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.

WLOG assume that $r>1$ (if $0<r<1$ then let $U' = rU$ and $r' = \frac{1}{r}$, so $r'U' = U$ and $r'>1$). Suppose that the intersection is nonempty. Then, there is some $x\in \partial (rU)$ such that $x\in\partial U$. Since $x\in \partial (rU)$, $y = r^{-1}x\in \partial U$. Thus, both $x$ and $y$ are on the boundary of $U$. We can draw a line segment from $0$ through $x$, which will also pass through $y$. If $0\in\text{Int}(U)$ there is an open neighborhood of $0$ also in $\text{Int}(U)$ since the interior is always open. Then, take the collection of line segments (not including endpoints) from points in that neighborhood to $x$. The union $I$ of all these line segments is an open set containing $y$. Since $U$ is convex, $I\subset U$. Thus, there is an open subset of $U$ containing $y$, contradicting that $y\in\partial U$. Thus, if $0\in\text{Int}(U)$ then the intersection must be empty.
If $0\notin\text{Int}(U)$ then $0 \in \partial U$ or $0 \in \text{Ext}(U)$. If $0\in\partial U$ then obviously $0 \in \partial (rU)$ so the intersection will be nonempty. If $0 \in \text{Ext}(U)$, then let the shortest distance between $0$ and $U$ be $R$ and the largest distance be $M$. If $0<r<\frac{R}{M}$, then $rU\cap U = \emptyset$. However, if $U$ is a line segment not through the origin, then $rU\cap U = \emptyset$ for all $r\neq 1$.
Thus, $\partial U \cap \partial (rU) = \emptyset$ for all $r$ if $0\in\text{Int}(U)$ and $\partial U \cap \partial (rU) \neq \emptyset$ for all $r$ if $0\in \partial U$.
