Post Number:#2 by Mehedi Hasan Nowshad » Tue Nov 01, 2016 7:51 pm
The main idea of my solution
Part (a) : WLOG, we may assume that the line segments are actually chords of a circle and all the chords intersect one another inside the circle. Now we shall label the endpoints one after another like $A_1, A_2,....A_{2n}$. Thus points $A_k$ and $A_{k+n}$ are the end points of same chord with $1 \leq k \leq n$. So if we place the frogs in odd indexed points, then geoff can fullfill his wish. also as n is odd, so $k$ and $k+n$ have different pairities ensuring that he didn't place the frogs in the endpoints of same chord.
Part (b) As $n$ is even here, so $k$ and $k+n$ have the same pairity. So eventually Geoff has to place at least two frogs in consecutive points. By which, he can never fullfill his wish.
"Failure is simply the opportunity to begin again, this time more intelligently."
- Henry Ford