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Posts Tagged ‘geometry’

Circles puzzle

October 13, 2016 Leave a comment

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: , , , ,

Circles and Triangles

July 15, 2016 5 comments

Regular readers will know that I love a good puzzle. I love all maths problems, but ones which make me think and get me stuck a bit are by far my favourite. The other day Ed Southall  (@solvemymaths) shared this little beauty that did just that:

I thought “Circles and a 3 4 5 triangle – what an awesome puzzle”, I reached for a pen an paper and drew out the puzzle.

I was at a bit of a loss to start with. I did some pythag to work some things out:

Eliminated y and did some algebra:

Wrote out what I knew:

And drew a diagram that didn’t help much:

I then added some additional lines to my original diagram:

Which made me see what I needed to do!

I redrew the important bits (using the knowledge that radii meet tangents at 90 degrees and that the line was 3.2 away from c but the center of the large circle was 2.5 away):

Then considered the left bit first:

Used Pythagoras’s theorem:

Then solved for x:

Then briefly git annoyed at myself because I’d already used x for something else.

I did the same with the other side to find the final radius.


A lovely puzzle using mainly Pythagoras’s theorem, circle theorems and algebra so one that is, in theory at least, accessible to GCSE students.

I hope you enjoyed this one as much as I did!

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:

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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:

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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.

Exact Trigonometric Ratios

November 8, 2015 4 comments

This morning I read this interesting little post from Andy Lyons (@mrlyonsmaths) which looked at teaching the exact Trigonometric Ratios for certain given angles (namely 0, 30, 60, 90 and 180 degrees). The post gave a nice little info graphic linked to the unit circle to show what was going on and then focused on methods yo remember the ratios.

While reading it I thought about how I introduce these exact Trigonometric Ratios. I first like to know that my students have a thorough and in depth understanding of right angled triangles and the trigonometry involved with them (including Pythagoras’s Theorem). I feel this is imperative to learning mathematics, the Triangle is an extremely important shape in mathematics and to fully understand triangles you must first fully understand the right angled triangle. The rest follows from that.

Once these are understood then you can move on to the trigonometric graphs, showing how these can be generated from right angled triangles within the unit circle, as shown in the info graphic on Andy’s post. Once the graphs are understood then the coordinates f the x and y intercepts and the turning points give us nice exact values for angles of 0, 90 and 180 degrees. This leaves us with 30, 60 and 45 to worry about.

At this point I introduced 2 special right angled triangles. First up is the right angled isosceles triangle with unit lengths of the short sides. This obviously gives us a right angled triangle that has two 45 degree angles (as the angle sum of a triangle is 180) and a hypotenuse of rt2 (via Pythagoras’s Theorem).

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Using our definitions of trigonometric ratios (ie sin x =opp/hyp, cos x = adj/hyp and tan x = opp/adj) we can clearly see that tan 45 = 1 and that sin 45 = cos 45 = 1/rt2. This aids the understanding more than just giving the values and allows students a method of working these values out easily if stuck.

The second triangle is an equilateral triangle of side length 2 cut in half. This gives us a right angled triangle with hypotenuse 2, short side lengths 1 and rt3 (again obtained through Pythagoras’s Theorem) and angles 30, 60 and 90.

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Again we can use our definitions of trigonometric ratios to conclude that sin 30 = cos 60 = 1/2, sin 60 = cos 30 = rt3/2, tan 30 = 1/rt3 and tan 60 = rt3.

This is again good for deeper understanding and for seeing why sin x = cos 90 – x, and cos x = sin 90 -x. This can lead to a nice discussion around complementary angles and that the word cosine means “sine of the complementary angle”. This triangle is also a good demonstration that tan x = cot 90 – x, when you come to higher level trig.

Parallelograms

October 21, 2015 Leave a comment

Parallelograms, you know, the weird quadrilaterals that look like a sheared rectangle. These:

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I’ve never rally thought that deeply about them, to be honest. They have some uses in angle reasoning lessons, and we need to be able to find their area in the GCSE, but I’ve not thought too deeply about them recently at all.

When teaching how to find the area I normally do this:

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It’s a fine method, and easy to show that it works by showing that you can cut the end off, stick.it in the other end and get a rectangle which is clearly of the same area.

But last week I marked a mock exam in which one of my year 11s had done this:

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I love this method, it’s much, much nicer than the other. I couldn’t wait to question him. When I did he said that he “couldn’t remember” how to do it, but knee how to find the area of a non right angled triangle so split it into two of them which were congruent using SAS.

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I asked him what would happen if you split the parallelogram across the other diagonal. He thought about it for a while, and eventually told me it would be fine because of “how the sine curve is” and because, “the angles add up to 180”.

I was impressed by his reasoning. He has clearly understood this method and generalised the area of a parallelogram in a way I’d never considered. I would have phrased is slightly differently though:

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The area of a parallelogram is equal to the product of two adjecent sides multiple by the sine of one of the angles. (Either will so as Sin x = Sin (180 – x) )

Quadrilateral Puzzle

January 10, 2015 Leave a comment

Yesterday the fantastic Ed Southall (@solvemymaths) tweeted this brilliant puzzle:

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It looked fun, so I thought I’d give it a try. First I sketched it out and gave all the vertices labels. A strategy I advise my students to take.

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I then considered triangle wzg, as it was the triangle I knew most about:

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My first thought was to find the length of the hypotenuse using Pythagoras’s Theorem. This was something that I didn’t use in the end, but I had yet to really formulate at strategy, and as I tell my students, you can never have too much information. I then thought I’d find the angles, but realised that it is this defaulting to trigonometry that often leads me to overcomplicate matters, so I thought I’d leave that til later, (plus I don’t know tan 2 or tan 0.5 off the top of my head.)

I considered the area of the triangle, then sketched the next triangle I knew stuff about fyz. It was here I saw my strategy.

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The right angles meant I could use congruent triangles.

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I could work out the area of the triangle yxk, which is congruent to fyz, as half the parallelogram area is 32, which is made up of this triangle and one which has area 8.

Thus the other leg of the right angled triangle must be 6rt2, and so a=12rt2 (as y is it’s midpoint!)

From there it was a question of Pythagoras’s Theorem to find b.

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A fantastic little puzzle, one that I enjoyed solving, and one which should be accessible to higher GCSE learners. If you haven’t already do check out Ed’s website.

Perimeter, the Hero’s way

December 28, 2014 2 comments

Forgive the title, but I do love Heron’s Formula, (named after Hero of Alexandria) it’s my favourite geometric formula and this is only the second time I’ve had need to mention it in a blog post. The first was Area the Hero’s way.

This post is a look at my solution to this beautiful puzzle which was set by Ed Southall (@solvemymaths) the other day:

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This struck me as a tricky one. The fact it’s a right angled triangle meant my brain was crying out “we’ve got to use trigonometry”, but I couldn’t see an obvious was to do it. We have the area of the triangle, and there is an incircle in the picture so I figured we’d need to use the relationship that the area of a triangle is equal to the semiperimeter multiplied by the radius of the incircle. It was possibly the fact that this relationship mentions semiperimeter that put Heron’s Formula in my mind.

I sketched the problem and filled in what lengths I could deduce using circle theorems, filling gaps with variables:

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From this the semiperimeter was nice to work out, so I went down Hero’s route.

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Which led me to:

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I had one equation, but two variables. I needed another equation in the same two variables, so I used the aforementioned relationship from incircle:

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I rearranged this for x (as it was the simplest rearrangement) and subbed it into the other equation:

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I then solved this equation for r, discounting the negative, as a radius can’t be negative. Once I had r I could use the relationship area/radius = semiperimeter so twice area/radius =perimeter:

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Which rounds to 44 cm. (The question asks to the nearest cm.)

I love this question, and I’m happy with my solution. I think it’s fairly elegant and uses a nice array of mathematics, but I can’t help but think that I’ve missed something blinding obvious that would have led to a simpler solution. If you solved it a different way, I’d love to hear how you did it. (I’d also like to hear if you did it the same way!) I asked Ed how he solved it, and he used the fact that the point where the incircle meets the longest side of a triangle splits the longest side into two segments, the product of whose lengths is the area. This is a nice relationship, and not one I’d known, so I will look to explore it. This could have saved be a bit if working, as I could have jumped to tge bit where I wrote x=6.

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