Metric on $\mathbb R^2$ in which each sphere $B((x,y),r)$ is an equilateral triangle Is there a metric on $\mathbb{R}^2$ in which each sphere 
$B\bigl((x,y),r\bigr)$ is an equilateral triangle centered at $(x, y)$, one of whose vertices has the form $(x',y)$ with $x'\geqslant x$?

[Note: if the triangle was replaced by a centrally symmetric convex polygon, the answer would be "yes", since such a polygon induces a norm. For a triangle we have a gauge function which, however, does not yield a metric.]
 A: I can prove the non existence under the following:

Assumption: for each point in $\mathbb R^2$ there are arbitrarily small balls not consisting of a singleton. 

This is implied for instance requiring that there aren't any isolated points (or if you don't consider points as degenerate triangles!).
Start from any two points on the $x$ axis, say $x_0=(0,0)$ and $x_1=(-1,0)$. They will have a certain distance, say $1$. I claim that the points
$$x_k=\left(-2+\frac{1}{2^{k-1}},0\right)$$
satisfy $d(x_k,x_{k+1})=1$.
Indeed by the symmetry of the distance function the sphere $S_1:=\{y:d(x_1,y)=1\}$ centered at $x_1$ will pass through $x_0$ (observe that this sphere could in principle be "thick"), and from the assumption on the shape of balls it will pass also through $x_2$. Consider now the sphere of radius $1$ centered at $x_2$ and repeat the same argument.

If $x=(-2,0)$ is the limit point, from the highlighted assumption above there is a ball of radius $r<\frac12$ and center $x$ which is not a singleton but actually a triangle. In particular it will have nonempty interior in $\mathbb R^2$, and will contain at least two consecutive points $x_k$ (in fact all of them eventually). This is impossible: a ball of radius strictly smaller than $1/2$ can not contain two points at distance one from each other, by triangle inequality: 
$$d(x_k,x_{k+1})\leq d(x_k,x)+d(x_{k+1},x)<1.$$
A: For $(x,y) \in \mathbb R^2$ consider the map $\Vert \cdot \Vert_\triangleright : \mathbb R^2 \to \mathbb R_{\ge 0}$:
$$(x,y) \mapsto \sup(-x,x+\frac{1}{\sqrt{3}}y,x-\frac{1}{\sqrt{3}}y)$$
You can prove that $\Vert \cdot \Vert_\triangleright$ is a norm. Moreover for $r \ge 0$, the subset $$B_\triangleright((x,r),r) = \{(x,y) \in \mathbb R^2 \ : \ \Vert (x,y) \Vert_\triangleright \le r\}$$ is a triangle having the properties you requested.
From this norm, you can define a distance having all the required properties.
