Posts Tagged ‘Algebra’

Algebraic sequences puzzle

June 18, 2020 Leave a comment

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

Which leads to

7x² – 168x + 1008 = 3x -2

7x² – 171x +1010 = 0

Which facorises quite nicely


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.

A short area problem

June 9, 2020 2 comments

At the weekend I saw this nice question:

It came from @lasalleed

And it looked interesting. Initially I didn’t know what to do, but I realized if I looked at the region with the area 95 I should be able to get an answer:

It was quite a simple form and solve a quadratic problem in the end.

When I got my answer I considered the other solution, the negative 19. If we moved the line from the left hand side do it was 19 away from the other side it would give us a rectangle area 95 on the right of the red rectangle. That’s quite an interesting thing.

19 is also the side length of the square. Which is also interesting.

It makes sense though, you are multiplying a by (a+b) to get the area shown (95) as b is 14 then you are multiplying side length -14 by the sidelength to get 19. A different but similar solution could be obtained.

Quite nice and interesting. How did you solve it?

Categories: #MTBoS, GCSE, SSM Tags: , , ,

Polygons, Area and some generalisations

May 7, 2020 Leave a comment

On Tuesday the maths teaching world recieved the terrible news that the great Don Steward has sadly passed away. I didnt know Don personally, but friends I have who did speak of him very highly. I do know his blog, median. It is an amazing source of lesson resources and some insightful musings that have often got me thinking and have helped me plan many lessons. Without knowing it, Don has helped me (and I imagine many others) become a better maths teacher and a better mathematical thinker with his posts, and I will be forever grateful to him.The news prompted Chris Smith (@aap03102) to share this:I thought that if Don had found this puzzle interesting enough to engage in email chat about it then it was definitely something worth looking at.I drew a sketch and sid some preliminary workings:I know the areas are the same. And I’ve got a load of trapeziums so that’s easy enough to work out:I can then set areas equal to each other:This brings about the answer that it will produce 3 sections with the same area every time k is 2 bigger than n. I thought about what this looks like, theres the simplest case of course:Which gives us 3 rectangles. This pair of solutions is probably the simplest to actually work out. But I think lots of students would not even consider it as it doesn’t look like the picture in the question. I also wondered about the examples where n was bigger than 2? I.e. if n = 3 then k = 5 and this would look inverted. Would people discount this also because it didn’t look like the diagram in the drawing?I then considered this one.For some reason, while I was looking at this one another route to the solution jumped out at me. If I consider one of the top trapezia, a or b in the first diagram:Then as the area of the square is 36, each section is 12 so the solution falls out from knowing that the sum of k and 6-n is 8 (from area of a trapezium formula.k + 6 – n = 8Simplifies to:k = n + 2A nicer solution.I considered the case where n = 0 as I thought it was quite nice:My thoughts turned to the general case offered in the question. Following my original working it falls out this way:So k need to be one third of the side length of the square larger than n.We could have just considered that the trapezium again:(1/2)(k + s – n)(s/2) = (s^2)/3Which reduces tok + s – n = 4s/3So againk = n + s/3Interestingly I think equating areas is a simpler solution in the general case, even though the area of the trapezium seemed simpler when we had numbers.All this, and the thoughts on how it looked earlier made me wonder what would need to happen if we wanted the angles at the “middle” vertex were all equal (so 120).I did some preliminary sketches:I know some ratios for right angled triangles with a 30 degree angle, looking first at s = 6:Then the general:Lots of interesting maths coming out of this puzzle. I wondered whether other angles would produce other ratios. I wondered what would happen if rather than equal areas we assigned the areas other ratios. There are plenty more directions to go in, but it’s getting late so they will have to be for another post and another day.

Power Puzzle and building resilience

May 4, 2020 Leave a comment

Today’s puzzle comes from “Britain’s Brainiest Dad” Chris Smith (@aap03102). Its a puzzle that was in his awesome newsletter, but that I spotted first on Twitter:

It caught my eye for a couple of reasons. Firstly it was a puzzle that wasn’t a geometry puzzle, and I’ve been a bit geometry heavy recently. Secondly when I saw it I couldn’t see an obvious path to a solution, perhaps I should have spotted the path a bit easier, but when I forst looked I was flummoxed, and that makes me want to do a puzzle.

I wrote out the problem:

Then I thought, “I can split that root 6”:

Then I realised that there were no roots on the left hand side so powers would be better. And noticed that the 3 and the 4 could also be expressed as powers of 3 or 2:

Typing this up now, I realise I should have definitely seen the answer from here, but I didn’t.what I did was simplify it:

Then spot the route to the solution;

A nice little puzzle with a neat solution. While I feel I should have spotted the solution quicker, I think that this process is good to demonstrate how to go about solving problems of this kind when you don’t see how straight away. I took the problem and applied bits of maths I know until the solution came to me. I feel that often this is something that students struggle with. I see it in my class and I have been seeing it with my daughter during the lockdown. Often they cannot spot a way forward so think they cannot solve the problem. I’ve been working on it lots over the last few years, and most of my classes are getting better at it. But each year brings us new classes and the same issues. I find modelling processes like this is a helpful way to help build that resilience into them.

I’d be very interested to hear how you approached this problem, and whether you have any different solutions. I’d also be interested to hear what strategies you use to build resilience in your students. Please let me know either in the comments or via social media.

A nice money puzzle

April 23, 2020 Leave a comment

Here is an interesting money puzzle I came across on twitter. It’s from @puzzlecritic and I came across it via Ed Southall (@edsouthall).

My first thought was that a barn owl shopping for books is a bit strange. But then I got over that and started to think about the puzzle.

It appeared to me that there were actually more than one way to get to an answer here. But knowing Ed, I figured only one would be correct. I assumed that some would give monetary values that are impossible (I.e. answers with part pennies involved).

I thought about the various ways. We have 3 books listed in order of price, a,b and c. The first way to get a solution is that b = 3a, and c = 8b = 24a. Thus the total price 10.50 divided by 28 would give the price of the cheapest book. This would give an answer of 37.5p so I discounted it.

The next was would be that b=8a and c=3b=24a. This would lead to 10.50/33 which is a shade less than 32p. Again, not a whole number.

The third way was where b=3a and c=8a. This would lead to 10.50/12, which would be 87.5p again, not a whole number.

The fourth way would be where c=3b and c=8a. This is a little less straightforward to solve. I set a=3x for ease meaning c=24x and b =8x. This leads to x=10.50/35 which is 0.3, meaning the cheapest book, a, cost 90p.

In hindsight, I should have known that it would be the final way that generated the solution!

I enjoyed this little puzzle, it got me thinking about ratios and algebra and it strikes me as a very good puzzle to show to my students. It made me wonder if there were other methods to solve it, if there was a way to do it without trialling different solutions. I can’t see any at the minute but will continue to think about it. I also wondered about the thought process behind setting it up, how easy would it be to create a question like this that gives answers that don’t work the other ways. I’m not sure, but I do intend to ponder it further. How did you solve it? Do you have any other questions or puzzles like this? I’d love for you to let me know in the comments or on social media.

Proof by markscheme

March 16, 2019 2 comments

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 comments 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…….

Simultaneous Equations

March 10, 2019 3 comments

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?

A lovely circle problem – two ways

December 7, 2017 5 comments

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

June 26, 2017 1 comment

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

March 24, 2016 4 comments

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)


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.