Brain Teaser No 1

2 + 9 * 1 =
3 + 9 * 12 =
4 + 9 * 123 =
5 + 9 * 1234 =

etc. obviously do the multiplication before the addition ("bidmas" or something...)

enjoy

ken

Nice one ken, you know what I mean!

Cheers

Don

quote:

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Originally posted by Don Atkinson:
ken c
_do the following multiplications and tell me what you notice_

Nice one ken, you know what I mean!

Cheers

Don

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numbers are fascinating things aren't they?

enjoy

ken

I too watched the Acropolis program and found its content fascinating. Unfortunately I also found its content could have been conveyed in about 1/10 of the time that the program makers strung it out. It is just me? And tonight I watched this documentary about some divers off the shore of Alexandria who were searching for Cleopatras lost palace in the bay. Great plot, potentially exciting story - BUT instead a great deal of repetition and slow, slow, slow, slow narrative. Very little technical detail and very much boredom. Elevator music filling the pregnant pauses. And I've noticed similar things about Horizon these days. Am I speeding up or is the great TV God's baud rate declining rapidly?

I have arrived at the following equation:-

sqroot(L>2 + 1) - 1 = D + 1/D

Just in case my notation is wrong, it can be described as follows:-

Square L and add 1, then take the square root of the result, then deduct 1. This value is equal to D plus the reciprocal of D.

So, once L is known, D can be solved reasonably quickly by quadratic equation.

If this is not the "elegant" solution, please will you now declare the correct one, as it is driving me crazy!!!!!!!!

STEVE D

If this is not the "elegant" solution, please will you now declare the correct one, as it is driving me crazy!!!!!!!!

The ladder problem has been around since the 21st Dec (page 13). You commented almost immediately. IMHO it is intriguing, annoying, irritating and driving me crazy also,etc(just like all maths and life in general). It is also tedious to re-read the past postings to see what has been tried. It is MOST irritating to be told " your current proposal was tried on 13th Jan, page 16, post No 10, and Bam wasn't impressed. He only gave us 5/10 for trying, in post No 11", so I'm sorry for irritating you. Really!

But please don't give up. Even if we don't have the solution, what we post or re-post might provide that spark of inspiration for somebody else who is just browsing/glancing through this thread.

But it is bl**dy irritating not being able to discover the answer and before long, we shall all be crazy!!

Cheers

Don

enjoy

ken

Firstly, sorry for not realising my equation had already been suggested. Pity, I was very pleased with it!! I feel very demoralised today, having gone to bed last night dreaming of that bottle of champagne.

It is virtually impossible to trace what has already gone before though.

Where D>1, then D = L - (1 plus "a bit").
The "bit" starts at 0.828427etc (where L = sqrt8), and rapidly tends to zero (as L tends to infinity).

BAM has already suggested we need to use less conventional maths (does this mean conventional maths but less of it, or maths which is less conventional?), so I agree with Ken that we should look elsewhere other than the standard equations.

I've already explored converging series (ie D converges to L - 1) and also calculus (ie rate of change of D with respect to L) but so far to no avail.

I would like a clue as to where to look next.

STEVE D

I could offer more clues but this would devalue the puzzle and it would no longer deserve M&C as a prize.

Of course, I must be careful about what I write lest it be misconstrued as a clue. There are one or two burried in this tomb already - which have been recognized, started upon and then abandoned.

And I should state for the record that although I went off on a bit of a tangent with my rant about TV documentaries I can assure you there are no hidden clues there.

For those attempting curve-fitting, this may be of some interest: http://www-groups.dcs.st-andrews.ac.uk/~history/Curves/Curves.html

BAM

1 large outer circle, inside which we draw
5 middle circles, each of which touch the outer circle and the two ajacent middle circles
1 inner circle which touches each of the 5 middle circles

overall picture looks a bit like a wheel bearing with 5 roller bearings!

We have already learned how to calculate the diameters of the middle and inner circles given the diameter of the larger circle (and the number of roller bearings)

Try calculating :-
(a) distance (centre to centre) between two ajacent middle circles
(b) distance between two non-ajacent middle circles

Compare ratio of a/b and b/a and post comments??

Cheers

Don

a = 2r
b = 4rsin0.3Pi

So b/a = 2sin0.3Pi = 1.618 etc.

We don't need circles here, the ratio is the length of the side of a regular pentagon to the distance between two non-adjacent vertices.

IIRC the 'Golden Ratio' is where a/b = (b-a)/a or more conveniently b^2 - b = 1. It happens to be 1.618....

It would be nice to demonstrate analytically that 2sin0.3Pi is the 'Golden Ratio', I don't see an obvious method at the moment.

Paul

It would be nice to demonstrate analytically that 2sin0.3Pi is the 'Golden Ratio', I don't see an obvious method at the moment.

No doubt Bam will supply an ELEGANT version of this proof???????

The rest of us can rest in blisful ignorance and content with the practical demonstration provided by Paul........or can you?

Cheers

Don

This is of the form: x=0.5(1+sqrt(4n + 1))

Thus x^2=x+n

Don's conjecture is that 2sin(0.3pi)=phi

Substitute x=2sin(0.3pi) and n=1
4(sin(0.3pi)^2=sin(0.3pi)+1

Now, 2(sinA)^2=1-cos2A

so 2-2cos(0.6pi)-2sin(0.3pi)=1
or cos(0.6pi)+sin(0.3pi)=0.5

This is true when calculated.

I haven't been able to proove it algebraically yet.

i am following the idea in my earlier posting:

http://forums.naim-audio.com/eve/forums?q=Y&a=tpc&s=67019385&f=58019385&m=9501935533&p=20

and various thoughts by none other than our Don.

will let you know how i get on -- i feel excited, hopely not for nothing...

enjoy

ken

cos(0.6pi)+sin(0.3pi)=0.5

I spent some time getting confused in infinite series and stuff....

But it's relatively simple to prove that the ratio we got from the pentagon is 'Golden', which obviously amounts to the same thing.

To save typing,
here's one somebody else did earlier.

Paul

The distance to from the ladder to the edge of the cube will be nil i.e. by definition touching??

[This message was edited by Happy Listener on SUNDAY 10 March 2002 at 21:06.]

Phew.

Paul

There are, I think, a number of possible mis-directions in the text.

Will think some more post watching '24'!!!

the trigonometry of the pentagon (of size 1, without loss of generality) dictates that:

2 * cos (0.2pi) * cos (0.4pi) = 0.5

this follows from the relationship between any diagonal and the equal sides of the pentagon.

by using the trig half-angle formulae, we can reduce this to:

cos (0.6pi) + cos (0.2pi) = 0.5

or

cos (0.6pi) + cos (0.5pi - 0.3pi) = 0.5

i.e.

cos (0.6pi) + sin (0.3pi) = 0.5

as was required.

the link to the golden ratio also comes directly from the identity:

2 * cos (0.2pi) * cos (0.4pi) = 0.5

which can be expressed as:

4 * cos (0.2pi) (2 cos^2(0.2pi)-1) =1

putting x = cos 0.2pi, we have:

8x^3 - 4x - 1 = 0

put z = 2x, and we have:

z^3 - 2z - 1 = 0

which factorizes to:

(z+1)(z^2 - z - 1) = 0

the second factor solves to:

z1=1.61803 and z2=0.61803

hence the familiar golden section again.

hope this is clear.

enjoy

ken

Near a small town with a big church in Wiltshire, an audio industry worker named Paul suplements his meagre salary with part-time farming. Paul owns a circular field of grass of unit radius and he also owns a very hungry goat from which he derives milk for cheese. Paul wishes to tie the goat to an exisiting post on the circumference of the field and wants to manage the growth of the grass so the goat doesn't decimate it too quickly. He therefore decides to use just enough rope so that the goat can only eat 50% of the grass.

What must the length of the rope be? Assume the distance between the jaws of the goat and the point of attachement of the rope is insignificant.

Don, OK what is the answer?

Have sent the following reply which incriminates myself and possibly one or two others so my apologies to one and all, including Bob!!

Bob,

Sorry, I didn't read my e-mails last night so just found yours a few moments
ago.

If you are looking for the answer to the 'golden ratio' question about five
circles inside a larger circle, then

Remember, the external angles of a five sided polygon (pentagon) are 72 deg
or 0.4pi radians. Similarly, the internal angle at the apex of each point of
a five sided star is 36deg, and the half angle is 18deg or 0.1pi. The base
angle of each major triangle in a star is also 72 deg.A quick look at a
calculator will show that 2*cos72deg=0.618.... and 2*sin18deg=0.618. I
think ken c and others have clearly demonstrated how to generate golden
ratios in a five sided or five pointed regular polygon from fairly basic
principles.

Its also worth noting that the horizontal line accross the top of a star
(and of course the other four similar 'long legs' of a star) are trisected.
The inner and outer length of these trisected lines are related by golden
ratios.

Whenever regular five sided figures are involved, there is nearly always a
golden ratio in there somewhere. In the case of the five circles, clearly
the primary golden ratio is the distance between centres of circles. As
others have pointed out, the circles quickly lead you into the pentagon and
the star.

If you are looking for the answer to the ladder problem, then I'm as baffled
as you about the elegant solution Bam refers to. I could only derive the
'neat' formula.

If you are looking for the answer to Bam's goat, eating 50% of the grass,
well, i've only just read that one and haven't had a chance to think about
it yet.

Of course, if you are the professor of pure mathematics at Trinity College,
Cambridge, and all this banter and pseudo analysis is getting too much, well
for sure you have found me, and probably a few others, out!!!! I doubt if
(m)any of us could derive any of our 'solutions' from the axioms of
arithmetic, and I would certainly struggle to pass A Level maths again!!!

Hope this helps, if not let me know again.

Cheers

Don

I forgot to add, "welcome to this thread" and if you can throw any light on Bam's ladder problem, "double welcome". And if you are the professor........perhaps you could just humour us with 'easy' brain teasers ??

Cheers

Don

Looking through the "curve" website which BAM suggested, the closest match seems to be a pursuit curve. I'm still trying to figure out the relevance to the ladder puzzle ie what are the pursuer and pursuee! The equation of the curve is
y = cx^2 - logx. (log means natural log)

Another thought:

In calculus, the tangent to a curve is the derivative of a function.

The ladder is a tangent to a curve, and is equal to 1/d. We need to integrate this to arrive at the original equation.

The indefinite integral of 1/d, is logd + c
(again log means natural log).

Again, I am struggling to put it in context, but you will see from my previous threads that I have a sneaky feeling that "e" is involved somewhere.

In the forlorn hope that this is of some help!

Steve D

1.1587285 to 7 d.p

Could be wrong though, it is one ugly equation!

0=4*ASIN(L/2)+2*L^2*(PI/2-ASIN(L/2))-(1-(L/2)^2)^0.5*2*L3-PI

cheers

Matthew

quote:

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Originally posted by Matthew T: 0=4*ASIN(L/2)+2*L^2*(PI/2-ASIN(L/2))-(1-(L/2)^2)^0.5*2*L3-PI

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i also got quite an ugly expression and decided that perhaps a different approach was required. my initial stab was to find the area in the intersection of the 2 circles as the sum of the chord segments thus created and then solving for the fact that this area is half the grazing circle area. but i got a mixture of angles and sin of angle and decided to rethink. your expression suggests you took the same approach, yes?

enjoy

ken
ps: funny sort of dejavu -- i seem to remember solving this problem in my former life??

I took the approach of splitting the area in half and then taking segments of the two circles (one of unit radius and one of unknown radius) and the subtracting the triangle for which these two circles over lap. Doing the chord approach is going to give one ugly equation to integrate!

Matthew