Fix three finite real numbers $p,q,r > 0$. Up to isometry, there is a unique 2-simplex $\Delta$ in the Euclidean plane bounded by a geodesic triangle with these three reals $p,q,r$ as side-lengths. The same holds for hyperbolic triangles $\Delta_{\mathrm H}$ bounded by three geodesic segments of lengths $p,q,r$ in the hyperbolic plane. There is clearly an isometry between the boundaries of $\Delta$ and $\Delta_{\mathrm H}$, unique up to rigid symmetries of the triangle, and I'd like to have an iron-clad proof such a map can be extended to a bi-Lipschitz map on the interiors.
Intuitively, it seems clear to me this should be possible because such triangles live in bounded regions of these planes, and the Euclidean and hyperbolic metrics $$dx^2 + dy^2 \quad \mbox{and} \quad \frac{dx^2 + dy^2}{y^2}$$ (taking the upper half-plane model for the hyperbolic plane) differ by a bounded factor, so it seems like any sufficiently smooth, non-stupid map between the regions should be bi-Lipschitz as a result of these metrics being "close."
On the other hand, the map $x \longmapsto \sqrt x$ is a non-Lipschitz self-homeomorphism of $[0,1]$, so it is probably possible to find a map sufficiently dumb that it isn't bi-Lipschitz.
It's possible to find explicit homeomorphisms between these regions of course, especially in polar coordinates, and one would like to bound the partial derivatives in such a way that it's somehow clear that the distances differ in a bounded fashion, but these expressions are just messes, and it feels like this degree of computation shouldn't be necessary. That is, it seems like this problem should be easier than it has proven.
What am I missing? How do I find such a bi-Lipschitz extension?
Here is an outline of how I would attack this problem. I am going to describe it for an equilateral triangle where $p=q=r$, but I guess that the method extends with little trouble.
The hyperbolic equilateral triangle $\Delta_H$ has a kind of "barycentric subdivision", where each side is subdivided at its midpoint, and the barycenter of $\Delta_H$ is the "centroid" $c_H \in \Delta_H$ which is the point fixed by the symmetry group of $\Delta_H$. Let $x_H,y_H,z_H$ be the midpoints of the sides. Note that $$\text{Length}[c_H,x_H]=\text{Length}[c_H,y_H] = \text{Length}[c_H,z_H] \le \log(\sqrt{3}) $$ (the right hand side is the extreme value that is acheived by the ideal triangle).
The Euclidean equilateral triangle $\Delta$ has an exactly similar barycentric subdivision, with centroid $c$ and side midpoints $x,y,z$. Note that $\text{Length}[c,x] = \text{Length}[c,y] = \text{Length[c,z]}$ increases linearly to infinity with $p$ (which is why no uniform bilipschitz constant exists independent of $p$).
Define a map $\Delta_H \mapsto \Delta$ in pieces as follows. First, map the geodesic segments $[c_H,x_H] \mapsto [c,x]$ stretching by a uniform manner, and similarly for the other three side endpoints. Next extend over each of the six triangles of the barycentric subdivision in a similar fashion. For example if $\overline{vw}$ is the side of $\Delta$ with midpoint $x$ then for each $u_H \in [x_H,c_H]$ with image $u \in [x,c]$ map $[v_H,u_H] \to [v,u]$ stretching length uniformly, et cetera.
You should be able to get concrete bounds on the bilipschitz constant for this map, expressed in terms of $p$, and my guess is that these bounds will be close to optimal. Roughly speaking your bilipschitz constant should be somewhere around the ratio of the "thinness" constants of the triangles, by which I mean $$\text{Length}[x,c] \,\, / \,\, \text{Length}[x_H,c_H] $$