## Constructions

One topics I have never been a fan of teaching is constructions. I think that this is due to a few factors. Firstly, there is the practical nature of the lesson, you are making sure all students in the class have, essentially, a sharp tool that could be used to stab someone. I remember when I was at school a pair of compasses being used to stab a friend of mines leg and this is something I’m always wary of.

Secondly, the skill of constructing is one that I struggled to master myself. I was terrible at art, to the point where an art teacher kept me back after class in year 8 to ask why I was spoken about in the staffroom as the top of everyone else’s class but was firmly at the bottom of his. I explained that I just couldn’t do it, although it was something I really wished I could do. He was a lovely man and a good teacher and he offered to allow me to stay back every Monday after our lesson and have some one to one sessions. I was keen and did it, this lasted all through year 8 and although my art work never improved my homework grades did, as he now knew I was genuinely trying to get better. I have always assumed the reason I am poor at art is some unknown issue with my hand to eye coordination, and I have always blamed this same unknown reason for struggling sometimes with the technical skills involved in constructions. Since coming into teaching I have worked hard to improve at these skills, and I am certainly a lot lot better than I used to be, but I still feel I have a way to go to improve.

For these reasons I chose to go to Ed Southall’s (@solvemymaths) session “Yes, but constructions” at the recent #mathsconf19. Ed had some good advice about preparation and planning, but most of that was what I would already do:

*Ensure you have plenty of paper, enough equipment that is in good working order, a visualiser etc.*

*Plan plenty of time for students to become fluent with using a pair of compasses before moving on.*

He then moved on to showing us some geometric patterns he gets students to construct while becoming familiar with using the equipment. Some of these were ones I’d not considered and he showed us good talking points to pick out and some interesting polygons that arise. The one I liked best looked like this:

*This is my attempt at it, I used different coloured bic pens in order to outline some of the shapes under the visualiser.*

The lesson was successful, the class can now all use a pair of compasses and we managed to have some great discussions about how we knew that the shapes we had made were regular and other facts about them.

Next week we need to move on to looking at angle bisectors, perpendicular bisectors, equilateral triangles, and the such. I hope to get them constructing circumcircles of triangles, in circles of triangles and circles inscribed by squares etc.

Here are some more of my attempts at construction:

“Constructing an incircle” – I actually did this one in Ed’s session!

“A circumcircle” – I drew the triangle too big and the circle goes off the page. Interesting to note the centre is outside the triangle for this one.

“A circle inscribed within a square” – this is difficult. Constructing a square is difficult and that is only half way there if that. This is the closest I have got so far and two sides are not quite tangent.

“A flower” – nice practice using a pair of compasses and this flower took some bisectors too.

*If you have any ideas for cool things I can construct, and that I can get my students to construct, please let me know in the comments or on social media.*

## Is one solution more elegant?

Earlier this week I wrote this post on mathematical elegance and whether or not it should have marks awarded to it in A level examinations, then bizarrely the next day in my GCSE class I came across a question that could be answered many ways. In fact it was answered in a few ways by my own students.

Here’s the question – it’s from the November Edexcel Non-calculator higher paper:

I like this question, and am going to look at the two ways students attempted it and a third way I think I would have gone for. Before you read in I’d love it if you have a think about how you would go about it and let me know.

**Method 1**

Before I go into this method I should state that the students weren’t working through the paper, they were completing some booklets I’d made based on questions taken from towards the end of recent exam papers q’s I wanted them to get some practice working on the harder stuff but still be coming at the quite cold (ie not “here’s a booklet on sine and cosine rule, here’s one on vectors,” etc). As these books were mixed the students had calculators and this student hadn’t noticed it was marked up as a non calculator question.

He handed me his worked and asked to check he’d got it right. I looked, first he’d used the equation to find points A (3,0) and D (0,6) by subbing 0 in for y and x respectively. He then used right angled triangle trigonometry to work out the angle OAD, then worked out OAP from 90 – OAD and used trig again to work out OP to be 1.5, thus getting the correct answer of 7.5. I didn’t think about the question too much and I didn’t notice that it was marked as non-calculator either. I just followed his working, saw that it was all correct and all followed itself fine and told him he’d got the correct answer.

**Method 2**

Literally 2 minutes later another student handed me her working for the same question and asked if it was right, I looked and it was full of algebra. As I looked I had the trigonometry based solution in my head so starter to say “No” but then saw she had the right answer so said “Hang on, maybe”.

I read the question fully then looked at her working. She had recognised D as the y intercept of the equation so written (0,6) for that point then had found A by subbing y=0 in to get (3,0). Next she had used the fact that the product of two perpendicular gradients is -1 to work put the gradient of the line through P and A is 1/2.

She then used y = x/2 + c and point A (3,0) to calculate c to be -1/2, which she recognised as the Y intercept, hence finding 5he point P (0,-1.5) it then followed that the answer was 7.5.

A lovely neat solution I thought, and it got me thinking as to which way was more elegant, and if marks for style would be awarded differently. I also thought about which way I would do it.

**Method 3**

I’m fairly sure that if I was looking at this for the first time I would have initially thought “Trigonometry”, then realised that I can essential bypass the trigonometry bit using similar triangles. As the axes are perpendicular and PAD is a right angle we can deduce that ODA = OAP and OPA = OAD. This gives us two similar triangles.

Using the equation as in both methods above we get the lengths OD = 6 and OA = 3. The length OD in triangle OAD corresponds to the OA in OAP, and OD on OAD corresponds to OP, this means that OP must be half of OA (as OA is half of OD) and is as such 1.5. Thus the length PD is 7.5.

**Method 4**

This question had me intrigued, so i considered other avenues and came up with Pythagoras’s Theorem.

Obviously AD^2 = 6^2 + 3^2 = 45 (from the top triangle). Then AP^2 = 3^2 + x^2 (where x = OP). And PD = 6 + x so we get:

*(6 + x)^2 = 45 + 9 + x^2 *

x^2 + 12x + 36 = 54 + x^2

12x = 18

*x = 1.5*

Leading to a final answer of 7.5 again.

Another nice solution. I don’t know which I like best, to be honest. When I looked at the rest of the class’s work it appears that Pythagoras’s Theorem was the method that was most popular, followed by trigonometry then similar triangles. No other student had used the perpendicular gradients method.

I thought it might be interesting to check the mark scheme:

All three methods were there (obviously the trig method was missed due to it being a non calculator paper). I wondered if the ordering of the mark scheme suggested the preference of the exam board, and which solution they find more elegant. I love all the solutions, and although I think similar triangles is the way I’d go at it if OD not seen it, I think I prefer the perpendicular gradients method.

*Did you consider this? Which way would you do the question? Which way would your students? Do you tuink one is more elegant? Do you think that matters? I’d love to know, and you can tell me in the comments or via social media!*

Cross-posted to Betterqs here.

## Hippocrates’s First Theorem

Over the half term I was doing some reading for my MA and I happened across Hippocrates’s First Theorem. (Not THAT Hippocrates, THIS Hippocrates!)

Here is the mention in the book I was reading (Simmons 1993):

It’s not a theorem I’d ever come across before, and it doesn’t seem to have any real applications, however it is still a nice theorem and it made me wonder why it worked, so I set about trying to prove it.

First I drew a diagram and assigned an arbitrary value to the hypotenuse of triangle A.

I selected 2x, as I figured it would be easier than x later when looking at sectors.

I then decided to work out the area of half of A.

A nice start – splitting A into two smaller right angled isosceles triangles made it nice and easy.

I then considered the area b. And that to find it I’d need to work out the area the book had shaded, I called this C.

Then the area of B was just the area of a semi circle with the area of C subtracted from it:

Which worked out as the area of the triangle (ie half the area of A**)** as required.

This made me wonder if it worked for all triangles that are inscribed in semi circles this way – ie the areas of the semicircles on the short legs that fall outside the semicircle on the longest side equal the area of the triangle.

My first thought was that for all three vertices to sit on the edge of a semi circle in this was then the triangle must be right angled (via Thales’s Theorem).

I called the length eg (ie the diameter of the large semi circle and the hypotenuse of efg) x and used right angled triangle trigonometry to get expressions for the two shorter sides ef and fg. Then I found the area of the triangle:

I then considered the diagram, to see where to go next:

I could see that the shaded area needed to be found next, and that this was the area left when you subtract the triangle from the semicircle.

I could now subtract this from the two semi circles to see if it did equal the triangle.

Which it did. A lovely theorem that I enjoyed playing around with and proving.

*I think there could be a use for this when discussing proof with classes, it’s obviously not on the curriculum, but it could add a nice bit of enrichment.*

*Have you come across the theorem before? Do you like it? Can you see a benefit of using it to enrich the curriculum?*

**Reference:**

Simmons M, 1993, *The Effective Teaching of Mathematics*, Longman: Harlow

## Problem solving triangles

Brilliant – a lovely puzzle app and a source of many little puzzlers if you follow their Facebook page. The other day, I came across this one:

It looked like it might be interesting so I screen shot it and thought, “I’ll have a go at that later, when I’ve got a pen. It’s bound to be nice using a bit of trigonometry and angle reasoning.”

But as I thought about it I realised I didn’t need paper. The hypotenuse of the large triangle is easy enough to find (6rt2) using Pythagoras’s Theorem. You can deduce the size of the green square is then 2rt2 as the big triangle is isosceles meaning the angles are 90, 45 and 45, as the square is only right angles then the little blue triangles in the 45 degree corners must also be isosceles. Thus the two blue and the green segments of the hypotenuse are equal.

The area of the square is then way to find (8) by squaring 2rt2. A nice easy puzzle.

My first thought had been that it would take a bit of working out, but it didn’t, it was a very straightforward question once I got going. That got me thinking, problem solving is something that I would love my students to get better at and I’m hoping to launch a puzzle of the month in January. This sort of puzzle is ideal. It will require then to build their perseverance skills as well as their problem solving skills and will give them a mental workout. I’m going to use this as a starter this week to warm them up.

*This post was cross posted to the BetterQs blog here.*

## The trouble with prisms

Wherever I see incorrect maths, it annoys me, whether it be in election material, newspapers or anywhere else. But the place where it annoys me most is the maths classroom. I don’t mean students getting the answers wrong, that’s an invaluable part of the learning experience. What I mean is when teachers get it wrong. This happens more than people would expect. I’ve written before about people teaching things wrong (ie rounding or the order of operations) but today is want to discuss a different annoyance.

Take a look at this:

It’s from a resource I downloaded from the TES website. The resource itself was pretty good, but this was one of a number of questions that infuriated me. Have you noticed why? Take another look.

Yes, indeed. The right angled triangle that forms the cross section of this triangular prism is that well known Pythagorean Triple the “4,9,10” triangle. Never heard of it? Neither have I! That’s because 4^2=16, 9^2=81 and 10^2=100. And 16+81 is very definitely 97, which in turn is very definitely NOT 100. It’s not even as though it’s hard to generate triples!

This sort of thing is lazy, if it had been put in front of me, as a student I’d have called a teacher out on it. The first time I saw something like this was during a micro teaching assignment while on my PGCE. The person in that case was rusty! I’ve seen it a couple of times with trainees or NQTs during observations, again these can be excused.

I even realise that experienced teachers can make innocent mistakes, but please, please, please check these things. Especially for triangular prisms, as this is **THE **area that I see this happening again, and again and again.

*Have you encountered something like this? Do you get as angry as me about it? Do you think it doesn’t matter and I’m being overly pedantic on this? Please let me know.*

## A surprising find

The other day I my timehop showed me this lovely little post from last year. It includes “Heron’s Formula” for calculating the area of a triangle, as I read it I remembered thinking it was a little strange that not many people had heard of it before.

Today I was looking through a number of textbooks trying to find a decent set of questions on area, perimeter and volume for my year nines as I wanted to consolidate their learning at the start then move onto surface area. *I’m not a fan of textbook misuse- ie “copy the example and try the questions” but I do sometimes use them for exercises as we have a very limited printing budget and some of them have superb exercises. For a fuller picture on.my view of textbooks, read this.*

I was looking in one of my favourite textbooks:

And I happened across this:

There it is! Plain as day! Heron’s Formula! In a KS3 textbook!

I was disappointed that its function was described and its name wasn’t and there was no mention of why this worked. It basically reduces the question down from a geometry one to a purely algebraic substitution task and I would question the appropriateness of including it in an exercise on area, but still, I was incredibly exciting to find it there!

*Are you a fan of Heron’s Formula? Had you even heard of it? Do you have a favourite textbook? I’d love to hear your views.*

## Circles and Triangles

Regular readers will know that I love a good puzzle. I love all maths problems, but ones which make me think and get me stuck a bit are by far my favourite. The other day Ed Southall (@solvemymaths) shared this little beauty that did just that:

I thought “Circles and a 3 4 5 triangle – what an awesome puzzle”, I reached for a pen an paper and drew out the puzzle.

I was at a bit of a loss to start with. I did some pythag to work some things out:

Eliminated y and did some algebra:

Wrote out what I knew:

And drew a diagram that didn’t help much:

I then added some additional lines to my original diagram:

Which made me see what I needed to do!

I redrew the important bits (using the knowledge that radii meet tangents at 90 degrees and that the line was 3.2 away from c but the center of the large circle was 2.5 away):

Then considered the left bit first:

Used Pythagoras’s theorem:

Then solved for x:

Then briefly git annoyed at myself because I’d already used x for something else.

I did the same with the other side to find the final radius.

A lovely puzzle using mainly Pythagoras’s theorem, circle theorems and algebra so one that is, in theory at least, accessible to GCSE students.

I hope you enjoyed this one as much as I did!

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