# A game of two halves, but maths will be the real winner

7 Nov 2016

Image: matimix/Shutterstock

Maths is an integral part of sport. So is competition. Maths can even be used to learn the outcome of certain competitions.

It’s the middle of the school year. Five football teams are due to face one another in an epic football blitz. The teams are very competitive.

Each team has its own strengths and weaknesses. The Animals have the best goalkeeper but a weak midfield. The Bulldogs have an excellent striker, but the rest of the team are mediocre. The Climbers have strong midfielders but terrible defenders. The Devils have decent strikers and defenders but a terrible goalkeeper. The Eliminators have an average team but their captain is awful.

The rules are set. Each of the teams have to play each other once and only once. A win would give a team three points, while a draw would give them one point. A loss would result in zero points.

At last year’s blitz, the Devils claimed victory over the other teams, while the Bulldogs were missing their star striker due to injury, and trailed in last place. But this year, they are prepared. The striker is back on the team and he’s been practising all year.

“We’re going to win this year,” said the Bulldogs captain. “We’re going to be so good, we’ll beat whoever comes second by two points.”

He was feeling confident, but the captain of the Devils just laughed at him.

“That’s not even possible,” he gloated. “In fact, no matter where you come in the final ranking, you can’t possibly beat the team that comes after you by two points. No one can.”

“You don’t know what you’re talking about,” said the Bulldogs captain. “I bet at least one of the teams will rank two points ahead of the next team. It’s probably going to be us anyway, but since you said no one can, I’m willing to take that bet.”

The Devils captain smiled. “You’re on.”

Who will win the bet? Can any of the teams end up with two points more than the team directly below it in the ranks? If so, what is the maximum number of teams in this tournament that could line up in such a way that each is 2 points ahead of the one coming just before it?

This question was part of the Croatian Mathematics Olympiad, 1st Round. The International Mathematical Olympiad 2017 will take place in Rio de Janeiro on 12-23 July 2017. Secondary school students in Ireland who wish to participate in Olympiads are encouraged to attend free mathematics enrichment programmes organised as part of the Irish Mathematical Olympiad.

Scroll down for the solution to this week’s puzzle.

The Bulldogs were victorious. Image: matimix/Shutterstock

Solution:

The Bulldogs gleefully celebrated their victory over the other teams, while the Devils captain kicked the dirt in anger. He was not used to being wrong.

But the Bulldogs played well and won three out of their four matches, drawing with the Climbers. The Devils won two and lost two; the loss to the Bulldogs was particularly painful. The Animals lost all four of their matches and came last. The Climbers drew with the Eliminators, as well as the Bulldogs, and they won the other two matches. The results and points can be seen here:

Final rankings:

Bulldogs=3+3+3+1=10

Climbers=1+3+1+3=8

Devils=0+0+3+3=6

Eliminators=0+1+0+3=4

Animals=0+0+0+0=0

On the way home, the Devils captain thought about the bet. How did he get it so wrong? But then he realised. Of course it was possible for some of the teams to beat the next team by two points – but it wasn’t possible for all five to do it.

From the ten matches, the captain knew the total sum of all scores is the smallest when all matches are draws, in which case, it is because each match results in 2 points for the two teams. He also knew the total sum of all scores is the largest when there are no draws, in which case, it is because the winner always takes 3 points.

Let ‘m’ be the smallest final score in the tournament. If the final scores of the 5 teams are at a distance of 2 apart from each other, then we can write them as: m, m+2, m+4, m+6 and m+8.

Summing up all scores, we get:

m+(m+2)+(m+4)+(m+6)+(m+8)=5m+20

If m=1, then the scores would be 1, 3, 5, 7, 9, and the sum 1+3+5+7+9=25.

He knew that, because this sum had to be at least 20, or at most 30. Hence ‘m’ can only be 0, 1 or 2.

However, the Devils captain knew ‘m’ couldn’t be 0 because in that case, the total sum would be 20 and all matches would be draws, hence all teams would have equal scores.

He also deduced that ‘m’ couldn’t be 2 because in this case, the total sum would be 30 and so, teams would only get points by winning. But then, all scores would be multiples of 3, so you can’t get two scores with a difference of 2.

Hence, if m=1, then the scores would be 1, 3, 5, 7, 9, and the sum 1+3+5+7+9=25.

Let ‘d’ be the final number of matches ending in a draw, and ‘w’ the final number of matches ending with a winning team. Then 25=2d+3w and d+w=10. Solving this, we get d=w=5.

But he knew this is impossible to achieve.

Indeed, to get 9 points, the top team would have to win 3 matches and lose 1. (It couldn’t win 2 and draw 3 because it only played 4 matches).

Similarly, to get 7 points, the second best team would have to win 2 matches, draw 1 and lose 1.

In between the two top teams, there already are 5 wins. But the third best team would also need a win, otherwise it couldn’t get 5 points in 4 matches. This contradicts w=5.

Therefore, no such scores could be obtained. The Devils captain skipped home, happy that he had solved the problem and secretly planned to make a better bet for the next tournament. Not to mention, he planned to beat the Bulldogs next time too.

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Jenny Darmody is the deputy editor of Silicon Republic

editorial@siliconrepublic.com