s51 - back, or there is another player thatis even farther...

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Solution to Problem 51 Congratulations to this week’s winners Nathan Pauli, Ray Kremer Correct solutions were also received from Bradley alumnus TIm Kelley, Robert T. McQuaid, Sergiu Stefanov, William Webb, [email protected], Burkart Venzke, Aaron David Kahn. Numerous incorrect submissions were received. The following solution is courtesy of Sergiu Stefanov, a student of Computer Science in Timisoara, Romania. First we have to make the assumption that there are no equal distances between players, so everyone will have only one person to throw to. The King throws to the player farthest away; let us denote the distance to that player by d 0 . Now there are two possibilities: the King is the farthest away from him and he throws the disk
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Unformatted text preview: back, or there is another player thatis even farther away. In the first case, a tea party ensues, so consider the second case. Let us not the distance by d 1 . It is obvious that d < d 1 . The game continues in the same way, the distance reaching the value of d n . We can write d < d 1 < d 2 < . .. < d n . Imagine that this person throws the disk back to the King, then that person is at disance less than d from the King, since that is the distance to the farthest person from the King. But each new distance must be strictly greater than the last! So the King can never get the disk back again....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.

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s51 - back, or there is another player thatis even farther...

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