Problem 6

For odd $n$, write $n=2m+1$ and let $C_r$ ($r=0,1,\dots,2m$) be the segment whose frog starts at the odd-indexed point $Q_r=P_{2r+1}$ (the indices of the $P$'s are read modulo $2n$); its other endpoint is $P_{2r+1+n}$. Thus the placement is explicit: the frogs start at every odd point $P_1,P_3,\dots,P_{2n-1}$, one on each of the $n$ segments, and face along $C_r$ from $Q_r$ to $P_{2r+1+n}$. Fix $C_0$. For $r=1,\dots,m$, the endpoint $P_{1+2r}$ of $C_r$ lies on the open circular arc from $P_1$ to $P_{1+n}$; for $r=m+1,\dots,2m$, the endpoint of $C_r$ on that arc is its antipode $P_{1+2r+n}$, whose relative index from $P_1$ is $2r-n$. Since $n=2m+1$, these relative indices are $2,4,\dots,2m$ and $1,3,\dots,2m-1$, respectively, so they are exactly all integers $1,2,\dots,n-1$ in circular order. Therefore, if $C_r$ meets $C_0$, the rank of their intersection when counted from the frog on $C_0$ is $\rho_0(r)=2r$ for $r=1,\dots,m$, and $\rho_0(r)=2r-n$ for $r=m+1,\dots,2m$. Rotating the same argument to any pair $C_a,C_b$, and writing $r$ for the cyclic difference of their starting-point indices, shows that the two ranks of their intersection, counted from the two frogs, are $\rho_0(r)$ and $\rho_r(0)=n-\rho_0(r)$: the second frog sees the same crossing order in reverse. This is the invariant/parity count: every pair's two ranks are positive integers in $1,\dots,n-1$ whose sum is $n$, so they cannot be equal because $n$ is odd (equality would give $2\rho_0(r)=n$). After clap $k$ ($1\le k\le n-1$), a frog is at the $k$-th intersection on its directed segment; hence two frogs could collide only if the two ranks for their pair were equal, which we have just ruled out. Thus the explicit odd-point placement avoids every collision.