convex quadrilateral. Suppose also that $C_1$ & $C_2$ are not tangent at the same point on the boundary of D. Does anyone know of a result which states that $C_1$ & $C_2$ must intersect in 4 distinct points inside D? I am not assuming that $C_1$ & $C_2$ are convex, though I would be interested in the result for that case. I'm also not assuming any smoothness assumptions on $C_1$ & $C_2$, though again I would be interested in the result if $C_1$ & $C_2$ are analytic curves.

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    $\begingroup$ Look at the union of the rigions bounded by $C_1$ and $C_2$ and note that its boundry contains at least $[\tfrac{n+1}2]$ arcs of $C_1$; here $n$ is the number of sides of the polygon. Each end of such arc is a point of intersection, so we will get at least $2{\cdot}[\tfrac{n+1}2]$ of them. $\endgroup$ – Anton Petrunin Oct 24 '14 at 22:49
  • $\begingroup$ That appears to be a nice way to prove what I was looking for, though I want to look it over more for now to make sure it does not implicitly assume anywhere what one wants to prove. Is the statement that the boundary contains at least $[\dfrac{n+1}{2}]$ arcs of $C_1$ just based on the fact that $C_1$ is tangent at the $n$ sides of D ? $\endgroup$ – Alan Horwitz Oct 25 '14 at 20:46
  • $\begingroup$ yes, yes, sure. $\endgroup$ – Anton Petrunin Oct 25 '14 at 21:00
  • $\begingroup$ OK, thanks Anton. I'll look at it more closely to convince myself. $\endgroup$ – Alan Horwitz Oct 25 '14 at 21:07

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