is every point of a Berkovich space a Shilov point? Let $k$ be an algebraically closed non-Archimedean valued field with the value group $\mathbb R$, and let $X$ be a variety over $k$. Is it true that for any point $x \in X^{an}$ of the Berkovich analytification of a variety $X$ over $k$ there exists an affinoid neighbourhood $Sp A$ of $x$ such that $x$ is the unique point in the Shilov boundary of $Sp A$ (i.e. the point on which every function $f\in A$ reaches the maximum of it's absolute value)?
(I am sorry for making probably superfluous restrictions but I am not sure what  the natural/most general statement of this question is)
 A: No. Saying that the point $x$ lies in the Shilov boundary of $\mathcal{M}(A)$ gives strong restrictions on the completed residue field $\mathscr{H}(x)$. In your case, its residue field $\widetilde{\mathscr{H}(x)}$ will be of transcendence $d$ over $\tilde{k}$, where $d$ is the dimension of the variety (see proposition 2.4.4 of Berkovich's book).
To handle the more general situation (with no assumptions on the field $k$), let me denote by $s(x)$ the previous transcendence degree and by $t(x)$ the dimension of the $\mathbb{Q}$-vector space $\sqrt{|\mathscr{H}(x)^\ast|}/\sqrt{|k^\ast|}$. Then you will find the condition $s(x)+t(x) = d$ (what is called an Abhyankar point).
If you want an explicit counter-example, pick any point of type 1 ($s+t=0$) when $d\ge 1$.
For a more interesting question, you could ask whether the result holds when you replace $X$ by the Zariski closure $Y$ of the point associated to $x$ in $X$. This is still not true, but you have to consider subtler things like points of type 4.
Let me also add that if the point is Abhyankar, then your result is true. You may find this in my paper "Les espaces de Berkovich sont angéliques" (Bulletin de la SMF 141 (2013), no. 2, 267–297).
