Brain Teaser No 1

Ah ha!! Forgotten how to find square roots without the old calculator or log tables, have we??

Remember long division? sqrts are somewhat similar but involve dropping down 'pairs' of digits each time (counting pairs outwards from the decimal point) ....memory comming back to life yet?

Bet ken c can remember... his French might be a bit rusty, but not his algebra!

Cheers

Don

(And I'm pretty sure I could still do a Maths A level and score respectably. It only started to get non-trivial at a more advanced level, this is wrt 20 years ago.....)

Anyway, I thought BAM's clue might be 'astrophysics', so I spent a minute or two thinking about ladders between orbits.

The equation Ken and Don have produced is correct, so any more elegant result must be equivalent.

For those prepared to deal with manual conversion into a PDF there is some 'professional mathematician' stuff available
here.

I can email the pdf to anyone who wants it. It's too big to attach here by about 100k zipped.

Paul

quote:

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Remember long division?

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Long division I can do. Square roots didn't stick. I think I was in one of the last years to be taught some slide rule stuff. I have one somewhere, bought from WHSmiths.

I was a bit into calculators in the late 1970s/early 1980s. The last proper one I had was a Hewlett Packard that both didn't have an equals key (way cool....) and did have a 'Normal' and its inverse function, which saved looking them up in the tables.

Paul

my thinking leads me to that the distance <d> is not determined by <L>; its more the other way round -- the appropriate <L> is determined by the height of the floor where the NAP500's are stored. given this height <H>, then <d> is determined from H = 1 + 1/d i.e. d = 1/(H-1). when H=1, there is no solution, i.e dont bother with the ladder. given H, and therefore <d>, you can determine <L>. (i.e. L is the dependent variable in the problem)

i guess my point is that our industrial espionage friends(?) will be more concerned about finding the right ladder length (given the constraints of the problem) rather than finding <d>, which follows immediately if they know <H>. simply arranging any ladder such that it touches the points mentioned in the problem description might not do -- the ladder may be too short to reach where the NAP500's are -- hence this will be a wasted calculation.

waffle waffle ramble ramble babble babble... the above probably doesnt make any sense whatsoever ... its too late in the night. its your fault bam. it will soon be my turn to post a brain teaser soon --- and, you wait, it will TEASE!!! but i will start with one or two easy ones.

i think we now need the elegant solution to bams teaser. don, paul, omer, are you getting any further with this??

enjoy

ken

yeah, rubbish. just read previous posting that the ladder is "long enough" for the job... told you it was too late last night.

enjoy

ken

enjoy

ken

We know that

d = (X-sqrt(X^2-4))/2

were X = sqrt(1+L^2)-1.

We can also see that another solution to one of Don's quadratics gives h and h=1/d

threrefore

d = 2 / (X+sqrt(X^2-4)).

These are both exact and neither elegant enough to keep Bam happy. Now maybe there is some way of simplifying these that eludes me but otherwise the solution would not be exact.

Matthew

It appears to me that if one can find A where

LsinAcosA = sinA + cosA

then d = tanA

But I don't know how to proceed....

('A' is the angle between the ladder and the wall)

This post may vanish mysteriously.

Paul

On page 17 post No 8 (or 9), Bam says

I hope you are not angling for a hint, Don.

Now I think this might be the HINT that Bam refered to again, just a few posts ago. ie angling

Paul, I wouldn't delete your last post just yet, Bam might be sweating!!!

Cheers

Don

sin(A) + cos(A) = L sin(A)cos(A)

may also be written:

sqrt(2) * sin(A+45) = L sin(2A)

that 45 degrees cropping up is interesting as it is the angle that gives the minimum ladder length L(min).

but i could not immediately see how this would lead me to a simple expression for <d>. substituting the values of trig functions in terms of <d> either leads to a tautology, or the expression for <d> that we already have.

enjoy

ken

quote:

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sqrt(2) * sin(A+45) = L sin(2A)

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That was my previous step....

Basically I've been laying the ladder on a horizontal with a coordinate origin at its centre and then rotating the floor and wall around it, I was interested in what curve the corner of the crate would then trace. The junction of floor and wall obviously traces a circle radius L/2.

The interesting corner of the crate traces a curve,

x = lcos2A - sqrt(2)cos(A + 45)
y = lsin2A - sqrt(2)sin(A + 45)

'A' is again the angle between the ladded and the wall, and 'l' is L/2.

We are interested in the case where y = 0.

I haven't the means to draw a picture, but I think the visualisation works.

It should be possible to unparameterise the equations and express the curve in terms of x and y. d can easily be expressed in terms of the l and x where y = 0. But the ugly square roots immediately raise their heads.

quote:

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but i could not immediately see how this would lead me to a simple expression for <d>. substituting the values of trig functions in terms of <d> either leads to a tautology, or the expression for <d> that we already have.

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Unfortunately the expression we have is correct, but that does mean that the angle stuff is correct too....

Paul

if d = 0.5 * ( x + sqrt(x^2 - 4) )

then (1/d) = 0.5 * (x - sqrt(x^2 - 4) )

which also leads to a nice expression for:

d + 1/d = sqrt(x^2 - 4)

where x = sqrt(L^2 + 1) - 1

interesting, but on close examination, didnt seem to lead anywhere...

enjoy

ken

The Sony engineers have used the ladder and now got over the wall (they used my 'neat' formula !! not Bam's elegant one) and are now in the warehouse.

Now remember its Xmas, so the warehouse is virtually empty 'cos most of the kit was all delivered to Grahams and other fine dealers in time to fill our Xmas stockings. Anyhow, there are eight 250s and two 135s scattered around the room (randomly). But the room is pitch black and the boxes are identical as are the weights of each package. So it’s a genuine game of chance. The Sony guys can only escape with TWO boxes. They really want a pair of 135s.

What's their chances of picking them?

Cheers

Don

BTW, they used that bloody ladder to climb over Blzebub's orchard wall, again using my 'neat' formula and that handy crate!

Cheers

Don

quote:

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What's their chances of picking them?

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2/10 * 1/9?

Paul

quote:

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How can they do this without opening any of the sacks?

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Six equations for nine variables looks much too exhaustive for me.

OTOH exhaustive is what computers do well, so I know the answer.....

Paul

[This message was edited by Paul Ranson on SATURDAY 09 February 2002 at 23:42.]

10*9 / 1*2 = 45

only 1 of those combinations has 2x135 (passive!!)

so probability required is 1/45

another approach, which produces the name answer is to find out how many seperate 2 digit numbers you can form from the 10 numbers 0,1,2,3,4,5,6,7,8,9 -- exclusing repetitions. this is clearly 100 - 10 = 90

assuming 8 and 9 are seq nos for 135's -- you can form 2 numbers with 8 and 9, that is 89 and 98.

so probability required is 2/90 = 1/45 as before.

enjoy

ken

I'm glad that Paul's pretending to be taking his time on the bunch of apples so as to give others a sporting chance!

Cheers

Don

took a bit of doing this one, linear equations with integer constraints. but things got a bit easier when i allowed 0 sacks, which i had ignored initially (and got no feasible solution ).

solution process is a bit awkward to explain, but here it is:

Marco 4 sacks(16kg), 0 sacks(14kg), 4 sacks(9kg)
Vuk 2 sacks(16kg), 1 sack (14kg), 6 sacks(9kg)
Mick 1 sack (16kg), 6 sacks(14kg), 0 sacks(9kg)

and the solution checks too...

its important that the each share in weight is obviously 100kg (i.e 300/3)

how very fair of you to suggest equal sharing amongst those 3...

enjoy

ken

[This message was edited by ken c on SUNDAY 10 February 2002 at 01:59.]

Your answers and updates are posted in the early hours. I hope these teasers aren't giving you sleepless nights!! Just in case they are, try counting sheepdogs!!

Bill, the shepherd and Shep, his dog, are going home after a day on the hills. Bill walks at a steady 4 mph. When they are half a mile from his cottage, Bill sends Shep on ahead to warn his wife he is on his way. Shep races to the cottage, barks, immediately returns to his master and continues to run back and forth between the cottage and Bill until Bill reaches home. Shep's running speed is 16 mph. How far did he run altogether, from when he was sent on ahead?

Perhaps if you try this one 'in your head', you'll fall asleep in no time at all.

Cheers

Don

they are, and i blame you!!!

no seriously, i "work" odd hours. rarely sleep before 0200 hrs -- the fact that the hifi sounds best (even at very low volume) around this time may have something to do with it.

enjoy

ken

The shepherd has half a mile to go at 4mph.

The dog is running 4 times faster, in the same time period, and therefore runs 2 miles.

SteveD

Well done, but remember, these are the sorbets, eaten between the main courses, which of course are prepared by Bam. They are designed to prevent us getting indigestion through eating ladders!

A hi-fi dealer sells Naim hi-fi which is priced at, CD player £160; amplifier £230 and speakers £390; and Sony hi-fi where the prices are CD player £170; amplifier £240 and speakers £400. A customer buys some of this hi-fi and it costs him exactly £1,000. What does he buy? Hint, like in real life, there is absolutely no rationale about what he buys!!

Cheers

Don

PS the original was based on chocolate bars at 16p etc and the litle girl spent £1....but then you would realise that this one had been lifted out of an 11+ paper and you would ignore it with all the comptent it deserved!