Is the unit circle "stretchy" with respect to its norm? Suppose we have a collection of metric spaces on $\mathbb{R}^n$, each of which has a different p-norm, $1\leq p \leq \infty$. ($p=2$ is Euclidean distance, $p=1$ is taxicab distance, etc.)  Then, suppose we have a point $x$ in $\mathbb{R}^2 $ that's within the unit circle of the metric space with the $\infty$-norm, but NOT within the unit circle of the metric space with the 1-norm.  I'd like to know if there's a $p$ for which $\|x\|_p$ is exactly 1.  Intuitively, I'm asking if, by adjusting the $p$-norm, I can "stretch" the unit circle to sit anywhere I like, as long as it's between the 1-norm and the $\infty$-norm, as suggested by this picture.
I feel like it must be true, and maybe I'm missing something obvious.

Yes, this is true by a simple continuity argument.  Note that for any fixed $x\in\mathbb{R}^2$, the map $p\mapsto \|x\|_p$ is a continuous function $[1,\infty]\to\mathbb{R}$ (it is obvious that this is continuous for $p<\infty$; continuity as $p\to \infty$ requires a little work but is not hard).  So by the intermediate value theorem, if $\|x\|_1>1$ but $\|x\|_\infty<1$ there is some $p$ such that $\|x\|_p=1$.
(Here are the details of the continuity as $p\to \infty$.  If $x=(a,b)$ with (WLOG) $a\geq b\geq 0$, then $$a\leq \|x\|_p=(a^p+b^p)^{1/p}\leq (2a^p)^{1/p}=2^{1/p}a.$$  As $p\to\infty$, $2^{1/p}a\to a$, so $\|x\|_p$ is squeezed and must converge to $a$.)
