Oxford University MAT 2018 Solved Papers

The particle's motion is restricted to the wedge bounded by the lines $\mathit{y}=2\mathit{x}$ and $\mathit{y}=\mathit{x}∕2$, and the particle could reach any grid point on the boundary on the wedge just by using a single repeated move. Since the point $\left(25,75\right)$ is outside this wedge, we want the closest point on the boundary of the wedge to this point. We must minimise ${\left(\mathit{x}-25\right)}^2+{\left(2\mathit{x}-75\right)}^2$, which occurs when $2\mathit{x}-50+8\mathit{x}-4\cdot 75=0$, that is when $\mathit{x}=35$. The distance from $\left(35,70\right)$ to $\left(25,75\right)$ is $\sqrt{{10}^2+5^2}=5\sqrt{5}$. The answer is (b).