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Let $\gamma$ be a simple closed plane curve. We know that a curve with constant curvature $\kappa$ will trace a circle in the plane. The radius of this circle is the inverse of its curvature. Now, let's say we had a lower bound $L$ on the curvature $\kappa(\gamma)$ of $\gamma$. Then intuitively it seems that the entirety of $\gamma$ should be contained in some circle of curvature $L$. However, I can't seem to prove this. This leads me to ask the following:

If $\gamma$ is a simple closed plane curve with $\kappa(\gamma)$ bounded below by some constant $L \in \mathbb{R}$, how is the minimum radius of a circle containing $\gamma$ related to $\kappa(\gamma)$? In other words, what bound (if any) does $L$ place on the area $A(\gamma)$ of $\mathrm{int}(\gamma)$?

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1 Answer 1

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We need two ingredients, the first one is the total curvature formula $$ \int_C k(s)ds= 2\pi $$ for every smooth simple closed planar curve $C$. Here $k(s)$ is the curvature function. If $k(s)\ge K>0$ for all $s$ (I do not like using $L$ for a curvature bound), we obtain $$ K L(C)\le \int_C k(s) ds=2\pi, $$ where $L=L(C)$ is the length of $C$. Thus, $$ L\le \frac{2\pi}{K}. $$ The second ingredient is the isoperimetric inequality: $$ 4\pi A(C)\le L^2(C) $$ for every simple closed planar curve $C$, where $A(C)$ is the area bounded by $C$. (The equality is attained if and only if $C$ is a round circle.) Combining these, we obtain: $$ A(C)\le \frac{L^2}{4\pi} \le \frac{4\pi^2}{4\pi K^2}= \frac{\pi}{K^2}. $$
This inequality is sharp, consider for instance $C$ which is a round circle. The converse is also true: if the equality is attained then the curve is a round circle.

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