The story begins with seemingly unrelating situation. I was trying to find out an elementary solution of the following problem.
A circle is called a separator for a set of five points in a plane if it passes through three of these points, it contains a fourth point inside and the fifth point is outside the circle. Prove that every set of five points such that no three are collinear and no four are concyclic has exactly four separators. (IMO Shortlist 1999 G2)
This problem can be solved by means of inversion or Möbius transformations. But I was stuck making elementary solution which does not makes use of those tools. I reached at the point that the following may be useful
For each three distinct points $P$, $Q$, $R$, let $\Gamma(P,Q,R)$ be the outer circle of $PQR$. Let $A_1A_2A_3 \cdots A_n$ be a convex polygon where no three of the consecutive vertices are colinear and $n \ge 3$. Prove that among the $n$ circles, the one with the maximum radius encloses the polygon; $$ \Gamma(A_1A_2A_3), \Gamma(A_2A_3A_4), \ldots \Gamma(A_nA_1A_2) $$
At the moment when I got this observation, I guess this can be proved easily. But after trying some quick treat, I realized that this is essentially the problem of curvature.(Anyway, I cannot prove this either)
So my question is the following
Let $\gamma$ be a smooth closed simple curve whose curvature never vanishes. Prove that the osculating circle at the point where the curvature is minimum encloses the curve.
I know that the question is so elementary that it is almost surely duplicate. However, I have searched several keywords without encountering similar question. So it would be great if someone gives links or the same questions or references in which the proof could be found. Thanks for your attention.