»  Solutions to puzzles in my National Review Online Diary

  October 2002

—————————

In my October diary I posed the following brain-teaser:

                    The Hockey Tournament

Three teams compete in a hockey tournament. Team A beats team B, team B beats team C, and team C beats team A. Fewer than 40 goals were scored in the tournament. Team A says they should win the tournament because they scored the most goals. Team B says they should win the tournament because they had the best goal differential (goals-for minus goals-against). Team C says they should win because they had the best goal ratio (goals-for divided by goals-against). While the judges are deciding who to give the trophy to, determine the score of each game.

—————————

Solutions

To my embarrassment (having confessed, when I posed the problem, that I didn't know how to do it), plenty of readers worked through it to get the correct game scores, which were: 13-12, 3-0, and 7-3.

The solutions fell into two broad categories, the "verbal" and the "algebraic." From an esthetic point of view, the "verbal" solutions — those that use a minimum of algebraic symbols — are of course superior. I show a couple of the most succinct ones below. For the sake of mathematical rigor, I also give a third, purely "algebraic," solution.

All solutions, of both types, rest on two key insights:  (1) that we seek the minimum possible answer to every question that arises in solving the problem, and (2) that the three goal differentials must net to zero, so that at least one of them must be positive and at least one negative.

Forcing everything to the minimum gives the minimum possible number of total goals, 38. There is no solution with fewer total goals than this. The next solution has 40 total goals, so the 38 total is the only one satisfying the problem as stated. ("Fewer than 40 goals were scored in the tournament.") If you remove the "fewer than 40 goals" condition, there is a solution with 40 goals, no solution with 41 goals, and then a solution for every number from 42 onwards. The smallest solution with no zero scores is 16-15, 4-1, 8-4 (I think), for a total 48 goals.

    • Verbal solution 1 (from DanFMarsh@aol.com)

The trick is to think in terms of scoring margins. Since we have a constraint on scoring, we're going to frequently look for the smallest possible values.

Since Team C has the highest winning percentage, it must have also have more wins than losses. So its scoring margin is at least +1. Team B has the largest scoring margin, so it must be at least +2.

For team B have a +2 winning margin in spite of losing to team A, it must have beaten team C by three points. For team A have more points than team B, they also have to score 3 points against team C. Since team C gave up at least 6 points, it must have scored at least 7.

Since any points scored against team B ruins team B's scoring differential, team C must have scored all their points against team A. That means team C beat team A, 7-3, but lost to team B, 0-3.

Now, team B must have a scoring ratio lower than 7:6, but a +2 scoring differential. The lowest such ratio is 15:13. That means they lost to Team A, 12-13. And sure enough, that means team A scored 16 points, 1 more than team B.

OK, that ratio works all the way around. We have a total of 38 points scored, and 38 points given up. The equation is balanced, and the total is under 40.

    • Verbal solution 2 (from Hei Lun Chan)

  1. Since C has the best goal ratio, their goal differential must be at least +1.
  2. Since B has the best goal differential, it must be at least +2. That means B must beat C by at least 2 more goals than A beat B.
  3. B must beat C by at least 3, since they lost to A by at least 1.
  4. C must beat A by at least 4, from (1) and (3).
  5. In the C-A game, A must score at least 3, from (2) and they must also score the most overall goals. (Since B's total score is at least two more than the sum of A's score in A-B and C's score in B-C, and A's total score is at least 1 higher than B's total score, it must follow that A's score in C-A must be at least 3 higher than C's score in B-C, which right now is at least 0.)
  6. C-A must be at least 7-3, from (4) and (5).
  7. B-C must be at least 3-0, from (3).
  8. If the exact scores from (6) and (7) are true, then A must beat B by only 1 goal, since any bigger margin would make B's goal differential fewer than 2.
  9. C's goal ratio is 7:6. B's goal ratio must be less. With a +2 goal differential, B must at least score 15 goals and allow at least 13.
  10. A-B must be at least 13-12 to satisfy (8) and (9).
  11. A-B could potentially be 14-13, 15-14, etc., but that would violate the "fewer than 40" condition.

So the scores must be:  13-12, 3-0, and 7-3.

    • Typical algebraic solution (from several readers)

  1. Rename the teams Alpha, Bravo, and Charlie for improved clarity.
  2. Use upper-case letters for winning scores, lower-case for losing scores.
  3. The scores of the three games are then A:b, B:c, and C:a.
  4. We have to find the six numbers A, B, C, a, b, and c.
  5. Note they are all whole numbers; and the winning scores A, B, and C are all greater than zero; and, since there were no ties, every game was won by at least one goal.
  6. For Alpha, total goals "for" are A+a, total goals "against" are C+b, differential is A+a-C-b, ratio is (A+a)/(C+b).
  7. For Bravo, total goals "for" are B+b, total goals "against" are A+c, differential is B+b-A-c, ratio is (B+b)/(A+c).
  8. For Charlie, total goals "for" are C+c, total goals "against" are B+a, differential is C+c-B-a, ratio is (C+c)/(B+a).
  9. If you add all three differentials, they net to zero. So at least one is positive and one negative.
  10. Since Bravo's is the biggest differential, it must be positive.
  11. I.e. B+b-A-c > 0
  12. I.e. B+b > A+c
  13. Since both A and B are positive, I can divide: (B+b)/(A+c) > 1
  14. But that is just Bravo's ratio …
  15.  … which is smaller than Charlie's ration (because Charlie's is the biggest ratio).
  16. So Charlie's ratio (C+c)/(B+a) is also bigger than 1.
  17. I.e. (C+c)/(B+a) > 1.
  18. So C+c > B+a.
  19. I.e. C+c-B-a is positive.
  20. That is just Charlie's differential.
  21. Being positive, and a whole number, it is at least 1.
  22. So Bravo's differential, which is bigger, is at least 2.
  23. Since the three differentials net to zero, Alpha's differential is at most -3.
  24. Since Alpha won the A:b game by at least 1, they must have lost the C:a game by at least 4, to get such a differential.
  25. I.e. C ≥ a+4.
  26. Alpha has scored the most goals, so A+a > B+b.
  27. I.e. A > B+b-a.
  28. From (22), Bravo's differential is at least 2.
  29. I.e. B+b-A-c is at least 2.
  30. I.e. (B+b-c)-A is at least 2.
  31. But (27) says that A > B+b-a.
  32. If, instead of subtracting A in (30), I subtract the smaller thing, the result must then be bigger than 2.
  33. I.e. (B+b-c)-(B+b-a) > 2.
  34. I.e. a-c > 2.
  35. I.e. a > c+2.
  36. So a is at least 3.
  37. And so, from (25), the minimum score for the C:a game is 7:3. On the minimizing principle, I shall suppose this is the actual score.
  38. Since the differential in the C:a game is then 4, if Charlie is to have his minimum net differential of 1 (see (21)), the differential in the B:c game must be 3.
  39. I.e. B = c+3.
  40. Since the differential in the B:c game is 3, for Bravo to get his minimum net differential of 2, the differential in the A:b game must be 1.
  41. I.e. A = b+1.
  42. Charlie's for/against ratio is (C+c)/(B+a).
  43. From (37) and (39), this is equal to (7+B-3)/(3+B).
  44. I.e. (B+4)/(B+3), …
  45.  … which is 1 + 1/(B+3)
  46. Bravo's for/against ratio is (B+b)/(A+c).
  47. From (39) and (41), this is equal to (B+A-1)/(B-3+A), …
  48.  … which is 1 + 2/(A+B-3).
  49. Since Charlie's ratio in (45) is greater than Bravo's in (48), 1/(B+3) > 2/(A+B-3).
  50. Cross-multiplying:  A+B-3 > 2B+6.
  51. So A > B+9.
  52. From (39), B is at least 3.
  53. So A is at least 13.
  54. From (40), the minimum possibility for the A:b game is then 13:12.
  55. In (37) we decided that the C:a score is 7:3. So to get the right differentials, the B:c score must be 3:0.
  56. These scores — 13:12, 3:0, and 7:3 — satisfy all the conditions of the problem, as shown above.
  57. Since we took minimum possibilities at every point, no smaller solution is possible.
  58. Since the problem requires less than 40 goals scored altogether, and this solution has 38, the only bigger solution possible is one with total 39 goals scored. Is there such a solution?
  59. To get a 39-goal total, either Alpha, or Bravo, or Charlie needs an extra goal.
  60. If Alpha in the A:b game, then Bravo loses the lead in differential:  -2, 1, 1.
  61. If Alpha in the C:a game, then Charlie loses the lead in ratio:  17/19, 15/13, 7/7.
  62. If Bravo in the A:b game, then Alpha no longer wins this game, it's a tie.
  63. If Bravo in the B:c game, then Alpha no longer has most goals (it's 16, 16, 7).
  64. If Charlie in the B:c game, then Bravo no longer leads in differentials (it's -3, 1, 2).
  65. If Charlie in the C:a game, same problem (it's -4, 2, 2).
  66. There is thus no other solution in less than 40 goals.
Top of this page