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 frogs will ever occupy the same intersection point at the same time.

(a) Prove that Geoff can always fulfill his wish if $n$ is odd.

(b) Prove that Geoff can never fulfill his wish if $n$ is even.

Part (a) — $n$ odd:

Step 1: Label the segments $1,2,\dots ,n$. Each segment $i$ crosses all other $n-1$ segments. The frog on segment $i$ visits intersection points with segments ${\sigma }_i(1),{\sigma }_i(2),\dots ,{\sigma }_i(n-1)$ in order (determined by the geometry and the chosen endpoint).

Step 2: The frog on segment $i$ is at intersection $i\cap {\sigma }_i(t)$ at time $t$. Two frogs collide if the frog on segment $i$ meets segment $j$'s frog at the intersection $i\cap j$ at the same time. This happens when ${\sigma }_i^{-1}(j)={\sigma }_j^{-1}(i)=t$ for some $t$.

Step 3: Choosing the opposite endpoint of segment $i$ reverses the order ${\sigma }_i$. So the choice of endpoint corresponds to choosing an orientation for each segment. A collision between frogs $i$ and $j$ at time $t$ means the frog on $i$ reaches $i\cap j$ at the same step as the frog on $j$ reaches $j\cap i$.

Step 4: Model this as a tournament: direct each pair $\{i,j\}$ based on whether the frog on $i$ reaches $i\cap j$ before or after the frog on $j$ reaches it. Reversing the orientation of segment $i$ reverses all arcs incident to $i$, changing the score (out-degree) of $i$ by $\pm (n-1)$. For odd $n$, $n-1$ is even, so score parities are preserved.

Step 5: For odd $n$, one can orient segments so that no two frogs meet at the same intersection at the same time, using a parity/coloring argument on the arrangement.

Part (b) — $n$ even:

Define a parity invariant: the total number of "before" relations has a fixed parity regardless of the choice of orientations. For even $n$, this parity forces at least one collision.

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