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Title: Football Tournament Post by Barukh on May 1st, 2007, 12:09am Mr. H. read an article on a local football tournament in a Sunday newspaper. Six teams A, B, C, D, E, F took place in a round robin tournament. The final table looked as follows: ----------------- A 11 8 0 B 11 7 2 C 9 4 0 D 6 2 10 E 3 8 9 F 1 1 9 Mr. H. tried to figure out the result of every single game, but failed to do so. Later, he recalled the score of one of A’s games. After that, he was able to restore all the scores. Are you? Notes: 1. “Pts” means Points scored, “GF” – Goals for, “GA” – Goals against. 2. The winner gets 3 pts, the looser – 0; in case of a draw both teams get 1 pt. |
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Title: Re: Football Tournament Post by ThudanBlunder on May 1st, 2007, 5:42am A start: [hide] There were 6x5/2 = 15 games worth a maximum of 45 points. As the total of points obtained is only 41, there must have been exactly 11 wins/losses and 4 draws. A dropped 4 points without conceding a goal. Therefore A drew 0-0 in 2 games and won 3 games. B dropped 4 points and conceded 2 goals.. Therefore B drew 2 games and won 3 games. C dropped 6 points without conceding a goal. Therefore C drew 0-0 in 3 games and won 2 games. So A,B,C drew 7 games. So (as there were only 4 draws in total) A,B,C must have drawn 0-0 all the 3 games between themselves. So (as there were 11 wins and 4 draws) D,E,F shared a total of 3 wins and 1 draw. This draw must have been 0-0 between C and F as D wins 2 games, draws none, and E wins 1 game, draws none. But D scored only 2 goals. Therefore D beat both E and F by 1-0 So F scored its only goal against E and E scored 6 goals against F and 2 goals against B A,B,C score all their 19 goals against D,E,F D,E,F concede a total of 28 goals. So D,E,F score a total of 9 goals against each other and 2 goals against B So F has an aggregate score of 0-2 against A and B So both A and B beat F by 1-0 So A scored 7 goals against D and E So B scored 6 goals against D and E So C scored 4 goals against D and E D conceded 10 goals against A, B,C E conceded 7 goals against A, B,C Also, B scores more than 2 goals against E. There seems to be 6 different solutions, so maybe I am missing something. ------ A ------B ------ C ------ D ------ E ------ F A ---- * ---- 0-0 ---- 0-0 ----? ------- ? -------1-0 B ---- 0-0---- * ----- 0-0 --- ? ---- ?-2 ------- 1-0 C ---- 0-0 --- 0-0 --- * ------ ? -----? ---------0-0 D ----- ? ------ ? ----- ? ------ * --- 1-0 ------ 1-0 E ----- ? ------2-? ---- ? -----0-1 ---- * --------6-1 F ---- 0-1 ---- 0-1 --- 0-0 --- 0-1 --- 1-6 ----- * [/hide] Any predictions for the Champions' League semi-finals? I say 1-1 to United, and Chelsea after extra time. |
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Title: Re: Football Tournament Post by ThudanBlunder on May 1st, 2007, 7:14am OK, I have it. [hide] B must have scored 5 goals against E (the game he remembered), as this allows for only 1 solution. 3 goals allows 3 solutions and 4 goals allows 2 solutions. ------ A ------B ------ C ------ D ------ E ------ F A ----- * ---- 0-0 ---- 0-0 ---- 6-0 --- 1-0 ----1-0 B ---- 0-0---- * ----- 0-0 --- 1-0 ---- 5-2 --- 1-0 C ---- 0-0 --- 0-0 ---- * ------ 3-0---- 1-0-----0-0 D ---- 0-6 --- 0-1 --- 0-3 ----- * ----- 1-0 ---- 1-0 E ----- 0-1 --- 2-5 --- 0-1-----0-1 ---- * ------ 6-1 F ---- 0-1 ---- 0-1 --- 0-0 --- 0-1 --- 1-6 ----- * [/hide] Nice puzzle, Barukh! |
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Title: Re: Football Tournament Post by Barukh on May 1st, 2007, 10:15am on 05/01/07 at 07:14:16, ThudanBlunder wrote:
Uhmm? So what was the A's game score that gave H. the solution? ??? |
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Title: Re: Football Tournament Post by ThudanBlunder on May 1st, 2007, 10:22am on 05/01/07 at 10:15:39, Barukh wrote:
Oh, A's score? I see. Not reading the question again. Many a slip 'twixt cup and lip. |
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Title: Re: Football Tournament Post by ThudanBlunder on May 1st, 2007, 11:04am on 05/01/07 at 10:15:39, Barukh wrote:
I don't see how fixing A's score forces only one solution. ??? A scored 7 goals against D and E B scored 6 goals against D and E C scored 4 goals against D and E D conceded 10 goals against A,B,C E conceded 7 goals against A,B,C Fixing d = 5 forces only one solution: b=1, a=6, f=1, e=3, c=1 Fixing a = 4,5,or 6 (or b = 1,2,or 3) does not. . . . D E A a b = 7 B c d = 6 C e f = 4 -------------- . . .10 7 where d = 3, 4, or 5 Well, it's 7.30 and time for the real football now, Liverpool-Chelsea! |
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Title: Re: Football Tournament Post by SMQ on May 1st, 2007, 11:52am on 05/01/07 at 11:04:05, ThudanBlunder wrote:
What's wrong with fixing a = 4? ;) 1 < a,b < 6; 3 < d < 5 (since B must defeat E by a score of d to 2); 1 < c < 3 (because d > 3); 1 < g, h < 3. Fixing a = 4 forces c = e = 3. Edit: fixed stupid B loses to E mistake with a = 2... --SMQ |
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Title: Re: Football Tournament Post by ThudanBlunder on May 1st, 2007, 2:58pm Yes, you are right. . . . D E A a b = 7 B c d = 6 C e f = 4 -------------- . . .10 7 Fixing a = 4, b = 3 forces . . . D E A 4 3 = 7 B 3 3 = 6 C 3 1 = 4 -------------- . . .10 7 Fixing c = 1, d = 5 forces . . . D E A 6 1 = 7 B 1 5 = 6 C 3 1 = 4 -------------- . . .10 7 Fixing e = 1, f = 3 forces . . . D E A 6 1 = 7 B 3 3 = 6 C 1 3 = 4 -------------- . . .10 7 So there is a different table depending on whose result Mr H happens to remember. :) |
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