, 2016

Solution:

Let points $O, A_{1}, A_{2}$ lie in a plane such that $\angle A_{1} O A_{2}=\frac{2 \pi}{n+2}$. We represent the mirrors as line segments extending between $O$ and $A_{1}$, and $O$ and $A_{2}$. Also let points $A_{3}, A_{4}, \cdots, A_{n+2}$ lie in the plane such that $A_{i+1}$ is the reflection of $A_{i-1}$ over $O A_{i}$.

If Meghal shoots a laser along line $l$ such that the first point of contact with a mirror is along $O A_{2}$, the next point of contact, if it exists, is the point on $O A_{1}$ that is a reflection of the intersection of $l$ with $O A_{3}$. If we continue this logic, we find that the maximum score for round $n$ is equal to the maximum number of intersection points between $l$ and $O A_{i}$ for some $i$. We do casework on whether $n$ is even or odd. If $n$ is even, there are at most $\frac{n+2}{2}$ spokes such that $l$ can hit $O A_{i}$, and if $n$ is odd, there are at most $\frac{n+3}{2}$ such spokes. Then we must sum $2+2+3+3+\cdots+1009+1009=1009 \cdot 1010-1-1=1019088$.