HMMO, 2020

Solution:

If two neighboring dorms are separated by a distance of more than $s$, we can move them slightly closer together and adjust the other dorms, increasing $s$. Therefore, in an optimal arrangement, the dorms form an equilateral 14-gon with side length $s$.

By scaling, the problem is now equivalent to finding the smallest $a$ such that there exist 14 vertices on the boundary of an $a \times a$ square that form an equilateral 14-gon with side length 1. Such a 14-gon must be centrally symmetric, yielding the following picture:

We know that $x_{1}^{2}+y_{1}^{2}=x_{2}^{2}+y_{2}^{2}=1$ and $x_{1}+x_{2}+3=y_{1}+y_{2}+2=a$. Moreover, if these equations are satisfied, then such a 14-gon exists. We now consider the vectors $\vec{v}_{1}=\left(x_{1}, y_{1}\right)$ and $\vec{v}_{2}=\left(x_{2}, y_{2}\right)$. These unit vectors are in the first quadrant and add to $(a-3, a-2)$, which lies on the line $y=x+1$.

Since $\vec{v}_{1}$ and $\vec{v}_{2}$ must lie on the first quadrant, from the above diagram we deduce that the minimum value of $a$ occurs when one of $\vec{v}_{1}, \vec{v}_{2}$ is $(0,1)$, meaning that $(a-3, a-2)=\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}+1\right)$. This means that $a=\frac{\sqrt{2}}{2}+3$, so the maximum possible value of $s$ is
$$\frac{17}{\frac{\sqrt{2}}{2}+3}=17 \cdot \frac{3-\frac{\sqrt{2}}{2}}{17 / 2}=6-\sqrt{2}$$