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Posts Tagged ‘Coordinate Geometry’

Area 48

July 10, 2019 Leave a comment

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.

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A lovely circle problem – two ways

December 7, 2017 5 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.

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.

Why sketching is important

November 15, 2015 3 comments

I wrote a while ago about how important diagrams can be using this Chessboard puzzle as an example. Last week I happened across another example while helping a year 12 student complete some exam questions on coordinate geometry.

The question was this:

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A relatively easy past paper question, but one that is quite fun nontheless. The student had completed a correctly and got the correct equation for l2 in part b but had gone wrong finding the point of intersection. She hadn’t noticed and hence had lost all the rest of the marks.

Part a was relatively straightforward:

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And part b should have been too:

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But she hadn’t got that. She’d made a transcription sign error and had the x value as -7. I knew her answer was incorrect as soon as I saw it because I had sketched the lines. She spotted that it must be wrong as soon as she saw my sketch:

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Had she sketched the lines out she would have known that the point of intersection was clearly in the first quadrant and couldn’t have had a negative x value. This self checking mechanism is just one of the any reasons I try to get my students to sketch everything. I don’t understand their reluctance to do it, it makes the questions so much simpler and allows you to spot your mistakes. I’ll just have to keep highlighting these examples to them and trying to get them to see how foolish it is to avoid the sketching.

Have you any ideas of how to instill the knowledge that a sketch is majorly important? If so, I’d love to hear them.

Incidentally, here’s the rest of my solution.

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And in full:

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Find the radius puzzle

December 13, 2014 Leave a comment

Yesterday I came across this photo on my phone:

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It’s a question that I found on a wall in one of the classrooms at school when I started in September and thought, “that’s a nice puzzle”, then promptly forgot about. When I found it, I thought “let’s explore this one.”

I sketched it out and had a look, my brain made its usually first stop at trigonometry.

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After spending a while deriving (accidentally) that cos x = adj/hyp for right angled triangles using the sine rule, various trigonometric identities and Pythagoras’s Theorem I decided there must be a better way. I sketched the problem on a coordinate grid so the centre of the circle was at (r,0) when r is the radius (ie so the y axis is a tangent) and looked at the equation:

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As I know the dimensions of the rectangle, I can deduce that the point (2,r-1) is on the circumference, sub these values in and solve for r.

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When I have r= 1 or 5 I can discount 1 as I know r>2, thus the radius is 5 cm . I was surprised by this answer, as it felt like the answer should be 4, not sure why.

Bizarrely, as I was working through this Jo Morgan (@mathsjem) posted this which included this similar puzzle:

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So I used the same method to solve that:

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I was happy with the solution, but I had a real feeling that I was missing something obvious that would lead to a much more concise solution, so I sketched again, this time I dropped a perpendicular from the point on the circumference which meets the rectangle:

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A right angled triangle, with side lengths r-2, r-1, r. It’s only the most famous RAT of all, a 3,4,5 triangle!

I followed the algebra:

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A much more concise method. This method made me think Chris Smith’s (@aap03102) puzzle (the one Jo shared) would be better to use in class, as it’s much more likely that students will know a 3,4,5 triangle than a 20,21,29 triangle!

Area of a semi-circle – the return

December 8, 2014 3 comments

Last week I wrote this piece in which I relayed my thought process when solving this problem from Ed Southall (@solvemymaths).

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After sharing it a lot of people tweeted me their solutions and I thought I’d look at a few here. I think the variety of solutions people used, and the amount of people who engaged with this puzzle is testament both to how interesting a puzzle this is and also how fantastic maths is. If you did it a different way, do let me know!

My solution was fairly long compared to many of the others. I think this maybe down to my heavy preference to Trigonometry, and my brain’s insistence on using it if possible! Here are some other, more concise methods:

Using the Tangent Ratio

In my solution I used right angled triangle trigonometry to find the lengths of the chords, then used Pythagoras’s Theorem to find the diameter. Steve Atkinson (@Small_Ears) also used trig, spotted that as tan a = 2 realised that tan b had to be 2 and thus the remainder of the diameter must be 12. A much more efficient and concise method, but not as much fun!

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Steve’s method made me realise we are dealing with three similar triangles, and that fact alone is enough to know the ratio of the sides are equal, and hence the rest if the diameter must be 12. This idea was the one used by Rob Rolfe (@robrolfemaths).

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Drawing a line from the centre

Jo Morgan (@mathsjem) tweeted this solution:

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Which I think is brilliant. Instead of calculating the rest of the diameter, Jo went for the rest of the radius, using a line from the centre to the point the perpendicular touches the circumference. Pythagoras’s Theorem and algebraic manipulation drop the radius out quickly implying the answer in great time.

The I only need Pythagoras’s Theorem method

This was tweeted to me by Mr Draper (@mrdrapermaths):

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As you can see, he has used Pythagoras’s Theorem on all three triangles and solved the simultaneous equations to find the necessary information. Possibly not the most concise method, but certainly no less concise than my approach, and one that has a lovely feel to it.

Constructing the geometric mean

Martin Noone (@letsgetmathing) tweeted this fantastic solution:

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It is, I believe, derived from an old method for constructing square roots that I think dates back to ancient Greece. This rule is illustrated in hospital top sketch, then he’s enlarged it to ensure the one is a three. He later tweeted the proof, which he uses as an exercise in lessons and one I may start to use too!

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Ed Southall then tweeted this pic which shows the general rule and how a semi-circle can be used to construct the geometric mean of two numbers, which itself could be used to find the diameter.

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And is a great visual proof that the geometric mean can never exceed arithmetic mean.

Coordinate geometry

Stuart Price (@sxpmaths) informed me that he’d enjoyed the puzzle and left it in his board for his students to complete, and that they used a variety of methods. He mentioned coordinate geometry and equations of circles, so I thought I’d have a crack at that method:

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I drew the axes so that the origin was at the point on the diameter where the perpendicular touches it. Ie 3 cm in from the circumference. I considered the general equation of a circle:

(x-a)^2 + (y-b)^2 = r^2

Then subbed b=0 in as the centre lies on the x axis.

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I know two points on the circumference, (-3,0) and (0,6) so I subbed these in to get two simultaneous equations which I then solved.

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Once I knew a (the x coordinate of the centre) I could work out the radius, and hence the area. A brilliant solution.

Calculus

Stuart also mentioned calculus, specifically implicit differentiation. I’m not sure how this would help, to be honest, and I’d love to see this solution. I did, however, consider integration. I rearranged the equation of the circle:

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I started to integrate but realised I’d need multiple substitutions, so I ran it on Wolfram Alpha instead.

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An nice, but incredibly long winded solution!

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