Brain Teasers ? or 50 Years On........... ?

Hi ken, hi sj, and no doubt Steved too.........

I think you three have cracked it We are down into the fourth significant figure now. I actually used the "approx =" sign with my ≈ 0.108

My values for cos α and cos β agree with ken (*) and my value for CD is ¼L√14

(*) I actually put cos α = (5/8)* √2 and cos β = (11/16)* √2

Its only because I like to see the √2 in the numerator rather than the denominator - absurd surds

As I said above, not elegant, but straightforward..................................bit like moving a 50 tonne stone ?

what i didnt like about the solution using cosine rule is that it smacks of brute force. there is quite a bit os symmetry in the problem that we are not exploiting, for example line CD is at right angles to the diagonal of the square, and also those 2 isosceles triangles.

if a flash of inspiration comes exploiting these symmetries -- i will report, but for now, i believe we can lay teh problem to rest...

enjoy

ken

Yes ken,

A high degree of symmetry and so far I have only gone for the brute force (and ignorance) solution.

But beware, some aspects look as if they might be symmetrical, but they aren't. For example, the shaded area in question CEDFC is not bisected by CD.

As you say, inspiration, or at least a flash of it might help. ...................Sleep tight

no, i wasnt falling into that trap

that would be trully folly as the circles have different radii anyhow, so CD cannot bisect that area...

The diagonal of the sq does, but this information is embedded in he process of finding tha angles -- may be that is indeed the best we can do.

Steve, did you get "nice" expressions (whatever that means!) for where the circles intersect?

enjoy

ken

OK, as i suspected there is definitely some symmetry we can exploit to solve this problem and i have found it  yipee!!!

Still involves solving for points of intersection of the two circles but in a transformed space, where the solutions more or less "drop out" (with a little manipulation of course) -- but somewhat "pleasingly"

after that the angles alpha and beta, can be obtained much more directly without the ugly cosine rule. 

can any of you guys see this simplifying symmetry?? otherwise i will publish this alternative approach 2moro...

enjoy

ken

Not sure about this as there are a couple of symmetries (the co-ordinates of the centres of the 2 circles and also about the diagonal x + y = 2L).

Would translating (i.e. flipping over) the centre of the smaller circle from (0,2L) to (L,L) retain the same intersection points but provide a more manageable formula of:  (x-L)^2  + (y-L)^2 = L^2.

y're thinking along the 'right line'/in the 'right area' 

but not quite there yet.

later...

enjoy

ken

By selecting a specific coordinate system, significant simplication results. The coordinate system i am referring to is one where the origin is the point A, and in addition, the x-axis along the line AB, i.e. along the diagonal of the square. Then the kyte shaped quad ADBC is symmetrically placed around the x-axis. The y-axis passes along line CD -- and again the kyte is symmetrically placed with respect to this too. Of course, we cannot assume CD is at right angles to AB without justification, but that justification is very simple.

This specific choice of coordinate system makes the equations of the 2 circles very easy to solve. So here we go:

enjoy

ken

A really imaginative and elegant solution, Ken.

thank you sjbabbey...

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ken

Hi ken,

I agree with sj, brilliant and very elegant !!.......................but..

Just for the benefit of anybody working through the figures you quoted.............

........the penultimate formula for sinα "...which implies 0.467707 radians" should read 0.48669 radians...........

The value 0.467707 is the value of sinα

Yes, I know, it was a typo and I am nit-picking...............and I really do appreciate the way you adjusted the value of "x" to put the √2 in the numerator............................ apologies.

A well deserved 10/10 and a round of applause ! really !

btw ken,

I really like the formatting of your maths. Do you have a special program for writing out formulae so clearly ?

Cheers

Don

thanks for the correction Don --  it was a copy/paste error from my 'J' immediate execution session on my PC.  :-(

   _1 o. (7^0.5)%4* 2^0.5

0.486695

enjoy...

ken

i use MathType within Word... its a bit of pain as i have to take a .PNG image into ImageShack using a 'Snipping tool' before hosting it here. there is probably an easier way -- i havent figured it out.

am i correct you use Visio for your very nice diagrams?

enjoy

ken

Just for the record. my use of the cosine rule was as follows

AB = (2√2)L

Using the cosine rule in triangle ABD

(2L)^2 = L^2 + ((2√2)L)^2 - 2(L)((2√2)L)cos α

Simplifying

4 = 1 + 8 - (4√2)cos α

Cos α = (5/8)√2

Use similar method to find Cos β =  (11/16)√2

CD/2 = Lsin α = L√(1 - cos² alpha) = (L/8)√(14)

Brutal !

I use Power-point, but its tedious ! eg, to draw those quadrant "arcs" I had to draw circles, then blank out three-quarters using a rectangle for half and a square for the third quarter, then ........tedious, hit-and-miss....you name it ! A brain-teaser in itself

I then save it as a "web-page" in a JPEG format (that is an option in Powerpoint) then upload it into flickr before loading it into the forum as an HTML link..........yet another brain-teaser

Doing the maths sometimes seems so much easier.............

Cheers

Don

Thank you ken,

I'll give it a try.

ah, it turns out further simplication is possible!!!

we actually do not need to solve for y at all, if we accept that CD is perpendicular to the diagonal.

we can use the value of x found, plus the radius to find cos(alpha) instead, ...etc etc

enjoy

ken

ah, "pleased" to hear you suffer as much as i do! 

enjoy

ken

Looks like I could be making progress ..............many thanks ken !

y're welcome Don. if you find any short cut to the cumbersome process of publishing here, please let me know.

enjoy

ken

They say a picture paints a thousand words or something like that..........so here is the picture

So how many rollers need to be placed in front of the 2 metre long stone that is being moved 1,000m towards Stonehenge.

OK, I know that fatcat has pointed out you only need two rollers, but think of it as if we had a few dozen rollers, and we spaced them out at "X" metre intervals on a repetitive basis, how far apart do the rollers need to be ? etc

Cheers

Don

A car pulls away from rest at a set of traffic lights. During the first 3 seconds its speed v is given by v = 6t - t² m/s.where t represents time in seconds.

How far does it travel during the first 3 seconds ?

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ken

Hi ken

Nice work, as usual

Of course the maths is spot on, but I especially enjoy looking at the "style" of your Math Type in Word.

I think I shall try to find a few equations to test your skills with Math Type rather than the maths itself