Optimal $\delta$ for Gromov's $\delta$-hyperbolicity of the hyperbolic plane

The answer is $\delta = \ln(2) \approx 0.693147181$.

Claim: The correct placement of the four points at infinity is at the corners of an ideal square.

With the claim in hand, we can compute $\delta$ in the upper half plane model. We place the points at $0, 1, \infty, -1$. We place identical horocircles at each of these points. These are cyclically tangent, and all have the same minimal distance $\delta/2$ from the point $i$. The points of tangency are cyclically permuted by the order four rotation about $i$. If we take boundary of the horosphere about $\infty$ to be the line $y = H$ then we discover that the order four element (fixing $i$) sends $1 + iH$ to $-1 + 2i/H = -1 + iH$. Thus $H = \sqrt{2}$. So $\delta$ is twice the distance from $i$ to $i\sqrt{2}$ and we are done.

The proof of the claim appears to be difficult. We have to prove that, given four material points, we can increase $\delta$ by first moving them "outward" to lie on a circle (tricky), then to lie symmetrically on the circle (medium), and then increase the radius of the circle to infinity (easy).

Indeed, the hyperbolic plane is $\log(2)$-hyperbolic (with the 4-point definition of hyperbolicity) and this is the optimal constant. The result is nontrivial and first appeared as Corollary 5.4 in

Nica, Bogdan; Špakula, Ján, Strong hyperbolicity, Groups Geom. Dyn. 10, No. 3, 951-964 (2016). ZBL1368.20057.