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A lovely simple trigonometry puzzle

May 8, 2016 4 comments

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!

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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.

May 8, 2016 Leave a comment

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!

image

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

April 27, 2016 7 comments

Here’s an interesting little question for you:

image

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 latitude and longitude question

November 17, 2015 Leave a comment

A friend of mine, a geography teacher, tweeted me this question earlier:

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A lovely set of questions that I thought I’d run through here. I hope you’ve had a go at them before you read on….

The first one appeared at first to be a very simple question, just looking at the proportion of the circumference you have travelled. I then noticed the sleight of hand with the units and realised it involved an extra step. Still relatively straightforward though. Convert the distance from miles to km and work out the proportion of the circumference you have travelled. Multiply that by 36o and you have your angle, in this case 50.57 degrees to 2dp.

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The second question is more tricky. The first thing we need to work out is the radius of the circle we would get if we cut a cross section parallel to the equator 30 degrees north. In order to do this I drew a diagram  (always imperative! ).

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This allowed me to find a right angled triangle with the hypotenuse being the radius of the earth. This allowed me to find the radius of the circle I was after.

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Once I had this I could work out the distance travelled easy enough, as we have travelled 180 degrees so half the circumference.

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Or 15349 km (to the nearest km.) Here I’m assuming we want the distance we have travelled around the earth, rather than the displacement which would obviously be the diameter of the cross section or 9772km.

UKMT Maths Challenge Q16

November 7, 2014 1 comment

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Above is a photo of question 16 from yesterday’s UKMT Senior Maths Challenge. And what I fantastic question it is!

It shows (incase you can’t make it out on the picture) a rectangle and a circle which have the same centre. The two shorter sides of the rectangle are tangents to the circle, and the rectangle is 6×12. The task is to find the area that is inside both.

After the challenge had been completed one if my Y13s came to ask about it, and it took me a little while to work it out. I played around with some similar triangles, thinking that I would need to calculate the bit of the rectangle outside the circle and subtract them from 72. Then I had the realisation that I could easily split it up into triangles and sectors that I could calculate the area of with great ease.

First I sketched:

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Showing the rectangle split into areas and sectors. I then considered the triangles, if I split then into right angled triangles I had a hypotenuse of 6 and a height of 3, so the cosine of the top angle must be 0.5, which means the angle must be pi/3.

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I then used the sine of this angle ((rt3)/2) to calculate the opposite side as 3rt3. Then thought “why didn’t I use Pythagoras’s Theorem?!” A quick check using Pythagoras’s Theorem shows us the side is rt27, which simplifies to 3rt3. The area of the triangle then must be 9rt3.

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Knowing that the angles at the centre in the two triangles are 2pi/3, that angles round a point add up to 2pi and that the sectors are congruent means we can deduce the angle if each sector is pi/3. We also know the area of a sector is (r^2(theta))/2, which gives the area of each sector as 6pi.

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Finally we sum the areas and get 18rt3 + 12pi.

The student had looked at the answers and estimated them using the knowledge that rt3 is around 1.7 and pick is a but more than three. This meant he could eliminate c,d and e, but he didn’t get to the final answer.

Concentric Circles Puzzle

July 31, 2014 5 comments

This morning I happened across this tweet from David Marain (@dmarain)

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As you may know, I have a penchant for puzzles, and find it hard to leave them unsolved, so I thought about it and came up with an answer. I thought I would jot down my thought process here.

For those who can’t see the picture, the puzzle is:

A point is chosen at random inside the larger of two concentric circles. The probability it lies outside the smaller one is 0.84. What is the ratio of the larger radius to the smaller radius?

It’s a lovely little puzzle that combines a bit of geometric thinking with probability theory, so do have a little go first.

You done? Good.

My thinking started as: “that 0.84 must be equal to the area of the big circle – the small circle all over the area of the big circle.”

I used a as the area of the big circle, b as the area of the small circle and formed the following equation:

(a-b)/a = 0.84

With a little rearranging I got:

0.16a = b

So a ratio of areas that is

1:0.16 (a:b)

Or

6.25:1

Which is equivalent to

625:100

Which simplifies to

25:4

As we are looking for the ratio of radii, we need to square root each side, which gives a ratio of

5:2

A nice little solution to a lovely puzzle. Thanks for sharing David.

Nb: no photo of envelope workings as I did it mentally.

Similarity, and other stories

November 15, 2013 1 comment

Recently I have been looking into a variety of things. One thing is “Inquiry Maths” and another is something I found on the #matheme site through the explore the MTBoS project called “Notice and wonder”.

These got me thinking about how I could introduce some of their elements into some of my lessons. I had just introduced similarity to my year elevens and I was going to move to similar area and volume problems. So I came up with this starter:

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I put it on the board and gave them ten minutes (I think) and let them get their teeth into it. A few were a little confused at first, but the discussions on each table enabled all pupils to make their own way to the correct answer. I didn’t know what they would notice or wonder, but I was pleasantly surprised to hear some of their comments:

“I notice that the area has gone up by four, not two. Does that mean you double the scale factor for area?”

-I loved this one, and refused to answer it, instead I asked him to enlarge the shape sf3 and see if the area was enlarged by six. I then got:

“It’s nine, not six. Why’s it nine? Stupid thing. Oh hang in, it’s squared. Oh, of course it’s squared! you times each side by it [the scale factor] and you times them together! Duh!”

Others I particularly liked were:

“I wonder if there’s a way of doing Pythagoras on triangles without right angles” (I told her that we would be meeting the cosine rule soon enough).

“the angles are the same! Wait, that’s how this SOHCAHTOA thing works isn’t it, cos it’s ratios an that.” (I said “very good, but can we use the proper name please!” then another pupil interjected with “Trigonometry”)

The lesson goes on to pose question prompts similar to those I’ve seen on inquiry maths in which we discussed similar volume and then I included a set of questions for then to attempt. I have uploaded the resource to TES:
http://www.tes.co.uk/teaching-resource/Similarity-and-other-stories-6374388/

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