### Archive

Posts Tagged ‘Puzzles’

## Area 48

Today I was looking at some of Ed Southall’s (@solvemymaths) puzzles on his website. I saw this one that I had not seen before: I thought I’d give it a crack. You should too….. go on…. Did you get an answer? Well here is how I approached it:

First I did a little sketch, as I always tell my students to do: I labelled the points with letters as this is normally quite a good way of keeping track of things.

I then decided to let AB = 1 (I chose that bit to be 1 as I knew a unit square would lead to lots of fractions, in hindsight this also let to fractions and AB = 2 would have been better.) This gave me a few lengths straight off the bat, and I could find BD by Pythagoras’s Theorem and hence had the area of the larger square – which I need to answer the question.

I also noticed I had a RAT (ABD) and I knew the perpendicular sides, and therefore could work out the area. I then looked at the triangle BCH. This looked like it would be similar to ABD but I took a couple of moments to justify it to myself before moving on, just in case….

If angle ABD is x then as DAB and BCH are both 90 and the angle sums of a triangle and on a straight line are both 180 then CBH and BDA must both equal 90 – x and CHB must equal x, hence they are similar. They are similar and the scale factor is 2 (as BC is half of AD and they are corresponding sides).

Hence the Area scale factor is 4 and the area pf BCH is a quarter of the area ABD. As Area ABD = 1 then Area BCH = ¼.

From here I took the area of the two triangles away from the area of the square ACED to get the shaded area and put it over the area of the larger square. (Well, after momentarily putting it over the area of the smaller square like a fool!). So here I had an answer, 11/20. I clicked on the comments on Ed’s website and saw some answers that were not what I had. This had me second guessing myself, so I thought about a different approach.

I went for a coordinate geometry approach (coordinate geometry seems to have taken over from trig as my brains go to method). I chose the origin as the common corner of the two squares and called the point where the vertex meets the horizontal point B. This mean B’s coordinates were (1,2). I called this line l1 and could spot its equation was 2x. Part of the shaded area is the area under this curve between x = 0 and x = 1 so I calculated that area to be 1. The perpendicular through B is the other line that bounds the top of our shaded region. I know the perpendicular gradiens multiply to -1 and I know it goes through point (1,2) so I could work out the equation of this line easy enough: Then calculate the area below it between the values x = 1 and x = 2. This gave an area of 7/4. So I had a total shaded area of 11/4 and could divide this by the area of the large square to get 11/20 again. I felt happier now that I had the same answer though two different methods, and I stress to my students that this is what they should be doing with any extra time in exams. Doing different methods and seeing if they get the same answer!

I hope you tried Ed’s puzzle, and if you did, please let me know how you approached it.

Categories: Maths

## Nice area puzzle

April 23, 2019 2 comments

Yesterday evening I came across this lovely area puzzle on twitter:  The puzzle is from Gerry McNally (@mcnally_gerry) he says its his first, and I hope that’s “first of many”.

I reached for the nearest pen and paper and had a quick go: As you can see, I misread the puzzle originally and thought the lower quadrilateral was a square. The large triangle is isosceles as given in the question. This allowed me to use the properties of similar triangles and the base lengths given to work out the areas of the square, both right angled triangles and the whole triangle. This then allowed me to calculate the area of the shaded quadrilateral and hence that area as a fraction of the whole.

Then I went to tweet my solution to Gerry and realised that nowhere does it say that the bottom quadrilateral is a square. I had added an assumption. This made me ponder the question some more. Instincts told me that it didn’t have to be a square, but that the solution would be the sane whether it was a square or not. But I didn’t want to leave it at that, I wanted to be sure, so I had another go.

I sketched out the triangle again: This time I called the height of the rectangle x.

This made it trivial to find the area’s of the rectangle and the triangle GCD. Triangle HAB was easy enough to find using similar triangle properties. and then I found the area of the whole shape again using similarity to discover the height. This allowed me to find the shaded area: Then when I put it as a fraction the xs cancelled and it of course reduced to the same answer. I really like this puzzle, and would be interested to see how you approached it, please let me know in the comments or on social media.

## 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.

Categories: GCSE, Maths, SSM, Uncategorized

## An excellent puzzle – alternate methods

July 19, 2017 2 comments

Yesterday I wrote this post looking at a nice puzzle I’d seen and how I solved it.

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: I had a moment of stupidity: 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.

## An interesting area puzzle

July 5, 2017 12 comments

Regular readers will know that I have a tendency to collect puzzles and I like to have a go at them. This evening I had a crack at this while my daughter was playing before bed. It looked interesting when I saw it on Facebook a while ago and so I thought I’d have a crack. If you haven’t yet, do it now. I want to know if you took the same approach!

I wasn’t too sure where to start, so I drew it out, labelled some stuff and came up with some equations: I thought if I multiples two of my equations together and rearranged I could get the yellow area as the subject: Then I needed cx: I thought “now I need ay”, then realised I had it: This meant I could.sub back in through the equations: So the area: I thought 27 was a nice answer, and I’m fairly sure it’s correct, however I have a feeling that I may have missed something blindingly obvious that would have gotten me there much quicker.

If you did it, I’d love to hear your approach, especially if you spotted something I missed!

Categories: #MTBoS, Maths, SSM, Starters

## 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.

Categories: Maths, Starters Tags: , , , ,

## Is one solution more elegant?

May 14, 2016 13 comments

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.