= Solution
Fix a hyperbolicity constant $\delta_0>0$ for $\mathbb H^2$. For a prescribed $\delta>0$, scale its distance by
$$
d_\delta=\frac{\delta}{\delta_0}d_{\mathbb H^2}.
$$
All distances in every geodesic triangle, including its thinness constant, scale by $\delta/\delta_0$. Hence
$$
X_\delta=(\mathbb H^2,d_\delta)
$$
is $\delta$-hyperbolic. Multiplication of a metric by a fixed positive constant is a <bilipschitz equivalence> and hence a <quasi-isometry>. Therefore $X_\delta$ is quasi-isometric to $\mathbb H^2$ and cannot be a quasi-tree. This supplies an example for every $\delta>0$.
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