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BdMO Online Forum • View topic - IMO 2016 Problem 6

IMO 2016 Problem 6

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IMO 2016 Problem 6

Post Number:#1  Unread postby Phlembac Adib Hasan » Sun Aug 07, 2016 1:07 pm

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 every occupy the same intersection point at the same time.

  1. Prove that Geoff can always fulfill his wish if $n$ is odd.
  2. Prove that Geoff can never fulfill his wish if $n$ is even.
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Re: IMO 2016 Problem 6

Post Number:#2  Unread postby 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."
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