This is a question that has been bothering me in the back of my head for quite some time.

Suppose we have a metric space $X$ with metric $\mathrm{d}$. By an isosceles triangle we mean a tuple of three points $a, b, c \in X$ such that $\mathrm{d}(a,b)=\mathrm{d}(a,c)$, with all three points distinct.

Now, the question is, what can we say of a space $X$ for which no such triangle exists?

I have only gotten some very weak results. Specifically, considering the non-isosceles property we are assuming the space has, we can define a function $\lambda_x(r)$ for every $x\in X$, such that $\mathrm{d}(\lambda_x(r),x)=r$ wherever it is defined. That is, we can define a function that finds *the* point at distance $r$ from $x$, whenever such a point exists. This $\lambda$ is continuous at 0, and is not continuous on any open subset of $\mathbb{R}$.

Through some simple manipulations involving this $\lambda$, we can show that $X$ must be totally disconnected.

That is just about the best result I have managed to get. A possible thread to continue on is a result found in Arkhangel'skii & Tkachenko (2008), saying that a locally compact hausdorff space is zero-dimensional iff it is totally disconnected.