2016 gets a bad rap!

2016 is sitting slouched in its chair like a tired and stroppy Yr 11 student.  “I know that Brexit and Trump were my fault, but you can’t blame me for David Bowie or Prince.  Or George Michael.”  And then 2016 shouts: “Everyone’s got it in for me.”

For example:

Dear 2016,

It's been a tough year, please don't take Carrie Fisher.

Sincerely,

Everybody.

— Darth Vader (@DepressedDarth) December 23, 2016

Is it fair to have a go at 2016? 

The people whose deaths have made it into the national or international media all have one thing in common: they are famous.  It does feel as if there were more deaths of famous people during 2016 than in previous years, so what is going on?  Well maybe there are just more famous people who are getting old round about now.

Some assumptions:

• We have noted the deaths of celebrities
• The median age nowadays is about 75
• People become famous in their mid-twenties

Using these assumptions, many of the celebrities will have come to prominence in the mid-1960s.  Let’s compare 1966 with 1956.

How can we find how many famous people there were in each year?  One proxy we can use is the number of movies that were released.

According to the Internet Movie Database, there were 10,460 movies released during 1956.  In 1966 there were 18,053.

This is a massive increase, presumably fuelled by the rise of the teenager, greater disposable income, a rise in population.  This will have required an increase in the number of movies, and also more actors and therefore more celebrities. 

If this has happened in movies, then presumably it also happened in popular music and on TV too.  Now some of the celebrities who have died during 2016 were not active during the 1960s, but many of them were and the rise in population after the second world war and the increase of celebrity since the 1960s are likely to have affected this.

And things are just going to get worse.  The number of celebrities in the world is increasing at a massive rate.  Think of the number of reality-TV show celebrities we see and read about each year.  In 50-odd years time, when the X-factor/Bake-off/Apprentice/Big Brother/etc contestants start to get old, 2016 will look like a quieter, more innocent time.

Maybe the reason we have seen so many more celebrity deaths in 2016 is that there are more celebrities nowadays and more of those celebrities are getting older.

So let’s stop blaming 2016; it really isn’t its fault (but hope you won’t be teaching his younger brother, 2017).

Refs:

http://www.imdb.com/search/title?release_date=1956

http://www.imdb.com/search/title?release_date=1966

Tokyo 2020 Logo

Olympic logos are often controversial. Lots of people didn’t like the London 2012 logo, for example. I thought the Rio 2016 one was great, though.

The controversy over the Tokyo 2020 logo was originally about plagiarism. The first logo that was chosen for the Olympics was subsequently withdrawn after it was found to be rather similar to that of the Théâtre de Liège. 

This is the new logo:

I find it a little unsettling. 

I think I have now worked out why. It feels as if it ought to have reflection symmetry. I think I start looking at the top of the logo, where my brain tells me it is symmetrical (roughly). 

As you go down the image this symmetry breaks down, and that is what makes it slightly uncomfortable to look at. 

There is, however, some symmetry in the logo. Can pupils spot it? 

It has rotational symmetry order 3. The red dots show three related points. 

Even though I now know about the symmetry I still find the Olympic logo ‘swims’ in front of my eyes. 

The Paralympic logo though; that works far better for me: 

Aah - nice to have reflection symmetry!

Sources: 

https://tokyo2020.jp/en/games/emblem/ 

https://www.rio2016.com/ 

Predictions are not necessarily bad if they don't turn out to be true

This was in The Guardian:

 Oh dear!  This again.  I have no drum to bang for pollsters and they may well have messed things up completely, but this paragraph is fairly hopeless. 

If we have a prediction, with a probability attached, it is natural to look at the biggest probability and to make use of this.  If there is a “15% chance that Donald Trump will win” then this is deemed to be unlikely.  But unlikely isn’t impossible.  By saying that the statement was ‘wrong’ we would also be saying that if a total of 6 comes up when you roll two dice then you must have cheated!

This crops up over and over.  An example is the (in-)famous prediction by the Met Office in 2009 of a ‘barbecue summer’.  Here is how the Daily Mail website refers to it:

The April statement says sun is “odds on” and that it is “likely” to be warmer.  That it didn’t happen doesn’t make the prediction necessarily “wrong”, just that a less likely scenario prevailed this time around.  (I am not sure it makes sense to make such a high profile prediction where you think it is only 2/3 likely to happen, though.)  

To be clear, I am not saying that either of these predictions are definitely “right”, but just that a scenario that is deemed to be unlikely can occur without the probabilities being wrong.

FFT

As teachers we also see this with FFT grades.  These grades used to be presented as percentages.  They would show that, for a pupil with a similar background (eg KS2 results, etc) a certain proportion of the pupils went on to get an A*, A, B, C, D, etc grade in each GCSE subject.  

Nowadays only the median version is given, and this is shown as _the_ FFT grade for the pupil.  In the new system these are given to the nearest third of a grade. 

If a  pupil has B+ as their FFT grade what this really means is that about half of children with a similar background to that pupil will get a high grade B or above, and half will get a high grade B or below.  It might well be the case that a small fraction actually get a high grade B.  If that pupil doesn’t get a high grade B that doesn’t mean that over- or under-achieved, or that their FFT grade was ‘wrong’, just that it didn’t happen. 

[The initial article also forms part of a Quibans.  Find it here.]

Sources:

https://www.theguardian.com/commentisfree/2016/nov/09/polls-wrong-donald-trump-election

http://www.dailymail.co.uk/news/article-1202982/Met-Office-left-red-faced-Britains-forecast-barbecue-summer-turns-washout.html

Aliens vs Barca

I usually enjoy the weekly puzzle on the Weekly Newsletter from Chris Smith (find him on Twitter @aap03102 – he will happily add you to his list of subscribers).

The puzzle in the most recent edition was a corker:

I loved this because there are so many ways of doing it.  I won’t quibble over whether two ways really are ‘different’ or not and will list as many as I and various students have come up with.  (Yr 11 thoroughly enjoyed it too!)

In all diagrams CN is labelled as a.  All explanations are sketched rather than presented as full proofs.

1) Similar Triangles

Triangles SNM and CNS are similar (look at the angles).  This means that a/180 = 180/135

Rearranging gives a = 240 and the diameter is 375.

2) Pythagoras 1

In triangle SNM the hypotenuse is 225. 

Angle MSC is a right angle (angle in a semi-circle), so we have two further right-angled triangles: CNS and MSC.  We can use Pythagoras in both of them:

(a+135)² = b² + 225²

b² = a² + 180²

Eliminate b² and solve to get a=240.

A brief excursus – 3,4,5 triangles.

The numbers that Chris chose had some meaning (this puzzle appeared in newsletter number 375 and the answer is … 375).  It is also nice that they are numbers I recognise (usually in the form of angles).  135 = 3x45 and 180 = 4x45, which means we have a 3,4,5 triangle, so the hypotenuse is 5x45 = 225.  In my handwritten notes for these solutions I made the arithmetic easier by using e = 45 and calling the sides 3e, 4e and 5e.

3) Pythag 2

Draw in a radius (labelled r):

Do Pythagoras on triangle ONS and work out r.  Then double it.

4) Pythag 3

This is really the same as version 3 but with different labels.

Use Pythagoras on the triangle ONS

5) Intersecting Chord Theorem

This is related to version 1.

The intersecting chord theorem states that CN x NM = SN x NU

This gives 135a = 180 x 180

6) Area of a triangle

Looking at the big triangle, the area is ½ (a+135) x 180

If we treat CS as the base then it is also ½ b x 225.

Equate these and then use Pythagoras on CNS (a^2 + 180^2 = b^2) as a second equation and solve for a.

7) More similar triangles – using cos.

OT is the perpendicular bisector of SM.

We know SM = 225, so TM is half of that, 112.5.  Cos(M) = 135/225 (from triangle SNM).  But in triangle OTM we have cos(M) = 112.5/r

8) Trig 1

Work out the size of angle α (inverse tan of 180/135).

Angle OSM = α (isosceles triangle)

Angle NSM = 90 – α
Angle OSN = α – (90 – α)

So angle OSN = 2α – 90.

Use this angle, trigonometry and the side SN to work out either ON or OS.

9) Trig 2

We know what angle α is (using trig in triangle SNM) and we know that SM is 225 (Pythag).  We also know that angle CSM is a right angle (angle in a semi-circle).  Use trig in that triangle to get the diameter.

10) Trig 3

We could also find the length of CS using trig (as in version 9) and can then use Pythagoras to find CM.

11) Similar Triangles again

Triangle SNM is similar to triangle CSM. 

225/135 = diameter / 225.

How many different bits of maths did we use?

Many of these methods overlap, but it is rather nice that there are so many different aspects of maths that are involved.  It was also good (with Yr 11) to be able to focus on methods and explanation, rather than on finding the answer.

I think this can involve:

• Circle theorems and other angle rules
• Pythagoras
• Trigonometry
• Area of a triangle
• Similar triangles

Many thanks to Chris for a great problem!

Danny Baker's Sausage Sandwich Game

On Radio 5Live on a Saturday morning the Danny Baker Show features the Sausage Sandwich Game.  There are two contestants and a sporting celebrity involved.  Three questions are asked (unless there is a tie in which case additional questions are used to break the deadlock) and are the sorts of things that only the celebrity is likely to know.

The third question ("the question that gives the game its name") asks whether the celebrity would prefer to have on their sausage sandwich "red sauce, brown sauce, or no sauce at all".  Disappointingly there are no stats available for this, although Danny and Lynsey (his co-host) were incredibly surprised a few weeks ago when 'no sauce at all' was the winner.

This week (30 July 2016) Audley Harrison (Olympic boxing gold-medallist) was the guest.  The second question asked how many letters there were in the number of the house he lived in at the age of 10 or 12.  Danny then gave an example (so house number eight would have 5 letters, and house number thirteen would have 8 letters).  The contestants had to choose from: 3-6 letters, 7-9 letters, or more than that.

So: what do you choose?
My thoughts are below.



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Here are my thoughts:

It's complicated!

The first twelve numbers all fall into the first group (3-6 letters).
But then thereafter there are very few others that have only 3-6 letters.  Multiples of 10 up to and including ninety (but not seventy) are the only others that fit.  This means that after a strong start (12 out of 12) this turns into 13 out of 20, 14 out of 30 and 19 out of 100.
There won't be any after that ("one hundred" has 10 letters and even if you get as far as 10^100 then I I think you would say it as "a googol", which is 7 letters).

The 'teens' are then all 7-9 letters.  After that, for the 2-digit numbers up to 100 (but not in the seventies), the numbers ending in one, two and six are in 7-9, while the rest are 10-or-more.

Forty-four is the point at which 7-9 overtakes 3-6 letters.
10-or-more overtakes 3-6 at fifty-seven.
The lead changes between 7-9 and 10-or-more a couple of times, but from ninety-four onwards it is all 10-or-more.  (Everything after that is 10-or-more.)

So what do we choose?   Almost every road will have at least 12 houses in it.  But many roads have more than that.  Not many have over 100 though.

Knowing a bit about the background to the person might help.  Danny must have been fairly confident that Audley didn't live in a house with a name, and knowing that Audley grew up in London will presumably help to tell us that there are more likely to be higher numbers in the streets, but realistically not higher than a couple of hundred.  (In the USA they seem to skip numbers with abandon, so house numbers of 2505 are not uncommon - we don't do that in England.)

FWIW Audley grew up in house twenty-seven, which gave 11 letters.

Singapore Mental Math (7) – Finding remainders when dividing by …

[Part of the mental math series…]  Here are the strategies for Week 17, 18 & 19 in Grade 6.

Good things:

• I usually see these as divisibility rules, and while it is obvious that the divisibility rules will give the remainder (perhaps with some tweaking), this isn’t commonly explored (in my classroom at any rate), so I found these interesting.
• There is some mental maths to do.
• There is proper use of vocabulary.

Bad things:

• These are incredibly restricted and over-specify the conditions. 
• There is a missed opportunity to look at why they work.
• Some of them are expressed very badly.

Let’s look at each in turn:

Dividing by 5

I hope many pupils would spot that there is a far easier way to deal with the Note!  Rather than dividing the last two digits by 5, why not just instruct pupils to look at the final digit.  In fact, pupils might come up with an easier way to do this.  Let’s ask them to explore how we can tell what the remainder is when we divide by 5.  They could try a few out for themselves, could have an idea and might then link it to the rule for divisibility by 5.

If an integers ends in 0 or 5 then it is divisible by 5.  If it ends in 1 or 6 then the remainder is 1, if it ends in 2 then …

Dividing by 8

This only helps if you have a number with 4 digits or more. 

Why does it work?  1000 is divisible by 8, so any 1000s will automatically be divisible by 8.  Given that we only want the remainder we could also get rid of anything that is clearly a multiple of 8.  To find the remainder when 4169 is divided by 8 we can ignore the thousands, leaving 169.  160 is divisible by 8, so we can ignore that, leaving 9.  The remainder is therefore 1.

(It is a little frustrating that in all of the questions on these three strategies none of the numbers are actually divisible by the divisor.  It would be nice to have a remainder of 0 once in a while.)

Dividing by 9

Many pupils will recognise the algorithm for determining whether a number is a multiple of 9. 

This is phrased rather oddly in that it is only used here for 4-digit numbers (all of the questions are of that form), when it works for any integer.  It also doesn’t explain what happens if the single digit is 9.

Can pupils explain why this algorithm doesn’t produce zero (unless you start with the number 0)?

Can they explain why this algorithm works?  Here is an algebraic way to show it.

If a number looks like “abcd” then the algebra is actually 1000a + 100b + 10c + d.

This can be rewritten as 999a + a + 99b + b + 9c + c + d

Rearrange this to give (999a + 99b + 9c) + (a + b + c + d).

The first bracket is clearly divisible by 9, so we only need to worry about the second one, which is the sum of the digits.

Singapore Mental Math (6) – multiplying mixed numbers with identical fractions when the numerator is 1

[Part of the mental math series…]  This is the strategy for Week 27 in Grade 6.

Good: Pupils will be following an algorithm and are using vocabulary, while also doing some mental maths.

Bad: This is incredibly specific!  Is it really worth having an algorithm for this particular scenario?  The questions that follow all have numbers such that step 2 gives an integer.  This needn’t always be the case.

Missed opportunity:

Why does this work?  What is going on?  Is there a better way to do it?

I think the most interesting thing here is working out how to express the numbers in an algebraic form.  For some pupils it will be strange to realise that when we write numbers next to letters as algebra then it means we are multiplying, yet when we write a number next to a fraction it often means we are adding!

The scenario we have got can therefore be shown as:

The numerators are always 1 and the denominators are the same.

This is a quadratic!  If we expand we get this:

Now go back to the original algorithm and see how each part of this algebra is linked to the different steps.

Further thoughts:

How would we ordinarily multiply two mixed numbers?  We would turn them both into improper fractions and multiply.  Then if we wanted to we could turn the answer back into a mixed number.

The benefits of this are that it will work for all mixed numbers. 

Are there any drawbacks?  Well, in the example given above that gives us:

This isn’t straightforward to do mentally, and the calculations aren’t as easy as the calculations given above. 

If we want easy sums to do then the method given works very well.  The difficult part is remembering the method!

Climbing St Paul’s Cathedral

Here's a stimulus for a brief question, perhaps useful for Core Maths.

What can we do with this?

To start with we could look at the link between the height and the number of steps:

Gallery	Height in metres	Number of steps	Steps/metre	Height of each step (m)
Golden	85	550	6.47	0.155
Stone	53	398	7.51	0.133

It isn’t the same.  First thought is that the steps for the route up to the very top are about 2cm taller than the ones to the Stone Gallery.

But maybe it’s the same route, and everyone goes to the Stone Gallery first, with only some people going on the top.  In this case we need to find the difference in the height and in the number of steps:

Height in metres	Number of steps	Steps/metre	Height of each step (m)
32	152	4.75	0.211

These steps are considerably taller than the earlier ones (over 1 and a half times as tall).

The Shard in London is the tallest building in the EU at 309.6 metres.  It has a viewing gallery on the 72^nd floor (out of 95), which is 244.3 metres up.  Presumably people usually go via a lift, but there must also be emergency stairs.  How many stairs? 

Image from:

https://www.stpauls.co.uk/visits/visits/sightseeing-times-prices/new-sunday-dome-climb 

Singapore Mental Math (5) – Subtraction: reverse 3-digit numbers

This one comes from the Grade 6 book (I assume this is the equivalent to Year 7 in England).

Week 4 has this:

Good things:  The pupils will be practising following an algorithm.  There is some mental maths involved.

Bad things:  The idea of having an algorithm (that you are expected to remember) to subtract a three-digit number from its reverse is a bit extraordinary!  There are potential problems with the way it is set out.  The opportunity to explore this, to explain why, to use some algebra, to create a better formula and to expand it to other numbers of digits has been missed.

The potential problem:  “Find the difference of the hundreds digits”.  This would mean that 598 – 895 has the same answer as 895 – 598. 

In the classroom:  We could try a few of these out to convince ourselves that this always works.  The pupils might wonder why the middle number isn’t mentioned in the algorithm.  Hold on: if we reverse a 3-digit number then the middle number is in the middle both times, so when you subtract them this gives zero.

Let’s try some algebra.  It might be natural for pupils to want to show this as ABC – CBA (particularly if they have previously worked on letter-substitution puzzles).  It would be good to explore why this doesn’t work with algebra.  (ABC means AxBxC.)

If we use their example of 895 and want to have a = 8, b=9 and c=5 then we actually need 100a + 10b + c. 

The reverse of this will be 100c + 10b + a.  When you subtract you get 100a + 10b + c – 100c – 10b – a, which simplifies to give 99a – 99c.  Factorising this gives 99(a – c).

This explains why we subtract the hundreds digits, and then why we multiply by 100 and subtract that difference again (leaving 99 of them).

Extensions:

We could try 2-digit reverses:

10a + b – 10b – a = 9(a – b), which means that for 2-digit reverses you subtract the digits and then multiply by 9 (which also means that the answer is always a multiple of 9).

We could then do a similar thing for 4-digit reverses…

Singapore Mental Math (4) – adding a series of odd numbers

Do read the previous few blog entries (here) to see the background to this.

This is the strategy for Week 1 in Grade 7 (which I assume is equivalent to Year 8 in England).

Good: Pupils will be following an algorithm and are using vocabulary, while also doing some mental maths.

Bad: Pupils might assume that in general all ‘series’ must start with 1.  It wasn’t made clear that for this algorithm to work it must start at 1.

Missed opportunity:

Why not mention that the answer is a square number?  (Step 3 could easily be: “Square the quotient obtained in Step 2”)

Why not explore why this is a square number?  There are so many ways to do this.  Here are a few I can think of immediately.]

 If there are an odd number of terms we can use the method from my previous post.  With 1 + 3 + 5 + 7 + 9 the middle number is 5 and there are 5 numbers.  The sum is therefore 5 squared.

The same is true for any odd number.

For an even number of terms we can still use the method from the previous post.  With 1 + 3 + 5 + 7 + 9 + 11 there are 6 terms and the middle is (1+11)/2= 6.  The sum is 6 squared.

Ah: so the algorithm is really adding the first and last terms, dividing by 2 and then squaring it.

We could also draw a picture:

Each colour shows a successive odd number, and after each new colour has been added it is still a square.

We can also explore further why this pattern must increase by an extra 2 each time.  If we just add on the same amount as last time, this is what happens:

To fill those gaps we need to add on an extra 2.

Singapore Mental Math (3) – adding a series of numbers in a pattern

See previous posts (here and here) if you are new to this.  Here is the Week 4 strategy.

Good thing: lots of practise of following an algorithm, and of multiplying by 5, 7 or 9 mentally.

Bad thing: while some questions have 5 numbers (‘addends’!), some 7 and others 9, all of the series go up by 2 each time (the method actually works for any linear sequence).

This is a missed opportunity if only used to answer these questions in the way stated.

Why does this method work?  This is a nice example of where algebra isn’t enough by itself: this is so much easier to see and to understand if you choose you algebra in a particular way.  Make the middle number in the series be n, and this is now straightforward.  (To be clear, it still works if you make the first number n, but this is more difficult to generalise from.)

With a 5-number series that increases by 2 we have:  (n-4) + (n-2) + n + (n+2) + (n+4)

When we sum this we get 5n, which is 5 times the middle number (as per the algorithm).  But we can see why this works: there is some symmetry in that the number to the left of n is 2 below n while the number to the right of n is 2 above n, so these cancel each other out.  If you have additional numbers then they will also cancel each other out, leaving only extra ns.

Can we do this if the series increases by 3?  Yes – that gives us: (n-6) + (n-3) + n + (n+3) + (n+6)

What if the series increases by a?  (n-3a) + (n-2a) + (n-a) + n + (n+a) + (n+2a) + (n+3a)

What if there is an even number of terms?
This still works (although the 'middle' is no longer an integer.

We can also use a visual method for this:

The first diagram shows 6+8+10+12+14.  In the second diagram that has been converted to 5 lots of the middle number.

This diagram also shows neatly that we are working out the mean of the numbers in the sequence (the middle number is the mean) and then multiplying by the number of terms. 

As a final different method, it would be nice to show this on a 100-square to illuminate further what is going on.

Singapore mental math (2) - special numbers

More from this book (see yesterday’s post for background information).

Here is the strategy from Week 33:

Good things: If used like this then there is some mental maths going on and the following of an algorithm.  It could keep mental maths sharp and settle pupils at the start of a lesson.  Vocabulary is used (product, factor, multiple).

Bad things: If used just like this then it seems to me to be very unlikely that anyone will remember what the special number is, or what fraction to use.  It doesn’t make it clear that this is only for 2-digit numbers.

Pupils could be asked to explore why this works (to be fair, perhaps this happens even though it isn’t mentioned in the book).

What is going on?

Multiplying by 429 somehow involves multiplying by 3/7 (and then doing some other jiggery-pokery).  What is the link between 429 and 3/7?

Let’s do 429 divided by 3/7 – we get 1001.  Aha, if we write down a 2-digit number, put a zero in the hundreds place and then write the 2-digit number down for the thousands then that is the same as multiplying by 1001.  This explains the final few steps.

Questions for pupils to consider:

• One of the questions is 7x429.  How does the algorithm need to be tweaked to make this one work?
• What happens if you do this with a 3-digit number?  Or a 4-digit number?
• What would be a better way to explain the algorithm?  (Maybe: “multiply by 3/7 and then multiply by 1001”?)
• What happens if you choose a number that isn’t a multiple of 7? 

The strategy for Week 34 is almost identical, except that the ‘special number’ is 715 and the initial instruction is to multiply by 5/7.

Week 36 has special number 858 and ‘multiply by 6/7’.

Both of these are fractions of 1001 too. 

What other ‘special numbers’ could there be?  Can pupils write an algorithm for those? 

(For example, 1/7 of 1001 is 143…)

Singapore Mental Math

I am exploring the Bar Model at the moment and have bought some books to help me get to grips with it.  Most of these books are very interesting and contain illuminating ideas.  When I have digested them I will post a list of the ones I found most helpful.

While browsing through books on Amazon I also bought a Singapore Math ‘Mental Math’ book. This is very different from the bar model materials.

The book describes itself as containing ‘challenging activities based on the world-renowned Singapore Math curriculum’.  It says it includes ‘must-know strategies for solving problems quickly and accurately’.  It states: ‘To help students build and strengthen their mental calculation skills, this book provides strategies that will benefit students as they learn tips to solve math problems quickly and effectively.  After acquiring such invaluable skills, students can apply them to their future, real-life experiences with math, such as in shopping and banking.’

There are 52 strategies in the book (one for each week of the year) and many of them are quite extraordinary. In the rest of this post I will explain one of them and will then discuss what I see as the pros and cons of this approach.  Future blog posts will include other strategies from the book.

Week 24 has this strategy:

The good things about this:

• Pupils will practise following instructions and using an algorithm.
• It might make a good ‘settler’ task at the start of a lesson.
• Some mental maths muscles will be flexed (with lots of adding and doubling going on).

The bad things about this:

• This strategy has very limited application.
• Pupils are being encouraged to learn lots of strategies of questionable usefulness.
• There could be confusion and it could be applied in the wrong setting.
• There is a much easier way to do this!
• There is nothing to explain why this works or to lead pupils to explore it.

Perhaps I am being unfair, and maybe pupils will want to know how and why it works.  Maybe teachers using this resource _do_ explore why this works.  If so then I think this could be a wonderfully positive resource.  For example:

Let’s use some algebra.  Call the two numbers n+1 and n.

We have (n+1)² – n².  We could expand the brackets, to give n² + 2n + 1 – n²

Alternatively we could use the difference of two squares:

This gives us (n+1)² – n² = (n+1 + n)(n+1 - n), which simplifies to (2n+1).

Going back to the original:

The original has (n+ ½ ) multiplied by 2, which is 2n+1.  

The easier way:

But 2n+1 is also (n+1) + n, so this is actually the two numbers added together!  11+10 = 21

An alternative method could involve drawing some diagrams:

This shows that the non-yellow part contains (n+1) + n squares, and properly demonstrates that we are adding the two numbers together.

I prefer the difference of two squares method because it seems to me to be the most easily generalizable:

If the numbers are two apart then we have (n+2)² – n², which is (2n+2)x2

If the numbers are three apart then we have (n+3)² – n², which is (2n+3)x3.

Every time (and using (n+a)² – n²) makes this obvious), we need to add the two numbers and then multiply by their difference. 

I can see some really fertile material here for KS3 and KS4 pupils to explore, so as a starting point this turns out to be an exciting resource.  It also leads back to the important idea that there is no such thing as a ‘good resource’ or a ‘bad resource’, but that it depends on how you use them.

Constructions: it's all about the rhombus

Constructions seems to be an ideal topic in which links can be made with other parts of geometry.  It also seems a good opportunity to teach for understanding (in a Skemp-type way) as well as leaving pupils with a method they can use.

I learned this when I was at school (in the 20th century).  I remembered where to put the point of the compasses and which arcs to draw so I could carry out the constructions accurately, but I think there is more to it than that.

First off: what counts as a 'construction' and what doesn't?

If you are asked to draw an "accurate" version of a triangle when given the three sides, it is usual to use a pair of compasses.  (Draw one side - then make an arc from each end - where the arcs cross is the third vertex.)  This will give a pretty good version of the triangle (assuming the compasses don't slip and you have a sharp pencil), but this won't ever be entirely accurate because measurement is involved.  This isn't usually considered to be a proper construction for this reason.

Euclidean constructions involve using a 'straight edge' (a ruler - but not using it for measuring) and a pair of compasses (for drawing arcs).  [These also won't be exactly accurate because the compasses might slip, there is the thickness of the pencil to consider, etc, but the idea here is that this method will, in an ideal situation, give an idealised answer, so it considered to be a purer construction.]

The main constructions involve bisecting an angle and creating the perpendicular bisector to a line segment.  Arguably, all of the others build on these.

The big idea here is that they both involve constructing a rhombus.

There are three properties of a rhombus that we need:  

1. A diagonal of a rhombus bisects the angle
2. The two diagonals of a rhombus bisect each other
3. The diagonals of a rhombus are perpendicular

Consider how to construct an angle bisector.  Here is an angle:

When we draw in the arcs using a pair of compasses then we are actually constructing a rhombus:

We know this is a rhombus because we have set the compasses to a particular length, and then each of the sides of the rhombus is the same as that length.

Because of the properties of a rhombus, if we join the opposite vertices we bisect the angle.

Now consider how to draw the perpendicular bisector of a line segment:

Again, we need to draw arcs and construct a relevant rhombus:

The original line segment becomes one diagonal of the rhombus, with the line we will draw in becoming the other diagonal.  The properties of a rhombus tell us that the diagonals bisect each other at right angles, so we have got the perpendicular bisector.

There are lots of other constructions, such as constructing the perpendicular to a line that passes through a particular point, but this can be done by putting the compass point on the point and drawing an arc that intersects the line twice.  Now use the line segment between the places where the arcs cross the line and use the second construction above.

One way to teach this is to start by getting pupils to construct lots of rhombi.  Then they can sketch the rhombus they need onto the diagram, and finally they can construct the relevant rhombus.

Remember: it's all about the rhombus!