Last week I was discussing the problem wall with a colleague and this jumped out at me, so I thought I would give it a try. It took rather longer than I’d care to admit, to be honest. I set off on a few false starts and came up with some incorrect solutions due to an incorrect solution I’d made. After a while I gave up and left it a few days before tackling it afresh. When I retackled it the process was much shorter and gave me a lovely concise solution. I will explain some of my incorrect thought processes first, and then I will explain how I got my solution. Before reading on, why not have a go yourself – I’d love to hear how you approached it either in the comments or via social media.

**The problem:**

I’m not one hundred percent sure where I sourced this question, I think it’s from the great solvemymaths website – but if it’s not please let me know.

The first thing I did was a sketch; this led me to see that what we were dealing with was a quarter circle inside a square with a triangle. It had to be a square due to tangents from a point being equal and radii being equal. I also spotted that said triangle was an 8, 15, and 17 Pythagorean triple. These were observations that would be key when I eventually got round to solving the problem.

Then I made my mistake that caused a lot of issues. I marked the point that the triangle was tangent to the circle as the midpoint of the hypotenuse. Looking back now this is such a daft thing to do. I was pretty tired and must have briefly confused tangents and chords I guess. Either way, a silly and costly mistake. Using this I tried a coordinate geometry approach and got numbers that didn’t make sense. I knew the radius would have to equal 15 + y, but I was getting values less than 15 and y could not be negative as it was just a scalar length. I tried this approach a few times from different angles but each came up the same. I was convinced my algebra was correct, so the mistake must have been somewhere else. I left it for another day. Here are some of my incorrect workings:

T

he correct way:

When I came back to the problem, I had a clearer head and as soon as I sketched it I could see the way to answer it.

The point where the hypotenuse was tangent to the circle was not the midpoint, but it could be defined in other ways. Using tangents from a point from the points where the tangent intersects the sides of the square we can see that the hypotenuse must be the sum of the distances from said intersection points to the corners that are also tangent points to the circle. I.e. the lengths I marked x and z. Thus we know that x + z is 17, as that is the length of the hypotenuse of an 8, 15, 17 triangle. We also know, as it’s a square, that x + 8 is equal to the radius (labelled y in my diagram) and that z + 15 is also equal to the radius (y). Thus we have three equations in three unknowns so an easily solvable system that gives us the answer of y = 20. And hence area = 400pi.

A nice concise solution in the end to a lovely problem that caused me far too much headache. I’m off to kick myself some more…..

]]>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.*

A couple of nice parallel lines questions that I might grow at y11 tomorrow.

Both are fairly straight forward to solve. I looked at the first one, imagines a third parallel line through the join if x and saw x must be the sum of 40 and 60 hence 100.

The second I saw an alternate angle to the 50 in the top triangle and used angle sum of a triangle is 180 to spot that x is a right angle. I glanced down at the responses and saw the vast majority had the same answers as me. That would probably have been the end of it but then I noticed this response:

The same thought process for the first one, but a significantly different approach to the second.

It made me wonder what approach others would take, and which approach my students would take. I wondered if the first problem had led this respondent into this solutions the second, and if so why it hasn’t had the same effect as me.

I don’t know if either approach is better, I just thought the differences were interesting. I’d love to hear your thoughts on it and how you would approach it.

]]>That’s interesting I thought, so I thought it have a go.

The radius is given to be 2. So we have an equilateral triangle side length 2. Using my knowledge of triangle and exact trig ratios I know the height of such a triangle is root 3 and as such so is the area.

Similarly, as the diagonal of the rectangle is 2 and the short side is 1 we can work out from Pythagoras’s Theorem that the longer side is root 3. And again it follows that so is the area.

Lastly we have the square, the diagonal is 2 and as such each side must be root 2, again this is evident from Pythagoras’s Theorem this gives us an area of 2.

Which leaves us a nice product of the areas as 6.

*I think* *that is correct, I’ve justvwoken up nd this post has been my working, so do about up if you spot an error. And I’d love to hear if youbsolved it a different way.*

Then one of them aid, “if I’m looking for two years ago, can’t I just times it by the multiplier to the power -2? Wouldn’t that work.” I thought this was an excellent thought process. The other student disagreed though, sating “no, it has to be divide.” So I thought at this point I’d better interject a little.

“Does it give you the same answer?” I asked. They both thought about it and tried it and discussed it and said yes. So I asked “does it ALWAYS give the same number?” they tried a number of scenarios using different amounts, different interest rates and different numbers of years. Eventually they had convinced themselves. “Yes, yes it is always the same.”

“So is it a valid method then?” I probed. Some more discussion, then one ventured “yes. It must be.”

“Why does it work?”, I then asked. And left them discussing it.

When I came back to the pair I asked if they could explain why it works and one of them said, “we think that it’s because multiplying by a negative power is the same as dividing by the positive version.”

]]>This set off a lengthy chain of thoughts in my head. Firstly, I was quite impressed by the fact a 5 year old could articulate all this about knowledge about shapes so well. Then I thought, does it really matter whether they call them oblongs or rectangles? Then I thought, wait a minute, why are we prohibiting the use of rectangle because it can also mean a square, but we are not prohibiting the use of oblong when it can also mean an ellipse? My chain of thought then jumped down a rabbit hole questioning whether we should actually be referring to regular or equilateral rectangular parallelograms and non – regular/equilateral parallelograms. Why are we allowing children to call a shape a triangle, when it is one possible type of triangle in a family of triangles, but not allowing them to call a shape a rectangle when it is only one possible rectangle in a family of rectangles. These thoughts stewed around in my head for a while and I thought I’d ask the twittersphere for their opinions on the matter.

These opinions fell into a couple of camps. The first cam thought that oblong was a nice enough word and they didn’t mind others using it but preferred not to themselves. The second camp felt that it was important to distinguish between an oblong and a square so important to use oblong not rectangle and the third camp thought that actually it was better to use rectangles due to the elliptical oblongs. I questioned some of the respondents from the second two groups a little further to see why they fell into these groups. Those in the second seemed unaware that the word oblong also meant ellipse and those in the third thought it was more important to excluded ellipses than squares. Stating that it was easy enough to explain away the special case that is the square.

I’ve spend rather a lot of time considering this, and am now not really sure what I think on the issue. I can’t see a problem with using a rectangle and explaining away the square as a special case. We call all triangles triangles and expand as and when required. No one bothers about calling a non-rectangular parallelogram a parallelogram, despite the fact that that could mean a rectangle. But again I’m not sure I’m massively strongly against the term oblong either. It could open up a good discussion about the term and how it could apply to ellipses, although this probably is a little too much for a year 1 classroom. I think I’m leaning towards rectangle as a preference though, as explaining away a special case is, for me, much more preferable than ignoring a whole class of oblongs.

*If you have views on this, whichever way you lean, I’d love to hear them, either in the comments or via social media.*

Immediately I called shinanegans. The 5% difference between the green and the blue looked far too big. Initially I thought it was just down to the scale starting from 25000 and the size, but looking deeper there are also 4 extra sets if 5 notches on the blue which further add a to the illusion.

All in all a terrible diagram. Poor form Microsoft. Poor form.

]]>The puzzle again:

Lovely, isn’t It?

After I published my previous post I wondered if I may have been better using a vector approach or a coordinate geometry approach. So I gave them a try.

**Coordinate Geometry**

I started by sketching the figure against an axis.

I place the origin at the centre of the circle, worked out the equation if the circle and the right leg of the triangle and solved simultaneously for x. Giving x =1 and x=1/3. These x values correspond to half the base of each triangle, which shows the scale factor from the large triangle to the small one is 1/3. As the area of the large one is rt2 this gives the area of the small as rt2/9.

I like this method, probably a little better than the one prior to it.

**Vectors**

First I sketched it out and reasoned I could work it out easy enough with 4 vectors.

I saw that I could write AC as a sum of two others:

I knew that the length of AC was 1 so I used Pythagoras’s Theorem to calculate mu. It left me with the exact same quadratic to solve. This time mu was the fraction of DB needed so was automatically the length scale factor. The rest falls out as it did before.

As well as this, Colin Beveridge (@icecolbeveridge), maths god and general legend, tweeted a couple of 1 tweet solutions. First he used trig identities:

**Trig Identities **

I assumed this was right, but checked it through to ensure I knew why was going on:

We can see beta is 2 x alpha and as such the tan value is correct. The cos value (although it is missing a negative sign that I’m sure Colin missed to test me) follows from Pythagoras’s Theorem:

This is again the scale factor as it is half the base of the small triangle and the base if large triangle is 2.

**Complex Numbers**

Then Colin tweeted this:

At first I wasn’t totally sure I followed so I asked for further clarification:

And then saw where Colin was going. I tried to work it through, by way of explaining here in a better manner.

I sketched it out and reasoned the direction of lines:

Then I normalised that and equated imaginary parts to get the same scale factor:

I am happy that is is valid, and that it shows Colin is right, but I’m not entirely sure this as the exact method Colin was meaning. He has promised a blog on the subject so I will add a link when it comes.

I like all these methods. I dontvthink I would have though of Colin’s methods myself though. I’d love to hear another methods you see.

]]>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.** *

The author, Rufus William, suggested that when teaching trig for the first time we should be solely teaching procedurally using SOHCAHTOA as a mnemonic, but then went on to say we shouldn’t be discussing ratio or similarity and how that links until later on. This confused me as the mnemonic SOHCAHTOA is designed to help you remember the trig ratios. I.e. Sine is the ratio of the opposite side over the hypotenuse. Just by teaching that you ARE teaching the trig ratios and purely by the fact that you are teaching the students that this will work for all right angled triangles you are telling the students that the ratios are the same for any triangle with the same angle no matter what the length of the sides are. THIS IS THE VERY DEFINITION OF SIMILAR TRIANGLES.

This perplexed me a lot and I spent a lot of time thinking about it and asking the author to elaborate on what he meant. The only way I can fathom to teach this without reference to ratio and similarity would be to say: ” “SOHCAHTOA” it gives you 3 triangles. Label the sides circle them to see which triangle you use. Put numbers in, cover the missing one, its either a divide or a times”. To me this seems like a backwards way to go about things. It feels like you are teaching them unnecessary procedures to avoid discussing the underlying concepts of trigonometry, and it doesn’t really make sense to me.

I find that by the time students reach right angled triangle trigonometry they have already met the concept of similarity, I like to use this a way in to discussing the topic and to show that ratio of two sides that are the same in relation to an angle will be the same for all similar triangles. Students will have always encountered simplifying fractions before they meet trig and as such can see why this is. This is when I specifically discuss the sine, cosine and tangent ratios and introduce the procedural manner in which they can solve the problems, although I do avoid the dreaded formula triangles (for many reasons which I have blogged about here). I will show them some common mnemonics, and SOHCAHTOA is one of them. I’m not a fan of mnemonics personally, I’ve never found them that useful except for musical ones, but I know a lot of people do.

Rufus does make some salient points in his post about teachers who refuse to allow students to memorise things and the dangers this will have on learning. Although I’m not entirely sure that they exist, and if they do I certainly don’t think there are many of them. I’ve certainly never met any.

He also suggests that students cannot have a full understanding of the ins and outs of trigonometry when they first meet it. I would very much agree with him in that respect, I know many people who have taught trigonometry for decades and still don’t, but I don’t think that means we have to bypass all information.

**Reference List:**

Cavadino, S.R. 2014. Formula Triangles. 12th October. *Cavmaths. *[online] accessed 14th July 2017. available: https://cavmaths.wordpress.com/2014/10/12/formula-triangles/

Cavadino, S.R. 2016. Catchy Mnemonics. 16th September. *Cavmaths. *[Online] accessed 14th July. Available: https://cavmaths.wordpress.com/2016/09/16/catchy-mnemonics/

William, R. 2017. The understanding paradox. 7th July. *No easy answers. *[online] accessed 14th July 2017. available: https://noeasyanswerseducation.wordpress.com/2017/07/07/the-understanding-paradox/

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