Carnival of Mathematics 224

Roll up, roll up, roll up. Come hither come all to the Carnival of Mathematics. This is the 224^th Edition of the longest running Maths Carnival.

For those of you who are unaware, a “blog carnival” is a periodic post that travels from blog to blog and has a collection of posts on a certain topic. In this case the topic is mathematics.

224 is an interesting number – you can make some simple number sentences out of it for a start 2 + 2 = 4 and 2 x 2 = 4, even 2 ^ 2 = 4, if you’re feeling a little spicy! It is an abundant number, and admirable number and is also the perimeter of a Pythagorean triple! (which one you say? I’ll leave that as an exercise for the reader……) For more fun facts about the number 224, try this post here!

It has been a long time since I have had the honour of hosting the carnival, in fact it was the the 126th edition I last hosted (available here). Lots has gone on in the world of mathematics since then, and the most recent carvinval, edition 223, is available here, hosted by the great George Shakan at his Data Science and Math blog.

This month we’ve had some quirky submissions, and none quite as quirky as the first one I recieved, submitted by Katie, that is the “Minimum Wage Clock“, this is not for the faint hearted, as it shows in real time the vast difference in earning between those on minimum wage, and those in high up positions.

Sticking with the quirky theme, next up we have another submission from Katie which looks at some excellent research and data collected by an eight year old girl, who really is a data scientist in the making.

If you, like me, get annoyed at various words and semantics, then this thread on mastadon may have you giggling, or up in arms!

For those of you, like me, who are secondary maths teacher, this post from Jo Morgan at Resourceaholic has some highlights to help you get your classes ready for this summers GCSE exams.

While Dave Gale, at reflective maths, has these fun questions based around the no 2024.

If it’s maths news you are after, then Katie Steckles and Peter Rowlett over at The Aperiodical (the custodians of this very carnival) have got you covered with this excellent news round up.

If you’ve after something lighter, the wonderful Ben Orlin has put out a few of his hilarious post this month at Math with Bad Drawings, always worth checking out.

Prof Nira Chamberlain, president of the mathematical association, has Dr Jeffery Quay as a guest on the latest on his excellent VLog series “What’s the point of maths?”

Well that’s it for this months carnival – I hope you enjoyed it! While you’re here, you could check out other carnivals you may have missed at the Carnival’s homepage.

Sine and Cosine – do we need a formula?

While teaching non-right angled trig recently it occurred to me that when doing the questions myself, I don’t actually thing about the formula all that much. Particularly with the sine rule. Yet I still start the teaching of it using the formula heavily. It made me wonder if this was really all that necessary.

Firstly the sine rule. Here in this question I have two “opposite pairs”:

And I know that the ratios of side:sine of angle for these opposite pairs are the same, so I can easily set up an equation and solve. Not once in that process have I thought about or used the sine rule as it comes, in terms of a,b and c. The triangle isn’t even labelled, but I can solve the problem. I can also talk about ambiguous cases as and when they crop up.

The cosine rule on the other hand requires a bit more remembering, and the formulae certainly comes in handy as memory aid (as does the sine rule) but sometimes students can become a little over focussed on the labelling. When I approach a question I don’t label the sides, I look for he pair – the side and opposite angle – and I know these both have to be in the place of a in the formulae.

It makes me wonder whether there’s a better way to approach this. Certainly I find students get to grips with it quicker if I spend less time on labels and more time on pairs, know that the pair is a no matter what the labels say. I’d be interested in others thoughts.

Triangles, Trig and Squares, oh my.

Over the weekend I happened accross this loevely puzzle on twitter. It was tweeted out by Diego Rattaggi (@diegorattaggi) and I saved it to my ever growing folder of puzzles to try on my phone. When i had a bit of time spare I thought I would give it a go.

Have you had a try? If not, you should. Its a nice one and I’d love to hear some different approaches.

Anyway, here is how I approached it. Initially, I wondered about finding an algebraic expression that might fall out with some nice numbers, but i got a nasty looking equation with z and y both raised to the power 4 so I thought I’d try a different tactic.

I did plenty of annotating on a diagram to visualise the information, and found that 180 = 180, so that wasn’t particularly useful.

Then I looked at a trigonometric approach. Which fell out nicely.

I has Tan 2b = 2, and I needed tan (90-2b), my insitinc was to use double angle formulae, but I quickly realised tan 90 was undefined so that wouldn’t work. I sketched a quick right angle triange, then realised I was being a tad silly.

While the triangle did give me the correct answer, it also made me remember that the definition of Cotangent is “The tangent of the complementary angle”, so cot 2b is quite literally tan (90-2b), so all i needed to do was take the reciprocal.

It was a nice little puzzle that incorporated a number of things, and it has me wondering how my year 13s would approach it. I’m gonna try it out with them later this week. I’d be intrested to hear how you approaced it, expecially if you soved it a different way.

1-9 math walk

Today I want to look at another puzzle I found on math walks (from Traci Jackson @traciteacher):

I love these 1-9 puzzles, and thought I’d have have a crack.

First I considered the 9, with the 1 gone already that means that the 9 must share a line with the 2 and the 3 to make 14.

That means that the 4 shares 1 line with a 2 and one with a 3. That means the 4 is one lines 4,2,8 and 4,3,7.

I considered these lines:

If I put the 7 on the left, I’d need the 6 at the intersection of the green lines. That would also mean that the 2 was above it, but I’d need another 6 on that line which doesn’t work.

So the 8 must be on the left and if we follow through we get:

I enjoyed this puzzle, if you have any cool 1-9 puzzles do send them on.

Step perimeters and infinite coasts

A few weeks ago I came across this picture on twitter:

I looked at it and thought “13cm, that’s a great question to use for perimeter.” I like questions like this, where you can show that if you invest the sides to the opposite sides of little rectangles the perimeter stays the same. I’ve seen them referred to as “step perimeter” and “cross perimeter” problems. Usually it’s the cross version I use, or a rectangle with a smaller rectangle cut from a corner. I liked this question as it had many different cuts and I feel would be great to open up a discussion on class.

I then thought about the problem a lot more, you could make many cuts. You could make them at irregular intervals like this, or you coukd make them at regular ones. You can look at it many ways. I started thinking about using more and more regular intervals and how it would start to look like a straight line. I then started to consider that. Intuitively it feels like as the intervals get infinitesimally small the limit should be the straight line. But that’s not what happens.

Bizarrely after I’d been thinking on this for a long time I was scrolling through twitter and saw this:

Someone else had obviously been having similar thoughts to me.

I started thinking about it again. I changed it slightly in my head, I changed it to a 5cm by 12 cm rectangle. This was so i could visualise the difference better.

Obviously due to Pythagoras’s Theorem the diagonal of a 5×12 rectangle is 13. But the distance we get by travelling in a series of steps that are parallel to the sides is the semi perimeter- in this case 17. That’s 4cm difference.

What’s going on here? It feels counterintuitive and almost paradoxical. What is going on is that no matter how small the steps are, it’s still longer to take them than it is to cover the distance diagonally. And the sum of all these differences will always be 4 for a 5×12 rectangle.

When you draw them so small it looks diagonal, it’s not actually diagonal. Its still steps. But you cant see them due to the width of the line and the inaccuracies in drawing and human vision. If you zoom in then the steps are their.

It made me think about the coastline paradox. Which states that you can never accurately measure a coastline as it is fractal in nature. The smaller the unit of measurement you use the longer the coastline will be. Meaning that the length of the coastline grows to infinity as your unit of measurement shrinks towards zero, giving coastlines infinite length of you use infinitesimal measurements.

This kinda feels the same as the case if the steps that create the illusion of a straight line. In reality you can’t physically measure an infinitesimal length and you will find a length for the coastline. And in reality you won’t be able to draw the steps that are that small. But it’s a nice interesting result. One I’ve enjoyed thinking about so far, and one I expect I’ll be thinking about some more

Tilings and areas

My daughter and I had another play around with pattern blocks. Firstly we played around and made some patterns. She made this one that was pretty cool:

We talked about tiling the plane and how shapes tesselate. Looking at which shapes fit together. Then I asked her if she could make a repeating pattern.

She came up with this one. Which wasn’t exactly what I meant but cool non the less.

Then she made this one that was more what I had meant. At this point we discussed which colour had more shapes and which took up more of the area.

We had similar discussions about these two tilings. We discussed how the red and yellow had the same amount if area in the red yellow and green one even though the yellow had twice as many squares. She showed this by making a hexagon put if the trapeziums.

She said the green, blue and purple one looked 3d.

I agree. I mentioned briefly that it was to do with the angles if the lines and that you can get dotted paper to help draw 3d which has dots at these angles. We talked briefly about rotational and reflective symmetry too.

She then made a hexagon:

We talked about how much bigger it was. She said it looked about 4 times bigger. We then discussed what this mean, and looked at the areas. Counting triangles.

I showed her that we could do it without counting triangles. We then looked at the side lengths of the hexagons and discussed how and why this scale factor was different to the area one. I think this photo of the hexagons is an excellent visual to use when looking at similarity in secondary school. Normally I just talk about squares and rectangles but can see an excellent set of visuals using these shapes.

We then started to look at fitting shapes together round points and on a line. And we found that if you put the thin blue rhombuses together on a line you can get some cool patterns:

We didn’t get into angles that much, but I can certainly see this could be a great entry point to those discussions in future. I can also see that as well as similarity there can be further discussions around area and perimeter that build from using these shapes and I hope to explore this more in future sessions.

This is the tenth post in a series looking at the use of manipulatives in maths teaching. You can see the others here.

Returning numbers

I’ve recently discovered the website “Math Walks“, by Traci Jackson (@traciteacher). The site is full of pictures taken on walks that have happened during quarantine where Traci has used chalk on a paving stone to create a maths prompt to aid discussion. Sometimes they are puzzles, sometimes they are sequences, and sometimes another maths picture and they always seem to get me thinking about maths. This one is one I’ve been thinking about today:

It’s a great visual, and I really like it on so many levels. Obviously the task if ti find a solution that when you follow the path round you get your original answer. It struck me as interesting that this can be accessed on a number of different levels.

I could give this to my daughter and she would be able to complete trial and error and eventually find solutions to each of these, but theres much more too it than that.

I thought about how I would tackle this problem and decided I would use algebra:

I quite liked this as a forming and solving equations exercise, I think it’s accessible but bit too easy. Many students may struggle with the notation around the forming and many may get confused with the order they need to do things in.

I considered how one might challenge a student who does just guess, and I feel that asking them to prove whether their answer is the only one or not would be a good follow up question in this case, I think it would be unlikely that many using trial and error would get both answers for the one which includes a square.

I then considered if there was a trick to generating these puzzles, presumably you start with your answer then you can make sure you always get back there.

I think these ones are lovely, and I hope to use them at some point when we get back. I’d love to hear your thoughts on them. How would you approach them? How would you generate them?

Circles and an octagon

Here’s an interesting puzzle that came via Diego Rattaggi (@diegorattaggi) and involves circles and octagon.

I started as always with a diagram:

I labelled some sides up, then changed my mind and changed labels as I was thinking about taking a coordinate geometry approach and didn’t want to have used x and y. As it happened I didnt use that approach anyway. While looking at the sketch I realised that the triangle AOC was a right angled isoceles. Due to an error a few weeks back I wanted to double check I wasn’t making an assumption here so did some work to justify this was the case:

I was using some similar reasoning to this hexagon puzzle, I could justify that to had to be an isoceles, and that extending the lines gave an isoceles, I could just that the vetex was definitely on diameter I’d drawn and was equidistant from both circles in x and y, but felt that wasn’t enough, and if it wasn’t definitely the centre there could be multiple solutions, then I saw a different version in my screenshot:

Once I had this information it was fairly straightforward using Pythagoras’s Theorem:

At this point I realised the ratio if the radii squared was the same as the area so that’s all I needed:

I got to the end and realised I had my fraction upside down so I flipped it over.

This was an interesting puzzle, and I think I will need to think further on the case where the centre being the exact of the right triangle wasn’t specified. I might need to look on geogebra.

Thinking about circles

A number of things over the last few weeks have got me thinking about circle theorems. I’ve been using them quite regularly to solve a number of the puzzles recently, and most of them work both ways. When I did this puzzle, I initially did it wrong:

So what I did was see that angle CBD was double angle CAD, using exterior angles theorem. And then at that point I thought, “well that makes B the centre of the circumcircle” then I followed the angles to get alpha as 30. A friend had sent me the puzzle and when I sent him my answer he said he had a different one, so I relooked at mine and realized that if 30 was the answer it wouldn’t work. Triangle DEC would have 2 right angles in it, and that’s impossible. I tried again and got the same answer as my friend (36).

But it got me thinking, and talking, about circle theorems. I didn’t know whether this one worked both ways or not, but assumed it did as all the other ones do. When I was discussing it I had a realisation though:

If you have a chord, and the centre of a circle you can always make a triangle (they can’t be collinear as that would make a diameter not a chord). If you have a triangle can always draw a circumcircle and the arc of that circle which falls within the original circle would always make the same angle from the ends of the chord as the centre.

It’s quite obvious when you think about it, but it wasn’t something I’d thought much on.

Then a few days later I was thinking about this puzzle and in particular the bit in the circle:

I was looking at it and thinking how its interesting that when you draw 2 tangents from a point the angle they make at the point is always 180 – the angle made at the centre when you draw radii from the points the tangents meet the circle. This is always true, as tangents always meet radii at 90 so the other 2 angles in the quadrilateral add to 180.

While I was thinking about how nice and interesting this was, it occurred to me that this means that the quadrilateral mounted by 2 tangents from a point and the radii they meet is always a cyclic quadrilateral (as that circle theorem does work both ways).

When I was thinking about this I thought “that’s weird, that’s the exact same circle I was thinking about the other day when considering the angle at the centre theorem”. So the circumcentre of the triangle OAB will always generate this circle.

It then occurred to me that as the radii meet the tangents at 90 the line OC is a diameter, so its midpoint, D, is the centre of the circle. So the circumcentre of OAB will be the midpoint of the line between the centre of the circle and the point where tangents from A and B meet.

It also struck me that alternate segment theorem falls out nicely from this:

I think these are cool properties of circles. It’s nice to just sit and ponder on maths sometimes, and investigate stuff you’ve not really thought about.

If you’ve been pondering anything recently I’d love to hear about it. Also, if you’ve got any cool circle or circle theorem properties o might not know I’d love to hear them too.

Circles on a line

I saw this lovely question from Mr Gordon (@MrGordonMaths) the other day:

I looked at it and even though it said it was GCSE maths only it didn’t look at all obvious how to find an answer. It did look interesting though, I wondered how my y11s would get on with it. I thought I’d give it a try:

As always I drew a sketch:

I was looking at straight line shapes I could draw and realised the trapezium was the better option in this case:

From here it was a bit of Pythagoras’s theorem:

Which gave me all I needed for the final answer, which is 1:4. (Obviously I discounted the trival R=0 as it doesn’t make sense in the context of the question).

A nice little puzzle that I can see could be rather taxing for students despite using only concepts they will have learned in KS3 and 4. It’s the sort of question that can really help with problem solving ability and is one i will definitely try on my year 11s when we are back fully.

I’d love to see how you solved this one, especially if you took a different approach.