## 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.

Categories: #MTBoS, A Level, Maths

## Revision, past papers and clarity of instruction

Recently I gave some of my year 12 students some practice papers as homework to aid their revision. When they brought them back in one of them had literally copied to  entire markscheme into to the answer paper, while another had watched a walk through and written up the answers with notes on how to do it next to each line of working. When I discussed with the class that there was absolutely no point in blindly copying the markscheme and handing it in as a) I already know the markscheme has the correct answers, and b) this isn’t getting them thinking about the maths and it really isn’t going to help them improve, they were all keen to point out that the other student had used a youtube walkthough.

This led to an interesting discussion with them. At first they couldn’t see a difference. I explained that actually, watching a walkthough and making notes on each line of working about what you are doing and why is actually a great way of revising. It is basically the same as me doing a walking talking mock with them. It allows them to see how a more experienced mathematician would tackle the questions and allows them to build their own thought process. So when a past paper is given to aid revision purposes, this is actually a decent use of it.

This led me to thinking about the instructions I had set. Those that copied the markscheme seemed to think the point of the homework was for them to get as many correct answers on it as possible. But actually, the point of the homework was for them to get as many correct answers on their upcoming assessment as was possible. A key difference, given that they will be doing their assessment in exam conditions rather than at home with textbooks and the internet freely available.

Perhaps I should have been more explicit with my instructions, using papers and exam questions for a variety of uses over the course of the year perhaps causes confusion if there is not additional information. I usually use exam questins and exam style questions in class and for home work to assess how well they understand toopics as we go along, I also use exams for data collection points, and then there is the use for revision. I think I need to work on explaining what each one is for when I’ve set it.

I’ve given them another paper to use as practice this week, I just hope that the discussion has an effect and that they use this one to help them prepare for the exam better, rather than to be more proficient copiers!

Categories: #MTBoS, A Level, Commentary, Exams, GCSE

## A pythagorean problem

Came across another nice puzzle from Mr Gordon (@MrGordonMaths): First thing I noticed was that it was a pythagorean triple. My initial thought was that there might be a solution involving circle theorems, but then I realised that as an area was given this might be the best route.

As angle QPS IS 90, then the area of triangle QPS is 24 (6×8/2). That means a perpendicular from P to QS must be 4.8 (as QS is 10 and would be the base on this instance.

The area PRS is given as 8 so the area of triangle QPR must be 16 (24 – 8).

This means that 4.8x/2 must be 16 (where x is the length QR. So x = 32/4.8 = 6 ⅔ so length RS is 3⅓.

At this point I realised that I’d actually gone a slightly longer than necessary way. As 4.8 is also the perp height of PRS when RS is the base I could have used that triangle. 4.8y/2 = 8 where y is length RS, so y = 16/4.8 = 3⅓.

I then realized QPR and PRS were both triangles with the same perpendicular height, and that as the area for each triangle is bh/2 then the ratios of their areas would be the same aa their bases. So as its 16:8 (or 2:1) all I really had to do was split 10 in the ratio 2:1 to get 6⅔:3⅓, and pick the smaller one as the length PR.

A lovely problem with a nice solution.

Categories: #MTBoS, KS3

## Accumulator maths

Earlier today I was discussing and thinking about football accumulators, and accumulators in general. In case you don’t know what one is here is a quick overview. You basically pick a set number of bets and put an initial stake on, then if your first bet wins the winnings and stake roll over to the next bet etc.

The idea behind them is quite interesting. The more bets within your accumulator, the more you can win and the growth can be exponential.

For instance, if you backed a number of football teams all at 2-1. If you put a quid on, and bet on the one match, you’d finish with 3 quid (£1 stake returned and £2 winnings). If it was 2 matches then that 3 quid would roll to the 2nd match and if they won to you’d end up with 9 quid. A third match and you’d have 27, a 4th and you’d have 81n a 5th 243, a 6th 729, 7th 2187 8 matches 6561 etc. Its a geometric series.

I thought about this and thought it might prove an interesting real live discussion on exponential growth and geometric series. You could see how quickly these things would grow. As most accumulators aren’t all the same odds you could discus how these models change with different amounts and this would lead to nice discussions around commutativity and the like.

I then wondered if this was something that should be discussed in a lessons. Gambling can become an addiction and it can ruin lives. It might have already affected the lives of students in our care, and discussion on it in lessons might be seen as promoting it.

I then started thinking about some of the topics we do teach, and the origins of it. Vast swathes of the maths we teach stems from mathematicians trying to get an advantage in some game of chance or other. And although we might not talk about the gambling we still teach the maths that came about from it.

The maths that comes from accumulators is very interesting, as is a lot of maths with roots in gambling. I would love to discuss it, but think it’s a topic to be wary of. I’d love to hear your views. Do you discuss this sort of thing in your lessons? Do you manage to do it in a way that doesn’t promote gambling? Do you think we should leave it out of lessons? Please let me know in the comments or via social media.

Categories: #MTBoS, Teaching

## Envelope puzzle and neat little square

I’ve been looking through my saved puzzles again and I found this nice little one in the maths newsletter from Chris Smith (@aap03102): It’s a nice little question that took me some thinking about.

First I considered the half squares with hypotenuse 2. As these are isoceles RATs, that means their side length is rt2 so each has an area of 1.

Then I thought about the half square with hypotenuse 3. Again it’s an isoceles RAT so pythagoras’s theorem gives us a sidelength of 3/rt2. So an area of 9/4.

These 3 half squares add to 17/4. The area of the rectangle is 6 so the part not covered by the half squares must be 7/4.

When thinking about the overlap, we need to consider what that means. I assume it means the bit that goes over the others, so in this case 9/4 (area of the halfsquare) – 7/4 (area of the gal). Which is 1/2 cm².

I think this is an amazing little question that I cant wait to try out in class. It also got me thinking about the square with diagonal of 2 and area of 2. It’s an interesting square really, and the only one where we see this. Consider a square, side length, x. The diagonal is (2x²)^½, the area is x². If we equate the. And square both sides we have 2x² = x⁴ , or x⁴ – 2x² = 0, so x²(x² – 2) = 0. This generates a solution when x = 0 (which is trivial and discountable as no square can exist without side length. We also get positive and negative square roots if 2. But we can ignore the negative as lengths in this case are scalar, so we have one answer. It’s a neat little square.

Categories: #MTBoS, KS3, SSM, Starters

## 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.

Categories: #MTBoS, KS2, KS3, Maths, Starters Tags: , , ,

## A surprising fraction?

The other day I saw this tweet from John Rowe (@MrJohnRowe): I looked at the picture and decided the answer was probably a half. And thought about it a bit more. The 4 small circles in the top left should be the same as the white quarter circle below, and the 4 quarter circles in the top right should fit over the white circle below them. This was interesting to me, and I thought I’d look at the algebra.

Looking at the top left corner, the circles involved there have the smallest radius, so we will call that radius r. That means each circle has a radius of pi r².

Below it we have a square the same size and a quarter circle radius 4r, so the white area is (16pi r²)/4 or 4pi r² this is the same as the pink area in the square above. We can see from this that half of the left rectangle is pink. Or we can continue with our algebra. The area of each square is 16r² (its sidelength is 4r) so the pink but here is 16r² – 4pi r², so if we add this to the bit above it we have 16r² shaded (and 16r² white).

The top right has 4 shaded quarter circles, each with radius 2r, so the total shaded area is 4pi r². Below it the white circle has the same radius so same area 4pi r², again that makes the shaded area 16r² – 4pi r² so the shaded area in the right rectangle 16r², and in the big square 32r². The total area of the big square if 64r² (radius is 8r) so the shaded fraction is a half. Which is nice.

Now I know I started by saying I thought it looked like a half, which is what I did think at the time I first saw it. But I still think it’s a surprise fraction. I’ve done a lot of geometry puzzles. And many have included shapes like this, so when I look at this that knowledge helps me see. Most students would not have done anywhere near the amount of puzzles I have so wont have that foresight, and I think it would be a very surprising result that could open many up to the wow factor. I also think that it might be a good starting point for some rich class discussions.

I think it’s a great visual and a lovely answer.

Categories: #MTBoS, Commentary, GCSE, KS3, SSM, Starters

## Step perimeters and infinite coasts

June 22, 2020 1 comment

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

Categories: #MTBoS, GCSE, KS3, Maths

## 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. 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.

## Algebraic sequences puzzle

I came across another Mr Gordon (@MrGordonMaths) brainteaser that I liked the look of:  It looked interesting to me. It’s not a type of sequence problem I’ve seen lots before and I thought it worth exploring.

I considered the first 2 terms. From this I can form an equation:

x² – 24x +144 = (3x -2)/7

7x² – 168x + 1008 = 3x -2

7x² – 171x +1010 = 0

Which facorises quite nicely

(7x-101)(x-10)

I expect that means x = 10 is our solution as surely the intention is an integer sequence.

This would give us -2, 4, 16 as our known terms. So we know that

10a+ 6 = 16²

10a + 6 = 256

10a = 250

a = 25. Which is a nice answer and a nice solution.

Although it doesn’t actually specify integer answers, so what happens if we use x = 101/7 ?

17/7 , 289/49 , 317/14 son the third term doesn’t work. So this x value doesn’t generate a valid solution.

I then wondered what would happen if I used the second and third term to generate an equation:

((3x-2)/7)² = (3x+2)/2

2(3x-2)² = 49(3x+2)

18x² – 24x + 8 = 147x +98

18x² – 171x – 90 = 0

(18x +9)(x-10) = 0.

This one again generates 2 solutions for x, but only one matches the other solution. So the common x value must be the only solution.

This was a lovely puzzle that I enjoyed thinking and working through. If you have a different solution I’d love to see it.

Categories: #MTBoS, GCSE, Starters