A man is walking in the north-east direction and wind appears to blow from north. If the man doubles his speed, wind appears at angle arccot 2 east of north. Find the actual direction of the wind.
The wind comes from the west. You can check that it works; anyway, here is how I found this out:
Let us denote by $\displaystyle \vec{w}={w_E\choose w_N}$ and $\displaystyle \vec{v}={\alpha \choose \alpha}$ the respective speeds of the wind and the man, in an orthonormal basis $\displaystyle (\vec{e}_E,\vec{e}_N)$ (E for east, N for north). Since the man goes NE, both components of $\displaystyle v$ are equal to some positive $\displaystyle \alpha$.
We know that the relative speed, i.e. the vector $\displaystyle \vec{w}-\vec{v}={w_E-\alpha\choose w_N-\alpha}$, is along the N direction, hence the E component is zero: $\displaystyle w_E=\alpha$.
Then we are told that the vector $\displaystyle \vec{w}-2\vec{v}={w_E-2\alpha\choose w_N-2\alpha}$ is (positively) colinear with $\displaystyle -{1\choose 2}$. This is the content of the statement about "arccot 2" (draw a right triangle with perpendicular sides of length 1 and 2; the least angle measures $\displaystyle {\mathop{arccot}}(2)$). As a consequence, $\displaystyle \frac{w_N-2\alpha}{w_E-2\alpha}=2$. Replace $\displaystyle \alpha$ by $\displaystyle w_E$ (as seen above), to get $\displaystyle 2-\frac{w_N}{w_E}=2$, hence $\displaystyle w_N=0$.
Finally, $\displaystyle \vec{w}={\alpha\choose 0}$ (the same $\displaystyle \alpha$ as in $\displaystyle \vec{v}$), and $\displaystyle \alpha>0$, which means that the wind comes from the west.
I hope my explanations are clear.
Laurent.