# Enclosing a convex plane domain in a disc

The following statement seems obvious to me:

Let $\gamma:S^1\to\mathbb R^2$ be a smooth injection such that $\dot\gamma$ and $\ddot\gamma$ never vanish. Then $\gamma$ encloses a strictly convex domain $\Omega\subset\mathbb R^2$. Let $\kappa>0$ be the minimum of the curvature of $\gamma=\partial\Omega$. Take any point $x\in\partial\Omega$ and draw a circle $C$ of constant curvature $\kappa$ which is tangent to $\partial\Omega$ at $x$ and curves in the same direction. Then $\Omega$ is inside the circle $C$.

Proving the statement seems far less obvious. I have spoken about this with other postdocs, and we have made no real progress despite several attempts. Assuming the statement is true, how could I go about proving it?

It suffices to show that $C$ and $\gamma$ only intersect at $x$ — or the intersection is the connected component of the set $\{y\in\partial\Omega;\text{curvature of }\partial\Omega\text{ at y}=\kappa\}$ containing $x$ in case $x$ is a point of minimal curvature. It may also be useful that if $C$ and $\gamma$ meet elsewhere, it has to happen transversally, not tangentially.

• I think that in Sec 1.7 of Toponogov's book Differential geometry of curves and surfaces. A concise guide you'll find the ideas you need to prove the statement. – Liviu Nicolaescu Jun 3 '16 at 10:31
• @LiviuNicolaescu, thanks! I got the book from my campus library and wrote up an answer based on it. – Joonas Ilmavirta Jun 3 '16 at 15:36

## 1 Answer

One method can be found in Toponogov's book Differential geometry of curves and surfaces as suggested by Liviu Nicolaescu in the comments. Let me adapt the proof given in problem 1.7.10 in the book to the present situation.

Consider the circle $C_\epsilon$ of curvature $\kappa-\epsilon$ instead. If we can show that $\partial\Omega$ does not meet $C_\epsilon$ for any $\epsilon>0$, we get the desired claim in the limit $\epsilon\to0$.

For simplicity, let us denote the circle by $C$ and assume it has constant curvature $k<\kappa$. Parametrize $\gamma$ by arc length $s$ starting from $x$, and denote the curvature function by $\kappa(s)$. Also parametrize $C$ by arc length in the same direction. Let $\alpha_\gamma(s)$ and $\alpha_C(s)$ be the angles of the tangent compared to those at $x$ (that is, $s=0$). Then $$\alpha_\gamma(s)=\int_0^s\kappa(t)dt\quad\text{and}\\ \alpha_C(s)=\int_0^skdt=ks.$$ Because $k<\kappa(s)$, we have $\alpha_C(s)<\alpha_\gamma(s)$ for all $s>0$.

It suffices to prove that there are no arc lengths $s_\gamma$ and $s_C$ so that $\gamma(s_\gamma)\in C$ and $\alpha_\gamma(s_\gamma)\leq\pi$. (If the turning angle of an intersection point is more than $\pi$, one can just go along the curves in the opposite direction.) We can also exclude the case $\alpha_\gamma(s_\gamma)=\pi$, since that would correspond to $\gamma$ meeting $C$ again only at the antipodal point of $x$, which is only possible by a tangential touch. Impossibility of tangential intersections was observed in the question.

Let $L$ be the tangent of $C$ and $\gamma$ at $x$. The orthogonal projection of the point $\gamma(s)$ to $L$ is at (signed) distance $\int_0^s\cos\alpha_\gamma(t)dt$ from $x$. A similar formula holds for points of points on $C$. At the hypothetical intersection point we have $$\int_0^{s_\gamma}\cos\alpha_\gamma(t)dt = \int_0^{s_C}\cos\alpha_C(t)dt.$$ It follows from the inequality $\alpha_C<\alpha_\gamma$ obtained above that $s_C<s_\gamma$.

It then remains to show that the arc $A=\gamma([0,s_\gamma])$ is shorter than the corresponding arc $B$ of the circle $C$ to obtain a contradiction. The arc $A$ lies between $B$ and the line segment $S$ joining $\gamma(0)$ and $\gamma(s_\gamma)$. Therefore $A$ is shorter than $B$.