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Two circles and a trapezium

June 4, 2020 Leave a comment

This puzzle is one that really got me thinking about a number of things, and had me stuck for a little while. I found it on twitter, where it was shared by Elyem Gercek Boss (@_eylem_99) . Here it is:

I started, as always, with a sketch:

I feel that I should mention here that the radius OF wasnt added until much later. In fact adding this line was my breakthrough moment to be honest. I had been struggling for a bit when I thought to add it in. But more of that later. The first thing I did was this:

I saw I come express AC in terms of the 2 radii, I then moved on from that and looked at the similar triangles I had and came up with x = ((r1)^2)/r2

I then worked out I had another similar triangle:

And decided this must be the route to the answer. I did various things:

Including this that was a load of work to get a result I already knew.

After some dead ends and obsessing over the similar triangles I looked back to my original triangle and that’s when I drew the line:

I realised that as I already had an equation in r1, r2 and x and a right angled triangle featuring these on the sides this would be a route to go down.

I realised I had a a quadratic:

At this point I could express the square of one radius over the square of the other, and as area is directly proportional to the square of the radius this was all I needed:

A really nice problem that had me thinking lots. I’m certainly there will be more concise solutions. If you have one, I’d love to see it! Please let me know in the comments or on social media

Coordinate integer puzzle

June 1, 2020 Leave a comment

Today’s puzzle is a nice question from the 2019 Ritangle competition. Ritangle is an integral maths competition aimed at students who study mathematics Level (or equivalent). They have tons of questions from the last few years on the website, many of which look very interesting. This one is one that I came across ages ago in Chris Smith’s (@aap03102) maths newsletter and it’s been sitting in my screenshots folder on my phone since. Here it is:

It’s an interesting question, and one I put a bit of thought into. First I considered what the midpoint was:

The mean of the x coordinates and the mean of the y coordinates.

Then I thought, what sums could I have for the coordinates?

Because all the coordinates, including the midpoint, had to be integers I could discount the odd sums:

Then I considered how I could get these sums. And what half of them would be:

Discounting any possibilities that would lead to repeating the same number in the sum and the midpoint it left limited possibilities. And when looking at which pairs involved all of the numbers 1-6 and nothing else it only left two possibilities:

Once I knew what the coordinates would be I could look at what their differences would be. Due to the nature of Pythagoras’s Theorem, it wouldn’t matter whether the coordinates were the x or y, or even which were paired, as they’d give the same length of line, so there was two possible lengths. 2rt2 and 4rt2. So the longest possible length is 4rt2.

I found this a nice puzzle to tackle, I suspect there are other ways to go about it, and I’d love to hear how you approached it, especially if you did it differently.

A lovely sequence proof problem

May 28, 2020 Leave a comment

Another puzzle I found in Chris Smith’s (@aap03102) newsletter, and had fun doing is this one:

It’s a nice little puzzle that relies on ones ability to use algebra, reasoning and to understand what an arithmetic sequence is.

I considered the problem and thought that as I know the difference between each consecutive term in an arithmetic sequence is always the same I should first start by considering the differences. I looked at each one and expressed it as a single fraction:

When I had the two defences I knew that I could then equate them:

At this point I simplified the equation:

And was surprised it had fallen out so nicely . This obviously show that the differences between x^2, y^2 and z^2 are also equal so this is also and arithmetic sequence. I think this is a lovely result.

It occurred to me later that I could have set the mean of the 1st and 3rd term equal to the second and the algebra may have been a little easier. I’d love to hear your approach.

Part whole division – why?

May 21, 2020 Leave a comment

I’ve been thinking a lot about division recently. I wrote this here a short while ago about dividing by fractions, then I was sent a document by Andrew Harris (2001) entitled “Multiplication and Division”, which I was asked to read as part of a series if CPD sessions from the local maths hub, then a number of different people have asked me questions about division recently too. I think probably for most this is due to helping their own kids with maths and meeting methods and structures that they aren’t familiar with, as they weren’t taught when they were at school themselves.

The main thing from friends that keeps popping up is using part whole models for division. And funnily enough it is one of the structures I was considering after reading the Harris document and looking at the distributive law and what higher level topics this underpins in later maths.

So what is it?

Using the part whole method for division is where you split a number into 2 or more parts before dividing then add your answers back at the end. For instance, if you want to divide 486 by 6 you can split it into 480 and 6. The benefit of choosing these numbers is that 48 is in the six times table. So you can see that 48÷6=8 so 480÷6=80, then you have 6÷6=1, add them together and you get 486÷6=81.

This structure, or method, is a very common mental strategy used by lots of people when dividing numbers in their head. Lots of those people will never have heard the term “part whole model” and will not have seen it laid out in a pictorial manner as students today will, but they will use that structure nonetheless. I myself was using it as a mental strategy a long time before I’d heard anyone refer to a part whole model or seen the visual representations.

What we are doing when we do this is using the distributive law of multiplication and division to break our problem into chunks that are easier to manage.

One of the questions I was asked was “is there a rule to how you split it up?” The person who asked me was wondering if you always split it up into hundreds, tens, ones etc or if you could do any. I explained that it didn’t matter, and that actually the divisor would normally be important in deciding. For instance if you were dividing 423 by 3 it wouldn’t make much sense to use 400 as this isnt divisible by 3. It would be more sensible to choose 300 (÷3=100), 120 (÷3=40) and 3(÷3=1).

But why not use short or long division?

This is a question I’ve seen a lot of times from a lot of people. They see the part whole method as a long and clumsy way to solve problems that they can solve easily using one of the two standard algorithms. I can see the point in asking, the algorithms are far more efficient as written methods. But that’s not why this model is taught. No one expects students to get to their GCSE and start drawing part whole models to solve division problems. The visual representations are their to help build an understanding of what is going on, an understanding of the relationship between numbers and mathematical operations. In this case it’s to build an understanding of how the distributive law works and to give a good mental strategy for division. It even helps understand how the long and short division algorithms work, as they are both based on splitting the dividend up into parts. There must come a point when these structures and representations are removed and students move to the abstract, but that doesn’t devalue their importance to that learning journey.

What else is the model used for?

The idea of a part whole diagram is introduced way earlier than this. Students get used to partitioning numbers into part whole models while working on addition and subtraction. It helps then see at that level that they have a relationship, that they are the inverse of each other. So when students come to meet this model for division it’s a small step on what they were already doing.

These are similar to some of the earliest part whole models my daughter did when she started school. They were being used to show place value, and also to show how addition and subtraction work and interact. For both these tasks this model is an excellent visual representation to help students understand the concepts.

Part whole models can also be represented as bar models. Here the one on the left can again be used to show either place value or addition/subtraction. The one on the right is an early algebraic model, and if we are told that x+2=9 we can use this representation to show why x must equal 7. This representation is more effective if students are familiar with it from their earlier mathematics.

Building on this we can show the distributive law when it comes to multiplication:

And show how that links to division:

As we go further into maths this idea of part whole division comes up again and again. One place that springs to mind is when calculus is first introduced at A level. One of the first things that we teach is how do differentiate and integrate polynomials with different powers of x. And a favourite style of question from examiners is this:

Or its derivative equivalent.

The easiest way to do this, when it comes to integration or differentiation, is to rewrite the fraction as separate terms:

What we have done here is used the part whole model to divide the expression on the numerator by x^2. We could draw that in our part whole model:

I wouldn’t advise that, its unnecessary, but having a secure knowledge of that model and how it works due to the distributive law is key to understanding how and why we can simplify this fraction in that way.

I’ve thought a lot about division recently, and I’m sure I will continue to do so, so if you agree,disagree or have anything else to add please get in touch either in the comments or via social media as I’d love to hear your views.

Proving sums and differences

May 20, 2020 Leave a comment

Today I want to look at a proof puzzle that I got from a previous addition of Chris Smith’s (@aap03102) brilliant maths newsletter. The puzzle is this:

It’s one I saw a few weeks ago, had a think about, got stuck and promptly forgot all about. But today I found the screenshot on my camera roll and thought I’d another go.

I tried a few things out that became fruitless endeavours, then I considers different ways to express the integers. As we were thinking I thought about modulo 100, what are the remainders we could have, as we are dealing with sums and differences we need to get the remainders to equal 0 for the sum or difference to be divisible.

Using mod 100 we would have 100 possible remainders. I only had 52 integers though, I I needed more thought. I considered this:

Using negatives I could express b as:

So I could get the magnitude of b to be one of 51 options. That meant that if I had 52 integers, they could be expressed this way and at least 2 of them would have the same value.

If the b values had the same sign, then their difference would be 0 hence the difference of the integers would be divisible by 100. Likewise if the signs were different then their sum would be divisible by 100.

I then considered how else I could write that.

I could express the integers as 100a + b (argh, why did I use a gain but for something different. Grrr.)

Then it follows the same way that at least 2 must have the same magnitude of b either the sum or the difference is divisible by 100. As shown here with the example b = 49:

A nice puzzle that I enjoyed exploring. I think there are probably nice proofs, certainly ones laid out better. I’d love to see yours.

An irrational triangle

May 15, 2020 3 comments

The sad news about Don Steward last week prompted many people to share blog posts and tweets about how he has helped and inspired them over the years. One of the people who blogged about him was Jo Morgan (@mathsjem) who wrote this piece. Jo had the pleasure of knowing Don personally and it was really nice to read the things she shared. If you missed it, do give it a read.Today I want to look at the problem she shared at the end of the post, which she says is her favourite Don Steward problem:It’s not one I remember seeing before, although I have spent so long on Don’s fantastic median website it is highly probably I have seen it. But I thought I’d have a go. Usually when working out an are of a triangle with 3 sides I will apply Heron’s formula, but I looked at this one and realised I would end up with a massive expansion of surds and quite possibly nested radicals and that would take a long time to work through by hand, so I considered other options. I thought a locigal first attempt would be to use trigonometry. I drew a sketch:

Labelled one if the angles x. I can then work out cos (x) fairly easily:

I did think I might get stuck at this point, with the limitation in the question prohibiting calculators, but actually it worked out really nicely. If I know cos (x) I can use a right angled triangle to work out sin (x):

Here I only have to square 1 (1) and the root of 26 (26) do a subtraction and take the square root of 25 (5) to complete my triangle using Pythagoras’s Theorem. All easily done without a calculator.Now I have sin x I can work out the area using area = absin(c)/2

When I started playing with the surds I wasn’t quite expecting it to fall out so nicely. Who’d have thought a triangle with 3 irrational side lengths would have such a nice integer area? I think I will explore this more when I get time. See how easy this type of triangle is to generate and if there is a rule to triangle of this sort occurring.

I said earlier that I didn’t think I’d seen this one before. But now I think about it, I think I may have. I think I remember giving it to a year 12 class maybe last year and some of them cheating and using the calculator but not getting the right answer as it had rounded the cosine and sine values, I will need to check this later on a calculator too. If this is the case, then I think this is a great discussion point when looking at rounding and why we shouldn’t round too early. I’m also left wondering how easy/difficult it would have been to get to the nice answer using Heron’s Formula, so I will give that a go in due course too.

4 squares and a rectangle

May 14, 2020 7 comments

Today’s puzzle comes from Catriona Shearer (@cshearer41):

It’s one that gave me a little trouble early on but finished up with a lovely solution. Have a go before you read on.

Here’s one of a few aborted efforts:

So my solution started, as usual, with a sketch. I worked on my sketch and came up with this:

Which proved to be the key. But I realise this is not a very easy to follow solution so I’ll try to explain a bit. Here is a second diagram, this time I have included some vertex labels:

Starting with triangle ABF I labelled side FB as a and FA as b, I know the side AB is 1 aa it is a side of a square of area 1. Angle BFA is a right angle so if angle FAB is x then angle ABF IS 90 – x. As angle ABC is 90 and FG is a straight line that means angle DBG is also x (180 – 90 – (90 – x)).

This means triangles ABF and BDG are similar (both have one angle 90, one x and the other 90-x). As the hypotenuse of BDG is 2 and the hypotenuse of ABF is 1 (from the side length of the squares being 1) then we know that GB is 2b, and GD is 2a. At this point I considered reflecting DBG to get BED but decided against that root.

I worked out that angle JDI (where J is directly above I such that JI and GK are perpendicular) is x. (Again 180 – 90 – (90- x)). IJD is 90 by design and the hypotenuse DI is 2 hence triangle IJD is congruent to BGD which means IJ is 2a.

However, angle JIK is also x (HIK is 90, HIJ is 90- x so JIK is 90 – (90-x)). IJK is 90 and the hypotenuse IK is 1 so triangle IJK is congruent to triangle ABF. So IJ is b.

This means as IJ is 2a, but IJ is also b we can say that b = 2a.

Back our original diagram now. Using these similar and congruent triangle I can see that the vertical side of the rectangle is a + 2b, and the horizontal one is 3a + 2b.

Subbing b = 2a in we get the horizontal side to be 7a and the vertical side to be 5a. So the area is 35a.

Almost there. We also have a number of right angled triangles we could use for the next be I chose the top left one BGD in the other diagram.

By Pythagoras’s Theorem we can see that a^2 is 1/5.

So the area is 7.

I think this is a lovely solution to a problem I very much enjoyed wrestling with. I imagine there are plenty of other approaches that you could use to solve it. Please let me know your solution either in the comments, by email or via social media.

Consecutive integer puzzle

May 12, 2020 Leave a comment

Today’s puzzle comes via Chris Smith’s (@aap03102) newsletter and was bet by Guney Mentes:

Chris’s newsletter is a lot of fun, full of great maths facts and great puzzles, if you aren’t a subscriber you definitely should be.

Anyway, this puzzle looked quite nice so I thought I’d have a quick go.

I decided I needed to think about it algebraically, rather than jump in an start guessing. So I considered a list of 8 consecutive integers starting with n.

n, n+1, n+2, n+3, n+4, n+5, n+6, n+7

This means the sum of the blue squares would be:

3n + 3

The yellow:

2n + 7

And the pink:

3n + 18

Equating the blue and yellow gives:

3n + 3 = 2n + 7

n = 4

That gives blue sum = 15, yellow sum = 15 and pink sum = 30.

I thought it interesting that the blue and yellow sums were equal to the sum of the first and last term. Then I realised that there was a good reason for this. The blue and yellow sums are each equal to a quarter of the total sum. The series is arithmetic with 8 terms. The sum of a series is equal to half the number of terms multiplied by the sum of the first and the last. It occurred to me then that this might be a good puzzle to use with a post 16 class to allow a discussion about summing series.

Terminal Exams

May 10, 2020 Leave a comment

Earlier this week I came across this post from 2014 in which I was thinking about the then upcoming switch from modular A Levels to linear A Levels. A move from short exams occurring throughout the course to 3 long ones at the end.

I was excited by the prospect, mainly because the earlier stuff is easier once you have learned the harder stuff, and on top of that their is more time to teach if you dont have to go into revision for exam mode every few months.

We’ve been through it a few times now, so I thought it might be a good time to revisit the subject.

What’s happened?

We switched from 6 short modular exams to 3 terminal exams, and the link between AS and A Level was severed so marks gained in Y12 no longer count toward your final grade. This to me made a lot of sense and I didn’t want to enter any for AS exams. The first year we didn’t enter them, but since the decision has been made at a whole trust level to enter, so that those who leave to go on to apprentices etc still get a qualification.

That first year we were the only subject in our school not to enter, and we gave them internal exams at the same time. We didn’t have anything to base grade boundaries on and it didn’t tell us much about the students abilities. The next year we struggled to get through all the year 12 content in time and didn’t have as much exam prep time as we would have liked before sitting AS exams. This year we re did the SFL and were done teaching new content by mid march, which was handy in the end.

I’m in two minds about AS exams, they don’t count for anything once they finish year 13 and it adds additional pressure. But it does give them additional impetus to revise and ensure the Y12 content is thoroughly stuck in their minds. When all the subjects but us sat them we found that the students revised a lot more for their other subjects which makes me think that it should at least be a school wide decision.

What about the other exams?

Losing the repeated exam windows has definitely helped with the scheduling if learning. Its given me the flexibility to teach in the order I feel is lost sensible, rather than teaching stuff in a certain order as that happens to be the module it was put in. The fact we dont need to go into exam mode and revision mode as often is also of benefit allowing more time to focus on understanding the mathematics and the concepts that students need to understand in order to succeed. For certain students though, I feel they would benefit from that “oh crap” moment when they don’t revise for the first exam and fluff it. But I think thise students are less common than some suspect.

How do the exams compare?

Obviously the spec has changed, so it’s not a direct comparison, but I feel that the linear approach allows a wider range of questions, different topics can be merged into different questions and there have been some really nice racing questions so far. My worries in 2014 about the length of the exams seem unfounded as I’ve not heard anyone complaining about the length, although have heard some of them say they needed more time.

What about the spec?

I like the new spec. It has lots of fun maths in it and I feel it’s a good broad range if maths to know for moving forward into higher education. I’ve very much enjoyed teaching it. I miss some of the stuff that isn’t there, but all in all I think it’s a good spec.

In summary

I still think that the linear model with terminal assessments is preferable over a modular model. I’m torn as to whether I think AS exams should be sat or not and I’m a fan of the new spec on the whole.

What are your opinions? Do you agree or do you think modular was/is better? Why have you come to this conclusion? I’d love to hear all views from both sides, please let me know in the comments, via email or via social media.

Flummoxing triangles

May 8, 2020 Leave a comment

Earlier on in the week I came across an interesting geometry puzzle from (@puzzleprime):

I had a quick try:

Many dead ends, I did work out that ABC is an isoceles triangle and I had the embryonic thought that I would need some more lines, but wasnt sure where they’d go. I then left it for a while. I came back to it today:

I started and realised I was mainly taking the same approach. So I thought about what other lines I could draw. At this point I spotted something. If I rotated ABC so the vertex B washes at A and CB ran along where AB is we would have 2 equal sides separated by 60 degrees. So if we joined their ends we would get an equilateral triangle:

At this point I still wasn’t sure what to do next. Then while I was looking at it I realised that, despite my terrible diagram, A’B’E and ABE are congruent triangles.

This meant that the line AE bisects the angle BAA’.

As angle BAA’ is the angle that makes up the 80 degree angle with x that means x is 70.

This was a lovely puzzle that had me thinking and flummoxed at different points. I’d love to hear of you have a different solution.

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