Source: IMO-2016-P6
Category: Existence and Construction

Problem:
There are $n\ge 2$ line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands $n-1$ times. Every time he claps, each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time.

\begin{enumerate}[(a)]
\item Prove that Geoff can always fulfill his wish if $n$ is odd.
\item Prove that Geoff can never fulfill his wish if $n$ is even.
\end{enumerate}

Solution:
Geoff can fulfill his wish if and only if $n$ is odd.

The preliminary geometric framework operates as established context. A large circle $\omega$ encases all $\binom{n}{2}$ intersection points. The boundary points $P_1, P_2, \dots, P_{2n}$ are ordered clockwise. Each line segment connects $P_i$ and $P_{i+n}$.

(a) For odd $n$, place the frogs on $P_1, P_3, \dots, P_{2n-1}$. Parity dictates that no two frogs intersect simultaneously.

(b) For even $n$, frogs cannot occupy consecutive $P_i$, requiring placement on alternating points. Frogs cannot occupy diametrically opposite points, yielding an immediate contradiction.


Grading guidelines:
(Partial) 1. Set up (or correctly used) the standard circular model: all intersections inside a circle, the 2n boundary points P1,…,P2n in order, and each segment is the chord PiPi+n; and interpreted a frog’s positions after k claps as the k-th intersection along that chord. 2. For odd n, gave the correct placement pattern (e.g. frogs start at P1,P3,…,P2n−1 or an equivalent alternating set) and proved a nontrivial invariant/observation implying two frogs cannot land on the same intersection at the same time (e.g. a parity argument on indices of endpoints/intersections). 3. For even n, proved a necessary restriction on any collision-free placement, such as: starting endpoints cannot be consecutive among the P_i (otherwise an immediate collision occurs), hence frogs must be placed on alternating boundary points. (Almost) 1. Solved both (a) and (b) in the circle/chord model with the correct constructions and contradictions, but left a small gap in justifying the key translation between “same intersection at same time” and the derived parity/index condition on the corresponding boundary labels. 2. Gave the odd-n construction and essentially complete no-collision proof, and for even n obtained the forced “alternating endpoints” condition, but did not finish the final contradiction (e.g. did not fully justify why some diametrically opposite pair must be chosen / why that forces a collision). 3. Solution is almost complete, but made minor mistakes which are not negligible.