HMMT February

For a construction, note that one can change $(1,0,0),(-1,0,0) \rightarrow(\sqrt{2}, 0,0),(0,0,0) \rightarrow(1,0,0),(1,0,0)$ and similarly for $(0,1,0),(0,-1,0)$ and $(0,0,1),(0,0,-1)$. Then $u=(2,2,2)$. For the bound, argue as follows: let the vectors be $v_{1}, \ldots, v_{6}, n=(x, y, z)$ be any unit vector, and $S=\sum_{i}\left(n \cdot v_{i}\right)^{2}$, where the sum is over all vectors on the board. We claim that $S$ is invariant. Indeed, we have $\left(n \cdot \frac{1}{\sqrt{2}}(v+w)\right)^{2}+\left(n \cdot \frac{1}{\sqrt{2}}(v-w)\right)^{2} =\left(\frac{n \cdot v+n \cdot w}{\sqrt{2}}\right)^{2}+\left(\frac{n \cdot v-n \cdot w}{\sqrt{2}}\right)^{2} =\frac{2(n \cdot v)^{2}+2(n \cdot w)^{2}}{2} =(n \cdot v)^{2}+(n \cdot w)^{2}$. Also, at the beginning we have $S=2x^{2}+2y^{2}+2z^{2}=2$. Therefore we must always have $S=2$. Thus, by the Cauchy-Schwarz inequality we have $n \cdot u=\sum n \cdot v_{i} \leq \sqrt{\sum_{i}\left(n \cdot v_{i}\right)^{2}} \sqrt{6}=\sqrt{12}=2\sqrt{3}$. But since $n$ is arbitrary, this implies that $|u| \leq 2\sqrt{3}$; otherwise we could pick $n=u /|u|$ and reach a contradiction.