Annual Harvard-MIT Mathematics Tournament

Solution:

Answer: $\frac{\sqrt{6}+\sqrt{2}}{2}$

Let the square be $ABCD$. First, suppose that all four vertices are closer to $X$ than $Y$. Then, by the triangle inequality, the sum of the scores is $AX + BX + CX + DX \geq AB + CD = 2$.

Similarly, suppose exactly two vertices are closer to $X$ than $Y$. Here, we have two distinct cases: the vertices closer to $X$ are either adjacent or opposite. Again, by the triangle inequality, it follows that the sum of the scores of the vertices is at least $2$.

On the other hand, suppose that $A$ is closer to $X$ and $B, C, D$ are closer to $Y$. We wish to compute the minimum value of $AX + BY + CY + DY$, but note that we can make $X = A$ to simply minimize $BY + CY + DY$. We now want $Y$ to be the Fermat point of triangle $BCD$, so that $\measuredangle BYC = \measuredangle CYD = \measuredangle DYB = 120^{\circ}$.

Note that by symmetry, we must have $\measuredangle BCY = \measuredangle DCY = 45^{\circ}$, so $\measuredangle CBY = \measuredangle CDY = 15^{\circ}$.

And now we use the law of sines: $BY = DY = \frac{\sin 45^{\circ}}{\sin 120^{\circ}}$ and $CY = \frac{\sin 15^{\circ}}{\sin 120^{\circ}}$. Now, we have $BY + CY + DY = \frac{\sqrt{2} + \sqrt{6}}{2}$, which is less than $2$, so this is our answer.