If $0Let $U\subseteq\mathbb{R}^n$ be convex and closed. I'm trying to find conditions for which the following happens:

If $0<r\neq 1$ then $\partial U\cap\partial (rU)=\emptyset$.

It seems to me that this is not true if $0$ is in the frontier of $U$, for instance $U=\{(0,y):y\in\mathbb{R}\}$ or $U=\{(x,y):y\ge |x|\}$. In fact, in both examples $U=rU$.
So, let us suppose that $0\in\text{Int} (U)$. Does this suffice to prove that $\partial U\cap\partial (rU)=\emptyset$? I couldn't find counterexamples so far.
What do you think?
Thank you.

I have an approach
that seems promising to me
but I can't push it through.
I'll enter it here
in the hope that
someone else can complete it.
$\partial U$
can be described as
$(t, f(t))$
for
$0 \le t \lt 2\pi$.
Suppose $(0, 0) \in U$ \
$\partial U$.
Suppose
$\partial U \cap \partial(rU) \ne \emptyset$
where $0 < r < 1$.
Then there are
$a, b \in [0, 2\pi)$
 such that
$(a, f(a)) = r(b, f(b))$
so that
$a = rb$ and
$f(a) = rf(b)$.
Consider the line from
$(b, f(b))$
to
$(rb, rf(b))$.
This has equation
$c(b, f(b))+(1-c)(rb, rf(b))$
for $c \in \mathbb{R}$
or
$c(b-rb, f(b)-rf(b))+(rb, rf(b))\\
=c(1-r)(b, f(b))+r(b, f(b))\\
=c(1-r)(b, f(b))+(a, f(a))
$.
Frustrating.
I can't see where
this leads to a contradiction.
What I would like to do
is show that this line
which passes through
two points on
$\partial U$
also passes through
a third point,
which would be a contradiction.
