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Consider going a small distance $h$ along the diagonal away from a corner. The probability of finding a point in the corner triangle with height $h$ is $n h^2$, and such a point is at least $x = \sqrt{2} - 2h$ away from a point in the same triangle at the opposite corner. If $D$ is the maximum distance, we have for small enough $h$ $$\Pr[D > \sqrt{2} - 2h] \approx n h^2 .$$ Rewriting $h = (\sqrt{2}-x)/2$, we hence get the approximate distribution function $$F(x) = \Pr[D \le x] \approx 1 - \frac{n}{4} (\sqrt{2} - x)^2$$ with the lowest acceptable $x_0$ where $F(x_0) = 0$ or $x_0 = \sqrt{2} - 2/\sqrt{n}$. Integrating then gives the expectation, $$E[D] \approx \int_{x_0}^{\sqrt{2}} \! x \, F'(x) \, dx = \sqrt{2} - \frac{4}{3\sqrt{n}}.$$ Here is a plot of the approximation and simulated distances:

Consider going a small distance $h$ along the diagonal away from a corner. The probability of finding a point in the corner triangle with height $h$ is $n h^2$, and such a point is at least $x = \sqrt{2} - 2h$ away from a point in the same triangle at the opposite corner. If $D$ is the maximum distance, we have for small enough $h$ $$\Pr[D > \sqrt{2} - 2h] \approx n h^2 .$$ Rewriting $h = (\sqrt{2}-x)/2$, we hence get the approximate distribution function $$F(x) = \Pr[D \le x] \approx 1 - \frac{n}{4} (\sqrt{2} - x)^2$$ with the lowest acceptable $x_0$ where $F(x_0) = 0$ or $x_0 = \sqrt{2} - 2/\sqrt{n}$. Integrating then gives the expectation, $$E[D] \approx \int_{x_0}^{\sqrt{2}} \! x \, F'(x) \, dx = \sqrt{2} - \frac{4}{3\sqrt{n}}.$$

The idea here is that there is a density $n_1$ of points for which the maximum distance is a.s. larger than unity. In that case, only points in opposing corners need to be considered. From the approximation above, $x_0 = 1$ implies $n_1 \approx 12 + 8\sqrt{2} \approx 23.3$, so that is where we can expect the result to become reasonable.

Here is a plot of the approximation and simulated distances, which seems to indicate that this is more or less what's actually going on:

Of course, there is then the further approximation of the true distance by the projection onto the diagonal. But the more we move into the corners, the better that approximation becomes.

Consider going a small distance $h$ along the diagonal away from a corner. The probability of finding a point in the corner triangle with height $h$ is $n h^2$, and such a point is at least $x = \sqrt{2} - 2h$ away from a point in the same triangle at the opposite corner. If $D$ is the maximum distance, we have for small enough $h$ $$\Pr[D > \sqrt{2} - 2h] \approx n h^2 .$$ Rewriting $h = (\sqrt{2}-x)/2$, we hence get the approximate distribution function $$F(x) = \Pr[D \le x] \approx 1 - \frac{n}{4} (\sqrt{2} - x)^2$$ with the lowest acceptable $x_0$ where $F(x_0) = 0$ or $x_0 = \sqrt{2} - 2/\sqrt{n}$. Integrating then gives the expectation, $$E[D] \approx \int_{x_0}^{\sqrt{2}} \! x \, F'(x) \, dx = \sqrt{2} - \frac{4}{3\sqrt{n}}.$$ Here is a plot of the approximation and simulated distances:

Consider going a small distance $h$ along the diagonal away from a corner. The probability of finding a point in the corner triangle with height $h$ is $n h^2$, and such a point is at least $x = \sqrt{2} - 2h$ away from a point in the same triangle at the opposite corner. If $D$ is the maximum distance, we have for small enough $h$ $$\Pr[D > \sqrt{2} - 2h] \approx n h^2 .$$ Rewriting $h = (\sqrt{2}-x)/2$, we hence get the approximate distribution function $$F(x) = \Pr[D \le x] \approx 1 - \frac{n}{4} (\sqrt{2} - x)^2$$ with the lowest acceptable $x_0$ where $F(x_0) = 0$ or $x_0 = \sqrt{2} - 2/\sqrt{n}$. Integrating then gives the expectation, $$E[D] \approx \int_{x_0}^{\sqrt{2}} \! x \, F'(x) \, dx = \sqrt{2} - \frac{4}{3\sqrt{n}}.$$ Here is a plot of the approximation and simulated distances:

Consider going a small distance $h$ along the diagonal away from a corner. The probability of finding a point in that triangle is $n h^2$, and such a point is at least $x = \sqrt{2} - 2h$ away from a point in the same triangle at the opposite corner. If $D$ is the maximum distance, we have for small enough $h$ $$\Pr[D > \sqrt{2} - 2h] \approx n h^2 .$$ Rewriting $h = (\sqrt{2}-x)/2$, we hence get the approximate distribution function $$F(x) = \Pr[D \le x] \approx 1 - \frac{n}{4} (\sqrt{2} - x)^2$$ with the lowest acceptable $x_0$ where $F(x_0) = 0$ or $x_0 = \sqrt{2} - 2/\sqrt{n}$. Integrating then gives the expectation, $$E[D] \approx \int_{x_0}^{\sqrt{2}} \! x \, F'(x) \, dx = \sqrt{2} - \frac{4}{3\sqrt{n}}.$$ Here is a plot of the approximation and simulated distances:

Consider going a small distance $h$ along the diagonal away from a corner. The probability of finding a point in the corner triangle with height $h$ is $n h^2$, and such a point is at least $x = \sqrt{2} - 2h$ away from a point in the same triangle at the opposite corner. If $D$ is the maximum distance, we have for small enough $h$ $$\Pr[D > \sqrt{2} - 2h] \approx n h^2 .$$ Rewriting $h = (\sqrt{2}-x)/2$, we hence get the approximate distribution function $$F(x) = \Pr[D \le x] \approx 1 - \frac{n}{4} (\sqrt{2} - x)^2$$ with the lowest acceptable $x_0$ where $F(x_0) = 0$ or $x_0 = \sqrt{2} - 2/\sqrt{n}$. Integrating then gives the expectation, $$E[D] \approx \int_{x_0}^{\sqrt{2}} \! x \, F'(x) \, dx = \sqrt{2} - \frac{4}{3\sqrt{n}}.$$ Here is a plot of the approximation and simulated distances: