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Lionel Messi sentimentally remembers a small soccer tournament in which he participated many years ago.

  • There were four teams in the tournamnent: Atleticolino, Barcelino, Chicorrotico and Diminuto.
  • Each team played exactly one match against each of the other teams. For a win/draw/loss a team respectively scored 3/1/0 points.
  • Chicorrotico scored twice as many goals as Atleticolino did.
  • Barcelino conceded twice as many goals as Atleticolino did.
  • At least one match ended in a boring draw without a single scored goal.
  • The total number of goals scored in the tournament was at most 17.
  • At the end of the tournament, Atleticolino had scored 7 points, Barcelino 4 points, Chicorrotico 3 points, and Diminuto was last with only 1 point.

Question: What were the results of the six matches? How many goals did each team receive and score in every single match?

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  • $\begingroup$ Are you sure this is possible? If "twice as many" has to be exact, I think I can prove that it requires more than 14 total goals. $\endgroup$ – f'' Oct 17 '15 at 17:04
  • $\begingroup$ Well, weirdly enough there's two solutions, and they are different. At first glance they both seem fine. Are we to elaborate on all possible solutions, or did you not anticipate multiple different answers being possible? $\endgroup$ – Tim Couwelier Oct 19 '15 at 6:24
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Answer:


A-B 2-1
A-C 0-0
A-D 1-0
B-C 0-0
B-D 1-0
C-D 6-6

Explanation:


There's six games:
A-B
A-C
A-D
B-C
B-D
C-D We know A has 2 wins and a draw, B has a win, a draw and a loss, D had a draw and two losses. If those 3 had a total of 3 wins, 3 draws and 3 losses, C must have had 3 draws (1 win and 2 losses would mean the number of wins and losses isn't equal)
So A wins vs B, draws vs C and wins versus D / B loses vs A, draws vs c and wins vs D / C ties all.
A-B 1-0
A-C 0-0
A-D 1-0
B-C 0-0
B-D 1-0
C-D 0-0
Those are the minimal scores to achieve the win results. Now as for the goal balances: C scored twice as much as A did. So C had a big score draw vs either B or D. Given there's no extra limitations on D, it's easiest to try first to just alter the score of the C vs D draw.
A-B 1-0
A-C 0-0
A-D 1-0
B-C 0-0
B-D 1-0
C-D 4-4
So now C has scored 4, vs 2 for A.
B conceded twice as many goals as A did. Without alterations, we'd be at one conceded vs none. If we change the results from the A-B game to 2-1, we end up at two conceded vs one.
A-B 2-1
A-C 0-0
A-D 1-0
B-C 0-0
B-D 1-0
C-D 4-4
Now to check all that's asked:
A ends up with 7 points, B with 4, C with 3, D with 1.
C scored twice as many as A => Not met. We need to retweak the C-D game.
A-B 2-1
A-C 0-0
A-D 1-0
B-C 0-0
B-D 1-0
C-D 6-6
Now to check all that's asked:
A ends up with 7 points, B with 4, C with 3, D with 1.
C scored twice as many as A => 6 vs 3
B conceded twice as many as A => 2 vs 1
No more then 17 goals => 17 goals exactly.
QED.

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  • $\begingroup$ After reading f'' his answer, his should probably be upvoted over mine, given it elaborates on all possible outcomes. $\endgroup$ – Tim Couwelier Oct 20 '15 at 8:13
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From the number of points, we can deduce that Atleticolino (A) won two games and drew one, Barcelino (B) won one, drew one, and lost one, and Diminuto (D) drew one game and lost three. Chicorrotico (C) could have one win and two losses, or three draws. However, the total number of draws must be even, so C drew all three games.

Now we know that C drew against A, B, and D, so A won against B and D, and B won against D.

If A won against B, then B must have conceded at least one goal. Then A must also have conceded at least one goal. If A conceded a goal while winning two games and drawing one, then A must have scored at least three goals, so C must have scored at least six goals.

In the games that C played, C's opponents scored the same number of goals as C. There must have been at least one goal scored in each of the other three games. Therefore C could not have scored eight or more goals, or else there would be at least 19 total goals. So C scored exactly six goals, and A scored exactly three.

For A to score exactly three goals, they must have conceded exactly one goal, and won against B and D by a margin of exactly one goal. There are three cases, all of which produce a solution:

  • A conceded a goal to B. Then A went 2-1 against B, 0-0 against C, and 1-0 against D. B has conceded two goals to A, so they conceded no goals to C or D: B went 0-0 against C and 1-0 against D. Then C went 6-6 against D, with a total of 17 goals scored.
A 2-1 B
A 0-0 C
A 1-0 D
B 0-0 C
B 1-0 D
C 6-6 D
  • A conceded a goal to C. Then A went 1-0 against B, 1-1 against C, and 1-0 against D. B conceded one goal to D (couldn't be to C, because it would be impossible to have a 0-0 draw). So B went 0-0 against C and 2-1 against D. C went 5-5 against D, with a total of 17 goals scored.
A 1-0 B
A 1-1 C
A 1-0 D
B 0-0 C
B 2-1 D
C 5-5 D
  • A conceded a goal to D. Then A went 1-0 against B, 0-0 against C, and 2-1 against D. B conceded another goal against C (couldn't be to D, because there would end up being too many goals scored). So B went 1-1 against C and 1-0 against D. C went 5-5 against D, again with a total of 17 goals.
A 1-0 B
A 0-0 C
A 2-1 D
B 1-1 C
B 1-0 D
C 5-5 D
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