Isosceles Triangles in Hilbert Spaces and Metric Spaces generally In what types of metric spaces $\langle X, d \rangle$ is it possible to do the following?
Task:  For any two points $x, y \in X$ such that $d(x,y) \leq  2\epsilon$, find a third point $z$ such that $d(x,z)=d(y,z) = \epsilon$.
I believe I can do this in a Hilbert space, but my calculations are embarrassingly long and I think likely contain an error.  So a second question is whether there is a short proof that this is possible in a Hilbert space.
 A: This cannot generally be done in a $1$-dimensional real Hilbert space.  In "higher" dimensions (i.e. $\ge2$), there must exist a point $w$ that is not a scalar multiple of $x-y$.
It's not hard to show that $\dfrac{w\cdot(x-y)}{\|x-y\|^2} (x-y)$ is the orthogonal projection of $w$ onto $x-y$, so $w$ minus that is orthogonal to $x-y$, and is not $0$ since $w$ is not a scalar multiple of $x-y$.  So let
$$
v = w - \frac{w\cdot(x-y)}{\|x-y\|^2} (x-y),
$$
and then $v\ne0$ is orthogonal to $x-y$.
We will seek the desired point $z$ on the perpendicular bisector of the segment from $x$ to $y$.  Such points are of the form
$$
z = \frac{x+y} 2 + tv
$$
where $t$ is a scalar.  The distance from $z$ to either $x$ or $y$ is $\sqrt{\dfrac{x-y}2 + t^2\|v\|^2}$.  Set that equal to the desired distance $\epsilon$ and solve for $t$.
A: It is also possible in general Banach spaces with dimension at least $2$. For brevity, let us consider $x = 0$ and $\epsilon = 1$. Since the dimension is at least $2$, there exists $v$, such that $\{y,v\}$ are linearly independent. Now, we define for $t \in [0,\pi]$
$$r(t) = \cos(t) \, y + \sin(t) \, v$$
and
$$z(t) = \frac{r(t)}{\|r(t)\|}.$$
Then, $\|z(t)\| = 1$ for all $t \in [0,\pi]$ and it remains to find $t$ with $\|y - z(t)\| = 1$. It is clear that $t \mapsto \|y - z(t)\|$ is continuous,
$$\|y - z(0)\| = \|y - y/\|y\|\| = \|y\| - 1 \le 1$$
and
$$\|y - z(\pi)\| = \|y + y/\|y\|\| = \|y\| + 1 \ge 1.$$
Now apply the intermediate value theorem.
