Competition problem — Combinatorics

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Combinatorics Difficulty 5.8 AIME, harder Prove it

Problem 9. In a volleyball tournament, bets are accepted on four teams. Bets on the first team are accepted at a ratio of $1: 2$ (if the first team wins, the player receives the amount they bet on this team plus twice the amount, i.e., they receive three times the amount of money they bet, and if they lose, the money is not returned). Bets on the second team are accepted at a ratio of $1: 4$ , on the third team $-1: 5$ , and on the fourth team $1: 6$ . Is it possible to place bets in such a way as to win regardless of the tournament outcome?

Solution

Answer: Yes, it is possible.

Solution. If the first team wins, the bet is returned tripled, so more than $1 / 3$ of all the money should be placed on it. Similarly, more than $1 / 5$ of all the money should be placed on the second team, more than $1 / 6$ on the third, and more than $1 / 6$ on the fourth. Since the sum of these fractions is less than 1 (indeed, $1 / 3 + 1 / 5 + 1 / 6 + 1 / 7 < 1 / 3 + 1 / 3 + 1 / 6 + 1 / 6 = 1$ ), there exists a set of numbers that sum to one, each of which is greater than the corresponding fraction. Any such set will do.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.