Three-person soccer

From the 2003 Flanders Junior Maths Olympiad:

To play soccer with three people, two field players try to score past the player in goal, and whoever scores stands in goal for the next game. Arne, Bart and Cauchy play the game, with A playing 12 times on the field, B playing 21 times on the field, and C playing 8 times in goal.

Who scored the 6th goal?

It seems like we have been given very little information, but let’s do some investigating to see what we can find.

Whenever one player is in goal, the other two players must be on the field. So the number of times A played in the field must equal the number of times B played in goal plus the number of times C played in goal. C played in goal 8 times, and A played on the field 12 times, so B must have been in goal 4 times.

Similarly, since B played on the field 21 times and C played in goal 8 times, A must have played in goal 13 times. So the number of times A, B and C played in goal was 13, 4 and 8 respectively.

This means A was in goal for 13 of the 25 games, just over half the time. But due to the rule about players swapping in and out, no player can play in goal during successive games. The only conclusion is that A played in goal for all the odd numbered games: game 1, 3, 5 and so on, up to game 25.

In particular, A was in goal for the 7th game, and thus it was Arne who scored the 6th goal.

So we did in fact (surprise!) have enough information to solve the problem. Not too difficult, but requires a bit of thinking; a very well designed puzzle and a nice riddle to give to family and friends over the dinner table.



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