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

Claim: the 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 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).

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).

Nice problem. The answer is (I think!) $\ln(2) \approx 0.693147181$.

Let's suppose that the correct placement of the four points at infinity is at the corners of an ideal square. In the upper half plane model we place these at $0, 1, \infty, -1$. We place identical horospheres at each of these points. So these are cyclically tangent, and all have the same distance 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 $y = H$ 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.

Not shown: That $\delta$ is maximised by taking the points to infinity and then placing them at the points of a square. But that is all reasonable - When the points are close together, $\delta$ is small (comparable to the diameter).

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

Claim: the 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 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).

Nice problem. The answer is (I think!) $\ln(2) \approx 0.693147181$.

Here is the idea. The most symmetric placement of four points at infinity is at the corners of an ideal square. In the upper half plane model we place these at $0, 1, \infty, -1$. We place identical horospheres at each of these points. So these are cyclically tangent, and all have the same distance 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 $y = H$ 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.

Nice problem. The answer is (I think!) $\ln(2) \approx 0.693147181$.

Let's suppose that the correct placement of the four points at infinity is at the corners of an ideal square. In the upper half plane model we place these at $0, 1, \infty, -1$. We place identical horospheres at each of these points. So these are cyclically tangent, and all have the same distance 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 $y = H$ 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.

Not shown: That $\delta$ is maximised by taking the points to infinity and then placing them at the points of a square. But that is all reasonable - When the points are close together, $\delta$ is small (comparable to the diameter).