## Circles puzzle

Here’s a lovely puzzle I saw on Brilliant.org this week:

It’s a nice little workout. I did it entirely in my head and that is my challenge to you. Do it, go on. Do it now….

Scroll down for my answer….

Have you done it? You better have…..

I looked at this picture and my frat thought was that the blue and gold areas are congruent. Thus the entire picture has an area of 70. There are 4 overlaps, each has an area of 5, so the total area of 5 circles is 90. Leaving each circle having an area of 18.

This is a nice mental work out and I feel it could build proprtional reasoning skills in my students. I am hoping to try it on some next week.

Did you manage the puzzle? Did you do it a different way?

*This post was cross posted to better questions *here*.*

## Is one solution more elegant?

Earlier this week I wrote this post on mathematical elegance and whether or not it should have marks awarded to it in A level examinations, then bizarrely the next day in my GCSE class I came across a question that could be answered many ways. In fact it was answered in a few ways by my own students.

Here’s the question – it’s from the November Edexcel Non-calculator higher paper:

I like this question, and am going to look at the two ways students attempted it and a third way I think I would have gone for. Before you read in I’d love it if you have a think about how you would go about it and let me know.

**Method 1**

Before I go into this method I should state that the students weren’t working through the paper, they were completing some booklets I’d made based on questions taken from towards the end of recent exam papers q’s I wanted them to get some practice working on the harder stuff but still be coming at the quite cold (ie not “here’s a booklet on sine and cosine rule, here’s one on vectors,” etc). As these books were mixed the students had calculators and this student hadn’t noticed it was marked up as a non calculator question.

He handed me his worked and asked to check he’d got it right. I looked, first he’d used the equation to find points A (3,0) and D (0,6) by subbing 0 in for y and x respectively. He then used right angled triangle trigonometry to work out the angle OAD, then worked out OAP from 90 – OAD and used trig again to work out OP to be 1.5, thus getting the correct answer of 7.5. I didn’t think about the question too much and I didn’t notice that it was marked as non-calculator either. I just followed his working, saw that it was all correct and all followed itself fine and told him he’d got the correct answer.

**Method 2**

Literally 2 minutes later another student handed me her working for the same question and asked if it was right, I looked and it was full of algebra. As I looked I had the trigonometry based solution in my head so starter to say “No” but then saw she had the right answer so said “Hang on, maybe”.

I read the question fully then looked at her working. She had recognised D as the y intercept of the equation so written (0,6) for that point then had found A by subbing y=0 in to get (3,0). Next she had used the fact that the product of two perpendicular gradients is -1 to work put the gradient of the line through P and A is 1/2.

She then used y = x/2 + c and point A (3,0) to calculate c to be -1/2, which she recognised as the Y intercept, hence finding 5he point P (0,-1.5) it then followed that the answer was 7.5.

A lovely neat solution I thought, and it got me thinking as to which way was more elegant, and if marks for style would be awarded differently. I also thought about which way I would do it.

**Method 3**

I’m fairly sure that if I was looking at this for the first time I would have initially thought “Trigonometry”, then realised that I can essential bypass the trigonometry bit using similar triangles. As the axes are perpendicular and PAD is a right angle we can deduce that ODA = OAP and OPA = OAD. This gives us two similar triangles.

Using the equation as in both methods above we get the lengths OD = 6 and OA = 3. The length OD in triangle OAD corresponds to the OA in OAP, and OD on OAD corresponds to OP, this means that OP must be half of OA (as OA is half of OD) and is as such 1.5. Thus the length PD is 7.5.

**Method 4**

This question had me intrigued, so i considered other avenues and came up with Pythagoras’s Theorem.

Obviously AD^2 = 6^2 + 3^2 = 45 (from the top triangle). Then AP^2 = 3^2 + x^2 (where x = OP). And PD = 6 + x so we get:

*(6 + x)^2 = 45 + 9 + x^2 *

x^2 + 12x + 36 = 54 + x^2

12x = 18

*x = 1.5*

Leading to a final answer of 7.5 again.

Another nice solution. I don’t know which I like best, to be honest. When I looked at the rest of the class’s work it appears that Pythagoras’s Theorem was the method that was most popular, followed by trigonometry then similar triangles. No other student had used the perpendicular gradients method.

I thought it might be interesting to check the mark scheme:

All three methods were there (obviously the trig method was missed due to it being a non calculator paper). I wondered if the ordering of the mark scheme suggested the preference of the exam board, and which solution they find more elegant. I love all the solutions, and although I think similar triangles is the way I’d go at it if OD not seen it, I think I prefer the perpendicular gradients method.

*Did you consider this? Which way would you do the question? Which way would your students? Do you tuink one is more elegant? Do you think that matters? I’d love to know, and you can tell me in the comments or via social media!*

Cross-posted to Betterqs here.

## A lovely simple trigonometry puzzle

Sometimes a puzzle can look complicated, but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing!

Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms, it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

*Sin^6 + sin^4 cos^2 – sin^4*

See it now? What if I rewrite it as:

*Sin^4 sin^2 + sin^4 cos^2 – sin^4*

I’m sure you have seen it now, but to be complete, take the common factor of the first two terms:

*Sin^4 (sin^2 + cos^2) – sin^4*

Obviously sin^2 + cos^2 = 1, so we’re left with:

*Sin^4 – sin^4 = 0*

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

*Cross-posted to Betterqs here. *

Sometimes a puzzle can look complicated, but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing!

Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms, it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

*Sin^6 + sin^4 cos^2 – sin^4*

See it now? What if I rewrite it as:

*Sin^4 sin^2 + sin^4 cos^2 – sin^4*

I’m sure you have seen it now, but to be complete, take the common factor of the first two terms:

*Sin^4 (sin^2 + cos^2) – sin^4*

Obviously sin^2 + cos^2 = 1, so we’re left with:

*Sin^4 – sin^4 = 0*

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

## An interesting area problem

Here’s an interesting little question for you:

*Have you worked it out? How long did it take you to see it?*

It took me a few seconds at least, I had screenshotted the picture and was reaching for the pencil when the penny dropped, and that’s why I thought it was an interesting question.

The answer is, of course, 100pi. This follows easily from the information you have as the diagonal of the rectangle is clearly a radius – the top left is on the circumference and the bottom right is on the centre.

*So why didn’t I spot it immediately? *

I think the reason for me not spotting it instantly might be the misdirection in the question, the needless info that the height of the rectangle had me thinking about 6, 8, 10 triangles before I had even discovered what the question was.

I see this in students quite often at exam time, they can get confused about what they’re doing and it links to this piece I wrote earlier about analogy mistakes. The difference is I wasn’t constrained by my first instinct but all too often students can be and they can worry that it must be solved in the manner they first thought of.

Earlier today a student was working on an FP1 paper and he was struggling with a parabola question, he had done exactly this, he had assumed one thing which wasn’t the right way and got hung up in it. When he showed me the problem my instinct was the same as his, but when I hit the same dead end he had I stepped back and said “what else do we know”, then saw the right answer. I’m hoping that by seeing me do this he will realise that first instincts aren’t always correct.

I’m going to try this puzzle on all my classes tomorrow and Friday and see if they can manage it!

*How quickly did you see the answer? Do you experience this sort of thinking from your students? I’d love to hear any similar experiences.*

Cross-posted to Betterqs here.

## A lovely angle puzzle

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)*

*And*

*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.*

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December 9, 2015
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## A puzzle with possibilities

Brilliant’s Facebook page is a fantastic source of brain teasers, they post a nice stream of questions that can provide a mental work out and that I feel can be utilised well to build problem solving amongst our students.

Today’s puzzle was this:

It’s a nice little question. But when I use it in class I will only use the graphic, as I feel the description gives away too much of the answer. Without the description students will need to deduce that the green area is a quarter of a circle radius 80 (so area 1600pi) with the blue semicircle radius 40 (so area 800pi) removed, leaving a green area of 800pi.

I find the fact that the area of the blue semi circle is equal to the green area is quite nice, and in feel that with a slight rephrasing the question could really make use of this relationship. Perhaps the other blue section could be removed or coloured differently and the question instead of finding the area could be find the ratio of blue area to green area.

Another option, one I may try with my further maths class on Friday, could be to remove the other blue section and remove the side length and ask them to prove that the areas are always equal, this would provide a great bit of practice at algebraic proof.

*Can you think of any further questions that could arise from this? I’d love to hear them!*

*This post was cross-posted to the blog Betterqs here.*