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Qetesuesi

Maths question

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A right-angled triangle has short sides of lengths 1 and 2.  What is the length of a square that fits inside the triangle such that one corner coincides with the triangle's right angle and the opposite corner lies on the hypotenuse?

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That takes me back!

Setting up the equation as being the biggest square that will fit in there will lie with its sides on the opposite and adjacent sides of the triangle (the 1 and 2).

Then putting the side 2 at the bottom and calling the side of the square "a" we have:

Tan (bottom angle) = opposite (a) / adjacent (2-a)

Tan = o/a in the big triangle so same again = 1 (opposite) / 2 (adjacent)

So Tan (bottom angle) = 1/2 = a / (2-a) or 2-a = 2a , so 2 = 3a and a = 2/3 which is the length of a side of the square.

I think.  Been a while.

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2/3     (two thirds)

Solve by dropping a perpendicular from the corner of the square on the hypotenuse to the side of length 1 and use similar triangles.

If x is the required length then (1-x)/x = 1/2, hence x=2/3

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See, schooling wasn't wasted for at least three of us when we can solve geometry problems when we should be working...

 

If 2/3 is right that is anyway.

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Just now, Frank Hovis said:

See, schooling wasn't wasted for at least three of us when we can solve geometry problems when we should be working...

 

If 2/3 is right that is anyway.

Only two of you.  I caught Chump putting 2/3 sqrt 2, without offering proof, before editing it to your answer and referring us to your proof!

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Just now, Qetesuesi said:

Why did you put 2/3 sqrt 2 before you edited it?

I guess because Chumpus went the long way round initially as I also did.

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1 minute ago, Qetesuesi said:

Tell me more....

Starting working out what the intermediate lengths are when they're not required, it doesn't matter for instance that the hypotenuse length is sqrt(5) when you don't need to use it.

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I never considered sqrts at all.  Or trig (cf. the use of tan).  However Spaniard's method is arguably a bit more straightforward than how I did it, as he compared one of the smaller triangles to the large triangle, whereas I compared the two smaller triangles which involves a bit more algebra.

But what's really eye-opening is the context of how I came by this question, a couple of weeks ago.

A year 13 further maths student said that someone at his school had gone for the Cambridge interview along with several others from other schools.  At this interview they were asked this question, and not one of them managed to get it - even though it needs no A level maths at all, nor even trig or Pythag.

I've yet to come across anyone offline who's been able to do it, but I'll keep trying.  Even on here, the question had at least 30 views before anyone posted the solution and proof.  So well done at least to you guys.

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I'd ask - do they teach schoolchildren to think in this way anymore?

My two main maths teachers were both superb.  Each had a delight of puzzles and tricks and would trot out mathematical problems for you to do (one - on the board in front of the class) and encourage you to think laterally rather just follow proscribed methods and formulae.  They would come up with absolute crackers for Christmas and it was genuinely enjoyable - which I doubt would be the case now in a world of SATS tests and pure curriculum teaching.

Marcus de Sautoy did a similar verison of this with a show, mixing maths and logic which was excellent.

I have tried to explain to people the difference between "surface" learning - learning the rules for set situations - and seeing that surface learning as merely an entry route into getting a feel for what lies behind it. 

An example would be mathematical set and group theory, one of the last things I did before I stopped maths, for which I am aware I never made it beyond the surface rules and descriptions as one person in particular that I spoke to had a real "feel" for it.

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I would also say that trig / tan is really just formalising relationships between angles and sides so all three solutions are doing the same thing with minor differences

I used tan as a relationship between sides in a right angled triangle rather than to deriev the angle.

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Part 2.

Consider all of the rectangles which fit inside the triangle in the same way as the square above.  What are the dimensions of the one with the largest area, and what is the area?

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32 minutes ago, Frank Hovis said:

I'd ask - do they teach schoolchildren to think in this way anymore?

My two main maths teachers were both superb.  Each had a delight of puzzles and tricks and would trot out mathematical problems for you to do (one - on the board in front of the class) and encourage you to think laterally rather just follow proscribed methods and formulae.  They would come up with absolute crackers for Christmas and it was genuinely enjoyable - which I doubt would be the case now in a world of SATS tests and pure curriculum teaching.

Marcus de Sautoy did a similar verison of this with a show, mixing maths and logic which was excellent.

I have tried to explain to people the difference between "surface" learning - learning the rules for set situations - and seeing that surface learning as merely an entry route into getting a feel for what lies behind it. 

An example would be mathematical set and group theory, one of the last things I did before I stopped maths, for which I am aware I never made it beyond the surface rules and descriptions as one person in particular that I spoke to had a real "feel" for it.

I had two maths teachers who were polar opposites. My GCSE teacher used to say 'I can't teach you to think', he was right - he was a useless *******. My A-level teacher really went out of his way to teach why something worked and in the process you started to get to grips with how to think about problems and how to approach them. 

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31 minutes ago, Scunnered said:

Part 2.

Consider all of the rectangles which fit inside the triangle in the same way as the square above.  What are the dimensions of the one with the largest area, and what is the area?

Nope. I have had a go but I get to a fiendishly complicated equation that I can't solve which, were I able to do so, would express a in terms of b (the two sides of the rectangle) which would then allow me to calculate the maximum.

I have just ground it out with Pythagoras to get me to that impasse so I await with interest the much neater solution.

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1 minute ago, Frank Hovis said:

Nope. I have had a go but I get to a fiendishly complicated equation that I can't solve which, were I able to do so, would express a in terms of b (the two sides of the rectangle) which would then allow me to calculate the maximum.

I have just ground it out with Pythagoras to get me to that impasse so I await with interest the much neater solution.

Off the top of my head I would guess you need calculus for this, I'm on holiday and don't have a pen and paper to hand but I'm guessing there's an equation for a and b that use the same variable (if a and b are the lengths of the sides), as a result you can express the area with a single variable related to the length of the side (lets call it x), then set the derivative da/dx = 0.

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Let A(0,0), B(0,b) and C(c,0) be a general right angled triangle, short sides lengths b and c.

Let P(x,y) be a point on the hypotenuse BC.

Then y/(c-x) = b/c, hence cy = b(c-x), (c/b)xy = cx-x'2     (c times x minus x squared)

Maximum xy occurs when d/dx(xy) =0, thus when x = c/2, when P is the mid-point of the hypotenuse.

The dimensions of the maximal rectangle are b/2 and c/2.

The area is bc/4, which is half the area of triangle ABC.

If b=1 and c=2 then the area is 1/2.

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7 minutes ago, The Spaniard said:

Let A(0,0), B(0,b) and C(c,0) be a general right angled triangle, short sides lengths b and c.

Let P(x,y) be a point on the hypotenuse BC.

Then y/(c-x) = b/c, hence cy = b(c-x), (c/b)xy = cx-x'2     (c times x minus x squared)

Maximum xy occurs when d/dx(xy) =0, thus when x = c/2, when P is the mid-point of the hypotenuse.

The dimensions of the maximal rectangle are b/2 and c/2.

The area is bc/4, which is half the area of triangle ABC.

If b=1 and c=2 then the area is 1/2.

That's a neater way than I would have used to get there, but pretty much the same approach as I gave above.

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10 minutes ago, ChumpusRex said:

I interpreted the question to say diagonal of the square. 

My first good maths teacher used to stand aside whilst indivdiuals had a go at working out on the board, he'd generally nod in encouragement but when he saw somebody steering off from the required answer his cry would ring out "Don't muck it!"

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3 hours ago, The Spaniard said:

Let A(0,0), B(0,b) and C(c,0) be a general right angled triangle, short sides lengths b and c.

Let P(x,y) be a point on the hypotenuse BC.

Then y/(c-x) = b/c, hence cy = b(c-x), (c/b)xy = cx-x'2     (c times x minus x squared)

Maximum xy occurs when d/dx(xy) =0, thus when x = c/2, when P is the mid-point of the hypotenuse.

The dimensions of the maximal rectangle are b/2 and c/2.

The area is bc/4, which is half the area of triangle ABC.

If b=1 and c=2 then the area is 1/2.

That'll do.  That's essentially how I did it as well, but I was hoping someone would come up with a more obvious geometric solution.

In the original case, you can observe that if the rectangle has width x (between 0 and 2) then it has height ½(2-x), so the area is  A = x(2-x)/2 = (2x-x²)/2, so dA/dx = (2-2x)/2 = 1-x, which is 0 at x=1. 

However, note that the equation for the area is symmetric in x and 2-x, so that every area occurs twice:  for instance, when x=0.3, A=0.3*1.7/2, and when x=1.7, A=1.7*0.3/2, which is the same.  The only time when an area occurs once only is when x=2-x, ie x=1.  This would show us that the maximum area occurs when x=1, if we knew that there was a unique maximum.  It's not entirely obvious that this is the case though (without using calculus): maybe the area could increase to a maximum as you increase x from 0 to 0.3, then decrease until x=1, then increase again as x approaches 1.7, then decrease to 0 as x approaches 2 (this would be so much easier with pictures). 

It feels like someone should be able to come up with a diagram that makes you go "Oh yes, that's obvious", but I can't come up with anything at the moment...

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