Quarter circle problem

March 2, 2018 2 comments

Here is a problem I have had on my classroom wall for a long time. I have a large display of problems on there that sometimes the students present me solutions to. This is one no one yet had done and I had not attempted.

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:


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…..

Categories: Maths

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.

Angle problem

December 5, 2017 Leave a comment

Today has been quite a geometric based day for me. I spent a couple of hours solving non-RAT trigonometry problems with year 10 and then a while with year 11 looking at various algebra angle problems. Then I went on Twitter and saw this from Ed Southall (@solvemymaths):

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.


Saturday puzzle

December 2, 2017 2 comments

One of the first things I saw this morning when I awoke was this post from solve my maths on facebook:

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.


Reverse percentages and compound interest

November 15, 2017 3 comments

The other day a discussion arose in my year 10 class that I found rather interesting. There was a question on interest which incorporated compound interest and reverse percentages. One student was telling the other how to find the answer to the reverse part, “you need to divide it, because it was that amount times by the multiplier to get this amount and divide is the inverse of times.” All good so far, then they discussed how to complete it if it was a reverse of more than one year, “so in that case it’s the new amount dived by the multiplier to the power of how many years.” I was pleased at the discussion so I didn’t really interject.

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.”



November 10, 2017 5 comments

Last week while we were waiting for a swimming lesson to start my daughter told me that one of her teachers had got “higgledy piggledy” about oblongs. I asked what she meant and she said that she’d accidentally called one a rectangle and had to correct herself and had informed the class that at her last school she’d had to call them rectangles but at this school had to call them oblongs and sometimes got higgledy piggledy about this. I asked my daughter why they couldn’t call them rectangles and she said that it was because squares can be rectangles too.

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.

Categories: #MTBoS, KS2, KS3, Maths, SSM, Teaching Tags: , ,

Dodgy Microsoft Graphics 

October 10, 2017 1 comment

So my new laptop arrived today and I quickly set about using it. It’s a Windows 10 laptop and as such has all the usual Microsoft stuff preloaded in it. I was going to set chrome as the default browser when it suggested I try Microsoft edge as it’s apparently faster and made for Windows 10. When I opened it it showed me this graphic:

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.

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