, 2019

Solution:

Note that we want the path drawn out by the lasers to come back to $P$ in as few steps as possible. Observe that if a laser is fired with an angle of $n$ degrees from the tangent, then the number of points it creates on the circle is $\frac{180}{\operatorname{gcd}(180, n)}$. (Consider the regular polygon created by linking all the points that show up on the circle—if the center of the circle is $O$, and the vertices are numbered $V_{1}, V_{2}, \ldots, V_{k}$, the angle $\angle V_{1} O V_{2}$ is equal to $2 \operatorname{gcd}(180, n)$, so there are a total of $\frac{360}{2 \operatorname{gcd}(180, n)}$ sides).

Now, we consider the case with both $n$ and $n+1$. Note that we wish to minimize the value $\frac{180}{\operatorname{gcd}(180, n)} + \frac{180}{\operatorname{gcd}(180, n+1)}$, or maximize both $\operatorname{gcd}(180, n)$ and $\operatorname{gcd}(180, n+1)$. Note that since $n$ and $n+1$ are relatively prime and $180 = (4)(9)(5)$, the expression is maximized when $\operatorname{gcd}(180, n) = 20$ and $\operatorname{gcd}(180, n+1) = 9$ (or vice versa). This occurs when $n = 80$. Plugging this into our expression, we have that the number of points that show up from the laser fired at $80$ degrees is $\frac{180}{20} = 9$ and the number of points that appear from the laser fired at $81$ degrees is $\frac{180}{9} = 20$. However, since both have a point that shows up at $P$ (and no other overlapping points since $\operatorname{gcd}(9, 20) = 1$), we see that the answer is $20 + 9 - 1 = 28$.