You can also find lots of lovely activities on the sport.maths.org pages.
Remember that our stages roughly refer to the English Key Stages in which the mathematical content is likely to be found and the number of stars indicate how difficult is is likely to be to get into the problem. All our problems are likely to contain enough depth to satisfy the most enthusiastic problem solvers!
May the best person win
Image
The Paralympics
The Paralympic Games include athletes with mobility impairments, visual impairments, brain damage and other disabilities. They started in the UK in 1948 with a small group of men who had been injured in the Second World War. The first post-war Olympic Games took place in London that year, and on the opening day Stoke Mandeville Hospital, which has a world-famous spinal injuries centre, organised a sports competition for some of the patients. The first official Paralympic Games, no longer restricted to war veterans, were held in Rome in 1960. The Paralympic Games now operate in parallel with both the summer and winter Olympics - and are called the 'Paralympic' Games from the Greek word 'para' meaning beside or alongside.
Team events and some individual events are for specific categories of athletes, such as wheelchair basketball, goalball and judo for the visually impaired, boccia for people with cerebral palsy (brain injuries), and rowing for people who only have use of their arms. In these events, people with other disabilities are excluded, and any variations in disability within the category are ignored.
In most individual events, athletes are divided into six broad categories - amputees, people with cerebral palsy, people with intellectual disabilities, wheelchair users, visually impaired people, and Les Autres. 'Les Autres' is the French term for 'others', and includes people with dwarfism, multiple sclerosis and congenital deformities. Within each of these categories, athletes are then further sub-divided into classes according to the degree to which their disability impacts on their performance.
Classifying athletes
It isn't only Paralympic sports where people are classified in some way. Golfers have their handicap as do race horses. Boxers and weight-lifters are classified according to their body weight, and many amateur sports classify people on the basis of what they have achieved so far.
Ideally each class should be as homogeneous as possible but different from other classes on the basis of certain criteria. In the case of the Paralympics, the criteria include sex, the nature of the event, the disability category and the degree to which the athlete's individual level of disability affects their performance. The classification needs to be seen to be fair, as far as possible. The Paralympics are a high stake competition, so the process also needs to be rigorous.
There are, however, many problems in grouping athletes into classes. It is difficult to apply handicapping in a consistent way across different classes. Racing conditions and changes in technology mean that classification frequently has to be revised. Whether a class is small or large may give athletes an unfair advantage or disadvantage. There have also been instances of cheating, with athletes claiming to be more disabled than they actually are in the hopes of being put into a class in which they would have an unfair advantage.
One way of avoiding such problems is to use a mathematical algorithm which focuses on athletes' performances, rather than the degree of their disability. This also avoids the need for committees to meet to debate which class an athlete should be put into and how classification should be updated as equipment or other factors change.
One proposed method involves calculating an overall median for an event, and also calculating median performances for each class within that event. The ratio of the median for a class compared to the overall median for the event tells us how good a particular class is relative to the others in the event. Competitors' performances are then scaled using this ratio to get their score. This process is illustrated at the end of this article.
Setting students
The classification of athletes for the Paralympics is a very similar process to that of classifying students for mathematics and other lessons. We might use criteria such as exam performance yet we are aware that this is a 'one off' snapshot of performance and therefore may not be entirely representative of a student's ability. To try and get a broader picture of performance we often add teacher assessment. Our attention may then turn to student confidence as we try and aim for conditions that will maximise a particular student's performance. Does the discussion of classification for the Paralympics above help further refine or inform our setting of students?
In the classroom - the school sports days
So what is the relevance here to school sports days and helping students to develop their mathematical thinking? The challenges we face when planning a sports day include:
- what events to choose
- how to allocate students to each event
- whether we need some sort of handicapping system or whether it is simply a 'flat playing field'?
The first challenge is taken up in one of this month's stage 1 problems, Our Sports. This encourages students to look at what events they might want to include in their sports day. This could be developed at stages 2 and 3, as a sampling problem. For instance, how many students should be asked to get a fair impression of the whole school view? Does it matter how good at sport those asked are? What would a representative sample look like?
The second challenge depends on the purpose of the sports day. For the Paralympics it is about peak performance and winning your event to gain the coveted gold medal. It is entirely competitive. For a school sports day it could be that everyone should take part, which raises the question of whether the most athletic should take all the 'medals' or not. There may be other issues which are as important as individual performances. The purpose will inform how students are allocated to each event. How important is it that we are modeling our school sports day on the real world of sport?
Students could be allocated to a particular event according to:
- previous performance
- their preferences
- ensuring everyone takes part in one event
- ensuring events are of equal size
- ??
By choosing one event - such as the 100m sprint - students could investigate the impact of different classification strategies. They could select participants using the their own classification strategies, try out the race with the different groups and then compare outcomes.
But how will they represent the results in order to make meaningful comparisons? For stage 2 and 3 students this could lead to the idea of an average as in some way typifying a particular group, with the range giving an indication of the variability within that group. Stage 3 and 4 students could think further about why the median might be preferable to the arithmetic mean.
The third challenge is informed by what we mean by 'fair'. This could lead to an investigation of different handicapping systems, such as that for golf or horse-racing. Students could explore how the system works and what it could be like in the school sports day. In school we may want the events to be fair yet perhaps not go as far as a handicapping system.
Questions to consider
How might the size of class affect your chances of winning?
- In a school setting, how might the size of the set/group affect a student's attainment outcomes?
- On what grounds do we/might we make a correlation between group size and ability?
What advantages are there in using median performances rather than mean, minimum or maximum performances for each class?
- How does this apply in a school setting in the way we may choose to group students?
Stage 3 students could also look at How Would You Score It?
Example of classification using median performances
Thanks to Professor David Percy, University of Salford, for his help with this section.
Here is a set of 50 random scores, between 0 and 100:
66 | 1 | 44 | 5 | 29 | 92 | 76 | 76 | 23 | 60 |
53 | 84 | 50 | 55 | 44 | 64 | 25 | 77 | 80 | 51 |
54 | 92 | 10 | 67 | 72 | 38 | 79 | 10 | 28 | 53 |
74 | 17 | 79 | 95 | 68 | 23 | 17 | 42 | 78 | 80 |
4 | 62 | 74 | 2 | 32 | 26 | 83 | 17 | 20 | 6 |
Suppose we want to divide the 'athletes', whose scores these are, into three equivalent groups so that we can judge who the overall medal winners should be. We could base our classification on the median, so that we make the median of each group the same, then having scaled all the scores, see whose score is best.
This process is illustrated below.
I start by subdividing the scores into three classes, according to whether the score is in the range 0 - 33 (Class A), 34 - 66 (Class B), and 67 - 100 (Class C). If a high score is best, this might indicate people who are severely affected by their impairment, those moderately affected, and those only slightly affected.
Class A: 1, 5, 29, 23, 25, 10, 10, 28, 17, 23, 17, 4, 32, 2, 26, 17, 20, 6 (18 members)
Class B: 66, 44, 60, 53, 50, 55, 44, 64, 51, 54, 38, 53, 42, 62 (14 members)
Class C: 92, 76, 76, 84, 77, 80, 92, 67, 72, 79, 74, 79, 95, 68, 78, 80, 74, 83 (18 members)
The overall median for the 50 scores is 53, and the three class medians are:
mA = 17, mB = 53, mC = 78.5
You could ask the question here whether the median of the central group is necessarily the same as the overall median (or is it simply a consequence of the members of the groups being symmetrically distributed?).
A simple scaling factor for each group can be calculated by dividing the minimum median by each class median, giving:
fA = 1, fB = 0.321, fC = 0.217
(17 x 1 = 17, 53 x 0.321 = 17, 78.5 x 17 = 17)
The raw scores are then scaled by these factors to give adjusted scores from which medal winners can be determined.
It turns out that all the medal winners would be in Class A, if a high score was best. These are the adjusted scores:
Class A: 1, 5, 29, 23, 25, 10, 10, 28, 17, 23, 17, 4, 32, 2, 26, 17, 20, 6
Class B: 21.2, 14.1, 19.3, 17.0, 16.0, 17.6, 14.1, 20.5, 16.4, 17.3, 12.3, 17.0, 13.5, 19.9
Class C: 19.9, 16.5, 16.5, 18.2, 16.7, 17.3, 19.9, 14.5, 15.6, 17.1, 16.0, 17.1, 20.6, 14.7, 16.9, 17.3, 16.0, 18.0
The ranges of classes with higher medians are reduced by the scaling process. Whereas for Class A 0 scales to 0, and 33 scales to 33, for Class B the minimum (34) and maximum (66) scale to 10.9 and 21.2 respectively, and for the minimum (67) and maximum (100) scale to 14.5 and 20 respectively. This favours the class with the lowest median, as we have seen.
This results from choosing data by randomly selecting integers from a range, which implies that, eg. 50 is twice as good as 25. Sporting data is not random and minimum scores are unlikely to be close to the bottom of the class, so such extreme results as this would be much less likely.
What difference would it make if class maxima were used, instead of class medians? If we assume that a high score is good, might this make for a fairer system of score adjustment?
You can also find lots of lovely activities on the sport.maths.org pages.
Remember that our stages roughly refer to the English Key Stages in which the mathematical content is likely to be found and the number of stars indicate how difficult is is likely to be to get into the problem. All our problems are likely to contain enough depth to satisfy the most enthusiastic problem solvers!
Alternative Record Book
Can you do it too?
Try some throwing activities and see whether you can throw something as far as the Olympic hammer or discus throwers.
Decathlon: the art of scoring points
FA Cup
Going for Gold
Looking at the 2012 Olympic Medal table, can you see how the data is organised? Could the results be presented differently to give another nation the top place?
Half Time
High Jumping
Little little g
Match the Matches
May the best person win
Medal Muddle
National Flags
Now and Then
Have a look at the results for some events at past Olympic Games. Can you use these to predict the results at the next Olympics?
Nutrition and Cycling
Olympic Logic
Olympic Measures
Olympic Records
Olympic rings
Can you design your own version of the Olympic rings, using interlocking squares instead of circles?
Olympic Starters
Look at some of the results from the Olympic Games in the past. How do you compare if you try some similar activities?
Olympic Triathlon
Olympic Turns
This task looks at the different turns involved in different Olympic sports as a way of exploring the mathematics of turns and angles.
Opening patterns
Performing beyond expectations - using sport to motivate students in mathematics lessons
Playing Squash
Secondary Sport Collection
Speed-time problems at the Olympics
Sports Equipment
Squash
Stadium Sightline
The animals' sports day
One day five small animals in my garden were going to have a sports day. They decided to have a swimming race, a running race, a high jump and a long jump.
The Fastest Cyclist
The Games' Medals
The Olympic Torch Tour
Tournament Scheduling
Track design
Training schedule
Twenty20
What's the point of squash?
Who can be the winner?
Who's the winner?
Twenty20
Several years ago I devised a set of rules to allow you to model a football match and even a World Cup competition. The rules are so simple you can play a full World Cup in one lesson, and they're accessible enough to appeal to primary as well as secondary schools. If football, why not cricket?
These two simple rules allow an innings to be played through in just a few minutes. However, it gives a pretty anaemic game, as neither the number of runs nor the number of wickets is sufficiently large to make things at all interesting. Worse still, it's purely mechanical. I want any simulation to require pupils to make decisions (so does Attainment Target 1, for that matter). The batting side should be able to decide when to play conservatively, scoring fairly slowly but with a relatively low risk of losing wickets. At other times a more aggressive mode may be needed, and the batsmen will take more chances in an attempt to score extra runs.
This is easily incorporated. We'll call the original style of play Style 1, but we'll introduce the possibility of Style 2: at any point the batting player may decide to accelerate, so s/he rolls the die not once but twice and adds their scores to get the total score for the over. Of course, more aggressive batting means more likelihood of losing a wicket, so the bowling player also rolls twice for that over. If the bowler rolls a 6 on either throw a wicket is lost (and two 6s mean that two wickets fall).
And since Twenty20 cricket often throws caution to the wind, there's Style 3 as well, where the batting player chooses to throw the die three times to get the total number of runs in the over. Likewise, the bowler throws three times, and each 6 means the fall of another wicket.
Now if you try this, you'll find that it's much better. However, it's still not 100% satisfactory. It's too easy for the batting side to use the super-attacking Style 3 throughout. They're unlikely to lose all their wickets and may well post a score of over 200 (in a game of Twenty20 cricket a total of below 100 is very poor, 150 might be reasonable, and the occasional 200+ marks an excellent total).
But a good simulation allows you to build in additional features to model the real situation, and in any cricket team the strongest batsmen are highest in the batting order and the later batsmen are weaker - less likely to score runs and more likely to lose their wickets. So to reflect the weakness of the lower batsmen we'll incorporate one more rule. Once the batting side has lost five wickets, further wickets fall whenever the bowling team rolls either a 5 or a 6.
Particularly if you're batting second you're going to have to evaluate the position each over, and I think the game is worth playing in its own right.
I imagine these simple rules will satisfy most people, but there's plenty of scope for taking things further. It is possible for a team to score no runs in an over - maybe you could reassign one of the numbers on the batting teams' dice to correspond to this, or say, for example, that if the bowler rolls a 1 then the batting team scores zero for that over, regardless of what they have rolled. You might like to use a slightly less blunt instrument for distinguishing between batsmen - perhaps wickets should fall when the bowler rolls a 6 for the top four wickets, then on a 5 or 6 for the next three, and on a 4, 5 or 6 for the final three. And wickets fall most easily in the last couple of overs, so you could introduce a modification to take care of that. We have batting Styles 1 to 3; is it worth trying a Style 4? And should the play be directed entirely by the decisions of the batting side, or should the bowling team have an input into the tactics?
For further information, a convenient first port of call is the Wikipedia Twenty20 entry
Performing beyond expectations - using sport to motivate students in mathematics lessons
I once made a school visit in just about the heaviest rain I've ever experienced. It came down with such force that it made you wet twice, once on the way down and again as it bounced up to waist height. It continued just as hard throughout the first half of the morning, and as morning break neared a deputation of worried-looking Y6 girls approached the teacher. "Please, miss, you're not going to
let a few drops of rain stop our rugby practice, are you?"
As the Olympics get closer and closer it seems to me that one of the healthiest developments in recent years has been the increased opportunities for girls to participate in sport. I've recently been doing One-To-One tuition in the middle school I taught at 25 years ago. Of course the new buildings and the computer rooms are a fascinating contrast to the days when our computer was a Sinclair
ZX81, but in 1985 a girls' football team was just as much pie-in-the-sky as a computer suite. The 2011 version is that we had to modify the timetable because one of my pupils wasn't prepared to miss her football practice.
Going back even further in my own experience, I spent the first couple of years in secondary school as a boarder, and we'd spend hours simulating our own Test Matches. All you needed was a pencil and an exercise book, and if they'd ever thought to look, my teachers would have been amazed by that exercise book. To call my exercise books in mathematics untidy would be to flatter them generously -
crossings out, ink blots, wrong answers and very few correct ones.
What a contrast there was in the appearance of my Test Match book - each game would take several days to play and there'd be perhaps 3000 separate pieces of information all impeccably and accurately recorded. On reflection there's no great insight needed to explain why I might have been at the bottom of the class in mathematics lessons and yet able to call upon all the mathematics I needed
to explore sporting situations - I spent two weeks in the sick bay and was perfectly happy that my only company was an old copy of Wisden Cricketers' Almanack.
A good stimulus can do much more than simply encourage children to work with a bit more enthusiasm. It can unlock levels of performance way beyond all expectations. I joined a Y6 teacher using a computer package about Grand Prix racing called "Cars - Maths In Motion". Now to my mind Formula One has not one single attribute that qualifies it as a sport, but I don't mind admitting that motor
racing does nevertheless stimulate some fine games. The package wasn't about cars whizzing around; indeed, there was no graphical component at all. Instead, children had to prepare a car for a race, making the best compromise between speed, power, brakes, etc, and to do this they had to do a vast amount of work involving estimating, decimals, and percentages.
The work lasted for several lessons, and the teacher told me that at one stage a visitor joined the class. She was unconvinced by the teacher's claims that a computer game about motor-racing was contributing to the pupils' mathematics and their understanding of percentages. The visitor buttonholed a child at random, choosing someone who was by no means a star in the class.
"What's 92% of 228?"
The child looked puzzled, and the teacher's heart sank.
"Go on Claire", she said, "I know you can do it. What's the answer?"
"About 209, I suppose", muttered Claire nervously.
When prompted, Claire explained her reasoning (all of which she had of course done in her head):
"10% of 228 is about 23, so if I do 228-23 I'll have 90%, but to get 92% I'll have to add back on another 2%. 1% is about 2, so the answer is about 228-23+2+2 = 209".
The visitor sat down to do some calculations on a piece of paper and several minutes later came up with a similar answer (the exact value is 209.76). When the visitor - somewhat more satisfied - had left, the teacher asked Claire why she hadn't given the answer immediately.
"It seemed too easy", she said, "I thought there must be a catch"!
This was, incidentally, another occasion when girls surpassed my rather bigoted expectations. Not only did two girls win the race, they overtook a further pair of girls whose car ran out of fuel on the final lap, and another girl left her sickbed not to miss the race. There are few computer packages which have been unsurpassed for more than twenty years, and you can find out more about this one
at Maths In Motion Challenge for Schools.
A little while ago I visited Arsenal Football Club's Double Club initiative. Arsenal-sponsored teachers are linked with local schools and teach lessons in basic subjects. The lessons have football-linked themes and the Arsenal teachers offer additional football lessons. I'm as keen on football as anyone, but even I could recognise there might be some downsides, so I visited with at least a
small amount of scepticism in my mind. I certainly recognised benefits - the pupils, many of whom had histories of little academic success, were enthusiastic and had a real commitment to the work. The DCSF evaluation (ref DCSF-RR101, ISBN 978 1 84775 422 6) was carried out by NFER, and reports:
Whether or not you want to build your teaching around football there seems a clear message here. Pupils who might not normally be motivated towards basic skills work were prepared to work hard because the work was presented in a way that they could relate to.
Yes, I do think the Cars - Maths In Motion program is a terrific stimulus, but I'm equally sure that perfectly ordinary ideas also can have significant effects. Back at my middle school of a generation ago the sports staff were so committed to athletics that I was convinced that they taught the children far more about decimals and measurement than the entire mathematics team. Equally, one lesson I drew from my Double Club visit was that the material itself didn't have to be particularly creative or exciting - simply presenting a standard task adapted to a football context ensured that the routine task was perceived as having a relevance and a meaning that's so often been lacking in maths lessons. Mathematics teaching is notorious for being totally divorced from anything real, but because sport is of interest to so many it may offer a motivating context for the most prosaic of activities.
Olympic Logic
Problem
These four problems require some logical thinking and a willingness to work systematically.
Can you deduce the missing information in each sporty situation?
Perhaps you might develop techniques in one that will help you to solve another.
1. Medals Count
Given the following clues, can you work out the number of gold, silver and bronze medals that France, Italy and Japan got in this international sports competition?
- Japan has 1 more gold medal, but 3 fewer silver medals, than Italy.
- France has the most bronze medals (18), but fewest gold medals (7).
- Each country has at least 6 medals of each type.
- Italy has 27 medals in total.
- Italy has 2 more bronze medals than gold medals.
- The three countries have 38 bronze medals in total.
- France has twice as many silver medals as Italy has gold medals.
2. Football Champ
Three points are given for a win and one point to each team for a draw.
The table below gives the total number of points and goals scored for and against each team.
Fill in the table and find the scores in each match.
Teams | Games Played | Won | Drawn | Lost | Goals For | Goals Against | Points |
---|---|---|---|---|---|---|---|
A | 2 | 5 | 3 | 3 | |||
B | 2 | 2 | 1 | ||||
C | 2 | 3 | 2 | 4 |
3. Fencing Tournament
Alice, Becky, Charlotte, Daphne, Elsie and Fran decide to compete in a fencing tournament. Each competitor has to fence against every other competitor. A match results in either a win or a loss.
- Image
- Daphne won her match against Becky.
- Alice and Elsie won the same, odd, number of matches, but Alice lost to Elsie.
- Becky and Fran won a total of seven matches
- Charlotte won only one match, against the only other person who also won only one match.
Can you deduce what all of the results were?
4. Hockey
In a hockey competition, four teams were to play each other once. 2 points were awarded for a win, and 1 point for a draw.
After some of the matches were played, most of the information in the results table was accidentally deleted.
Team | Played | Won | Drawn | Lost | For | Against | Points |
---|---|---|---|---|---|---|---|
A | 4 | 4 | |||||
B | 5 | 5 | |||||
C | 0 | 4 | 2 | ||||
D | 0 | 3 | 0 |
Can you work out the score in each match played?
Getting Started
Work out what each part of the problem tells you, and write it down.
What can you work out from what you know? Write down any new information that you deduce.
Can you think of any clear ways of representing the information?
Student Solutions
Lots of people got the answer to Medals Count right. Rachel, at Milton Keynes Academy, gave us a great explanation of her method:
There are 38 bronze medals in total for the 3 countries. France has 18 of them, so there are 20 bronze medals between Italy and Japan. These are the possible combinations, needing to be over 6:
Italy: either 6, 7, 8, 9, 10, 11, 12 , 13 or 14 bronze
Japan: 14, 13, 12, 11, 10, 9, 8, 7 or 6 bronze
Italy must have at least 8 gold, because they have more than France (who have 7). So, they need to have at least 10 bronzes as they have 2 more bronzes than golds. This means that Italy has at most 9 silvers.
Japan has at most 10 bronzes because there are 20 bronzes between Japan and Italy. They have one more gold than Italy, giving Japan at least 9 golds. Japan also have 3 less silvers than Italy, giving them at most 6 silvers. France has twice as many silvers as Italy has golds (8x2 = 16), so France has at least 16 silvers.
Now let's explore the possibilities for Italy's bronze medals. Italy can have 10, 11, 12, 13 or 14 bronzes. But if Italy has 11 or more bronzes, then it has 9 or more golds, and so it has at most 7 silvers. This means that Japan has at most 4 silvers, which isn't allowed. So Italy must have 10 bronzes exactly.
In conclusion:
Italy: 10 bronzes 9 silvers 8 golds
Japan: 10 bronzes 6 silvers 9 golds
France: 18 bronzes 16 silvers 7 golds
Excellent! Luke, from Cottenham Village College, did things more algebraically:
First, give the number of medals names:
gold | silver | bronze | |
France | a | b | c |
Japan | d | e | f |
Italy | g | h | i |
Now apply the rules given:
From rule 1: g=d-1, h=e+3
From rule 2: c=18, a=7
From rule 5: i=d+1
From rule 7: b=2d-2
The table now looks like this:
gold | silver | bronze | |
France | 7 | 2d-2 | 18 |
Japan | d | e | f |
Italy | d-1 | e+3 | d+1 |
Apply rule 6: f+d+1+18=38, and the table looks like:
gold | silver | bronze | |
France | 7 | 2d-2 | 18 |
Japan | d | e | 19-d |
Italy | d-1 | e+3 | d+1 |
As France has the fewest golds, d-1 must be at least 8 so d is at least 9.
As 2d+e+3 = 27 (rule 4), and e must be at least 6 (Japan must have at least 6 silver medals: rule 3), we see that d = 9.
This makes e = 6.
The table finally looks like this:
gold | silver | bronze | |
France | 7 | 16 | 18 |
Japan | 9 | 6 | 10 |
Italy | 8 | 9 | 10 |
Great - thanks! How about Football Champ? The Year 9 class at Ounsdale High School showed us their method:
- Josh said that to work out the wins, draws and losses we looked at the points and the only possibilities are given below.
- B and C must have drawn because they each had 1 draw.
- A must have beaten B, and C must have beaten A.
- Jordan said: the maximum that team B could have scored in their draw was 2, so it was either 2 - 2 or 1- 1.
- As C only let in 2 goals in total, and C must have scored at least one goal against A, the same between B and C must have ended up 1-1.
- Total number of goals for and against must be equal, because if A scored against B it would count for A and against B, for example.
And their results were as follows:
Teams | Games Played | Won | Drawn | Lost | Goals For | Goals Against | Points |
---|---|---|---|---|---|---|---|
A | 2 | 1 | 0 | 1 | 5 | 3 | 3 |
B | 2 | 0 | 1 | 1 | 2 | 5 | 1 |
C | 2 | 1 | 1 | 0 | 3 | 2 | 4 |
Nicely done!
For the Fencing Tournament problem, Szymon from Earl Mortimer College recommends making a table. He explains his method as follows:
Tip 2 says 'Daphne won her match against Becky'. Mark this in the table as follows:
A | B | C | D | E | F | |
A | ||||||
B | Image
|
|||||
C | ||||||
D | Image
|
|||||
E | ||||||
F |
- Tip 3: 'Alice and Elsie won the same, odd, number of matches, but Alice lost to Elsie'. You know that Elsie won against Alice, so you can plot these straight away. Now, they won either 1, 3 or 5 games each, but it can't be five, because only one person could have won all their matches.
- Tip 5 says Charlotte won only one match, against the other person who won only one as well. We don't know who this is, but there is only one of them, so it can't be Alice and Elsie, so they must have won 3 of their matches...
A | B | C | D | E | F | |
A | Image
|
Image
|
Image
|
Image
|
Image
|
|
B | Image
|
|||||
C | ||||||
D | ||||||
E | ||||||
F |
Thanks! Finally, the Hockey problem - this was tough! A bit of careful reading tells you that the results given are only a partial list of results - some games may not have been played yet. Samantha, from Laxey Primary School, gives her answer as follows:
Team | Played | Won | Drawn | Lost | For | Against | Points |
A | 3 | 1 | 1 | 1 | 4 | 4 | 3 |
B | 3 | 2 | 1 | 0 | 5 | 2 | 5 |
C | 2 | 0 | 2 | 0 | 4 | 4 | 2 |
D | 2 | 0 | 0 | 2 | 0 | 3 | 0 |
and the individual results:
A v B: B wins 2-0
A v C: draw 2-2
A v D: A wins 2-0
B v C: draw 2-2
B v D: B wins 1-0
C v D: not played yet.
Thanks to everyone who sent in solutions this month!
Teachers' Resources
Why do this problem?
These four problems require some logical thinking and a willingness to work systematically. Routes to a solution are not immediately obvious so working on the problems could help students to develop their resilience, an important quality for mathematical problem-solvers.
Possible approach
Before introducing the task:
"Mathematicians need to be resourceful problem-solvers who don't give up, who talk to each other and share good ideas, who work strategically and systematically. The answers won't be immediately obvious, so these problems will test your ability to work as mathematicians."
The problems could be used in several ways:
Hand out this worksheet and invite students to work on all four problems in pairs.
Cut the worksheet into the four separate problems and invite different groups to work on different problems.
Work on one or two of the questions together as a class and then invite students to use each other's useful insights to solve the remaining problems.
While students are working, circulate and listen for useful insights. Where appropriate, bring the class together so that students can share successful strategies with the rest of the class.
The following key questions or prompts could be offered to students who are stuck:
What do you know? Write it down.
What can you deduce from what you know? Write it down.
Is there a good way of representing what you know, to make it clear?
Possible extension
For other problems that require systematic and logical thinking see Two and Two, Product Sudoku and Cinema Problem.
Possible support
Of the four problems, Football Champ and Hockey use very similar techniques, so one could be solved as a class before students are given the second to apply any useful methods they have devised.
Olympic Measures
Problem
Olympic Measures printable worksheet
Olympic Measures additional support printable worksheet
Below are some interesting measurements and records from events at the Olympic Games. Unfortunately they have been muddled up.
Can you regroup them correctly? You can print off and cut out this set of cards.
Getting Started
Which quantities are times? Which are speeds? Which are distances?
There are some standard quantities, and some world records or times. Think about whether each of these is likely to be a whole number or a decimal.
Once you've identified all the distances, try ranking them in order of magnitude.
Here is an interactive version with fewer quantities for you to rearrange. Once you've reorganised them you'll be able to check which ones you have got correct.
Student Solutions
We tried to work out the problem. We knew we needed to make sensible estimates. To work out Usain Bolt's record, for example, we ran the distance ourselves and found out what times we scored. We thought that he'd be able to do it a little bit faster than us, so that helped make our guesses. Two groups guessed really well with 10 seconds. So we were using what we already know to help us solve a problem.
Fantastic! This is a great way to estimate the kinds of numbers you're looking for and use the data you have sensibly.
The final table of answers, which loads of people got right, is given below. Thanks to everyone for the submissions!
Usain Bolt's 100 metres World Record |
9.58 | seconds |
Women's Long Jump World Record |
7.52 | metres |
Men's High Jump World Record | 2.45 | metres |
Basketball hoop diameter | 18 | inches |
Basketball hoop height | 10 | feet |
Mass of Women's Discus | 1 | kg |
Women's Discus World Record | 76.8 | metres |
Diameter of Archery target | 122 | cm |
Archers' distance from the target | 70 | metres |
Average speed of World Record |
8.77 | mph |
Men's Shot put World Record | 23.12 | metres |
Height of Diving platform | 10 | metres |
Men's 10km Swimming 2012 Olympics winner's time |
01:49:55 | hr:min:sec |
Men's 50m Freestyle Swimming 2012 Olympics winner's time |
21.34 | seconds |
Paula Radcliffe's Women's Marathon World Record |
02:15:25 | hr:min:sec |
Triathlon Swim (distance) | 1500 | metres |
Triathlon Bicycle ride (distance) | 40 | km |
Triathlon Run (distance) | 10 | km |
Men's Triathlon 2012 Olympics winner's time |
01:46:25 | hr:min:sec |
Men's Pole Vault World Record | 6.16 | metres |
Teachers' Resources
Why do this problem?
This problem invites students to engage with units of measurement and orders of magnitude, by presenting a variety of records and measurements from events at the Olympic Games. Some will be familiar to students, others may lend themselves to estimation or a little research.
Possible approach
These printable resources may be useful: Olympic Measures,
Olympic Measures Support.
"Here are some interesting measurements and records from events at the Olympic Games. Unfortunately they have been muddled up. Can you regroup them correctly?"
Hand out this set of cards and invite students to work in pairs together. The first page of cards is self-contained, so if you wish to make the task easier, you can just use the cards from the first page.
If students are stuck, here are some key questions to help them:
Which quantities are likely to be whole numbers? Why?
Which quantities are lengths? Which are times? Which are speeds? Which are masses?
Which units might belong with the lengths... times... speeds... masses...?
Can you rank the different lengths... times... speeds... masses in order of magnitude?
Finish by bringing the whole class together to agree on a class ordering for the cards. Students will need to convince each other of their own ordering by explaining what they are certain of, and justifying their educated guesses.
The final ordering could be checked with the interactivity, which will give feedback when all the cells are occupied.
The videos of impressive world record performances available on this site might be of interest to students. (Link is to an external site, so NRICH is not responsible for the contents.)
One teacher created a PowerPoint to use with an earlier version of this task (some of the world records have since changed); you can download it from the TES Resources website.
Possible support
This set of cards only requires students to sort the measurements, as the units are included with each event.
Possible extension
Invite students to do some research to create a set of similar cards of their own to swap with a friend.
Students could also have a go at All in a Jumble where the measurements have again been all jumbled up.
Tournament Scheduling
Scheduling games is a little more challenging than one might desire.
There are 2 well-known types of tournament formats that sport schedulers use.
- Single Elimination Tournaments.
- Round Robin Tournaments.
Single Elimination Tournaments.
In this format, one defeat is enough to eliminate a team from the tournament.
Scheduling a single elimination tournament is relatively easy. The first step is to get the number of teams to a power of two: 2, 4, 8, 16, 32...
The schedule for 8 teams is shown below.
With 20 teams, you might select eight of those teams to have a preliminary round. The four winners would then join the other twelve to fill out the sixteen-team field.
If you have 24 teams, you could have everyone compete in a preliminary round; this would leave 12 teams. Then, 4 of those teams (randomly selected) could get a bye and the other 8 teams would play to find the other 4 contestants for the next round.
Round Robin Tournaments.
In a Round Robin tournament every team plays every other team .
There is a systematic approach to scheduling a Round Robin tournament. This method assumes that there are enough fields / pitches / courts so that all the games in a round can be played simultaneously. The technique is called the polygon method .
Round Robin scheduling: Even number of teams.
Let N = number of teams in the tournament. There will be N -1 rounds (each team will play N-1 games). Since each team will play every other team once, no team will be idle during any of the rounds.
Let us schedule a round-robin tournament for 8 teams numbered from 1 to 8.
Draw a regular (N -1) sided polygon (i.e., a heptagon for 8 teams). Each vertex and the centre point represents one team.
Draw horizontal stripes as shown below. Then, join the vertex that has been left out to the centre. Each segment represents teams playing each other in the first round.
So (7, 6), (1, 5), (2, 4) and (3, 8) play in the first round.
Rotate the polygon 1/(N-1)th of a circle (i.e. one vertex point). The new segments represent the pairings for round two.
So (6, 5), (7, 4), (1, 3) and (2, 8) play the second round.
Continue rotating the polygon until it returns to its original position.
One more rotation will bring the polygon back to its original position.
If A, B, C and D are the fields / pitches / courts, the schedule could look like this:
Round | A | B | C | D |
---|---|---|---|---|
I | 7, 6 | 1, 5 | 2, 4 | 3, 8 |
II | 6, 5 | 7, 4 | 1, 3 | 2, 8 |
III | 5, 4 | 6, 3 | 7, 2 | 1, 8 |
IV | 4, 3 | 5, 2 | 6, 1 | 7, 8 |
V | 3, 2 | 4, 1 | 5, 7 | 6, 8 |
VI | 2, 1 | 3, 7 | 4, 6 | 5, 8 |
VII | 1, 7 | 2, 6 | 3, 5 | 4, 8 |
We can also rotate the teams around so that each team plays in every field / pitch / court at least once (at present team 8 always plays in D).
Round | A | B | C | D |
---|---|---|---|---|
I | 7, 6 | 1, 5 | 2, 4 | 3, 8 |
II | 6, 5 | 7, 4 | 1, 3 | 2, 8 |
III | 1, 8 | 6, 3 | 7, 2 | 5, 4 |
IV | 4, 3 | 5, 2 | 7, 8 | 6, 1 |
V | 3, 2 | 4, 1 | 5, 7 | 6, 8 |
VI | 2, 1 | 5, 8 | 4, 6 | 3, 7 |
VII | 1, 7 | 2, 6 | 3, 5 | 4, 8 |
Round Robin scheduling: Odd number of teams.
Let N = number of teams in the tournament. There will be N rounds (since each team will play every other team once, and will be idle for exactly one round ).
Let us work out the schedule for 7 teams, numbering the teams from 1 to 7. Draw a regular N-gon (heptagon for 7 teams). Each vertex represents one team.
Draw horizontal stripes as shown below. The vertex that has been left out gives the idle team. Each segment represents teams playing each other in the first round.
So (7, 6), (1, 5) and (2, 4) play in the first round.
Rotate the polygon 1/Nth of a circle (i.e. one vertex point.) The new segments represent the pairings for round two.
Continue rotating the polygon until it returns to its original position.
One more rotation will bring the polygon back to its original position. Therefore, the schedule could look like this:
Round | A | B | C |
---|---|---|---|
I | 7, 6 | 1, 5 | 2, 4 |
II | 6, 5 | 7, 4 | 1, 3 |
III | 5, 4 | 6, 3 | 7, 2 |
IV | 4, 3 | 5, 2 | 6, 1 |
V | 3, 2 | 4, 1 | 5, 7 |
VI | 2, 1 | 3, 7 | 4, 6 |
VII | 1, 7 | 2, 6 | 3, 5 |
Why does this work?
The restriction that no vertex has more than one segment drawn to/from it ensures that no team is scheduled for more than one game in each round.
Restricting ourselves to horizontal stripes ensures that no segment is a rotation or reflection of another segment. This means that no pairing will be repeated in a future round.
Notice that in the case where N (no. of teams) was odd, by having only one idle team in each round, the tournament can be completed in the minimum number of rounds.
Arunachalam Y. is a member of the HeyMath! team
Who's the best?
Problem
Who's the Best? printable sheet
Who's the Best? medal table CSV file
Which countries are most "sporty"?
Below, we have provided the London 2012 Olympics medal table.
Do you think the medals table alone offers enough information to answer this question fairly?
What other information do you think could be taken into account?
Make a plan of the information you will collect and the calculations you will perform. Then carry out your plan to decide which countries are most sporty.
Send us your top-rated countries, together with your justification for the data and criteria you chose to include in your analysis.
In the London 2012 Olympics the final medal table was as follows (spreadsheet CSV form)
Country | Gold | Silver | Bronze | Total |
---|---|---|---|---|
United States | 46 | 28 | 29 | 103 |
China | 38 | 28 | 22 | 88 |
Great Britain | 29 | 17 | 19 | 65 |
Russia | 24 | 25 | 32 | 81 |
South Korea | 13 | 8 | 7 | 28 |
Germany | 11 | 19 | 14 | 44 |
France | 11 | 11 | 12 | 34 |
Italy | 8 | 9 | 11 | 28 |
Hungary | 8 | 4 | 6 | 18 |
Australia | 7 | 16 | 12 | 35 |
Japan | 7 | 14 | 17 | 38 |
Kazakhstan | 7 | 1 | 5 | 13 |
Netherlands | 6 | 6 | 8 | 20 |
Ukraine | 6 | 5 | 9 | 20 |
New Zealand | 6 | 2 | 5 | 13 |
Cuba | 5 | 3 | 7 | 15 |
Iran | 4 | 5 | 3 | 12 |
Jamaica | 4 | 4 | 4 | 12 |
Czech Republic | 4 | 3 | 3 | 10 |
North Korea | 4 | 0 | 2 | 6 |
Spain | 3 | 10 | 4 | 17 |
Brazil | 3 | 5 | 9 | 17 |
South Africa | 3 | 2 | 1 | 6 |
Ethiopia | 3 | 1 | 3 | 7 |
Croatia | 3 | 1 | 2 | 6 |
Belarus | 2 | 5 | 5 | 12 |
Romania | 2 | 5 | 2 | 9 |
Kenya | 2 | 4 | 5 | 11 |
Denmark | 2 | 4 | 3 | 9 |
Azerbaijan | 2 | 2 | 6 | 10 |
Poland | 2 | 2 | 6 | 10 |
Turkey | 2 | 2 | 1 | 5 |
Switzerland | 2 | 2 | 0 | 4 |
Lithuania | 2 | 1 | 2 | 5 |
Norway | 2 | 1 | 1 | 4 |
Canada | 1 | 5 | 12 | 18 |
Sweden | 1 | 4 | 3 | 8 |
Colombia | 1 | 3 | 4 | 8 |
Georgia | 1 | 3 | 3 | 7 |
Mexico | 1 | 3 | 3 | 7 |
Ireland | 1 | 1 | 3 | 5 |
Argentina | 1 | 1 | 2 | 4 |
Serbia | 1 | 1 | 2 | 4 |
Slovenia | 1 | 1 | 2 | 4 |
Tunisia | 1 | 1 | 1 | 3 |
Dominican Republic | 1 | 1 | 0 | 2 |
Trinidad and Tobago | 1 | 0 | 3 | 4 |
Uzbekistan | 1 | 0 | 2 | 3 |
Latvia | 1 | 0 | 1 | 2 |
Algeria | 1 | 0 | 0 | 1 |
Bahamas | 1 | 0 | 0 | 1 |
Grenada | 1 | 0 | 0 | 1 |
Uganda | 1 | 0 | 0 | 1 |
Venezuela | 1 | 0 | 0 | 1 |
India | 0 | 2 | 4 | 6 |
Mongolia | 0 | 2 | 3 | 5 |
Thailand | 0 | 2 | 1 | 3 |
Egypt | 0 | 2 | 0 | 2 |
Slovakia | 0 | 1 | 3 | 4 |
Armenia | 0 | 1 | 2 | 3 |
Belgium | 0 | 1 | 2 | 3 |
Finland | 0 | 1 | 2 | 3 |
Bulgaria | 0 | 1 | 1 | 2 |
Estonia | 0 | 1 | 1 | 2 |
Indonesia | 0 | 1 | 1 | 2 |
Malaysia | 0 | 1 | 1 | 2 |
Puerto Rico | 0 | 1 | 1 | 2 |
Chinese Taipei | 0 | 1 | 1 | 2 |
Botswana | 0 | 1 | 0 | 1 |
Cyprus | 0 | 1 | 0 | 1 |
Gabon | 0 | 1 | 0 | 1 |
Guatemala | 0 | 1 | 0 | 1 |
Montenegro | 0 | 1 | 0 | 1 |
Portugal | 0 | 1 | 0 | 1 |
Greece | 0 | 0 | 2 | 2 |
Moldova | 0 | 0 | 2 | 2 |
Qatar | 0 | 0 | 2 | 2 |
Singapore | 0 | 0 | 2 | 2 |
Afghanistan | 0 | 0 | 1 | 1 |
Bahrain | 0 | 0 | 1 | 1 |
Hong Kong | 0 | 0 | 1 | 1 |
Saudi Arabia | 0 | 0 | 1 | 1 |
Kuwait | 0 | 0 | 1 | 1 |
Morocco | 0 | 0 | 1 | 1 |
Tajikistan | 0 | 0 | 1 | 1 |
Notes and Background
The Plus article Harder, better, faster, stronger explores some of the same ideas as this problem.
Getting Started
How might the population size of a nation affect its chances of winning a medal?
What about the wealth of a nation?
Student Solutions
In this problem, we needed to find a way to use the Olympic medal tally in order to work out which country is the most sporty.
Gianluca from Salesian School in Chertsey sent us his excellent analysis of the Olympic tally. He made use of a scoring system for each type of medal and also took into account the population of each country. You can find his analysis here, thank you Gianluca!
Teachers' Resources
Why do this problem?
This problem offers an engaging context for planning and carrying out a data-handling activity. Many problems that students meet are "tidy"; this activity prepares students for the messy nature of real-world data-handling. Students will need to make decisions about the data they need, and to search for that data.
Possible approach
Show the medal table from the 2012 Olympics.
When groups present their findings, encourage the other groups to offer constructive feedback on the decisions they chose to make, the evidence they chose to include in their analysis, and the parts of the presentation they found most convincing. One way to do this is for each group to make a short presentation with time for comments and questions afterwards. Alternatively, groups could prepare a
poster presentation and then time could be allowed for everyone to go round and look at each poster, writing feedback on post-it notes.
Key questions
Does the Olympic table rank countries fairly?
Possible support
Some simple and quantifiable criteria that students could use in their analysis could be population size and wealth of the nation. It may be worth finding out some of this data in advance in order to guide the students' searches.
Possible extension
Students could use data from previous Olympics to see the extent to which their criteria give a constant list of most sporty nations over time.
Medal Muddle
Problem
If you are a teacher, click here for a version of the problem suitable for classroom use, together with supporting materials. Otherwise, read on...
Thirteen nations competed in a sports tournament. Unfortunately, we do not have the final medal table, but we have the following pieces of information:
1. Turkey and Mexico both finished above Italy and New Zealand.
2. Portugal finished above Venezuela, Mexico, Spain and Romania.
3. Romania finished below Algeria, Greece, Spain and Serbia.
4. Serbia finished above Turkey and Portugal, both of whom finished below Algeria and Russia.
5. Russia finished above France and Algeria.
6. Algeria finished below France but above Serbia and Spain.
7. Italy finished below Greece and Venezuela, but above New Zealand.
8. Venezuela finished above New Zealand but below Greece.
9. Greece finished below Turkey, who came below France.
10. Portugal finished below Greece and France.
11. France finished above Serbia, who came above Mexico.
12. Venezuela finished below Mexico, and New Zealand came above Spain.
Can you recreate the medal table from this information?
Can you describe an efficient strategy for solving problems like this?
Extension
The following year, twice as many teams entered the tournament. Can you use your strategy to sort out the medal table from these clues?
Perhaps you might like to try creating a similar problem of your own.
You will need to consider the following:
Although there are twelve statements above, there are more than twelve pieces of information, because some sentences compare more than one pair of teams.
What is the minimum number of pieces of information needed to order the teams?
Which information, if any, is redundant?
Getting Started
You may find it useful to print off and cut out these cards. You could arrange the countries randomly and then read through the clues adjusting the order as you go. Here is a set of cards for the extension.
Alternatively, you could begin by figuring out which teams couldn't have come first.
If the GeoGebra applet does not load correctly you can save the GeoGebra file and open it using the free to download GeoGebra software.
Student Solutions
Thank you to everyone who participated! The correct answers, which almost everyone got, were:
1. Russia
2. France
3. Algeria
4. Serbia
5. Turkey
6. Greece
7. Portugal
8. Mexico
9. Venezuela
10. Italy
11. New Zealand
12. Spain
13. Romania
Congratulations if you got the order right! Let's have a look at some of the ways of doing it.
Alex, from Winnersh Primary School, had the following interesting idea to find the countries in order, one at a time:
My method was to choose a random country and then go through the clues until I found a country that was higher up. I carried on until I found a country where I could go all the way through the clues without finding another one that was higher up. I then put that country (Russia) in 1st place. I would then do the same but ignoring Russia, and found the 2nd, then 3rd, etc.
Rebecca, from Woodchurch, had a similar idea:
Try to count how many times one country came above each other country. Then repeat this thirteen times. Then put your answers in order. Ta-da!
Daniel, from Wilson's School, wrote down at each step what he got from the hints:
From hint 1, you can get:
- Turkey / Mexico
- Italy / New Zealand
From hint 2, you can get:
- Portugal
- Venezuela / Mexico / Spain / Romania
From hint 3, you can get:
- Greece / Spain / Serbia / Algeria
- Romania.
etc.
Many people thought it was a good idea to write the names of countries on bits of paper or card and swap them round - this saves a lot of writing! For example, Michelle, from Globe Academy, wrote:
I started by putting the country names in a random order. Then I read through the clues and started swapping around the countries. When I got to the end of the clues I went back through the clues and checked again.
Mrs. McGuire's class at Lakewood Catholic Academy were another one of many who followed this approach - they say it took them about 45 minutes and lots of trial and error. Could it have been speeded up, do you think?
Jade, at Oakmeeds, sent us the following comments on the card idea:
I wrote some of the infomation to do with the country on the country's card, e.g. "above Spain and Algeria and below New Zealand".
Pros:
- Easy to read and clear
- Enjoyable when arranging the cards
- Pretty quick if you have an idea in your head
- Makes you happy when you complete it!
Cons:
- The writing process is slightly tedious
Tips:
- When writing the notes on the cards write short phrases and clearly so easily read.
- If you write the wrong infomation on the cards then it's not going to be pretty...
Charlie, from Wentworth Primary, had this interesting idea:
I used a mathematical method allocating points for each one above and subtracting a point for below, to eventually work out where each country should go by adding up the points I had allocated.
Mrs. Gale's class, from Churchill Academy, had a trick to speed things up slightly:
Colour coding the countries to make them stand out more easily. This made it clear there was most information about France.
On a similar note, Alastair from Richmond CoE sent us lots of flags that he printed out and cut up while constructing his solution. Nice!
A few people moved onto the extension problem using the same sorts of techniques as above. The correct answer was:
Sri Lanka, Great Britain, Brazil, Spain, Turkey, Austria, Romania, Finland, Mexico, Germany, Serbia, Italy, Canada, Algeria, New Zealand, Australia, Norway, France, Portugal, Greece, Japan, Sweden, Venezuela, USA, Russia, Denmark.
Thanks to Brain Academy at St. Peters CEVC Primary, Mrs. C's class at Court Moor School, and Ms. Troup's class at Prior's Field School for sending in their answers to the extension problem - this one was tough!
(Finally, Stefan from Afghanistan said: "this is so cool"! Thanks, Stefan!)
Teachers' Resources
Why do this problem?
This problem is an exercise in strategic thinking, accessible to lower Stage 3 students but hinting at work on sorting algorithms that they might meet at Stage 5 in Decision Maths.
Possible approach
"I'm going to give you a problem to solve, and while you work on it, I'd like you to think about the strategies you are using. Imagine you had to solve lots of problems like this one. How would you ensure that you found the correct answer accurately and efficiently?"
Hand out this worksheet for students to work on in pairs (or individually at first if they wish).
These cards could be printed and handed out to students so they can manipulate the order as they work their way through the different clues.
Once students have had a chance to discuss the merits of different approaches, hand out this worksheet with the extension challenge, so that they can test how their chosen strategy works on a longer problem with more information to consider. Here is a set of cards for the extension activity.
Key questions
Which representations or ways of organising your thinking help you to use the information given to solve the problem efficiently?
Possible extension
Challenge students to create their own versions of the problem, which could be shared on the blog.
Possible support
The visual representation shown in the hint is a very clear way of seeing the relationship between the different countries.
Olympic Records
Problem
Olympic Records printable sheet - graphs
Here are ten graphs showing how Olympic records have changed over time in ten athletic events.
Can you deduce which event each graph represents?
You can click on the thumbnails above to view each graph, or download this pdf or this powerpoint with all the graphs.
Here are some questions to consider, which may help you to make sense of the graphs:
- Can you determine what the units might be on the vertical axis for each graph?
- Why do some graphs show a decreasing trend and some an increasing trend?
- Are there any unusual features in any of the graphs? Can you think of a plausible explanation for them?
Thanks to Alan Parr for suggesting this problem.
Getting Started
This PDF contains a list of the ten events.
Can you match the events to the graphs?
Student Solutions
Patrick, Elliot, Oscar and Peter, Jaa, Lucas and Jerry, Prinn, Xander and Ryan, Pair, Korya, Sasha and Ysabel and Pat and Andrew from Rubgy School Thailand, Edward from Plumstead Manor in England, Seokwon (Andy) from Suzhou Singapore International School in China, Mahdi from Mahatma Gandhi International School in India and Angel from England all identified this event correctly.
Isaac, Mr Robinson and Miss Morgans from St. Joseph's RC Primary School, Newbury in the UK wrote:
First, we looked at all the graphs and deduced that the x-axis was the year of the event and the y-axis was the score/time achieved. We then thought about which events the score was increasing in and which ones decreased. We
thought about the types of events each graph could possibly represent and made some initial suggestions.
Graph 1: Speed - running
Jaa, Lucas and Jerry explained:
The measurement gets less and less which could maybe show that it is a running event.
Patrick, Elliot, Oscar and Peter wrote:
We believe that the units are in seconds.
We think it is the 100m men's event.
The world record is 9.58 and was set in 2009, but Usain Bolt didn’t do it in the Olympics, but he did it in the World Championships. The OLYMPIC record is 9.63 which was set in 2012, which corresponds to the graph so therefore we know that this is the men's 100m.
Patrick, Elliot, Oscar and Peter, Jaa, Lucas and Jerry, Prinn, Xander and Ryan, Pair, Korya, Sasha and Ysabel and Pat and Andrew, Edward, Seokwon (Andy), Mahdi and Angel all identified this event correctly.
Pat and Andrew wrote:
I think the unit is a meter because the record keeps going up.
Prinn, Xander and Ryan wrote:
At first we thought about the sports which are apt for the integers given on the y axis, firstly, I thought that it would be javelin, conversely it was brought to light after doing some research that the olympic record for the javelin is over 90 metres, then this alluded us to think about the different categories for jumps, resulting in doing a little bit of research ont he high and long jump.
Angel wrote:
It is because the longest jump was 8.9m and nobody has beaten that ever since which is why the graph finishes in 1968.
Mahdi thought about the y axis units:
Unit is time, which can suggest that it is either a marathon (if in hours) or a run of 1 or 2 km for 3.5 minutes. I reckon Men's or Women's 1500 metres.
Jaa, Lucas and Jerry made some suggestions based on the x axis:
A new sport due to there only being records from not long ago or it could be a women’s sport due to the Olympics not allowing women to participate a while ago.
Angel, Seokwon (Andy) and Edward identified the graph. Angel and Seokwon (Andy) used the same idea. Angel wrote:
It is [women's 500 metres] because women’s 1500 metres started in 1972 and that’s where the graph starts.
Seokwon (Andy) added:
Identified via the approximate times, and the few number of Olympic records.
Peter, Elliot, Patrick and Oscar, Jaa, Lucas and Jerry, Prinn, Alexander and Ryan, Pair, Korya, Sasha and Ysabel, Edward, Seokwon (Andy), Mahdi and Angel identified this event. Pair, Korya, Sasha and Ysabel wrote:
y = meter
High jump - men
The y value increased over time, so it could measure length. A short length such as 2 metres may have meant the event was a jumping event including height would be most reasonable because 2 metres is too short for specific jumps. Therefore, we think it is a high jump.
Lucas, Jaa and Jerry added:
High jump because the world record in 1994, it was 2.45 meters for men.
Prinn, Xander and Ryan, Pat and Andrew, Edward, Seokwon, Mahdi and Angel correctly identified this event. Prinn, Xander and Ryan wrote:
We also think that this is the men's javelin olympic record, as there was a major rule change to do with javelin just before the 1988 olympics, which you can see on the graph; the record dropped during the 1988 olympics, and slowly climbed back up. The current men's olympic record was set in 2008 and is 90.57 metres.
Seokwon (Andy) investigated the change:
Interesting data point when Olympic Record decreases. Redesign of javelin when almost reaching 100m as danger of javelin going closer to the seating area. Centre of mass altered, so that the javelin sticks into the ground, rather than hoizontally [travelling].
Mahdi, Angel, Seokwon (Andy), Edward, Pat and Andrew and Jaa, Lucas and Jerry correctly identified this event. Angel wrote:
Women's High Jump, because it was similar to the Men’s High Jump and it started later than the Men’s.
Mahdi wrote:
No clue. Confused about the fluctuations and drastic changes too. No trend as well, and hard to guess the units. They don't seem like time or distance, something else completely.
Pat and Andrew, Seokwon (Andy) and Angel worked out the event. Seokwon (Andy) wrote:
Mens Decathlon, on the y-axis points
Jaa, Lucas and Jerry, Pat and Andrew, Edward, Seokwon (Andy), Angel and Mahdi correctly guessed the event. Jaa, Lucas and Jerry wrote:
Women’s 10km run because women have recently been able to participate in the olympics and the scores get lower and lower. The record is also 29:17.45 (held by Almaz Ayana). This applies to the graph.
Unit: Minutes:seconds
Mahdi wrote:
Some sprint, because of the decreasing time, probably? A 40 second record could be for men's 400 metres?
Pat and Andrew, Edward, Seokwon (Andy) and Angel identified the event as Men's 4$\times$100 metres Relay.
Pat and Andrew, Edward, Seokwon (Andy), Angel and Mahdi correctly identified this event. Mahdi wrote:
Probably women's 100 metres. The declining time which is lowest at around 10 seconds can be a good reason for women's 100 metres. However, I wonder if the record isn't broken since 1990 or somewhere around that year.
Angel wrote:
Women’s 100 metres, because it is similar to the men’s and the record hasn’t been broken since 1988.
Teachers' Resources
Why do this problem?
This problem gives students the opportunity to make sense of graphical data and challenges them to apply their own knowledge about athletics to explain and interpret key features of the graphs.
Possible approach
Arrange the class in groups of three or four, and give each group the ten graphs.
While they are working, circulate and listen to their reasoning. If some groups are stuck, here are some useful prompts:
- What can you deduce about the event if the record increases over time?
- What about events where the record decreases over time?
After they have had sufficient time to consider the graphs, ask them to write down a list of what events they think each represents.
Possible support
You may wish to hand out this list of the events and ask students to work out which belongs with each of the ten graphs.
Alternatively, this card sort activity contains the ten graphs and the events on cards for students to match up.
Possible extension
Ask students to carry out some extra research in order to explain unusual features in the graphs, such as:
- Drastic changes
- Long gaps
- Absence of data in the early twentieth century
- Records that go both up and down
Students could create their own set of graphs and challenge the rest of the class to work out which Olympic athletics events they relate to.
Olympic Triathlon
Problem
Olympic Triathlon printable sheet - data
The Olympic triathlon consists of a 1.5km swim, a 40km cycle, and a 10km run, completed in sequence. The first person to finish wins.
Do you think the triathlon will be won by someone who is very strong in one event and average in the other two, or someone who is strong in all three disciplines?
Take a look at this spreadsheet showing the results from the 2008 Beijing Olympics Men's Triathlon.
What do you notice?
You may find it helpful to sort the results in various ways, work out averages and measures of spread, or plot some graphs to test correlations between times for individual events and overall finishing positions.
Can you come up with any explanations for what you have noticed?
This spreadsheet contains results from male and female Olympic Triathlons since the introduction of the event in 2000.
Do the results from the other years mirror what you noticed about the 2008 Men's Triathlon?
Are your explanations for the 2008 results plausible for the results from other years too?
Are there any events with unexpected results, or outliers?
Send us anything interesting that you notice, together with graphs or statistics that highlight what you have noticed and your suggested explanations.
Triathlon swimming, and in particular cycling, is affected by a phenomenon called drafting. Click below to read more about drafting. Does it help to explain the spread of times in the different events?
In cycling, the majority of the work done at racing speeds will be to battle air resistance. This means that drafting has a large effect: a rider in a peloton (a large cluster of cyclists) can use over 30% less energy to move at the same speed as a cyclist riding alone. Drafting can be both co-operative and competitive: a small group of cyclists can work together to maintain a high speed in a paceline, rotating the lead position (who must work hardest) between them; alternatively, a lone rider can try to sit on the wheel of a competitor, allowing them to do the harder work and conserving energy for later. To try and get ahead of a peloton is called a break or a breakaway, and is difficult for the lone rider. This makes teamwork a very important part of cycling.
Drafting also has an effect on swimming, because the main work is done against the drag from the water. Because of the much lower speeds involved, the slipstream of each swimmer is more spread out. This means that to draft in swimming, one can be adjacent and slightly back from the swimmer in front, instead of directly behind as in cycling. In swimming events without lanes, such as the triathlon, competitors frequently form groups and lines just like pelotons and pacelines.
If you have access to YouTube, you can try to observe the effects of drafting in the following videos from the 2008 Beijing Olympics: Men's Triathlon and Women's Triathlon.
Getting Started
Sort the data according to times for the swimming leg, cycling leg, and running leg.
How well do each of the individual event times correlate with the overall times?
Here are some ideas to consider when trying to explain what you notice:
How long do athletes spend on each leg of the Triathlon?
How might the order of the events in the Triathlon affect the final times?
What are the advantages of staying with other athletes over breaking away early?
How do the energy requirements differ for each leg of the Triathlon?
Student Solutions
Adam and Eva, at Ratoath Senior School, made a comparison of the various spreadsheets of data for men's and women's events since 2000:
We figured out that the first 3 winning countries did not always have the highest score in any certain task. We also figured out the the overall winners in the men's events are very often Germany, Canada and New Zealand, and in the women's events from 2000 till 2008, Australia always comes within the top three.
The Pythagoreans Club at All Saints Catholic School, on the other hand, commented on the men's results in 2008 in more detail:
Firstly, we sorted the data in terms of the swimming results. We found out that none of the top three medallists was the fastest at swimming; in fact the gold medallist came 16th. Therefore we think that the swimming does not have a major impact on the final positions. For example the fastest swimmer actually finished 34th overall.
Secondly, we resorted the data in terms of fastest cycling times. We noticed that the cycling also did not have a big impact on the final positions as the gold medallist came 33rd. Also the person who came 1st in swimming actually came last in cycling.
Lastly, we found out that the running time was the most important result, as the ranking in the running event was almost the same as the eventual ranking with only a few changes. So whatever position you place after running really determines where you will place overall.
We plotted a scatter graph of the running time against the total time. We wanted to find out whether these two variables were related, and they turned out to be very closely related.
To conclude we would say that in order to have the best total time, you have to be good at running. If you want to be an Olympic gold medallist, you don't have to get on swimmingly or get on your bike but you do have to be Forrest Gump.
Great! Thanks for all your responses.
Teachers' Resources
Why do this problem?
This problem offers a context to analyse some real world data and invites students to use their knowledge of sport to explain what they notice.
Possible approach
"The triathlon consists of a 1.5km swim, a 40km cycle and a 10km run. Do you think the triathlon will be won by someone who is very strong in one event and average in the other two, or someone who is strong in all three disciplines?"
Give students some time to share their ideas.
Display the spreadsheet.
"We have some data showing the times for each leg and the overall times from the Men's 2008 Olympics. What are you going to look for in the data to see if you are right about what makes a good triathlete?"
Give students time to plan what analysis they will carry out. Suggestions might include scatter graphs to look for correlation between individual event times (or rank) and overall time (or rank), or an investigation of the average and spread of times for each leg.
If a computer room is available, students could work in pairs with the spreadsheet. If no computers are available, the data is on this worksheet.
Finally, invite students to share what they found in the data, including any unexpected results, together with their explanations drawn from their knowledge of swimming, cycling and running.
If time allows, further work could be done with the results on the second spreadsheet, so that comparisons can be made with the Triathlons that took place in previous years.
Key questions
How well do each of the individual event times correlate with the overall times?
Your explanations might want to take into account the following:
How long do athletes spend on each leg of the Triathlon?
How might the order of the events in the Triathlon affect the final times?
What are the advantages of staying with other athletes over breaking away early?
How do the energy requirements differ for each leg of the Triathlon?
Possible support
Who's the Best? offers a simpler context for exploring Olympic data.
Possible extension
David and Goliath invites students to look for correlations and explain what they find in the data in the context of the men's shot put.
The Fastest Cyclist
Problem
In Nutrition and Cycling, you were invited to work out Andy's calorie needs on a ride from Land's End to John o'Groats.
If you haven't already done so, print out a copy of these cards.
Andy is competing in a Land's End to John o'Groats race, and YOU are his coach!
You are determined that Andy will win the race, as long as he is willing to cycle for more than 7 hours a day.
What is the shortest time in which Andy could complete the ride, if we ignore his 7 hour daily cycling limit, while making sure his energy needs are met?
You may wish to consider minimising the cost, and/or the weight that Andy needs to carry, together with any other appropriate constraints that occur to you.
You might find this article, describing the diet of Tour de France cyclists, of interest.
Cycling image from Wikimedia Commons, uploaded by Evdcoldeportes, released under the Creative Commons Attribution-Share Alike 2.5 Colombia license.
Getting Started
What is the maximum calorie intake available to Andy, based on the snacks and energy drinks he can carry?
If he cycled for $7\frac{1}{2}$ hours, how many calories would he need? How many calories could he take in?
What if he cycled for $8$ hours?
Student Solutions
This spreadsheet shows the limitations assuming all other information stays the same when the 7 hour per day limit is relaxed:
This leads to the conclusion that 8 hours per day cycling would give him a very small calorie deficit.
Algebraic solution (where h = hours/day cycling):
Calories available to consume: 4000 + 350h
Calories used: 2500 + (14 x 1.61 x 100h)/4.19
4000 + 350h = 2500 + (14 x 1.61 x 100h)/4.19
h = 7.98 hours per day cycling
If Andy cycles 8 hours per day, total race time:
1407 / (1.61 x 14 x 8) = 7.8 days
= 7 days + 6.422 hours cycling on Day 8
=7 days + 6 hours 25.34 minutes cycling on Day 8
Can you think of any other approaches?
Teachers' Resources
Why do this problem?
This problem offers a challenging extension to Nutrition and Cycling. Whereas the former problem requires students to make sense of information and engage in proportional reasoning, this is an optimisation task which students could tackle using spreadsheets.
Possible approach
Students will need to have worked on Nutrition and Cycling first, and will need the same set of cards.
"What if Andy could cycle for longer than seven hours a day? Imagine you are his coach, and you need to come up with a plan that makes sure his energy needs are met, but helps him to complete the race in the fastest possible time. At the end of the task, you will each get a chance to present your race plan to the rest of the group. Who can come up with the winning plan?"
Arrange the class in small groups. Students will need access to calculators or spreadsheets. While you circulate, ensure that students are recording their calculations clearly so that they will be ready to present them.
There is no 'correct' answer to this task - students will have to make decisions about whether a calorie deficit can be permitted, and whether additional off-road snacks could be allowed.
Plenty of time should be allowed at the end of the task for groups to present their plans, the decisions they made, and their reasons, with clear justifications for their results.
Key questions
What are Andy's calorie needs?
What are Andy's calorie allowances?
How do these change as he cycles for a longer time?
Possible extension
Students may wish to represent the situation algebraically, and apply different constraints before solving the resulting equations.
Possible support
Encourage students to use trial and improvement - choose a time greater than 7 hours for Andy to cycle, calculate his calorie intake and calorie needs, and then adjust up or down as necessary.
Track design
Problem
Imagine you are building a new Olympic stadium and you are responsible for designing and marking out the running track. The track needs to fulfil the following specifications:
- The distance around the inside edge of the inner lane should be 400m.
- There should be 8 lanes.
- Each lane should be 1.25m wide.
- The track should consist of two straight sections joined by two semi-circular sections.
- The straight sections should each be 85m in length (a straight section is extended over the curve for the 100m race, as shown below).
Can you work out the radius of the curved sections in order to produce an accurate scale drawing?
For the 200m race, runners start on the curved section at the right of the diagram and run anticlockwise to the finish line at the top left.
As the outer lanes are longer than the inner lanes, a staggered start is needed so that at the finish line all runners have run the same distance.
Can you work out where each runner should start so that they all run 200m in total?
For the 400m race, the runner in lane 1 does one complete lap of the track, so the start line is the same as the finish line. The runners in lanes 2 to 8 again have a staggered start.
Can you work out where each runner should start so that they all run 400m in total?
Getting Started
If you know the arc length, how do you find the radius?
Student Solutions
Well done to Daniel from Kings School, New Zealand, Josh from Chatham Grammar School for Boys, and Anna from the Sandon School, who all correctly worked out the dimensions of the running track. Here is Josh's solution:
Radius of the curved sections:
As the straight sections of the track are both 85m long, 170m of the track is straight. That means that the remaining 230m of the track is comprised of the two semicircular sections. As there are two semicircular sections, each of equal radii, the sum of their inside edges (of the inside lanes) is equal to the circumference of a circle that would fit neatly into the inside curves of the track.
The circumference of a circle, $c$, is given by the equation: $c = 2 \times \pi \times r$, where $r$ is the radius.
Therefore, as we have deduced that the circumference of this imaginary circle is equal to the sum of the inside curves of the track, we can write: $230 = 2 \times \pi \times r$, so $r = \frac{230}{2\pi} = 36.606$ metres (to 3 d.p.)
This measurement is only the measurement of the radii of the semicircles which comprise the inside edge of the inside lane on the curved section. We want to find a measurement for the radii of the semicircles which comprise the whole of both curved sections; there are 8 lanes that we must consider in order to do this. As each lane is a constant 1.25 metres, the sum of the thickness of the lanes is equal to 8 x 1.25 = 10 metres. Therefore, the radius of the curved sections, in order to produce a scale drawing, is equal to: 36.606 + 10 = 46.606 metres (to 3 d.p.)
The 200m Staggered Start:
As it is mentioned in the problem for the 400m start that the measurement for the length of a given lane is that of its inside edge, I will assume that it is the inside edges of each lane that must be made constant in the 200m race in order to make the race fair. The runner in lane 1 starts at the curved section on the bottom right of the track, so I will use that lane for comparison with the others. The straights do not need to be compensated for in the staggered start, as the runners would all run the same distance here. Therefore, we only need to consider the 115m of track that the runners will run on before they hit the straight; in this time, all runners will be running on the curved section of track.
The runner in lane 2 is running on a track with an inside edge that is 1.25m further outwards than that of lane 1. Therefore, the radius of runner 2's track is equal to $36.606 + 1.25 = 37.856$ metres. Thus, the circumference of the semicircle that makes up the curved section is equal to: $\frac{2 \pi r}{2} = \pi r = 118.927$ metres This is 3.927 metres longer than the inside edge of lane 1, so the rune in lane 2 will start 3.927 metres in front of the runner in lane 1.
We can also generate an nth term sequence for the lanes. The nth term for this sequence is: $(36.606 + 1.25(n-1)) \times \pi$ I arrived at this conclusion because the formula for the arc of a semicircle is pi x r, hence the reason for multiplying the sequence by pi, and the section in brackets is the way to determine the radius of the inside edge of any given track. Each runner, as we move from lane 1 to lane 8, will start 3.927 metres (3 d.p.) in front of the previous runner.
The 400m Staggered Start:
The obvious conclusion for this problem would be to say that each runner starts twice as far behind the runner in the next inside lane as they did in the 200m race, but I will investigate this mathematically.
Each runner is running 400 metres, but 170 metres of the track they will run (85 x 2) is made up of straight sections, where the runners will run the same distance regardless of the lane they are in, so this does not need to be compensated for. Thus, 230m of the track they will run on is comprised of a curved section, which must be compensated for. When the runner in lane 2 reaches a curved section, the radius of the inner edge of his/her lane is equal to $36.606 + 1.25 = 37.856$ metres. Thus, the arc lengths of both semicircles together is equal to: $2 \times \pi \times 37.856 = 237.854$ metres (3 d.p.) This is 7.854 metres longer than the length of curved track in lane 1, so the runner in lane 2 will start 7.854 metres in front of the runner in lane 2.
We can also generate an nth term sequence for the lanes. The nth term for this sequence is: $2 \times ((36.606 + 1.25(n-1)) \times pi)$. I arrived at this conclusion using the nth term formula I generated before to calculate the arc length of one of the semicircular pieces of track, then multiplied it by 2 (for the two semicircles involved in the 400m. As we move from lane 1 to lane 8, each runner starts 7.854 metres in front of the runner on the next inside lane to them.
Rajeev, from Haberdashers' Aske's Boys' School, sent us his calculations for the staggers presented clearly in a table. You can see it here.
Teachers' Resources
Why do this problem?
This problem offers an authentic context within which to calculate arc lengths and requires students to present their findings in a convincing manner.
Possible approach
This printable worksheet may be useful: Track Design
Arrange the class in groups of three or four, and hand out this worksheet to each group.
Give the students plenty of time to discuss and work on the problem in their groups. For groups not used to working collaboratively, these roles may be useful to guide students in organising themselves to work together.
To finish off, results could be presented as a list of the dimensions/angles that a groundskeeper would need in order to paint the lines for the running track, together with explanations of how students worked them out.
Additionally, students could be asked to produce a scale drawing of the track design to an agreed scale (1cm to 2.5m would fit on a large flipchart sheet) and then each group's drawing could be overlaid on another group's to check to see if they coincide.
Key questions
Possible support
By working in groups, students are encouraged to support each other in making sense of the problem and working towards a solution.Possible extension
David and Goliath
Problem
In the tables below are some of the top performing shot putters of all time, their weight in kilograms, and the distance they threw in metres.
Look at the two data sets. Do any results stand out?
What questions occur to you?
Choose appropriate statistical analyses to answer some of your questions.
Are there any other data you will need to collect?
You may wish to download this spreadsheet containing the two data sets.
It is suggested that heavier competitors are at an advantage in the shot put.
Do the data support that claim?
What would be the effect on the competition of introducing weight categories (like those found in boxing or weightlifting), or giving lighter shots to lighter competitors?
Name | Weight (kg) | Distance (m) |
Randy Barnes | 137 | 23.12 |
Ulf Timmermann | 118 | 23.06 |
Alessandro Andrei | 118 | 22.91 |
Werner Günthör | 130 | 22.75 |
Kevin Toth | 144 | 22.67 |
Udo Beyer | 135 | 22.64 |
John Brenner | 127 | 22.52 |
Adam Nelson | 115 | 22.51 |
Reese Hoffa | 133 | 22.43 |
Sergey Smirnov | 126 | 22.24 |
John Godina | 129 | 22.20 |
Sergey Gavryushin | 95 | 22.10 |
Sergey Kasnauskas | 126 | 22.09 |
Aleksandr Baryshnikov | 130 | 22.00 |
Gregg Tafralis | 132 | 21.98 |
Mikhail Kostin | 120 | 21.96 |
Tomasz Majewski | 132 | 21.95 |
Remigius Machura | 118 | 21.93 |
Cottrell J. Hunter | 135 | 21.87 |
Mike Stulce | 122 | 21.82 |
Women's Shot Put:
Name | Weight (kg) | Distance (m) |
Natalya Lisovskaya | 105 | 22.63 |
Ilona Briesenick | 95 | 22.45 |
Helena Fibingerová | 95 | 22.32 |
Claudia Losch | 84 | 22.19 |
Meisu Li | 92 | 21.76 |
Natalya Akhrimenko | 90 | 21.73 |
Vita Pavlysh | 90 | 21.69 |
Xinmei Sui | 90 | 21.66 |
Verzhinia Veselinova | 95 | 21.61 |
Margitta Pufe | 90 | 21.58 |
Ines Müller | 90 | 21.57 |
Nunu Abashidze | 105 | 21.53 |
Zhihong Huang | 100 | 21.52 |
Larisa Peleshenko | 95 | 21.46 |
Heike Hartwig | 95 | 21.31 |
Liane Schmuhl | 90 | 21.27 |
Astrid Kumbernuss | 90 | 21.22 |
Kathrin Neimke | 90 | 21.21 |
Helma Knorscheidt | 90 | 21.19 |
Heidi Krieger | 95 | 21.10 |
Data source: IAAF
Mass of shot for male competitors: 7.260 kg (16 lb)
Mass of shot for female competitors: 4 kg (8.82 lb)
Getting Started
Teachers' Resources
Why do this problem?
This problem encourages students to engage with a data set, pose their own questions about it, and perform some statistical analysis to try to answer their questions. The real-life sporting context provides a 'hook' to draw students in to speculation and discussion about how to compete on a level playing field.Possible approach
Key questions
Possible extension
Consider other athletic events where attributes such as height or weight might confer an advantage. Collect some data and analyse it to see whether there is a case to be made for some sort of 'handicapping' system.Possible support
Make it clear to students that there are no right or wrong answers in this task, but that you are looking for answers with good statistical justifications.High Jumping
If you are training to be good at any sport then you are in the business of optimisation - doing all you can to enhance anything that will make you do better and minimise any faults that hinder your performance. This is one of the areas of sports science that relies on the insights that are possible by applying a little bit of mathematics. Here we are going to think about two athletics events where you try to launch the body over the greatest possible height above the ground: high jumping and pole vaulting.
This type of event is not as simple as it sounds. Athletes must first use their strength and energy to launch their body weight into the air. If we think of a high jumper as a projectile of mass M launched vertically upwards at speed $U$ then the height $H$ that can be reached is given by the formula $$U^{2}=2g H$$ where $g$ is the acceleration due to gravity. Alternatively we can think in terms of energy conservation. The kinetic energy of the jumper at take-off is $\frac{1}{2}M U^{2}$ and this will be transformed into the potential energy $M g H$ gained by the jumper at the maximum height $H$ when he is instantaneously at rest at the highest point. Equating the two gives $ U^{2}=2g H$ again.
All this sounds straightforward but the tricky point is the quantity $H$ - what exactly is it? It is not the height that is cleared by the jumper. Rather, it is the height that the jumper's centre of gravity is raised, and that is rather a subtle thing because it makes it possible for a high jumper's body to pass over the bar even though his centre of gravity passes under the bar. When an object has a bendy shape it is possible for its centre of gravity to lie outside of the body. One way to locate the centre of gravity of an object is to hang it up from any point on the object and drop a weighted string from the same point, marking where the string drops. Then repeat this by hanging the object up from another point. Draw a second line where the hanging string now falls. The centre of gravity is where the lines of the two strings cross. If the object is a square then the centre of gravity will lie at the geometrical centre but if it is L-shaped or U-shaped the centre of gravity will not lie inside the boundary of the body at all.
It is this possibility that allows a high jumper to control where his centre of gravity lies and what trajectory it follows when he jumps. The aim of our high-jumper is to get his body to pass over the bar whilst making his centre of gravity pass as far underneath the bar as possible. In this way he will make optimal use of his explosive take-off energy.
The simple high-jumping technique that you first learn at school, called the 'scissors' technique is far from optimal. In order to clear the bar your centre of gravity, as well as your whole body, must pass over the bar. In fact your centre of gravity probably goes close to 30 centimetres higher than the height of the bar. This is a very inefficient way to clear a high-jump bar. The high-jumping techniques used by top athletes are much more elaborate. The old 'straddle' technique involved the jumper rolling around the bar with their chest always facing the bar. This was the favoured technique of world-class jumpers up until 1968 when the American Dick Fosbury amazed everyone by introducing a completely new technique which involved a backwards flop over the bar and won him the Gold Medal at the 1968 Olympics in Mexico City. This method was only safe when inflatable landing areas became available. Fosbury's technique was much easier for high jumpers to learn than the straddle and it is now used by every serious high jumper. It enables a high jumper to send their centre of gravity well below the bar even though their body curls over and around it. The more flexible you are the more you can curve your body around the bar and the loweryour centre of gravity will be. The 2004 Olympic men's high-jump champion Stefan Holm, from Sweden, is rather small by the standards of high jumpers but is able to curl his body to a remarkable extent. His body is very U-shaped at his highest point. He sails over 2m 37 cm but his centre of gravity goes well below the bar.
The photographer Peter Kjelleras captures the Olympic high-jump champion Stefan Holm jumping at the World Athletics Championships in Paris, in 2003. Holm dramatically demonstrates his ability to send his centre of gravity far below the bar he is clearing 2.32 metres above the ground.
Let's see if we can get a rough estimate of how well we might expect them to do. Suppose they manage to transfer all their horizontal running kinetic energy of $\frac{1}{2}MU^{2}$ into vertical potential energy of $MgH$ then they will raise their centre of mass a height of:
$$ H=\frac{U^{2}}{2g}$$
If the Olympic champion can reach $9\mathrm{\ ms^{-1}}$ launch speed then since the acceleration due to gravity is $g=10\mathrm{\ ms^{-2}}$ we expect him to be able to raise his centre of gravity height of $H=4$ metres. If he started with his centre of gravity about $1.5$ metres above the ground and made it pass $0.5$ metres below the bar then he would be expected to clear a bar height of
$1.5+4+0.5=6$ metres. In fact, the American champion Tim Mack won the Athens Olympic Gold medal with a vault of $5.95$ metres (or $19^{\prime }6\frac{1}{4}"$ in feet and inches) and had three very close failures at $6$ metres, knowing he had already won the Gold Medal, so our very simple estimates turn out to be surprisingly accurate.
John D. Barrow is Professor of Mathematical Sciences and Director of the Millennium Mathematics Project at Cambridge University.
Alternative Record Book
Problem
In the Olympic and Paralympic Games, medals are awarded for the best performances in each event. But what if we used different criteria for writing an alternative record book?
Investigate some of the questions below to write your own Alternative Record Book.
You will need to do some research to answer these questions, and you will need to make some estimations and approximations. You may also need to refine the questions to make them more precise. If you can't determine an exact value for certain contexts, you could try to compute upper or lower bounds to allow you to compare.
In which Olympic or Paralympic event does:
- A human travel fastest? How fast?
- An object travel fastest? How fast?
- An object travel highest? How high?
- A human expend the most calories? How many?
- A human react fastest? How fast?
- A human experience the greatest acceleration or deceleration? How much?
- An object experience the greatest forces, stresses or strains? How much?
Send us your thoughts about which events should be included in the Alternative Record Book, together with your evidence and calculations.
Can you think of any other categories to include in your Alternative Record Book?
You may wish to do some experiments and create an Alternative Record Book for your own school and its athletes.
Getting Started
The 100m sprint isn't the only event where humans move fast! Here are some suggestions of other speedy competitors you could consider:
A cyclist in a sprint
A wheelchair racer in the Paralympics
A diver as he enters the water
A long jumper as he takes off
or perhaps including the Winter Olympics too...
Student Solutions
Here are some of our ideas about this problem. Do you agree with our answers? Do you have any other ideas?
1. A human travels fastest in the men's 100m race. The record time is 9.58 s and if we subtract an average reaction time 0.15 s then the average speed would be $38.2 km/h$. If we let a human use a bicycle then the event where they travel fastest would be track cycling. The speed is about $70 km/h$.
2. A human reacts fastest again in the men's 100m race. Reaction times are about 0.09 - 0.15 s.
3. An object travels fastest in a shooting event. The speed of a bullet travelling out of a rifle is about $3500 km/h$. Moreover, a ping pong ball can travel about $110 km/h$, a tennis ball travels about $150 km/h$, a javelin travels at $90 km/h$ and in badminton shuttlecocks travel at over $400 km/h$.
4. An object travels highest in a javelin throwing event. A javelin reaches a height of 50m or it could be a missed bullet which can reach a height of a few kilometres.
5. A human's head experiences about 60g acceleration in a boxing event.
6. A human expends the most calories in a cycling - road event. Cyclists are racing for about 7 hours and they approximate burn 9000 kcal.
7. Force. For example a hammer could be thrown for 80 meters. This means it has a speed of about 28 m/s just before being released. The hammer's mass is 7.257 kg and measures 1.215 m. Thus the force on the hammer $(7.257 \times 28^2)/(1.215) = 5 kN$. In a high jump event, the current record is $2.45 m$ which means that the jumper had a speed of $6.9 m/s$. In the high jump, the acceleration off the ground takes about $0.15 - 0.2 s$. Thus, the force a jumper exterts on the ground is $(80 \times 6.9)/0.16 = 3.5 kN$.
Stress is a measure of the average force per unit area. We calculated that the force on a hammer is about 5 kN and we know that the diameter of standard hammer throw wire is 4mm. Thus, the stress in wire is $5000/(1.25 \times 10^{-5}) = 400 MPa$. Moreover, there are high stresses in a pole vault pole about 1 GPa.
Strain is a measure of deformation. It is highest in a pole which is used in a pole vault event. It is about 2.5%. Read more.
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Teachers' Resources
Why do this problem?
This problem invites students to take a look at the amazing limits that athletes push themselves and their equipment to in search of Olympic glory.
In order to come up with answers, students will need to undertake some research, calculate with measurements and different units, and make comparisons.
Possible approach
For younger students, concentrate on the first few questions dealing with speed, distance and time. Older students studying mechanics and physics may be interested to explore questions on acceleration, stresses and strains.
"What Olympic records do you know about? Do you know who holds the record, and what it is?" Students will share their knowledge of Olympic events and records.
"What if we wanted to make an Alternative Olympic or Paralympic Record Book? For example, we might want a record for the event where a human travels the fastest. Talk to your partner and suggest which event might hold this record."
Give students a couple of minutes to discuss, and then collect their suggestions on the board. If everyone suggests that it must be e.g. Usain Bolt running the 100m, invite them to consider some other possibilities:
- A cyclist in a sprint
- A wheelchair racer in the Paralympics
- A diver as he enters the water
- A long jumper as he takes off
- or perhaps including the Winter Olympics too...
If students reject cycling/wheelchair as they include extra equipment, suggest that they might have two categories in their record book, for fastest unaided human and for fastest aided human.
"You'll need to do some research to determine for sure which one of these would get the record."
Perhaps invite students to suggest some other categories for their record books, or offer them the ones from the problem. Then allow them time for research in groups, either to research one question of their choice or a range of questions.
Finally, each group can produce their own 'Alternative Record Book' or if each group has worked on one question, their work could be compiled into a class Record Book.
Key questions
Which events are likely candidates?
Have you taken a look at the list of Olympic events to check that you haven't missed any out?
What research and calculations will you need to carry out in order to compare the speeds and distances in different events?
Possible extension
The later questions on forces, stress, strain and acceleration require the use of more advanced mathematics, and may prompt students to think of their own challenging follow-up questions.
Possible support
To get students thinking about measurements and units at the Olympics, start with Olympic Measures.
Decathlon: the art of scoring points
The decathlon consists of ten track and field events spread over two days. It is the most physically demanding event for athletes. On day one, the 100m, long jump, shot putt, high jump and 400m are contested. On day two, the competitors face the 110m hurdles, discus, pole vault, javelin and, finally, the 1500m. In order to combine the results of these very different events - some give times and some give distances - a points system has been developed. Each performance is awarded a predetermined number of points according to a set of performance tables. These are added, event by event, and the winner is the athlete with the highest points total after ten events. The women's heptathlon works in exactly the same way but with three fewer events (100m hurdles, high jump, shot, 200m, long jump, javelin and 800m).
Originally, the points tables were set up so that (approximately) 1000 points would be scored by the world record for each event at the time. But records move on and now, for example, Usain Bolt's world 100m record of 9.58s would score him 1202 decathlon points whereas the fastest 100m ever run in a decathlon is 'only' 10.22s for a points score of 1042. The current world record that would score the highest of all in a decathlon is Jürgen Schult's discus record of 74.08m, which accumulates 1383 points.
All of this suggests some important questions that bring mathematics into play. What would happen if the points tables were changed? What events repay your training investment with the greatest points payoff? And what sort of athlete is going to do best in the decathlon - a runner, a thrower or a jumper?
The decathlon events fall into two categories: running events where the aim is to record the least possible time and throwing or jumping events where the aim is to record the greatest possible distance. The simplest way of scoring this would be record all the throws and jumps distances in metres, multiply them together and then multiply all the running times in seconds together and divide the product of the throws and jumps by the product of the running times, T. The Special Total that results will have units of $(\mathrm{length})^6 \div (\mathrm{time})^4 = \mathrm{m}^6/\rm{s}^4$ and spelt out in full it looks like this:
$$\mbox{Special Total, ST} = \frac{\mathrm{LJ} \times \mathrm{HJ} \times \mathrm{PV} \times \mathrm{JT} \times \mathrm{DT} \times \mathrm{SP}}{T(100\mathrm{m}) \times T(400\mathrm{m}) \times T(110\mathrm{mH}) \times T(1500\mathrm{m})}$$
If we take the two best ever decathlon performances by Šebrle (9026 pts) and Dvořák (8994 pts) and work out the Special Totals for the 10 performances they each produced then we get
Šebrle (9026 pts): ST = 2.29
Dvořák (8994 pts): ST = 2.40
Interestingly, we see that the second best performance by Dvořák becomes the best using this new scoring system.
In fact, our new scoring system is not a good one. It contains some biases. Since the distances attained and the times recorded are different for the various events you can make a bigger change to the ST score for the same effort. An improvement in the 100m by from 10.6s to 10.5s requires considerable improvement but you don't get much of a reward for it in the ST score. By contrast reducing a slow 1500m run by 10 seconds has a big impact. The events with the room for larger changes have bigger effects on the total. The actual scoring points tables that are used incorporate far more information about comparable athletic performances than the simple ST formula we have invented.
The points awarded (decimals are rounded to the nearest whole number to avoid fractional points) in each track event - where you want to give higher points for shorter times (T) are given by the formula
$$\mbox{Track event points} = \rm{A} \times (\rm{B} - \rm{T})^\rm{C},$$
where T is the time recorded by the athlete in a track event and A, B and C are numbers chosen for each event so as to calibrate the points awarded in an equitable way. The quantity B gives the cut-off time at and above which you will score zero points and T is always less than B in practice -- unless someone falls over and crawls to the finish! For the jumps and throws - where you want to give
more points for greater distances (D) - the points formula for each event is
$$\mbox{Field event points} = \rm{A} \times (\rm{D} - \rm{B})^\rm{C}$$
The three numbers A, B and C are chosen differently for each of the ten events and are shown in this table. You score zero points for a distance equal to or less than B. The distances here are all in metres and the times in seconds.
Most importantly, the points achieved for each of the 10 events are then added together to give the total score. In our experimental ST scoring scheme above they were multiplied together. You could have added all the distance and all the times before dividing one total by the other though.
Event | A | B | C |
---|---|---|---|
100 m | 25.4347 | 18 | 1.81 |
Long jump | 0.14354 | 220 | 1.4 |
Shot put | 51.39 | 1.5 | 1.05 |
High jump | 0.8465 | 75 | 1.42 |
400 m | 1.53775 | 82 | 1.81 |
110 m hurdles | 5.74352 | 28.5 | 1.92 |
Discus throw | 12.91 | 4 | 1.1 |
Pole vault | 0.2797 | 100 | 1.35 |
Javelin throw | 10.14 | 7 | 1.08 |
1500 m | 0.03768 | 480 | 1.85 |
In order to get a feel for which events are 'easiest' to score in, take a look at this table which shows what you would have to do to score 900 points in each event for an Olympic-winning 9000-point total.
Event | 900 pts |
---|---|
100m | 10.83s |
Long jump | 7.36m |
Shot put | 16.79m |
High jump | 2.10m |
400m | 48.19s |
110m hurdles | 14.59s |
Discus throw | 51.4m |
Pole vault | 4.96m |
Javelin throw | 70.67m |
1500m | 247.42s (= 4m 07.4s) |
There is an interesting pattern in the decathlon formulae that change the distances and times achieved into points. The power index C is approximately 1.8 for the running events (1.9 for the hurdles), close to 1.4 for the jumps and pole vault and close to 1.1 for the throws. The fact that C > 1 indicates that the points scoring system is a 'progressive' one, curving upwards in a concave way;
that is, it gets harder to score points as your performance gets better. This is realistic. We know that as you get more expert at your event it gets harder to make the same improvement but beginners can easily make large gains. The opposite type of ('regressive') points system would have C < 1, curving increasingly less, while a 'neutral'; one would have C = 1 and be a straight line. We can
see that the IAAF tables are very progressive for the running events, fairly progressive for the jumps and vault, but almost neutral for the throws.
In order to get a feel for how the total points scored is divided across events, the Figure below shows the division between the ten events for the averages of the all-time top 100 best ever men's decathlon performances.
Figure : Average points spread achieved across the 10 decathlon events in the 100 highest points totals
It is clear that there has been a significant bias towards gathering points in the long jump, hurdles and sprints (100m and 400m). Performances in these events are all highly correlated with flat-out sprinting speed. Conversely, the 1500m and three throwing events are well behind the other disciplines in points scoring. If you want to coach a successful decathlete, start with a big strong sprint hurdler and build up strength and technical ability for the throws later. No decathletes bother much with 1500m preparation and rely on general distance running training.
Stop Press! A new world record was set for this event in the US Olympic trials during the weekend 23rd June 2012 by Ashton Eaton. The performances for each event and the points accrued by them can be found at http://www.usatf.org/events/2012/OlympicTrials-TF/Results/Summary-39.htm
It is an interesting little project to compare the pattern of points obtained across each of the events with the ideal 'average' decathalete in the graph in the article. Eaton is actually very different, doing far better in the running events and worse in the throws.
Stop Press! Kevin Mayer broke the decathlon world record at the Decastar meeting in Talence on 16th September 2018, scoring 9126 at the final IAAF Combined Events Challenge fixture of 2018.
$^i$ See http://www.iaaf.org/mm/Document/Competitions/TechnicalArea/ScoringTables_CE_744.pdf
Training schedule
Problem
- 200m sprint
- 800m run
- 100m hurdles
- high jump
- long jump
- shot put
- javelin
The values for a, b, and c in each event are given below:
Event | a | b | c |
200 meters | 4.99087 | 42.5 | 1.81 |
800 meters | 0.11193 | 254 | 1.88 |
100 metres hurdles | 9.23076 | 26.7 | 1.835 |
High Jump | 1.84523 | 75 | 1.348 |
Long Jump | 0.188807 | 210 | 1.41 |
Shot Put | 56.0211 | 1.5 | 1.05 |
Javelin Throw | 15.9803 | 3.8 | 1.04 |
In the table below are the best times and distances of an Olympic hopeful in training, as well as the World Records for each heptathlon event (as of April 2011).
Event | Olympic hopeful | World records |
200m | 25.34s | 21.34s |
800m | 2min 13.00s | 1min 53.28s |
100m hurdles | 13.65s | 12.21s |
High jump | 1.43m | 2.09m |
Long jump | 5.67m | 7.52m |
Shot put | 12.45m | 22.63m |
Javelin | 45.05m | 72.28m |
Getting Started
Why not use a spreadsheet to calculate quickly the score for each event given the athlete's times or distances?
Then try tweaking her times and distances to see how her scores can be affected by different training schedules.
Student Solutions
Joseph from Park View Community School sent us the following solution:
Find the percentage away from the world record for each event. Increase the 3 lowest events by 20% and accordingly maintain current levels for the 3 greatest events. increase performance by 10% on the remaining event. I calculated all points including all percentage increases as well using excel.
You can download Joseph's spreadsheet to see how he calculated this solution here.
Teachers' Resources
Why do this problem?
When we watch sports coverage of the multi-discipline events such as the heptathlon, the scoring mechanism is not usually made explicit. This problem explores some of the maths behind the scoring system and invites students to optimise an athlete's performance by choosing a suitable training schedule. Along the way, students can practise substituting into formulas, make sense of functions, and use spreadsheets to repeat routine calculations quickly.
Possible approach
Set the scene by introducing the seven heptathlon events (perhaps asking students if they can name the events). Then display the two equations:
Key questions
Possible support
Students may opt to use a calculator to solve this problem, but it is much more efficient to use a spreadsheet. In order to make the most of the task, it may be worthwhile spending some time with the whole class talking about how to set up formulas in a spreadsheet and how to make changes to investigate different training schedules.Possible extension
Who's the winner?
Problem
Who's the Winner? printable sheet
Why are they not all equally likely?
This mathematical model assumes that when a goal is scored, the probabilities do
not change. Is this a reasonable assumption?
Alison suggests that after a team scores, they are then twice as likely to score the next goal as well, because they are feeling more confident. What are the probabilities of each result according to Alison's model?
Charlie thinks that after a team scores, the opposing team are twice as likely to score the next goal, because they start trying harder. What are the probabilities of each result according to Charlie's model?
The models could apply to any team sport where a small number of goals are typically scored.
You could find some data for matches between closely matched teams that finished with two goals and see which model fits most closely to what happened.
You will need to make some assumptions about what it means for teams to be "closely matched".
Send us your conclusions, and explain the reasoning behind the assumptions you chose to make.
Getting Started
Simulate a few games where two goals are scored and record how many "wins", "draws" and "losses" occur.
Student Solutions
Thank you to Anna from Mt Eliza North Primary School, Australia, who sent us this solution:
1. What are the possible results if 2 goals are scored in total?
2-0, 0-2, 1-1.
2. Why are they not all equally likely?
They are not all equally likely because there are two ways you can get the end result, 1-1, by the teachers scoring first and then the students or the students scoring first and then the teachers. There is one way for the end result, 2-0, the teachers get the 2 goals without the students getting 1. And there is only one way you can get the end result 0-2, by the students getting both the goals. Therefore the end result 1-1 is more likely.
3. Is this a reasonable assumption?
Yes. Even if the teachers are more confident and the students are trying harder, those equal out so the chance would stay at 50-50.
4. What are the probabilities of each result according to Alison's model?
In Alison's model the team who scores first is twice as likely to score the next goal.
The chance of ending 2-0, with the teachers winning, is:
$\frac{1}{2}$$\times$$\frac{2}{3}$$=$$\frac{1}{3}$
so there would be a 33.3% chance of that occurring.
The chance of ending 2-0, with the students winning, is also 33.3%:
$\frac{1}{2}$$\times$$\frac{2}{3}$$=$$\frac{1}{3}$
The chance of ending 1-1, with the teachers scoring the first goal and the students scoring the second goal, is:
$\frac{1}{2}$$\times$$\frac{1}{3}$$=$$\frac{1}{6}$
The chance of ending 1-1, with the students scoring the first goal and the teachers scoring the second goal, is also:
$\frac{1}{2}$$\times$$\frac{1}{3}$$=$$\frac{1}{6}$
So altogether the chance of ending 1-1 is also 33.3%:
$\frac{1}{6}$+$\frac{1}{6}$=$\frac{1}{3}$
Therefore, using Alison's model there is a 33.3% chance of any of the three results occurring.
5. What are the probabilities of each result according to Charlie's model?
In Charlie's model, after a team scores, the opposing team are twice as likely to score the next goal, because they start trying harder.
The chance of ending 2-0, with the teachers winning, is:
$\frac{1}{2}$$\times$$\frac{1}{3}$$=$$\frac{1}{6}$
so there would be a 16.7% chance of that occurring.
The chance of ending 2-0, with the teachers winning, is also 16.7%:
$\frac{1}{2}$$\times$$\frac{1}{3}$$=$$\frac{1}{6}$
The chance of ending 1-1, with the teachers scoring the first goal and the students scoring the second goal, is:
$\frac{1}{2}$$\times$$\frac{2}{3}$$=$$\frac{1}{3}$
The chance of ending 1-1, with the students scoring the first goal and the teachers scoring the second goal, is also:
$\frac{1}{2}$$\times$$\frac{2}{3}$$=$$\frac{1}{3}$
So altogether the chance of ending 1-1 is 66.6%:
$\frac{1}{3}$+$\frac{1}{3}$=$\frac{2}{3}$
Therefore, using Charlie's model there is a much greater chance of ending up with a 1-1 result.
Krystof from Uhelny Trh, Prague, used tree diagrams to work out the probabilities.
Teachers' Resources
Why do this problem?
This problem offers the opportunity to use probability in an authentic sporting context. It introduces the idea of using a probability model to make predictions, and then to refine the model using real data.
Possible approach
Introduce the idea of two equally matched teams (teams who are equally likely to score the next goal).
Key questions
Why are the results "win", "draw" and "lose" not all equally likely?
Possible support
Ask pairs of students to simulate the different models using dice, and collect together all the results. Then introduce tree diagrams to explain the results of the simulation.
Possible extension
Students could find data (perhaps from school or local teams) and decide on criteria for identifying "closely matched" teams. Then matches between closely matched teams where exactly two goals were scored can be analysed to see which model best fits the data.
Stadium Sightline
Problem
One of the challenges in designing an Olympic stadium is to make sure that spectators can see the event without their views being blocked by the spectators in front.
Your task is to design the tiered seating for the stadium. The back of each seat is 80cm behind the back of the seat in front, but raised up so that each row can see over the row in front.
Make some sketches showing what the cross-section (side view) of the seats might look like. Will the seats lie on a straight line, or a curve? How steep?
What factors do you think you will need to take into account when working out how high each seat needs to be?
The Olympic organisers have stipulated that all spectators must be able to see clearly a point 10m in front of the front row of seating:
The spectator in the second row needs to have line of sight to the same point as the spectator in the first row, as seen in the diagram above. Notice that the spectator in the second row needs some extra clearance in order to see comfortably over the first spectator's head.
Here is a zoomed-in diagram:
The first spectator's eye-level is 1.2m above the ground.
There is an extra 0.2m of clearance from his eye-level to the second spectator's line-of-sight. The second spectator is a further 0.8m away from the point on the pitch.
How high above ground level does spectator 2's seat need to be?
Now draw a similar diagram with the dimensions and unknowns for spectators 2 and 3.
How high above ground level does spectator 3's seat need to be?
Finally, imagine there were 40 rows of seating in the stadium.
Can you work out the heights above ground level of each of the 40 rows, and hence plot a side view of the seating?
It is very helpful to use a spreadsheet to perform the repeated calculations and plot the results.
Getting Started
What lengths do you know?
Look for similar triangles.
Student Solutions
Our first step should be to generalise the dimensions shown, for the nth person in the stadium. We shall let the point which must be visible to all be denoted as P. The distance, d, from P to the nth person's seat is therefore given by: $d_n = 10 + 0.8(n - 1) = 10 + 0.8n - 0.8 = 9.2 + 0.8n$. Furthermore, let $\theta_n$ represent the angle between the horizontal and the line of sight of the n-th person. We assume every person has a torso height of 0.8m. Let $x_n$ represent the height above ground level of the n-th person's seat. We can consider the following:
$\theta_{n+1} = \arctan\left(\frac{1 + x_n}{d_n}\right)$.
This formula is much easier to consider when we break it down step by step. Consider the zoomed in picture from the original problem. Consider the right-angled triangle formed with the perpendicular at person 1's seat extended to the line of sight of person 2. That triangle will enclose the angle between person 2's line of sight and the horizontal, $\theta_2$. Its height is given by adding the overhead clearance (0.2m), the torso height (0.8m) and person 1's chair height ($x_1$). Its base length is given by the distance from person 1's seat to the point P ($d_1$). This is true generally with 1 and 2 replaced by n and n+1, by considering similar diagrams. Therefore by use of the expression $\tan\theta = \mathrm{opp}/\mathrm{adj}$, we arrive at the above general formula.
Now we can replace the $d_n$ in the formula with $9.2 + 0.8n$, to give:
$\theta_{n+1} = \arctan\left(\frac{1 + x_n}{9.2 + 0.8n}\right)$.
The last step now is to generate a recurrence relationship between $x_{n+1}$ and $x_n$, by eliminating $\theta$. We can do this by getting an expression for $\theta_{n+1}$ in terms of $x_{n+1}$. By again studying the zoomed in diagram, we shall get an expression for $x_2$ and then generalise. Consider the triangle formed between the perpendicular at person 2's seat up to their eye-level and the horizontal along to P. This horizontal distance is $d_2$ and the vertical distance is $x_2 + 0.8$. Therefore, by use of the tangent formula:
$\theta_{n+1} = \arctan\left(\frac{x_{n+1} + 0.8}{d_{n+1}}\right)$.
By eliminating $\theta_{n+1}$ from these two equations, we eventually end up with:
$x_{n+1} = \frac{2.64 + 0.16n + (10+0.8n)x_n}{9.2 + 0.8n}$
and we can now set up a table using this formula to calculate all of the values from $x_2$ to $x_{40}$, as we know that $x_1 = 0.4$.
The required values are:
$x_2 = 0.712$
$x_3 = 1.0388\dots$
$x_{40} = 19.0444\dots$
Fantastic!
Teachers' Resources
Why do this problem?
This problem encourages students to model a real-life situation mathematically, that of tiered seating design in sports stadia. The problem begins with some straightforward geometry but leads into some spreadsheet modelling. Although the problem can be solved using trigonometry, the most elegant solution simply uses similar triangles.
Possible approach
"Your task is to design the seating layout for the stadium. The seats will be tiered so that the back of each seat is 80cm behind the back of the seat in front, but raised up so that each row can see over the row in front. Start by making some sketches of how you think the seats might need to be arranged."
Give students time to try out some ideas. Perhaps sketch on the board (or invite students to sketch) a variety of possibilities - straight lines of different gradients, and convex and concave curves. Then allow some time for students to share their intuitions about which will best model a successful tiered seating design, together with any reasons they have to support it.
Next, introduce the diagram:
Set the next task, to work out the missing height. Students may wish to draw or plot accurate diagrams before making calculations.
Move on to working out the height above ground level of the third seat, and allow some time to discuss whether students are surprised that the extra height is not constant.
Finally, invite students to explore what happens for subsequent steps, perhaps using spreadsheets to iterate and calculate the next step height from the previous.
Key questions
What sort of a view would spectators have if each seat was higher than the one before by the same distance?
What sort of a view would spectators have if the angle between their line of sight and the line of sight of the spectator in front was constant?
Possible extension
Invite students to consider other factors that might affect stadium seating and perhaps adapt their models accordingly.
Possible support
Track Design looks at a simpler design context at the Olympics.
Speed-time problems at the Olympics
Problem
Speed-Time Problems at the Olympics printable worksheet
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At the 2012 Olympic games, the qualifying standards for the women's 100 metres race was 11.29s. How does this compare with the speed of a bus? |
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At the 2012 Olympics Shelly-Ann Fraser-Pryce won the women's 100m in a time of 10.75s. If she had continued running, how much further would she have run by the time an athlete running at the qualifying speed (11.29s) would have crossed the line? |
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In the 2009 IAAF World championship, Usain Bolt ran the 100m in 9.58s. Estimate how far he would have been ahead of the gold medallist from Lane 2 had they been racing together. |
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Imagine that you raced in the 200m with Usain Bolt. By what length would he beat you?
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Imagine that a 2km rowing race took place on a rowing lake with two separate legs of 1km. How would the total race time vary from a race on a river where one leg is upstream and the other downstream? |
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Imagine that cyclist A completes a lap following the blue line on the velodrome track. Cyclist B completes a lap 1m inside the blue line and cyclist C completes a lap 2m outside the blue line. How do the distances travelled vary between the cyclists? |
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In the past, the start of a 100m race was indicated by a pistol shot next to lane 1. Did this give a significant advantage to the runner in lane 1? Would it have given a significant advantage to anyone if this pistol was fired from the end of lane 4? |
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Imagine an announcement is made from a podium in the centre of a stadium. As the speaker talks into her microphone the sound is simultaneously sent to speakers which project the sound into the stadium and up to satellites which transmit the signal as digital radio. Who might hear the sound first: someone listening on the radio or someone listening in the stadium? |
Note: In this problem you might not have been given all of the required data and you might need to make estimates or approximations. You should be clear in your mind as to any assumptions that you make whilst constructing your answers.
Getting Started
It is important to be aware throughout that these questions are (deliberately!) not as 'precisely' stated as typical textbook questions. For example, the phrase 'If she had continued running ...' from lane 2 requires an assumption to be made before computation of an answer.
There is no absolutely 'right' way to make these assumptions, although assumptions need to be made clearly.
One possible assumption might be that a runner runs at their average speed for the full distance.
Student Solutions
Lane 1:
The required speed to qualify is $\frac{100\mathrm{m}}{11.29\mathrm{s}}=8.86\mathrm{ms}^{-1}.$ This is $8.86\times\frac{3600}{1600}\mathrm{mph} = 19.9\mathrm{mph}$. This is a similar speed to a bus travelling in a built-up area, but of course the bus will continue at this speed for much longer!
Lane 2:
Shelly-Ann Fraser-Pryce travelled at a speed of $\frac{100\mathrm{m}}{10.75\mathrm{s}}=9.30\mathrm{ms}^{-1}.$ If she had continued at this speed until the athlete running at the qualifying speed had finished (an extra $11.29\mathrm{s}-10.75\mathrm{s} = 0.54\mathrm{s}$), she'd have ran an additional $5.02\mathrm{m}.$
Well done to Annabelle from Saint Martns, Brendan from Colfe's School and Zoe, Abigail, Ella and Amy from Queen Elizabeth High School who found this!
Lane 3:
When Usain Bolt crossed the finish line in $9.58\mathrm{s}$, Shelly-Ann Fraser-Pryce would still have $10.75\mathrm{s}-9.58\mathrm{s} = 1.17\mathrm{s}$ to run, meaning she would be $1.17\mathrm{s}\times9.30\mathrm{ms}^{-1}=10.88\mathrm{m}$ behind.
(These calculations assume they run at their average speed throughout the race, which is certainly not the case, as the athletes take several seconds to get to full speed. However, these figures should give an acceptable approximation.)
Lane 4:
Your $200\mathrm{m}$ time depends on lots of factors including age, gender and fitness, but a good ballpark figure would be $30\mathrm{s}$. After Usain Bolt finishes in his world record time of $19.19\mathrm{s}$, you'd still have $\frac{10.81}{30}\times200\mathrm{m}=72\mathrm{m}$ left to run! (Again, we're assuming constant speed throughout the race.)
Well done to Richard from Wilson's School who managed to get up to this part!
Lane 5:
Suppose on the flat rowing lake, the crew take $210\mathrm{s}$ $(3\colon 30)$ to row $1\mathrm{km}$. Overall, the time for their $2\mathrm{km}$ race is $420\mathrm{s}$ $(7\colon 00)$. This corresponds to a speed of $4.76\mathrm{ms}^{-1}$, which we're assuming is the same in both directions and throughout the race.
Suppose on the River Thames, the tide makes the boat go $x\; \mathrm{ms}^{-1}$ faster than before on the downstream section, and $x\; \mathrm{ms}^{-1}$ slower than before on the upstream section. The overall time for this race is therefore $\left(\frac{1000\mathrm{m}}{(4.76-x)\mathrm{ms}^{-1}}+\frac{1000\mathrm{m}}{(4.76+x)\mathrm{ms}^{-1}}\right)\mathrm{s}$.
If $x=1$, the overall time on the Thames is $439.3\mathrm{s}$, $19.3\mathrm{s}$ slower than the race on the lake. However, if $x=0.2$, the overall time is $392.8\mathrm{s}$, $27.2\mathrm{s}$ faster than the race on the lake.
Lane 6:
Velodromes for the Olympics are allowed to measure $250\mathrm{m}$, $333.3\mathrm{m}$ or $400\mathrm{m}$ in length. We'll assume $400\mathrm{m}$ to simplify the calculations. Assuming half of this distance is along the home and back straights along which there's no difference which line you take, there's $200\mathrm{m}$ around the $180^{\circ}$ turns where the position on the track is important. If we assume the cyclist on the blue line travels $400\textrm{m per lap}$, then the radius of the semicircular sections is $\frac{100}{\pi}\mathrm{m} =31.8\mathrm{m}$. Cyclist B will therefore travel $200\mathrm{m}+2\times\pi\times(31.8-1)\mathrm{m}=393.7\mathrm{m}$ and cyclist C $200\mathrm{m}+2\times\pi\times(31.8+2)\mathrm{m}=412.6\mathrm{m}$.
Track cyclists try to cycle as close to the inside of the track as possible in order to minimise the distance travelled. However, if they do this they can end up 'boxed-in' by other cyclists; this is one of the other factors that should be considered.
Lane 7:
Assuming a lane is $1\mathrm{m}$ across, the sound from the starting pistol only has to travel $8\mathrm{m}$ to reach the athlete in lane 8. This takes approximately $\frac{8\mathrm{m}}{340\mathrm{ms}^{-1}} = 0.02\mathrm{s}$, much less than an average human reaction time, so doesn't give anyone a significant advantage. Firing the gun from the centre of the track by lane 4 would halve the time before the sound has reached everyone, but this isn't necessary.
Lane 8:
The speed of sound is around $340\mathrm{ms}^{-1}$ (amongst other things, it depends on altitude and air temperature). Estimating that the crowd is $100\mathrm{m}$ from the podium, the sound will take $\frac{100\mathrm{m}}{340\mathrm{ms}^{-1}}=0.29\mathrm{s}$ to reach them.
A geostationary satellite used for communications has an altitude of around $36000\mathrm{km}$. Radio signals (which move at the speed of light) will take around $2\times\frac{3.6\times10^7\mathrm{m}}{3\times10^8\mathrm{ms}^{-1}} = 0.24\mathrm{s}$ to travel to and return from the satellite.
It seems the two delays are similar, although this calculation neglects any delays in the process of transmitting the signal, either on earth or by the satellite, and also neglects any time spent by the digital radio decoding the signal.
Teachers' Resources
Why do this problem?
This task provides an engaging context for students to explore speed, distance and time problems. Some of the questions require students to make assumptions or find out extra information.
Possible approach
These resources may be useful: Speed-Time Problems at the Olympics,
Here are some ways the questions in this problem could be used:
- Display one question at the start of a lesson, give students some time to work out their response, and then discuss as a class different ideas and methods.
- Give all eight questions out and invite students to work on them in pairs or small groups before bringing the class together to share their answers and debate any disagreements.
- Give out different questions to different pairs and then invite each pair to present their answer, with the rest of the class acting as critical friends insisting on clear reasoning.
There may be opportunities for cross-curricular links with P.E. where students may have collected their own data about their best times for 100 and 200m. It may be appropriate to adapt some of these questions and use students' own times.
It is important to be aware throughout that these questions are (deliberately!) not as 'precisely' stated as typical textbook questions. For example, the phrase 'If she had continued running ...' from lane 2 requires an assumption to be made before computation of an answer. There is no absolutely 'right' way to make these assumptions, although assumptions need to be made clearly. You might need to encourage or reassure the class that they are 'allowed' to make their own sensible assumptions on which to base their calculations if they are unused to working in this way. You might find that rich mathematical discussion emerges from the discussion of the modelling assumptions made on certain parts of the question.
Key questions
What assumptions do you need to make?
Is there any extra information you need to know?
Possible support
Possible extension
The challenging task Speedo invites students to think about questions of speed, distance and time where acceleration plays a part.
The Olympic Torch Tour
Problem
If you are a teacher, click here for a version of the problem suitable for classroom use, together with supporting materials. Otherwise, read on...
In May 2012, the Olympic torch will arrive at Lands End for a 70 day tour of the UK, ending in London. The plan is to cover towns and villages so that 95% of Britons will be within 10 miles of the torch relay.
Imagine a mini-Olympic torch tour running between 4 cities in the UK, with the following constraints:
- The torch starts and finishes in London
- The torch should pass each city once and only once
- The following table lists the distance between each city
(in miles as measured by Google Maps)
London | Cambridge | Bath | Coventry | |
London | 0 | 50 | 96 | 86 |
Cambridge | 50 | 0 | 120 | 70 |
Bath | 96 | 120 | 0 | 80 |
Coventry | 86 | 70 | 80 | 0 |
- What is the shortest route?
- How can you be sure it is the shortest?
- How many different routes are there?
London | Cambridge | Bath | Coventry | Oxford | |
London | 0 | 50 | 96 | 86 | 60 |
Cambridge | 50 | 0 | 120 | 70 | 65 |
Bath | 96 | 120 | 0 | 80 | 54 |
Coventry | 86 | 70 | 80 | 0 | 46 |
Oxford | 60 | 65 | 54 | 46 | 0 |
- What is the shortest route now?
- How many different possible routes did you need to consider?
Is there an efficient way to work out the number of different possible routes when there are 10 cities? 15 cities?...
Suppose a computer could calculate one million routes per second. How long would it take to find the optimal route for 10 cities? 15 cities? 20 cities?
The type of question we have explored above is a famous problem in computation complexity theory known as the Travelling Salesman Problem. Perhaps a better question for the torch tour is not to find the shortest or longest route, but to find the maximum number of cities the torch can visit whose route length is at most, say 2000 miles, or visit as many populated towns as possible. These are variants of the original problem known as the 'Orienteering Problem' and the 'Prize Collecting Travelling Salesman Problem'. It is an active area of research among mathematicians and has a wide range of applications.
Getting Started
You could try to list all the possible routes and calculate the distance the torch would need to travel for each.
Student Solutions
Niharika was the only one to send us a solution this time round:
The possible routes are:
1. London-Cambridge-Bath-Coventry-London
2. London-Cambridge-Coventry-Bath-London
3. London-Bath-Cambridge-Coventry-London
4. London-Bath-Coventry-Cambridge-London
5. London-Coventry-Bath-Cambridge-London
6. London-Coventry-Cambridge-Bath-London.
I worked out the distances for each route, and they came to:
1. 336 miles
2. 296 miles
3. 372 miles
4. 296 miles
5. 336 miles
6. 372 miles
The shortest routes are routes 2 and 4.
Now imagine that each city along our route is a box. We have five boxes to line up. The first and last box must be 'London', so there is only 1 way to fill those boxes. The second box can be filled in one of 3 ways. Once we've used that city up, the third box can be filled in 2 ways. Finally when we come to the fourth box there is only 1 way to fill it in. So there are 1*3*2*1*1 = 6 routes.
I guessed that, when extending to 5 cities, we should probably start with the shortest route for 4 cities and then add the extra city in. In this case, we should start with routes 2 and 4, and add Oxford in in all the different places, and see which distance is smallest. In each case, least distance is covered if we add in Oxford between Coventry and Bath. In this case there are 6 boxes and 1*4*3*2*1*1 = 24 routes.
When there are n cities, there are n+1 boxes, and so there are $1\times (n-1)\times (n-2)\times \dots \times 2 \times 1\times 1$ routes - just think of how many ways there are to fill in each box. But if you know the shortest routes for the case of n-1 cities, my guess is that you should be able to add the n-th city to those.
Great - thanks, Niharika!
Teachers' Resources
Why do this problem?
As well as encouraging students to work systematically, this problem introduces ideas from Decision Maths and Computer Science such as the Travelling Salesperson problem and the efficiency of algorithms.
Possible approach
Hand out this worksheet and encourage students to work on their own or in pairs on the first part of the problem with four cities. Bring the class together to discuss the sort of strategies they used to find a short route, and to discuss the checking that was necessary to be sure that their route was the shortest.
Once students are happy with the idea that a 'brute force' algorithm that checks every possibility is the ONLY way to be certain of finding the shortest route, set them the second challenge to find the shortest route when there are five cities.
"You need to be able to convince everyone that the route you find is the shortest. Think about how you could record your work to make sure you don't miss any possible routes."
While students are working, circulate and notice anyone who is using a particularly useful representation such as a systematic way of listing possible routes, or a tree-diagram approach. Bring the class together and invite those students to share their representations.
"Why does it take longer to check all the routes for five cities than four?"
"Is there a quick way to work out how many different possible routes there will be, for six cities?"
"How will the number of possible routes continue to grow?"
Finally, as students begin to get some appreciation of the way the number of routes grows as the number of cities increases, challenge them to work out the time taken to implement the brute force algorithm by computer.
Key questions
How can you represent the different routes in a way that makes sure you don't miss any?
If I add in an extra city, how does that affect the number of different possible routes?
Possible extension
There are rich opportunities for students to go and research topics in Decision Mathematics and Computer Science and to learn about the challenges involved in seeking solutions to problems using computers.
Nutrition and Cycling
Problem
Nutrition and Cycling printable worksheet part 1 (questions)
Nutrition and Cycling printable worksheet part 2 (cards)
Andy is preparing to cycle from Land's End to John o'Groats.
He will undertake some training rides before the big ride.
These cards contain some information about his training schedule, details about the big ride, and his nutritional needs when he is cycling.
Have a look at the cards and try to make sense of the information.
Then use the information to help you to answer the questions below.
- Andy is planning a short training ride.
He wants to take either bananas or cheap cereal bars with him as on-the-road snacks.
How many bananas would he need to take, to minimise the calorie deficit at the end of his ride? How many cheap cereal bars?
(The calorie deficit is the difference between the calories Andy uses during his cycle ride, and the calories he consumes before and during the ride.)
- After his training rides, Andy is ready to cycle from Land's End to John o'Groats.
How many days will it take?
Work out some of Andy's different options for carrying and consuming on-the-road snacks and drinks.
How can he maximise his consumption while cycling?
Together with his meals, can he consume enough calories each day so that he doesn't lose any weight?
How much of his calorie intake will need to be provided each day through off-the-road snacks?
Possible extension
The Fastest Cyclist follows on from this problem and challenges you to devise a winning cycling and nutrition plan if Andy is racing to reach John o'Groats.
Getting Started
How could you organise the cards?
Are there any pieces of information you haven't used yet?
Are there any cards with useless information?
Could you combine the information on several cards to generate new pieces of information?
Student Solutions
The ride is 38 miles long, which is 38 x 1.61 = 61.18 km. Multiplying by the 100 kJ of enengy used to cycle each kilometer, Andy will use 6118 kJ by cycling. This is converted to kcal using the fact that 1 kcal = 4.19 kJ, so 6188 kJ = 6188 / 4.19 = 1460 kcal.
Since Andy had a big meal of 800 kcal for breakfast he has 1460 - 800 = 660 kcal to make up on the journey.
Even though the ride is short, Anna, Millie, Deena, Lizzie and Izzy wanted to account for the 2500 kcal that Andy needs on top of his exercise requirements. They assumed that Andy burns these 2500 calories at a constant rate during the 24 hour day.
On top of the exercise, during the 2 hour ride Andy needs 2 x 2500 / 24 = 208.3 kcal anyway, which means a total of 1460 + 208.3 = 1668.5 kcal need to be consumed. He has a big meal (800 kcal) for breakfast, so he needs to make up 868.5 kcal on the ride. If he eats 7 bananas, he can make up 840 kcal; if he eats 8 cereal bars, he can make up 800 kcal.
Good! But, didn't Andy only want to limit himself to 250kcal per hour on short trips? What happens here? Jim, from Strathallen, has the answer:
If he wishes to eat bananas, he can eat four of them and consume 480 kcal. If he wants to eat cheap cereal bars, he can eat five of them and consume 500 kcal. So the cereal bars minimises his defecit.
Great - thank you!
Ewan, from King Edward VII School in Sheffield, tackled the second problem with the following method:
The distance from Land's End to John o' Groats is 1407 km, which is 1407 / 1.61 = 874 miles. We know that Andy cycles at 14 mph on multi-day trips so the cycling will take him 874 / 14 = 62.4 hours. Since he wishes to cycle for no more than 7 hours per day, and 62.4 / 7 = 8.9, Andy will take 9 days to complete the trip.
Ewan assumed that Andy would cycle for 7 hours on the first 8 days and have a slightly shorter last day, then worked out the calorie defecit.
In a 7 hour day Andy travels 7 x 14 = 98 miles, which is 98 x 1.61 = 157.78 km. This needs 15778 kJ of energy which is 3766 kcal. Adding the 2500 kcal needed in addition to these calories, Andy requires 6266 kcal per day.
The three really big meals provide 3000 kcal, leaving 3266 kcal from snacks. He can have 350 kcal per hour for 7 hours in on-the-road snacks which is 2450 kcal. This leaves 816 kcal to get from off-the-road snacks.
Another student, Emily, worked out several ways of achieving 2450 kcal (or more!) from snacks and energy drinks:
Andy can carry 3l of energy drink which provides a total of 6 x 190 = 1140 kcal (and he can then refill with water to provide his fluid needs for the rest of the ride).
He can carry
8 bananas = 960 kcal
or 8 energy bars = 1760 kcal
or 16 cheap cereal bars = 1600 kcal
or 24 energy gels = 2640 kcal (though he wouldn't want to consume more than 2450kcal worth)
So he can meet his energy needs with any of these supplemented by energy drink, except bananas.
You might like to think of letting Andy carry combinations of different snacks, he does have 8 pockets after all. See if you can find a way of getting exactly 2450 kcal so that Andy doesn't carry spare food.
Teachers' Resources
Why do this problem?
This problem requires students to make sense of a wealth of information in order to analyse the nutritional needs of a long-distance cyclist. As well as handling data, students will also gain practice in converting units and proportional reasoning.
Possible approach
Arrange the class into twos or threes, and hand out these cards together with this worksheet.
As the activity is a sense-making task, there should be little teacher input, other than to explain that all the information they need to answer the questions is on the cards, and the expectations for justifying and communicating their solutions. While students are working, circulate and make a note of any insights that are worth sharing with the whole group.
Solutions could be presented in a variety of ways:
Groups could prepare a poster
Groups could present their solution to a part of the task to the rest of the class, with other students acting as 'critical friends'
Each group could present their solution to another group
The following key questions or prompts could be offered to groups who are stuck:
How could you organise the cards?
Are there any pieces of information you haven't used yet?
Are there any cards with useless information?
Could you combine the information on several cards to generate new pieces of information?
Possible support
The first question is much less demanding than the second, so you may initially want to hand out this smaller set of cards that just contains the information needed for the first question.
Zin Obelisk could be used to introduce this type of task with easier mathematical content.
Possible extension
The Fastest Cyclist follows on from this problem and challenges students to devise a winning cycling and nutrition plan if Andy is racing to reach John o'Groats.
What's the point of squash?
Problem
In the game of squash the serve passes from one player to another only when the serving player loses a rally. A player wins a point when, and only when, they win a rally on their serve.
Usually the winner is the first player to reach 9 points, but if the score becomes 8-all then the game can be played to either 9 or 10 points: the person who first reached 8 points makes this decision.
Suppose that the score in a game is 8-all and you reached 8 points first and you have a probability of $p$ of winning any particular rally. Under which circumstances is a 9 point game a good idea?
Sportspeople often have very clear strategies in their minds when playing different opponents and sometimes make shot decisions based on their chances of winning points in different circumstances: sometimes it is best to 'play it safe' and on other occasions more risky play is called for.
This problem is based on the Traditional International rules of squash, taken fromhttp://www.squashgame.info/squashlibrary/2
Pole vaulting
Problem
The pole vault event consists of these phases:
1. The vaulter stands at the end of the runway holding the pole vertically.
2. The runner sprints along the runway as the top of the pole is moved towards the ground.
3. The pole is planted in the box.
4. The pole flexes and absorbs the energy of the runner.
5. The pole straightens and the runner is propelled up and (hopefully!) over the bar.
This is a fascinating mechanical process!
Can you draw a sequence of simple pictures which represent the stages of a pole vault?
Then consider these questions, using appropriate data for the athlete:
1. What is the locus of the centre of mass of the athlete during this process?
2. How much kinetic energy does the runner and pole contain just prior to planting in the box?
3. How efficiently is this converted into potential energy?
There are various levels of sophistication at which this can be considered - analyse with as much depth as you feel is relevant and you can use real data (some are provided below) or approximations. Either is fine, provided that your assumptions and estimations are clearly stated.
You might wish to use the following data:
Pole vaulting world record holders as of January 2011:
Male = 6.14m (Sergey Bubka), Female = 5.06m (Yelena Isinbayeva)
Lengths of poles vary between 2.3m to 6.4m, with weight rated individually
Height and weight of vaulters: Sergey Bubka 1.83m/80kg, Yelena Isinbayeva 1.74m/65kg
Length of runway: 40m
Student Solutions
This investigative task can be approached at various levels, and there is no specific 'solution' as such.
Teachers' Resources
Why do this problem?
Possible approach
Possible variation
Key points of note
Little little g
Problem
This problem involves four different parts which you can either discuss, just think about or analyse with various levels of detail.
The effective weight of an object at any place on earth mainly depends on three things:
1. The gravitational pull of the earth on the object.
2. The centripetal acceleration on the object caused by the rotation of the earth on its axis.
3. The mass of the object.
The gravitational acceleration is typically quoted as $g$ is $9.80665\mathrm{ms}^{-2}$ and the weight $W$ of an object as $W=mg$.
Part 1: The figure quoted in this question for $g$ assumes that the earth is a sphere. Newton's law of gravitation says that the gravitational acceleration felt at a distance $R$ from the centre of a uniform sphere is given by
$$
g =\frac{GM}{R^2}\;,\quad G = 6.67300\times 10^{-11} \mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}.$$
What radius does this imply for the earth?
Part 2: In 1968 the Olympic Games were held in Mexico city, at an altitude of $2240\mathrm{m}$ above sea level. At these games, Bob Beamon jumped a staggering $8\mathrm{m}$ $90\mathrm{cm}$ in the long jump, smashing the previous record by $55\mathrm{cm}$. This record survived until 1991 when it was broken by a small amount in Tokyo (altitude $17\mathrm{m}$), by Mike Powell.
Do you think that the unusually high altitude of Mexico City contributed to the longevity of Bob Beamon's record? Back up your thoughts with an analysis.
Part 3: Is the rotational effect of any significance on your weight? Do as much analysis as seems necessary to determine 'significance'.
Part 4: Is there anything else that might have a tiny effect on your weight?
The earth is usually modelled as a uniform sphere of mass $5.9742 \times 10^{24}$ $\mathrm{kg}$.
Assume that the earth spins around its axis once every 24 hours (if you think that this statistic is 'obvious' then you might like to read http://en.wikipedia.org/wiki/Earth's_rotation !!)
Light weights
Problem
The weight $W$ of an object on earth depends on the mass $m$ of the object and the force of gravity. The weight is usually given by the expression
$$
W = 9.8 m.
$$
However, the actual weight decreases the further you get from the centre of the earth. Newton worked out that weight can be measured more accurately as
Here $M$ is the mass of the earth, $m$ is the mass of the small object you are trying to weigh in $\mbox{kg}$ and $R$ is the distance from the centre of the earth in metres; $W$ is the weight in Newtons, which have units of metres kilograms per second per second.
In Olympic weightlifting the biggest competitors can sometime lift $200\mbox{kg}$ masses overhead. Sometimes weight lifting events take place in high altitude cities and sometimes at sea-level. The question that you are asked is this:
Does the variation in gravity provide a significant effect for weightlifters?
Something else to think about: How high in an airplane or rocket would you have to go before you could lift a $200\mbox{kg}$ mass overhead?
$G$ is called Newton's gravitational constant, which you can read about on Wikipedia.The universal law of gravitation expressed here gives extremely accurate predictions for the orbits of suns and planets. It is eventually superseded by the difficult theory of general relativity.
Student Solutions
This problem is best tackled by using the formula given to you carefully.
Christopher from Sale Grammar sent us his working here. He comments:
Using the values of 1956N and 1975N for the extremities of the earths surface we can see that the range of values would be 19N which equates to around 2kg. so, if a weightlifter was to do some weightlifting at the bottom of the Arctic, and then the top of Mount Chimborazo, they would feel a slight difference. However, when it comes to the main aim of the olympics, winning, it is clear that of course, everyone will be under the same stresses, so in terms of competition, everything will be fine. And there are other considerations at various altitudes too, such as air quality and density, and there are some serious considerations to be had at the bottom of the arctic. So in conclusion, yes, it would make a difference to the appeared strength of the weightlifter, but only by about 2kg.
Patrick from Otterborne included a table with his solution, showing the weight varying with altitude. How high would you have to go to lift 200kg?
Altitude (km) | Weight (%, relative to 200kg) | Weight (kg) |
0 | 100 | 200 |
10 | 99.69 | 199.37 |
50 | 98.45 | 196.90 |
100 | 96.93 | 193.87 |
500 | 85.98 | 171.95 |
1000 | 74.71 | 149.41 |
5000 | 31.39 | 62.78 |
10000 | 15.14 | 30.29 |
50000 | 5.84 | 11.67 |
Teachers' Resources
Why do this problem?
Possible approach
Key questions
Possible extension
Possible support
Any win for tennis?
Problem
A mathematician tennis player said:
"In tennis you win a game if you score 4 points before your opponent scores 3 points. Or, if you both score 3 points at some stage you win if you manage to score 2 points in a row after the 3-all stage before your opponent does."
This sentence is quite a mouthful to say, so first think about what it means! If you play tennis, think about how this mathematically represents the scoring system.
Suppose that you have a fixed chance of $0.6$ of winning any given point. What is your chance of winning a game?
Numerical extension
In reality a fixed chance of winning a point is not a good assumption. Suppose that Ahmed has a 60% chance of winning the first point if he serves, 80% chance of winning a point if he has just won a point and a 40% chance of winning a point if he has just lost a point. Suppose that Bryoni's chances are 85%, 80% and 30% respectively if she serves.
What chance would each player have of winning a service match?
Explore.
In the 2010 Wimbledon Championships, Isner and Mahut played the longest match in tennis history: the match went on for three days and finished with a score of 70-68! (You can read about it in the Plus article here) After the match, Isner said that a match like this will never happen again.
I wonder if Isner was correct in this statement. The famous Cambridge mathematician Tim Gowers thought about this question on his blog.
Getting Started
Consider the difference between games which hit 3 points each and those which don't.
Note that some sequences of games occur with the same probability as others (e.g. LWWWW and WLWWW).
For the numerical parts, you will need to decide the accuracy to which you wish to compute the answers. You might like to verify numerically any answer that you make to the first part of the question.
Student Solutions
Steve writes
Once I had understood the scoring systems I realised that I needed to look at the chances of winning based on whether I reached deuce or not. Let $V$ be the event of victory in the the overall game and $D$ the event of reaching deuce ($3$ points all) and $P_0, P_1$ and$P_2$ the probability of winning the game whilst losing $0$, $1$ and $2$ points respectively.
$$P(V) = P_0+P_1+P_2+ P(V|D)P(D)$$
$P_0, P_1$ and$P_2$ correspond to games of $4$, $5$ and $6$ points respectively. Each of these must end in a winning point (denoted W). The number of such games equals the number of ways of distributing $0$, $1$ and $2$ lost points (denoted L) amongst $3$, $4$ and $5$ points. The game reaches deuce if $6$ points are played with $3$ $W$s and $3$ $L$s. The number of such games equals the number of ways of distributing $3$ $L$s into $6$ points. Since each point is won or lost independently of the other points the probabilities multiply. Therefore,
$$P(V) = (0.6)^4\cdot 1 + (0.6)^4(0.4)\cdot 4 + (0.6)^4(0.4)^2\cdot \frac{5\cdot 4}{2}+P(V|D)\cdot(0.6)^3(0.4)^3\cdot\frac{6\cdot 5\cdot 4}{3\cdot 2}$$
I now need to use conditional probability to work out the chance of winning given that I reach deuce
The red dots on this tree indicate branches where the probability of winning equals the probability of winning given that I am at deuce $P(V|D)$ and the blue dot is a victory $V$.
$$P(V|D) = (0.6)^2 + (0.6)(0.4)P(V|D)+(0.4)(0.6)P(V|D)$$
which can be rearranged to
$$P(V|D) = \frac{(0.6)^2}{1-2(0.6)(0.4)}$$
Putting all of this together yields a probability of victory of
$$P(V) =(0.6)^4\left[ 1+ 4(0.4)+10(0.4)^2\right] +\frac{20(0.6)^5(0.4)^3}{1-2(0.6)(0.4)}$$
This equals $0.736$ (3dp). I left the probabilities in so that I could check the calculation by replacing the $0.6$ and $0.4$ by $0.5$. In this case, I get $0.5$, which makes sense (even players have an even chance of winning in this set of game rules). I also get a guaranteed victory by changing the $0.6$ to $1$ and setting the $0.4$ to $0$, which also makes sense (you will win if you win every point!)
--------------
For the second part of the problem I used a spreadsheet. It was inefficient, but displayed the probabilities clearly. After a bit of thought it is clear that the $P(V|D)$ depends on whether deuce was reached on a $W$ or on a $L$. I'll need two new events $D_W$ and $D_L$ for these cases.
We now have
$$P(V)= P(V|\bar{D})P(\bar{D})+P(V|D_L)P(D_L)+P(V|D_W)P(D_W)$$
There is a fun pair of linked trees for the $P(V|D)$ - in the diagram the red circles represent $P(V|D_W)$ and the yellow circles represent $P(V|D_L)$. Reading off the probabilities gives a pair of simultaneous equations:
$$P(V|D_W) = (0.8)^2+(0.8)(0.2)P(V|D_L)+(0.2)(0.4)P(V|D_W)$$
$$P(V|D_L)=(0.4)(0.8)+(0.4)(0.2)P(V|D_L)+(0.6)(0.4)P(V|D_W)$$
This gives $P(V|D_W) = 0.792$ (exact value $\frac{80}{101}$) and $P(V|D_L) = 0.554$ (exact value $\frac{56}{101}$)
Using this piece of conditional probability to simplify the numerical analysis I then used a spreadsheet (Tennis.xls ) to calculate that
$$P(D_L) = 0.0755, \quad P(D_W) = 0.0947$$
Using this I calculate the chance of victory as
$$P(V) = 0.7364$$
For Bryony the numbers are
$$\begin{eqnarray}P(V|D_W) = 0.753\\P_(V|D_L)= 0.423\\P(D_L) = 0.0857\\P(D_W)= 0.0646\\P(V) =0.7383\end{eqnarray}$$
So, the players are very evenly matched.
You can check the results on this spreadsheet: Tennis.xls
Teachers' Resources
Why do this problem?
This is an interesting and complicated scenario which will require clear probabilistic thinking. There will be various different ways in which the first part of the problem can be attempted, and it would be informative to compare and contrast these. It is quite a difficult problem and could be set to keen students individually or discussed collectively. It would be good as revision of a module on probability or to motivate the advanced study of probability.
Possible approach
- There are different sorts of winning sequences, such as WWWLW, WWLLW.
- Each winning sequence ends with a W.
- There are arbitrarily long winning sequences WWWLLL(WL)....(WL)WW.
- Not all winning sequences are of the same probability.
Key questions
Possible extension
Possible support
Playing Squash
The rules of squash say that if the position is reached when the score is $(8,8)$ (we give A's score first), and $B$ is to serve, then $A$ must choose between the game ending when the first player reaches $9$ or when the first player reaches $10$. We want to decide what is the best choice for $A$ to make (when $p$ and $q$ are known).
If $A$ chooses $9$, then he wins the game with probability $\varphi$. Suppose now that $A$ chooses $10$; then the sequence of possible scores and their associated probabilities are as follows:
Sequence of scores | Probability | Probability in terms of $p,q$ |
---|---|---|
$(8,8) \to (9,8) \to (10,8)$ | $\varphi \times \theta$ | $p^3/(1-pq)^2$ |
$(8,8)\to (8,9)\to (9,9) \to (10,9)$ | $\mu\times\varphi\times\theta$ | $p^3q/(1-pq)^3$ |
$(8,8) \to (9,8) \to (9,9) \to (10,9)$ | $\varphi \times \lambda\times \varphi$ | $p^4q^2/(1-pq)^3$ |
Angle of shot
Problem
When a projectile is fired, it travels along a parabola (in the absence of wind and air resistance).
Part 1: A shot putter will release the shot from arms length. Estimate the optimal angle that the shot should be released from to make it travel furthest, assuming the the shot putter can launch the shot at the same speed from any angle. (Note: the shot is launched from around head height rather than ground level)
Part 2: In reality it is not possible to launch the shot at the same speed from any angle: the body is naturally able to put more power into certain angles. Linthorne (2001) constructed a mathematical model in which the velocity is related to the projection angle as follows (Linthorne has written about the model on the Brunel University site; the published reference is given at the foot of the problem)
$$v= \sqrt{\frac{2(F-a\theta)l}{m}}$$
where $F$ is the force (in newtons) exerted on the shot for a horizontal release angle, $a$ is a constant that characterizes the rate of force decrease with increasing release angle, $l$ is the acceleration path length (in metres) of the shot during the delivery and $m$ is the mass of the shot $\left(7.26 \mathrm{kg}\right)$. A typical set of values for these parameters might be
$F=450\mathrm{N}$, $a=3\mathrm{N/degree}$, $l=1.65\mathrm{m}$ and $m=7.26\mathrm{kg}$.
Determine approximately the angle the shot putter should choose, to maximise the length of the shot put.
You may wish to do a little research and study some video footage or stills showing shot putters releasing the shot, to see how close your theoretical optimum angle is to the angle of release used in practice.
The published reference for the paper is Linthorne, N. P. (2001). Optimum release angle in the shot put. Journal of Sports Sciences, 19, 359-372.
Getting Started
Use a spreadsheet to work out different launch velocities, and hence range, for each angle.
Student Solutions
Assuming for the time being that the athlete can launch the shot at the same speed at any angle, I plotted a graph of range against projection angle for a variety of speeds:
Looking at the graph, it appears the optimal angle for maximising range is slightly smaller than $45^{\circ}$.
We now need to incorporate how the maximum launch speed varies with the projection angle. Here's a graph of projection velocity against projection angle. Although the projection velocity decreases as the projection angle increases, the range previously had a maximum at $45^{\circ}$, so the optimal angle will now be a tradeoff between angle and velocity.
We can now calculate the range of the shot using this projection velocity, and plot a graph of range against projection angle (using the same value of the parameters as before):
For these values of the parameters,we can see the optimal angle is now around $35^{\circ}$, as opposed to $45^{\circ}$ in the previous model.
Solving this problem using differentiation is quite messy, but well done to Joshua who had a go; you can see what he did here.
Teachers' Resources
Why do this problem?
This problem gives students the opportunity to investigate projectile motion in a real-life context.
Possible approach
Pose the first question, to work out the optimum angle of release for the shot, bearing in mind that the shot is released from above ground level - perhaps discuss and agree as a group what height to assume the release takes place.
Once students have tackled this first question, ask them to identify the different assumptions they have made. The key assumption is that equal force can be applied at all release angles. Invite students to speculate what variables affect the force (and hence velocity) with which the shot can be launched.
Introduce the equation for release velocity varying with angle, and then invite students to calculate the range for different launch angles. This is best done numerically, as a spreadsheet task.
Key questions
What are our modelling assumptions?
Are our assumptions valid?
Possible extension
Students may be interested in reading the article Modelling Assumptions in Mechanics.
Possible support
Cannon Balls could provide a good introduction to thinking about the mechanics of motion under gravity.
10 Olympic Starters
Problem
Consider some of these questions concerning the mechanics of sport. You might need more data in some cases or need to make an approximation to allow for mathematical modelling. You might be able to give precise answers or answers bounded by some reasonable error range. Be as precise as you can in your assumptions so as to convince yourself or others of the answers.
1. What if a long jumper could launch him or her self from the platform at 45 degrees with the same speed as at their standard launch angle? How far would they jump?
2. In pistol and rifle events, competitors aim at a 10-ringed target from the set distances of 10m, 25m and 50m. Do you think that marksmen need to alter their angle of aim by a measurable amount between these targets?
3. Imagine that a wind of speed 1ms$^{-1}$ is blowing parallel to the straight parts of the athletics track. Do you think that this would help or hinder a 400m sprinter?
4. What if a shot-putter could launch the shot at an angle of 45 degrees at the same speed as their usual launch angle?
5. At what speed does a pole-vaulter hit the crash mat?
6. In football, a penalty is taken 12 yards away from the goal. How good do the goalkeeper's reactions have to be?
7. A basketball free throw is taken 4.6m from the hoop. The hoop is 45.7cm in diameter, and 3.05m high. The basketball is 24cm in diameter. How precise does a player's shot have to be to ensure the ball goes in the hoop?
8. A trampolinist can jump to a height of 10m. They perform a double somersault. How quickly must they be able to rotate in order to land safely on their feet and not on their head?
9. A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.90m. At the top of the swing his speed is momentarily 0ms$^{-1}$. Calculate his speed at the bottom of the swing.
10. Assuming the ball travels at a constant speed throughout, how much longer does a tennis serve to the edge of the court take to reach the baseline than a serve 'down the T'?
Student Solutions
Many of these problems involve an object moving under the influence of gravity and friction forces (such as air resistance). A first step in modelling is usually to neglect friction. In this case, the acceleration downwards due to gravity is about $9.8\mathrm{ms}^{-1}$. This leads to the simplifying modelling assumption that the downwards motion of a projectile is the reverse of the upwards motion. This can be used to calculate, for example, the speed of lift off by working out how fast an object will be landing when it hits the ground falling from a specified height (such as a pole-vault bar).
One of our summer students came up with these numbers:
1) Ignoring air resistance, if a long jumper could launch themselves at $45^{\circ}$ at a speed of $10.0\mathrm{ms}^{-1}$, they'd travel $\frac{10^2}{g}\mathrm{m}$ = $10.2\mathrm{m}$. (See here for a derivation). In reality, long jumpers would probably struggle to convert enough of their horizontal speed into the required vertical speed to travel on this trajectory. There's an interesting article here discussing how good a long jumper Usain Bolt could be.
2) Assume the gun is fired from exactly the same height as the target (though any height difference could be incorporated fairly simply into the model), and the bullet travels at a constant speed of $v= 400\mathrm{ms}^{-1}$. Suppose the bullet is fired at an angle $\alpha$ to the horizontal. The coordinates of the bullet are $$\left(v\cos(\alpha)t, v\sin(\alpha)t - \frac{g}{2}t^2\right)\;.$$ At time $t=T$ say, the bullet hits the target. If L is the distance to the target from the shooter, we find $\sin(2\alpha) = \frac{Lg}{v^2}$ (using $\sin(\alpha)=2\cos(\alpha)\sin(\alpha)$.) This gives the following values of $\alpha$ for the different distances:
L (m) | $\mathbf{\alpha (^{\circ})}$ |
10 | 0.035 |
25 | 0.088 |
50 | 0.18 |
These seem to be very small numbers. However, if we included air resistance, the bullet would slow down during the motion and consequently hit the target lower than predicted here, so in reality there might need to be a noticably different aim for the different distances.
3) Suppose the athlete is only affected by the wind on the back and home straights, which are around $85\mathrm{m}$ in length, and they run exactly $1\mathrm{ms}^{-1}$ faster with the wind and $1\mathrm{ms}^{-1}$ slower when running against the wind. If the athelte's time in still conditions is $50\mathrm{s}$, then their average speed during the race is $8\mathrm{ms}^{-1}$. Their wind affected time using these assumptions would be:
$$\frac{(400 - 2\times85)\mathrm{m}}{8\mathrm{ms}^{-1}} + \frac{85}{8+1}\mathrm{s} + \frac{85}{8-1}\mathrm{s} = 50.3\mathrm{s}$$
i.e. their time is predicted to be $0.3\mathrm{s}$ slower.
We could incorporate other factors into the model. Air resistance, for example, is proportional to the speed squared, so could affect the time significantly. The wind would also affect the athlete around the bends of the track; here the wind strength would vary continuously. We've assumed the wind would affect the speeds along the straights symmetrically, in general this is likely to not be the case, as the athelete's performance may be reduced more than expected running into the wind.
4) A launch speed of $13\mathrm{ms}^{-1}$ seems a reasonable approximation. Neglecting air resistance, a throw with the velocity angled at $45^{\circ}$ to the ground would travel $\frac{13^2}{9.8}\mathrm{m} = 17.2\mathrm{m}$. The speed at which a shot putter can launch the shot depends on the launch angle. For more information, see here.
5) Suppose the athlete has just equalled the world record of $6.14\mathrm{m}$. Assume the athlete has zero vertical speed at the top of the jump. Their speed at the bottom of the descent is therefore $\sqrt{0^2+2\times9.8\times6.14}\mathrm{ms}^{-1} = 11.0\mathrm{ms}^{-1}.$
6) A footballer can kick the ball at a speed of around $75\mathrm{mph}$. Assuming the football remains at this speed, it'll take $11\mathrm{m}\times\frac{1}{75\times\frac{1600}{3600}\mathrm{ms}^{-1}} = 0.33\mathrm{s}$ to reach the goal. The goalkeeper needs to have moved into position and be ready to save it in this time!
7) Assume for the time being, the vertical velocity and trajectory result in the ball landing in the centre of the hoop looking from above. In the horizontal plane, the athlete can shoot at an angle $\alpha$ off the centre, where $\alpha = \tan^{-1}\left(\frac{0.23}{4.6}\right) = 2.8^{\circ}$. Can you work out any restrictions on the angle to the horizontal of the initial velocity?
8) Using $s=ut+\frac{1}{2}at^2$ shows that the time (for a point particle) to fall from $10\mathrm{m}$ is $\sqrt{2\times 10/9.8}\mathrm{s} = 1.43\mathrm{s}$. Thus, the time taken to travel to a height of $10\mathrm{m}$ and back is $2\times1.43\mathrm{s} = 2.86\mathrm{s}$. If they need to rotate $720^{\circ}$ in that time, they need to rotate at $252^{\circ}\mathrm{s}^{-1}$.
9) If we assume all his mass is $0.9\mathrm{m}$ from the bar, then conservation of energy gives: $mg(2\times0.9) = \frac{1}{2}mv^2$, so $v \approx 6\mathrm{ms}^{-1}$.
10) A 'singles' tennis court measures $8.3\mathrm{m}$ by $23.8\mathrm{m}$. Suppose the player can serve at $50\mathrm{ms}^{-1}$. The straight serve takes $\frac{23.8}{50}\mathrm{s} = 0.476\mathrm{s}$ to get to the baseline. The diagonal serve travels $\sqrt{4.15^2 + 23.8^2}\mathrm{m} = 24.2\mathrm{m}$, so takes $0.49\mathrm{s}$.
Teachers' Resources
Why do this problem ?
Mathematical modelling is an important skill and mechanics courses are often the first places in which modelling can be explored. These interesting questions will allow you to practice these skills whilst developing awareness of the key concepts of mechanics.
Possible approach
There are several parts to this question. The individual pieces could be used as starters or filler activities for students who finish classwork early. Enthusiastic students might work through them in their own time. Since there is no absolutely 'correct' answer to many of these questions, they might productively be used for discussion: students create their own answers and then explain them to the rest of the class. Does the class agree? Disagree? Is there an obvious best 'collective' answer?
Key questions
- What assumptions will you need to make in this question?
- What equations of mechanics will you need?
- How accurate do you think you answer is?
- What order of magnitude checks could you make to test that your answer is sensible?
Possible extension
Can students make up similar questions? Can they put any upper or lower bounds on the numbers?
Possible support
Students might struggle with the 'open' nature of the questions. To begin, they might like to read the Student Guide to Getting Started with rich tasks
Squash
Problem
In the game of squash, players gain "points" by winning "rallies". If the "server" wins a rally, he or she wins a point; otherwise the service changes hands with no points gained.
Normally the game is won by the first player to reach 9 points -- typically by 2 or more points. But if the score reaches 8-8 then the person due to receive serve can call "9" (in which case the first to reach 9 wins) or call "10" (in which case the first to reach 10 wins).
I'm playing a game against Ivana Slogovitch, the Russian squash champion. I estimate that my chance of winning any particular rally against her, regardless of whether I serve or not, is p. The score gets to 8-8 and I am due to receive serve. Should I call "9" or "10"?
Getting Started
Student Solutions
In the game of squash, the player who serves is the only player who can win the next point. If the server loses the rally, no points are scored and the other player serves next. Note that if in two consecutive rallies both servers lose the rally, then the situation is exactly the same as it was before the two serves. This is not taken into account below where the tree diagram and solution from Allan Ling apply to the simpler problem where the player who wins a rally wins the point whether serving or receiving. Can you see how to re-draw the tree diagram to give the solution for the game of squash?
If I call '9', then I win with probability p.
If I call '10', then there are 6 possible outcomes, shown by the tree diagram.
The winning outcomes are shown in red. Calculating the probabilities of these three outcomes, we find that the total is p 2 + p(1-p)p + (1-p)p 2 = 3p 2 - 2p 3 .
Now we compare this to p to see which is larger.
When p< 0.5, then p> 3p 2 -2p 3 .
When p=0.5, then p=3p 2 -2p 3 .
When p> 0.5, then p< 3p 2 -2p 3 .
Therefore, in this simplified situation, if p is smaller than 0.5, I should call '9', and if p is greater than 0.5, I should call '10'. If p=0.5, I have an equal chance of winning on both choices.
Now what happens playing according to the rules of squash?
Teachers' Resources
Why do this problem?
A real world problem involving conditional probability.Possible approach
Half the class might work through the problem with a probability of 0.4 of winning rallies against Ivana to see whether calling 9 or 10 isadvantageous, and the other half might work on the problem with a probability of 0.3. Then a class discussion, and drawing tree diagrams of the possibilites, might help everyone to solve the problem for a general value of p and find the critical value.Key question
What are the possible outcomes of the next rally, which is served by Ivana, if my probability of winning the rally is p?What happens if Ivana loses that rally?
See Also
There is a detailed explanation of the solution to this problem in this article.The Olympic LOGO
Problem
You can do this theoretically, use logo or even try to do it in a limited number of characters making use of the twilgo http://twilgo.com environment. Using this small number of characters is a challenge!
LOGO is a simple programming language which instructs a pointer (or 'turtle') to move on a flat surface. Mathematicians often become very skilled computer programmers and learning how to create beautiful, efficient algorithms is a part of this.
Student Solutions
If you go to twilgo then
RT 90 REPEAT 3 [REPEAT 360 [FD 1 RT 1] PU FD 125 PD] PU RT 139 FD 245 RT 41 PD REPEAT 2 [REPEAT 360 [FD 1 RT 1] PU FD 130 PD]
will approximately produce the desired image. Perhaps you can produce a 'better' algorithm to display this? If so, let us know!
For trademarking reasons we are unable to display this image, so why not try this for yourself?
FA Cup
Problem
In the FA Cup 64 teams play a knockout tournament. Over the four years 2001, 2002, 2003 and 2004 Arsenal have been drawn against Chelsea each year and have beaten Chelsea every time. What is the probability of that happening?
Let's say that whenever Arsenal and Chelsea play the probability of Arsenal winning is 0.6 and otherwise, throughout the tournament, both these teams have a probability of winning of 0.7 in the first round, 0.6 in the second round and 0.5 in the subsequent rounds.
Getting Started
You'll need to work out, for each round, the probability that Arsenal play Chelsea and also the probability that they do not play against each other but both survive to play in the following round. A tree diagram is useful in thinking through this problem.
Student Solutions
Well done Roy M. for the following excellent solution.
To calculate the probability of Arsenal and Chelsea playing each other in the FA cup in four consecutive years and Arsenal winning each time we have to make reasonable assumptions about the probabilities of these teams winning their games. So we assume that whenever Arsenal and Chelsea play the probability of Arsenal winning is $0.6$ and otherwise, throughout the tournament, both these teams have a probability of winning of $0.7$ in the first round, $0.6$ in the second round and $0.5$ in the subsequent rounds.
To go about this problem I decided to find the chance of Arsenal playing and beating Chelsea round by round.
Round One: The chance of Arsenal being drawn against Chelsea and beating them is: $(1/63)\times(3/5)= 1/105$.
Round Two: The chance of Arsenal not being drawn against Chelsea in round one, and both teams winning their round one matches, multiplied by the chance of them playing each other in round two and Arsenal winning is: $(62/63)\times(7/10)\times(7/10)\times(1/31)\times(3/5)= 7/750$
Continuing in this fashion, the probability of Arsenal beating Chelsea in any given round is equal to the product of the chances of the two teams not being drawn against each other in any former round, times the chances of the two teams beating any team they were drawn against in all former rounds, times the chances of Arsenal being drawn against Chelsea in the given round, times the chance of Arsenal winning.
The odds for Arsenal beating Chelsea in rounds 1 to 6 (6 being the final) is as follows:
Round | Probability A plays C and wins | $\quad$ Probability A, C don't meet, both win $\quad$ |
1 |
$\frac{1}{63}.\frac{3}{5}=\frac{1}{105}$
|
$\frac{62}{63}.{\frac{7}{10}}^2$
|
2 |
$\frac{1}{31}.\frac{62}{63}.{\frac{7}{10}}^2.\frac{3}{5}=\frac{7}{750}$
|
$\frac{30}{31}.\frac{62}{63}.{\frac{7}{10}}^2.{\frac{3}{5}}^2$
|
3 |
$\frac{1}{15}.\frac{30}{31}\frac{62}{63}.{\frac{7}{10}}^2.{\frac{3}{5}}^2.\frac{3}{5}$
|
$\frac{14}{15}.\frac{30}{31}.\frac{62}{63}.{\frac{7}{10}}^2.{\frac{3}{5}}^2.{\frac{1}{2}}^2$
|
4 |
$\frac{1}{7}.\frac{14}{15}.\frac{30}{31}.\frac{62}{63}.{\frac{7}{10}}^2{\frac{3}{5}}^2.{\frac{1}{2}}^2.\frac{3}{5}=\frac{21}{6250}$
|
$\frac{6}{7}.\frac{14}{15}.\frac{30}{31}.\frac{62}{63}.{\frac{7}{10}}^2.{\frac{3}{5}}^2.{\frac{1}{2}}^4$
|
5 |
$\frac{1}{3}.\frac{6}{7}.\frac{14}{15}.\frac{30}{31}.\frac{62}{63}.{\frac{7}{10}}^2.{\frac{3}{5}}^2.{\frac{1}{2}}^4.\frac{3}{5}=\frac{21}{12500}$
|
$\frac{2}{3}.\frac{6}{7}.\frac{14}{15}.\frac{30}{31}.\frac{62}{63}.{\frac{7}{10}}^2.{\frac{3}{5}}^2.{\frac{1}{2}}^6$
|
6 |
$\quad \frac{2}{3}.\frac{6}{7}.\frac{14}{15}.\frac{30}{31}.\frac{62}{63}.{\frac{7}{10}}^2.{\frac{3}{5}}^2.{\frac{1}{2}}^6.\frac{3}{5}=\frac{21}{25000} \quad$
|
Adding all these probabilities together (i.e. finding the chances of one of them happening) you get: $${1101\over 35000}$$ which is roughly $3.15$ per cent. So for the odds given, there is a $3.15$ per cent chance of Arsenal playing and beating Chelsea in the FA cup.
However the question asks for the probability of it happening four years in a row. So you must multiply this result by itself four times $$\left({1101\over 35000}\right)^4=9.792 \times 10^{-7}$$ (giving the result to 4 s.f.) which gives roughly 1 in a million chance of it happening or a probability of $0.000098$ per cent.
So for a bet of £1 you could be laughing all the way to the bank with a sum of just over £1m
Teachers' Resources
This requires an understanding of the simple rules of probability and a lot of calculating with fractions.