## Proof by markscheme

While marking my Y11 mocks this week I came across this nice algebraic proof question:

The first student had not attempted it. While looking at it I ran through it quickly in my head. Here is the method i used jotted down:

I thought, “what a nice simple proof”. Then I looked at the markscheme:

There seemed no provision made in the markscheme for what I had done. *(Edit: It is there, my brain obviously just skipped past it)* *How did you approach this question? Please let me know via the comm*ents *or social media.*

Anyway, some of my students gave some great answers. None of them took my approach, but some used the same as the markscheme:

And one daredevil even attempted a geometric proof…….

## A lovely circle problem – two ways

So, I was working with some year 12s on a few problems around circles out of the new Pearson A Level textbook. (Incidentally, it’s this book, and I think it’s probably the best textbook I’ve come across. I would certainly recommend it.)

This question appears in a mixed exercise on circles:

It’s a lovely question. Before reading on, have a go at it – or at least have a think about what approach you’d take – as I’m going to discuss a couple of methods and I’d be interested to know how everyone else approached it.

Method 1:

I looked at this problem and saw right angled triangles with the hypotenuse root 52. I knew the gradient of the radii must be -2/3 as each radius met a tangent of radius gradient 3/2. From there it followed logically that the ratio of vertical side : horizontal side is 2 : 3.

Using this I could call the vertical side 2k and the horizontal side 3k. Pythagoras’s Theorem then gives 13k^2 = 52, which leads to k^2 is 4 and then k is 2 (or -2).

So the magnitude of the vertical side is 4 and of the horizontal side is 6.

From here it follows nicely that p is (-3,1) and q is (9, -7).

Finally there was just the case working out the equation given a gradient and a point.

L1: y – 1 = (3/2)(x +3)

2y – 2 = 3x + 9

3x – 2y + 11 = 0

L2: y + 7 = (3/2)(x -9)

2y + 14 = 3x – 27

3x – 2y – 41 = 0

I thought this was a lovely solution, but it seemed like a rather small amount of work for an 8 mark question. This made me wonder what the marks would be for, and then it occurred to me that perhaps this wasn’t the method the question writer had planned. Perhaps they had anticipated a more algebraic approach.

Method 2:

I had the equation of a circle: (x – 3)^2 + (y + 3)^2 = 52. I also knew that each tangent had the equation y = (3/2)x + c. It follows that if I solve these simultaneously I will end up with a quadratic that has coefficients and constants in terms of c. As the lines are tangents, I need the solution to be equal roots, so by setting the discriminant equal to zero I should get a quadratic in c which will solve to give me my 2 y intercepts. Here are the photos of my workings.

As you can see, this leads to the same answer, but took a lot more work.

*I’d love to know how you, or your students, would tackle this problem.*

## Proving Products

Just now one of the great maths based pages I follow shared this:

So naturally I figured I would have a go. I thoughts just get stuck in with the algebra and see what happens, normally a good approach to these things.

My first thought was that if I use 2n – 3, 2n -1, 2n + 1 and 2n +3 then tgere would be less to simplify later. I know that (2n + 1)(2n – 1) = 4n^2 – 1 and (2n – 3)(2n + 3) = 4n^2 – 9 so I multiples these together.

(4n^2 – 1)(4n^2 -9) = 16n^4 – 40n^2 + 9

I thought the best next move would be to complete the square:

(4n^2 – 5)^2 – 16

This shows me that the product of 4 consecutive odd numbers is always 16 less than a perfect square and as such that the product of 4 consecutive odd numbers plus 16 is always a square.

(4n^2 – 5)^2 – 16 + 16 = (4n^2 – 5)^2

A nice little proof to try next time you teach it to your year 11s.

## A lovely angle puzzle

I’ve written before about the app “Brilliant“, which is well worth getting, and I also follow their Facebook page which provides me with a regular stream questions. Occasionally I have to think about how to tackle them, and they’re excellent. More often, a question comes up that I look at and think would be awesome to use in a lesson.

Earlier this week this question popped up:

What a lovely question that combines algebra and angle reasoning! I can’t wait to teach this next time, and I am planning on using this as a starter with my y11 class after the break.

The initial question looks simple, it appears you sum the angles and set it equal to 360 degrees, this is what I expect my class to do. If you do this you get:

*7x + 2y + 6z – 20 = 360*

*7x + 2y + 6z = 380 (1)*

I anticipate some will try to give up at this point, but hopefully the resilience I’ve been trying to build will kick in and they’ll see they need more equations. If any need a hint I will tell them to consider vertically opposite angles. They should then get:

*2x – 20 = 2y + 2z (2)*

*And*

*3x = 2x + 4z (3)*

I’m hoping they will now see that 3 equations and 3 unknowns is enough to solve. There are obviously a number of ways to go from here. I would rearrange equation 3 to get:

*x = 4z (4)*

Subbing into 2 we get:

*8z – 20 = 2y + 2z*

*6z = 2y + 20 (5)*

Subbing into 1

*28z + 2y + 6z = 380*

*34z = 380 – 2y (6)*

Add equation (5) to (6)

*40z = 400*

*z = 10 (7)*

Then equation 4 gives:

*x = 40*

And equation 2 gives:

*60 = 2y + 20*

*40 = 2y*

*y = 20.*

From here you can find the solution x + y + z = 40 + 20 + 10 = 70.

A lovely puzzle that combines a few areas and needs some resilience and perseverance to complete. I enjoyed working through it and I’m looking forward to testing it out on some students.

*Cross-posted to Betterqs here.*

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October 9, 2015
5 comments
September 23, 2015
2 comments
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## A bizarre solution

On twitter earlier I saw this picture:

It was tweeted by “Mathster”, an app which claims to be “a complete solution for UK and IB curricula”. The picture had me scratching my head. Why would you add the unnecessary step one into preceding? Why would you refer to “moving terms”? Why? Why? Why?!

By saying “move terms” there is no explanation for changing sign, so learners may just move the term as is, which makes it perhaps even worse than the fabled “magic bridge”! There’s no mention of what’s actually going on! The final step is explained well, and I don’t know if that makes it better or worse!

## Area Puzzle 21

I came across the following area puzzle on Ed’s (@solvemymaths) site.

I found it quite an interesting idea and had a little go at solving it.

First I considered an equilateral triangle, all the angles are 60 degrees so we can see the area would be (x^2 sin (60))/2. As sin 60 = 3^(1/2)/2 it was easy enough to solve.

I then realised I could have used Heron’s Formula, so did it that way too. Luckily I got the same answer:

The square was a tough one:

Then I considered a pentagon, I wasn’t sure how to approach it at first, so I reverted to my favourite shape, the triangle. Split the pentagon into 5 congruent isosceles triangles and solved with trig.

Wolfram alpha gave me the lovely exact answer:

I then considered the hexagonal case, which is really just 6 equilateral triangles.

I approached the decagon in the same manner I had the Pentagon.

And got an equally lovely exact answer:

I enjoyed working through these, and thought it would make a nice lesson to build resilience and cognitive activation. I also thought about what else could be done. What does the sequence of side lengths generated look like? Could an equation be formed to describe it? What if we were looking at the areas of regular polygons with side length six, what would the sequence look like then? All these would make nice investigations.

## Simultaneous Equations

It’s been a while since i last wrote anything here. Which says more about how busy I’ve been than my desire to write, but I hope to start writing more regularly.

This week I was teaching simultaneous equations and a student asked a question that made me think about things so I thought i would share.

I was teaching elimination method and I had done some examples with the coefficients of y having different signs and I put one on the board with the same signs and asked the class to think how we may go about solving. One of the students in the class put uo his hand after a while and said he thought he had solved it.

5x + 4y = 13

2x + 2y = 6

I asked hime to talk us through his thinking and he said “first I multipled the bottom equation by -2”

5x + 4y = 13

-4x – 4y = -12

“then I added the equations as before”

x = 1

“Then I subbed in and solved.”

2 + 2y = 6

2y = 4

y = 2

“so the point of intersection is (1,2)”.

This wasn’t what I was expecting. I was expecting him to have spotted we could subtract instead, but this method was clearly just as correct. It wasn’t something I had considered as a method before this, but I actually really liked it as a method and it led to a good discussion with the class after another student interjected with her solution which was what I expected, to multiply by 2 and subtract.

It was a great start point to a discussion where the students were looking at the two methods, and understanding why they both worked, the link between addition of a negative and subtracting a positive and many more.

I was wondering, does anyone teach this as a method? Have you had similar discussions in your lessons? What do you think of it?## Share this via:

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