IMO 2016 Shortlisted Problems

(a) For odd $n$, we travel along the circumference of the disk and mark each of the points $A_i$ or $B_i$ 'in' and 'out' alternately. Since every pair of lines intersect in the disk, there are exactly $n-1$ points between $A_i$ and $B_i$ for any fixed $1⩽i⩽n$. As $n$ is odd, this means one of $A_i$ and $B_i$ is marked 'in' and the other is marked 'out'. Then Jeff can put a snail on the endpoint of each segment which is closer to the 'in' side of the corresponding line. We claim that the snails on $l_i$ and $l_j$ do not meet for any pairs $i,j$, hence proving part (a).



Without loss of generality, we may assume the snails start at $A_i$ and $A_j$ respectively. Let $l_i$ intersect $l_j$ at $P$. Note that there is an odd number of points between $\mathrm{arc}{A}_i{A}_j$. Each of these points belongs to a line $l_k$. Such a line $l_k$ must intersect exactly one of the segments $A_{i}P$ and $A_{j}P$, making an odd number of intersections. For the other lines, they may intersect both segments $A_{i}P$ and $A_{j}P$, or meet none of them. Therefore, the total number of intersection points on segments $A_{i}P$ and $A_{j}P$ (not counting $P$) is odd. However, if the snails arrive at $P$ at the same time, then there should be the same number of intersections on $A_{i}P$ and $A_{j}P$, which gives an even number of intersections. This is a contradiction so the snails do not meet each other.

(b) For even $n$, we consider any way that Jeff places the snails and mark each of the points $A_i$ or $B_i$ 'in' and 'out' according to the directions travelled by the snails. In this case there must be two neighbouring points $A_i$ and $A_j$ both of which are marked 'in'. Let $P$ be the intersection of the segments $A_i{B}_i$ and $A_j{B}_j$. Then any other segment meeting one of the segments $A_{i}P$ and $A_{j}P$ must also meet the other one, and so the number of intersections on $A_{i}P$ and $A_{j}P$ are the same. This shows the snails starting from $A_i$ and $A_j$ will meet at $P$.