A particle moves in the $xy$-plane, starting at the origin $(0, 0)$. At each turn, the particle may move in one of the two ways:
- it may move two to the right and one up
- it may move one to the right and two up
What is the closet distance the particle may come to the point $(25, 75)$?
This article can only move within the region bounded by the two lines $y=2x$ and $y=x/2$. Then, we realize the point $(25, 75)$ is "above" the region. Hence we need to find the smallest $k$ such that the article will land on the circle of $(x-25)^2 + (y-75)^2=k^2$, or minimize the value of $(x-25)^2 + (y-75)^2$.
Our first attempt is to let $y=2x$ (as this is on the boundary line). In this case, $$(x-25)^2 + (2x-75)^2 = 5x^2 - 350x + 6250$$
This quadratic function has a minimal value at $x=35$ which corresponds to $y=70$. We can verify the point $(35, 70)$ can be reached by the article by repeating the first motion for $35$ times. And the line connecting $(25, 75)$ is perpendicular to the line $y=2x$. Therefore, we confirm the desired closest point is $(35, 70)$ and the corresponding distance is $$\sqrt{(25-35)^2 +(75-70)^2}=\boxed{5\sqrt{5}}$$