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12 players took part in a chess tournament and any two of them played exactly one match against each other (1 point for victory, 1/2 point for draw, 0 point for loss). After the tournament the top-ranked three players together gained three times more points than the bottom-ranked five players together.

How did the match between the seventh and the eighth player end?

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    $\begingroup$ I haven't looked at the math of this at all yet, but is it within the constraints of the setup (not necessarily the mathematics) for two players to end up tied in the end (i.e. tied for a ranking)? $\endgroup$ – dpwilson Feb 23 '16 at 20:37
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Answer:

7th beat 8th.

Explanation:

There are 66 matches ($\frac{12 \times 11}2$). Each match results in a total of 1 point gained, so there are 66 points to be gained total. (This isn't necessary information, but is nice for checking our work.)

Only one person can score a zero, and only one person can score an 11. (You can't have more than one person lose every match or more than one person win every match.)

The highest possible score for the top 3 is 30. (Examples: [11, 10, and 9] or [11, 9.5, 9.5 ] or [10.5, 10.5, 9]) The important thing is that they each beat all the other players (9 $\times$ 3 = 27) and give up 3 points amongst themselves (27 + 3 = 30).

That means the maximum cumulative score the bottom 5 can achieve is 10 ($\frac{30}3$).

Note that this is also the minimum score that the bottom 5 can achieve. (They play 10 games against each other ($\frac{5 \times 4}2$), and someone has to win each match. So there must be at least 10 points between the five of them.) So these must be the exact cumulative scores for the bottom 5 and the top 3 (10 points and 30 points).

This leaves 26 points (30 + 10 + 26 = 66) for the middle 4, which is consistent with [5, 6, 7, 8]. So this is a valid scoring distribution given the premises. This distribution is possible if everyone beat all the people below them in ranking, which means that 7th would have beat 8th.

It doesn't really matter what 7th scored anyway, since if 8th would have lost or tied to anyone outside the bottom 5, the cumulative score of the bottom 5 would be above 10, which would put the cumulative score of the top 3 above 30, which is impossible.

TLDR:

The 'top-three-equals-three-times-the-bottom-five' restriction and the round-robin nature of the tournament requires that the top section be total winners and the bottom section be total losers (with respect to the other brackets). 8th had to lose to everyone above him.

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Let say 8th to 12th players are the last 5 and they lost their all matches except against each other. So 8th has to win or draw

(does not matter since total point would not change and I will go for win)

against 9,10,11 and 12th, 9th would win against 10,11,12th etc and lost against 8th. So The total point of the last 5 players would be 8,6,4,2 and 0 at worst. That makes total 20. So the first three, 1st, 2nd and 3rd has to have 60 point at least. This is only possible if 1st, 2nd and 3rd would win all matches against other players except the matches between them. They can draw or win against each other. The total point would be 60.

As a result,

4th, 5th, 6th, 7th games has to be lost against 1st, 2nd and 3rd. and win against 8th, 9th, 10th, 11th, 12th. In between 4th to 7th cannot be known and does not change the fact of the given in the question.

and

8th has to lose against 1st to 7th as mentioned before. and win against the rest to become 8th.

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    $\begingroup$ As a small quibble, there are a number of possible scenarios where Player 8 does not beat each of Players 9-12. The important thing is that 8-12 must all lose to 1-7. $\endgroup$ – Matt Feb 23 '16 at 21:44

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