## Ed’s Infamous Area Problem

Yesterday a colleague asked if I’d seen the maths problem that was going round and featured in the national press. I hadn’t, but was not surprised to see that it was Ed Southall (@solvemymaths) who had posed the problem that had got the world (well, the nation at least) talking maths. My initial thought was that it was great to see a positive discussion of maths in the press. Then I figured I’d need to solve it.

Here is the problem:

*What fraction of the area is shaded?*

What follows is my solution. Please attempt the problem before reading on, I’d love to see your approach.

Firstly,I did a sketch (of course I’d did. If you didn’t then why on earth not!)

I labelled the base of the rectangle 2x and the height b (it looked like a square, but I didn’t want to assume and figures if it was necessary to Ed would have told us). I realised that I was looking at 2 similar triangles (proof can be made using opposite and alternate angles), with a scale factor of 2 (the base of the bottom is double the base of the top). I know that when working with areas the scale factor is squares so using an area scale factor of 4, a for the height if the top triangle and (b – a) for the height of the lower triangle intake up with this equation:

Which solved to tell be b was 3a, thus b-a was 2a.

From here it was simple, I worked out the area of the shaded triangle and the whole rectangle put it as a fraction and simplified.

How did you do it?

## Quarter circle problem

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:

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…..

## A lovely circle problem – two ways

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

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

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

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.”

## Eid and Exams

Today is the day that all the students in year 11 at my school, and I believe a majority across the school, sat their final GCSE paper. It was a physics paper. Today also happens to be Eid al-fitr. Eid is a holy day in the Islamic faith and marks the end of the holy month of Ramadan.

Eid al-Fitr is an important day in the Islamic faith. Muslims start it out by attending the mosque for prayers, before sitting down to share a meal with their families, which will be the first time they have done this during daylight in a month.

I teach in a school where, I believe, around 30% of the student body practice Islam. This year has been particularly hard for year 11, as many have been observing the fast of Ramadan during their exam period and have had to miss important parts of their Eid rituals and celebrations in order to sit their final exams. Eid al-Fitr, and Eid al-adha – the most important Islamic holiday, follow the Islamic calendar and as such move year on year. Next year, Eid al-Fitr falls on June the 4

^{th}, the same day as one of the English Language GCSE exams, which ALL y11 students will sit, along with Business and music exams. This date will also feature A Level papers for English Language, English Lang and lit, art, RS and Chemistry.Personally, I would advocate for all holidays of all the major religions to be made bank holidays as the UK becomes an increasingly wonderful multi-cultural and diverse place, but I understand we are a long way from that dream becoming a reality. However, I am certain that a more achievable goal is becoming a society that manages to schedule GCSE and A-Level exams around such an important event.

Many students this year have been disadvantaged this year because they have had an exam on a day that is massively important to them, their families and their religion. GCSE exams that fell today cover science, taken by the vast majority of students, and Citizenship, an exam I would argue was extremely important. A level exams that fell today were PE, Economics, English literature, Mathematics, Further Mathematics and Chinese.

In the 2011 census Islam was the second biggest religion in the UK behind Christianity, and also the fasted growing religion. There are many local authority areas in the UK where more than 25% of the population follow the Islamic faith. These included Bradford, which is where I work, Blackburn, Luton and Birmingham. Some areas are over thirty, which includes Tower Hamlets, which has around 35% of its population following Islam.

GCSE and A Levels are important examinations; they massively affect the future of those that sit them. The stresses on students at this time is massively high and I feel that it is hideously unfair to make this more difficult on one subset of students purely based on their religion. To have a few fallow days during the exam period would mean what? Lengthening the session by a few days?! Surely it’s time we stopped punishing students for what they believe in.

## Share this via:

## Like this: