$\mathbb{P}$ is a regular polygon with 2018 sides. What is the minimum number $k$ of vertices that 4 of them form a convex quadrilateral with 3 common sides with $\mathbb{P}$.
My idea is to color 1010 vertices to make sure 2 consecutive ones are colored. Then color 506 of the remaining to make sure 3 consecutive ones are colored. Then color 264 of the remaining to guarantee that 4 consecutive ones are colored. And they for a quadrilateral sharing 3 sides with the 2018-gon
To make a quadrilateral with three common sides with $\Bbb P$, you need four vertices in a row. To avoid that, you can pick three, skip one, pick three, skip one, and so on. You get $\lfloor \frac {2018}4 \rfloor\cdot 3 +1=1513$ that way without four in a row. The next one gives four in a row.