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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.

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An excellent puzzle

July 18, 2017 3 comments

Today I saw this tweet:

The puzzle looked grand. Thanks to those people that tweeted at me to make sure if seen it, it’s much appreciated.

The puzzle itself is:

I drew it out and labelled a few things:

But soon realised that it’s impossible unless you make assumptions. 

With the assumption that the vertex of the triangle is at the midpoint if the line I was in a position to have a good crack at it. My first thought, as is often the case, was to run at it using right angled triangles:

My initial thought was to use right angled trigonometry, but I realised I’d probably need to approximate or use some maths software and that would take a bit of the fun out of it. I presumed I’d be able to find an exact answer in a better way.

I realised the big triangle and the green triangle were similar and I could easily work out the area of the big triangle.

I then realised I didn’t have the scale factor. I went back to rats.

Then realised I had another similar isoceles triangle to play with:

Using similarity I found the “base” of DFG and used that to find length EG. Thus giving me a scale factor between the blue and the green triangle.

As mentioned previously I knew that rt2/(SF^2) was the green area so using the scale factor of 3 I got the required area to be rt2/9. 

I would like to say that’s what I did. That’s what I see I should have done while writing this up. But it’s not what I did at the time. I took a longer way round. I got giddy with triangles:

Used  Pythagoras’s Theorem to find the peep height and found the area that way.

Luckily I got the same answer.

I then saw the same tweeter had tweeted this:

This is the same question but altered slightly in the information given and what is required as the final solution. If you make the same assumption it follows from the tan ratio that all the distances are the same, so you need to do the same to that point and then find the ratio green area / blue area. I’d done most of it above, so I finished it off:

I love this puzzle, and I hope to use it in my classes next year. I may give it to year 12 tomorrow and see if they can crack it. I think I prefer the second variation. I’d love to hear your thoughts on it, and how you solved it. Let me know in the comments or via social media. 

A little circle problem 

June 28, 2017 Leave a comment

I’ve just seen this post from Colin Beveridge  (@icecolbeveridge) answering this question:

Naturally I had a go at it before reading Colin’s solution. When I read his I found a lovely concise solution that we slightly different to mine, so I thought I’d share mine.

I started by just drawing a right angled triangle from the centre of the circle like so.

I seem to have cut off the denominator of 6 on the angle there. I know that the hypotenuse is 12-r (where r is radius) and the side opposite the known angle is r.

This means I can use the sine ratio of pi/6 to get  

r/(12-r)  = 1/2

Which leads to:

2r = 12 – r

Then 

3r = 12

r = 4

Which is the same as Colin got. 

I’ve seen questions like this on A level papers before and I know they often throw students, so I make sure I explore lots of geometry based problems and puzzles to combat this. I’d be interested to know which way you would approach this. Colin and I used a very similar approach, just differing in the point at which we introduced the 12. Which way did you do it? 

Categories: A Level, Maths Tags: , , ,

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.

Infuriatingly impossible exam questions

February 9, 2017 4 comments

Today I was working on some Vectors exam questions with my Y13 mechanics class and I came across this question:

A student had answered it and had gotten part d wrong. What he had done was this:

I have recreated is incorrect working.

Obviously he had found out when the ship was at the lighthouse, instead of 10km away. I explained this to him and started to explain how he should have tackled this when a sudden realisation angered me.

Now for those if you that didn’t work through the question, here is the actual answer:


Can you see what had me infuriated?

This is an impossible answer! If the lighthouse is on the trajectory of the ship and it will hit said lighthouse at t=3 then that would stop the ship! At the very least it would slow it down!!!! In reality it would have to avoid the lighthouse and change trajectory. Meaning the second answer, T=5, would not happen under any circumstances!

My initial thought was: “are they expecting students to spot this and discount the second answer? That’s a bit harsh.”

So I checked the markscheme:


Nope, they are looking for both answers. Argh! I can understand using a real life context in mechanics, I really can. But why not check for this sort of thing!

What do you guys think? Is this infuriating or am I just getting get up over nothing? I’d love to hear your views in the comments or via social media.

Catchy Mnemonics 

September 16, 2016 3 comments

I find most memory aids a little silly. Why learn a rhyme about horses when you can just learn the trig ratios? Why learn a rhyme about the duke of York when you can just remember the order the colours come in?

However, I find that music is a good way of remembering things. For some reason music is good for us to remember words. I can, for instance, remember the words to a great deal of 90s pop songs even though I didn’t like them and never chose to listen to them because I heard them out places and on TV so often that they got lodged in my brain forever. 

This is something I have seen work well in learning maths facts. Year on year I hear pupils sing “mean is average, mean is average…” etc in lessons to remember the averages. And I also hear a great many variations on the circle song.

Last year when I was teaching kinematics one of the students said “Sir,play the SUVAT song.” I’d not heard of the SUVAT song and he found it on you tube and we listened to it. It’s simple and it’s catchy and it really helped him and his class remember those equations. So on Tuesday I played it to one of my mechanics classes. By the end of the leson I’d heard three people sing it and it has been stuck in my head all week.

What do you think about mnemonics? Do they have a place? Have you any songs or rhymes that you use to remember things or that you encourage students to use? And do they help?

Circles and Triangles

July 15, 2016 5 comments

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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