TITLE: Intersections and curvature in the plane
QUESTION [10 upvotes]: Let $D$ be a nonempty compact convex plane region whose boundary is a smooth curve whose radius of curvature is at most 1 everywhere. Can the boundary of $D$ intersect a circle of radius 1 in more than two points?

REPLY [4 votes]: David Speyer writes "the only way this strategy could work
is if we improve the $2$ in the previous paragraph to $\pi$."
That improvement follows.
Suppose $f < 0$ and $f + f'' \leq 0$ on $(0,c)$, with $f(0)=f(c)=0$.
We show $c \geq \pi$, with equality if and only if
$f(\cdot)$ is a negative multiple of $\sin(\cdot)$.
Let $\alpha = \tan^{-1}(f'/f)$, so $\alpha(0) = \pi/2$ and $\alpha(c) = -\pi/2$.
Note that if $f(x) = A\sin x$ then $\alpha(x) = \pi/2 - x$.
In general we compute
$$
\alpha' = \frac{f\!f''-{f'}^2}{f^2+{f'}^2},
$$
whence
$$
\alpha' + 1 = \frac{f(f+f'')}{f^2+{f'}^2} > 0.
$$
Therefore $\alpha' \leq -1$ on $(0,c)$.  Since $\alpha$ decreases by $\pi$
on that interval, the interval must have length at least $\pi$.
Moreover if $c=\pi$ then $f+f''=0$ on all of $(0,\pi)$, which together with
$f(0)=f(\pi)=0$ makes $f$ a multiple of the sine function. $\Box$