Can you solve the maths question for Singapore schoolkids that went viral?

[wrighty]

Having done more maths exams than most, I answered the question as if it were on a maths paper, rather than (with China's addition) on a semantic logic module of a degree level philosophy exam. I suspect, in the test for kids, that '44' would get the marks rather than 'no, because we can't assume the figures are in a plane, or that the space within which they are embedded is flat Euclidean'.

[Tempus Fugit]

I thought it was pretty obvious they were both 44 and another of these things where someone on FB posts a puzzle and there's zillions of 'likes' if you got it right

 

(with it being perimeter does it matter if the angles are 90º as long as the sides are straight and parallel - we're not calculating area where it would matter !)

[Chinahand]

Anyone tried the A4 puzzle?

Edited June 29, 2016 by Chinahand

[Chinahand]

With it being perimeter does it matter if the angles are 90º as long as the sides are straight and parallel - we're not calculating area where it would matter !

Agree - as long as the sides are straight and parallel, and the shape is on a plain.

[pongo]

Anyone tried the A4 puzzle?

Solved it earlier. Thanks for posting. Enjoyed it.

 

I'm still not certain I can explain WHY you get the answer you do

Is that a philosophical question? Pythagoras was a philosopher.

 

The bit I was weakest on - I ended up with a rather long 4 part equation - and the bit for the shorter hypotenuse was particularly long since it also included shorter equations for the other two sides (of the shorter hypotenuse). I felt I should have been better at reducing it all. In which case my hunch is that I would not have needed a calculator except to verify the answer. I realised how lazy code has made me about wrapping everything in parentheses.

Edited June 29, 2016 by pongo

[Chinahand]

 

I'm still not certain I can explain WHY you get the answer you do

Is that a philosophical question? Pythagoras was a philosopher.

 

The bit I was weakest on - I ended up with a rather long 4 part equation - and the bit for the shorter hypotenuse was particularly long since it also included shorter equations for the other two sides (of the shorter hypotenuse). I felt I should have been better at reducing it all. In which case my hunch is that I would not have needed a calculator except to verify the answer. I realised how lazy code has made me about wrapping everything in parentheses.

 

I think it was a philosophical question. The answer really surprised me - doing it algebraically requires calculations involving root 2 and the square root of root 2 and I've been thinking how to explain how the actual answer drops out.

 

If you do it via an all encompassing equation it becomes, as you say, quite complicated and the answer just drops out - even if you do it algebraically it can be far from obvious why you get the result.

 

I've been thinking about it and have come up with a couple of methods which explain it, hopefully, more elegantly - one using geometry and one using ratios, but I'll wait a bit longer to see if anyone else wants to give it a go.

 

As the guy in the video said once a solution is up people can look through it without trying it for themselves and just go - yeah I get that, I'd have done it that way too.

 

The wonderful thing about maths is that there are lots and lots of ways of showing why things are the way they are and different methods can be insightful in different ways. Trying it for yourself is so much better than following someone else's working.

[Gladys]

Lots and lots of ways of showing why things are the way they are? So graph paper would have been equally valid?

[pongo]

Lots and lots of ways of showing why things are the way they are? So graph paper would have been equally valid?

 

Graph paper, or working out the answer, doesn't show the why of a thing.

[ManxTaxPayer]

 

Why only 1/2 marks, he got the same answer as your video.

Because the correct answer to the original question as posted here is "no".

 

The original question posted here was "can you work out what the perimeters of these two figures are with the information available?"

 

That's not the same test as the one in the video - where the shapes are described as rectilinear.

 

Alleluia.

[Chinahand]

Well ... here's my go at explaining the A4 sheet of paper problem found here.

I have to admit that a part of me thinks I shouldn't be putting it up and rather should be leaving it for people to solve without any hints, but the solution is really interesting so I'd like to discuss it.

Please, if you are going to read the below, give the problem a go first.

I really did mean it - don't look until you've tried it yourself.

It is quite an easy problem to set up. All it takes is an A4 piece of paper.

Fold the piece of paper as in the picture:

 

If the piece of paper represented by the third drawing had a width of 1 and a height of √2. What is the perimeter of the 1st drawing? Please try to do it yourself first before looking.

 

Any ideas?

 

 

Don't forget √2 is an irrational number it literally goes on for ever. Don't round your earlier results too much before adding them up to get a final answer or that might create a rounding error.

 

So what do you think the perimeter's length is? This is your last chance to do it yourself.

 

 

 

OK, I lied, but this is it ... if you haven't tried it yourself by now, I give up.

... the perimeters length is ...

 

 

 

4.

 

You did try it yourself, didn't you?

 

But the question is why?!

 

 

 

Math's isn't just about getting an answer it is all about explaining why and how you got your answer and this problem has lots of different ways to solve it, some of which will explain why you get such a nice round answer more or less clearly than others.

 

So do you want to know why the perimeter is length 4?

 

 

You did do the problem yourself, didn't you?

 

 

So here is my explanation:

 

 

 

Bit small isn't it ... I hope clicking on it makes it bigger!

... there's more ... come on this is a Chinahand post ... I always go on and on ...

 

 

When I first solved the problem I solved it via the second method shown in the picture, and was really really surprised when the sum √2 + (2 – √2) + √(6 – 4√2) + √2 came to 4.

 

I still get a sort of childish delight in understanding that (2 – √2) = √(6 – 4√2)

 

OK I'm a bit weird that way!

 

... there's more

 

 

 

The realization that side FE is the same length as side EC is really the key to solving the problem - the final shape ACEF is a figure called in geometry a kite and it is symmetrical along its long axis A-E in my picture.

 

In the first method shown in the picture above I managed to show that the figure is a kite using congruent triangles. I used Euclidean geometry to show triangle AFE is congruent to triangle ACE.

 

That is all well and good, but after solving it that way I realized there was another even simpler way to do it.

 

Do you want to know what that was?

 

 

 

It involves ratios.

 

 

 

 

In the figure there are two squares ABCG and HCDE.

 

ABCG has side 1 and HCDE has side (√2 – 1)

 

To find the perimeter of ACEF we need to know what AF + FE + EC + CA equals.

 

We already now AF = √2 and it is basic subtraction to show FE = (2-√2) and a trivial application of Pythagoras to show AC also equals √2 the problem length is EC.

 

But in the square HCDE the ratio of the side EC to the square side DC is the same ratio in the square ABCG to the sides CA to BA.

 

ie CA/BA = EC/DC

 

But we know CA = √2 and BA = 1 and DC = √2 – 1; the only length we don't know is the length we want EC. So ...

 

√2 / 1 = EC / (√2 – 1)

 

Rearranging the equation EC = √2 x (√2 – 1) = (2 - √2)

 

And so we have solved the problem now knowing all the sides of the perimeter: AF + FE + EC + CA.

 

That is a really simple solution - we are given that AF = √2, it takes some simple subtractions to find FE = 2 – √2, you need to do one application of Pythagoras' theorem to show AC = √2, and EC simply drops out of the ratios between CA & BA and EC & DC.

 

I admit this is really nerdy, but that is far simpler than the multiple applications of Pythagoras or Euclidian geometry I used in my first two attempts to solve it.

 

Maths is full of different ways of explaining the same problem.

 

That is a very important part of its power - the structure of maths is interlinked in incredibly fascinating and complex ways.

 

I recently read Rozsa Peter's book Playing with Infinity and in it she wrote:

 

Man created the natural number system for his own purposes, it is his own creation; it serves the purposes of counting and the purposes of the operations arising out of counting. But once created, he has no further power over it. The natural number series exists; it has acquired an independent existence. No more alterations can be made; it has its own peculiar properties, properties such as man never even dreamed of when he created it. The sorcerer’s apprentice stands in utter amazement before the spirits he has raised. The mathematician ‘creates a new world out of nothing’ and then this world gets hold of him with its mysterious, unexpected regularities. He is no longer a creator, but a seeker; he seeks the secrets and relationships of the world which he has raised.

 

That sums up a lot of what is amazing about maths. We created maths, but have no idea just how complex it is.

 

And through its wonders we can understand the world. Ah wonderful!

 

Sorry for being nerdy. But maths is cool!

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[pongo]

I really like this a lot.

 

But still don't think that it constitutes a why.

[Chinahand]

It is fascinating isn't it.

 

Here are 118 proofs of Pythagoras's theorem.

 

Do these proofs explain why right angle triangles are this way?

 

That is really an very deep question and I'm honestly not sure.

 

In one way, definitely, but in another way I'm not certain.

 

Why, when you have a series of axioms and apply them logically, do you get a certain result?

 

Because that is what you get when you are consistent in your thought. I think that is the only answer you can give to these sorts of questions, but I suspect some people will find it unsatisfying - maybe I'm one of them!

 

My main task was trying to understand how the root twos, and terms including the square roots of root 2 cancelled out. Have I achieved that? In a similar way to the proofs in the link, but yes I agree my only answer as to why (2 – √2) = √(6 – 4√2) is because it does.

 

And maybe that is unsatisfying!

[wrighty]

My abstract algebra is a little rusty, but I think we're straying into fields and field extensions here. A bit like in complex numbers - you don't need a super-complex number to take a square root of a complex, so a square root of a surd involving root 2 will also be a surd involving root 2. Don't know if helps the specific question of why though.

[pongo]

@chinahand (this is not an attempt to address the why).

I got to the answer using the same method as you did initially. If I had been answering that as an O level exam question I would have been able to state that I was applying a²+b²=c² - I would not have needed to explain that.

I suspect that in a similar way and at a more advanced level there is a similar rule which describes the ratio method as a given.

I also suspect that at a more advanced level (than me) (2 – √2) = √(6 – 4√2) is as obvious as 2 + 2 = 4.

That brings me back to what I was saying about my hunch that my long equation would reduce to [the answer] without me having to calculate it - ie that there would likely have been no need to be careful not to round √2. That the equation would simply reduce to that. A hunch. Not something I am able to demonstrate.

Edited July 1, 2016 by pongo

[Chinahand]

I do agree with you Pongo - Wrighty has reminded me of Surds - goodness I'm back in Mr Manning's class long long ago!

 

This video explains why (2 – √2) = √(6 – 4√2) is as obvious as 2 + 2 = 4.