Posts Tagged ‘Algebra’

A lovely circle problem – two ways

December 7, 2017 3 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.

A bizarre solution

October 9, 2015 5 comments

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

September 23, 2015 2 comments

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.

Bicycle Puzzle

September 12, 2015 5 comments

This week’s puzzle from Chris Smith is a nice contextualised simultaneous equations puzzle I intend to use next week.


It boils down to:

2X + B + T = 135
2X + 2B + 3T = 269
X + B + T = 118

Via elimination using 1 and 3 we can see that X (number of tandems) is 17.

That gives:

34 + B + T = 135
34 + 2B + 3T = 269
17 + B + T = 118

1 and 3 rearrange to the same leaving:

B + T = 101
2B + 3T = 235

Elimination gives T (number of Tricycles) to be 33 which leave B (number of Bicycles) to be 68.

A nice little problem.

Indices Help

February 16, 2015 Leave a comment

This is the second post in a series of posts which I’m writing to appear on a website for our A level students. You can see the first one, on fractions, here. After sharing the first one I got some great feedback and have some areas to improve it, if you see anything missing from this, or feel I need to clarify / remove anything, please let me know.

Indices, or index numbers. That’s those little numbers that appear raised just after a number or a letter. You know, the “2” that means squared, the “3” that means cubed, etc. They are quite often referred to as “powers”. They are sometimes written like this 3^2 means 3 squared. The power is called the index number (in this case 2) and the number being raised is the base number (in this case 3). Indices are pretty important in mathematics, and you will need to be very well versed in them to be successful on the A level maths course.

What are they all about?

In the first instance, they are the numbers of times a number is multiplied by itself. 2^2 means 2 x 2, 2^3 means 2 x 2 x 2, etc. This means that 2^1 is just 2. Which is something some people forget, don’t be one of those people. Repeat after me: “anything to the power 1 is itself, anything to the power 1 is itself….” 

But what do we do with them?

We need to be able to handle indices. We will save ourselves a lot of time later if we can simplify expressions using them, and it is essential for calculus. So here are “The Rules”, but remember, they only work with the same base number (or letter)….

Rule 1

The first rule of indices is, you don’t talk about indices…. Sorry, I couldn’t resist. Is Fight Club even a film that people still watch?

Rule 1 is: “When multiplying, add the powers“. (I say rule 1, that’s what I call it, not its official name…)

This rule is fairly intuitive. If you have 2^2 x 2^3 then you have (2 x 2) x (2 x 2 x 2) which is the same as 2 x 2 x 2 x 2 x 2, which is by definition 2^5. Try some yourself and see.

Rule 2

Again, this is fairly intuitive. The rule is: “When dividing with indices, subtract powers.

So 3^3/3^2 = 3

This is because we are “cancelling” common factors. Taking this as a fraction we’d have a numerator of 3 x 3 x 3 and a denominator or 3 x 3, so we divide top and bottom to give 3/1 which is just 3.

Rule 3

This rule involves raising a power to a power. Consider the problem (x^3)^2,  this means x^3 multiplied by itself, which gives x^6 (using rule 1). The “shortcut” here is to notice that because of the way multiplication works, “when raising a power to a power you multiply“.

A real common mistake on this type of problem occurs when you get something of the form (2y^4)^3. Often people will evaluate that as 2y^12, but that’s wrong. Don’t be one of those people.

(2y^4)^3 = 2y^4 x 2y^4 x 2y^4

So you get 8y^12.


Rule 4

Negative powers are the reciprocal of positive powers

This follows from rule 2, 3^4/3^6 gives 3^(-2) but if we cancel common factors it gives 1/3^2, hence they are the same. This one is extremely important when we get to calculus.

Rule 5

Fractional powers are roots

Think of it like this, 9^(1/2) x 9^(1/2) = 9^1 (using rule 1). Well 9^1 = 9 so 9^(1/2) multiplied by itself is 9. And we know 3 multiplied by itself is 9. This follows for all square roots. By the same logic we can see that a power of 1/3 is a cube root, a power of 1/n is an nth root, etc. This is quite important for calculus.

Another thing you need to be able to do with fractional roots is evaluate them, and I don’t mean just unit fractions. You need to be able to evaluate stuff like 32^(3/5).

This isn’t as hard as it looks, you just need to tackle it in stages. Split the fraction up using rule 3:

32^(3/5) = (32^(1/5))^3 always do the root first, it makes the number easier to deal with.

In this case 32^(1/5) = 2 and 2^3 = 8 so 32^(3/5) = 8. Questions like this do come up in a level papers, and they come up in the non calculator c1 paper, so it’s handy to know the first 10 powers of 2, 3 and 5. If you are struggling to remember them in an exam, you can work them out and write them down.

The final note on fractional indices is that when they are involved in problems using the other rules you deal with then in the same was as any fraction problem. See this page for help in fractions.

Rule 6Anything to the power 1 is itself” – as mentioned before.

Rule 71 to any power is 1

This is straightforward. 1×1=1 no matter how many times you repeat it.

Rule 8 – “Anything, except 0, to the power zero is 1

This is one people sometimes struggle to get their heads around. An nice way to think of it is this: 2^2/2^2 = 2^0 but 4/4=1. You can try with any base and any power, this always works. The only exception comes when the base is 0. 0^0 is undefined (or indeterminate), in the same way that dividing by zero is.