Brain Teaser No 1

Posted by: Don Atkinson on 16 November 2001

THE EXPLORER

An explorer set off on a journey. He walked a mile south, a mile east and a mile north. At this point he was back at his start. Where on earth was his starting point? OK, other than the North Pole, which is pretty obvious, where else could he have started this journey?

Cheers

Don

Posted on: 23 February 2002 by ken c

the impression i get from bam's various posts is that we are not using all the information that we can to solve this problem. the equations we have so far are definitelyt correct, but perhaps not elegant enough. i guess elegant means not involving sqrt of a sqrt for the <d> expression. bam has also dropped other hints like "astrophysics" or "astronomy" which may suggest something to do with shapes such as hyperbola's etc...

i have given up on this problem and i am now looking fwd to the elegant solution.

don, i find andrew wiles solution using a clever combinationIwasawa Theory and Kolyvagin-Flach method to prove the Taniyama-Shimura conjecture (from which Fermat's follows) somehow untimately "unsatisfying". i would have preferred a "simpler" solution. its very elegant (i dont fully understand it) but boy, all that heavy mathematical machinery to prove what on the surface is a rather "simple" theorem. but whatever i say is totally irrelevant -- the theorem has been proved. i just hope that his proof may in fact suggest a simpler alternative. but who am i...

i first encountered this problem when i was doing A levels. i thought it was very simple, proceded to spend a long time getting absolutely nowhere.my main approach was factorizing: f(x,y,z)=(x^n) + (y^n) - (z^n)

ah well...

enjoy

ken

Posted on: 23 February 2002 by ken c

the goldbach conjecture. even euler failed to prove it so i am not sure i want to try. there have been various limited "constructive" proofs using a computer, but, AFAIK, no general proof. do you know different??

put your proof in a word document, publish it and become famous...

enjoy

ken

Posted on: 23 February 2002 by Don Atkinson

which may suggest something to do with shapes such as hyperbola's etc

ken c, y = 1/x is a rectangula hyperbola and d = 1/L is a half decent approximation for the ladder when L>(say)5, so I agree that some sort of hyperbolic equation is a distinct possibility.

However, a quick look through any maths book shows a hundred equations that plot as hyprbolae. Not one of them seems to give that magic gem that Bam is looking for!!

Mind you, I have also plotted the curve using the 'precise' formula that you published. If L is plotted on the X-axis and d is plotted on the Y-axis, then L is asymtotic to the X-axis at y=0 BUT d ISN'T asymtotic to the Y-axis at X=sqrt(8), because at this point the tanget to the curve is parallel to the Y-axis.

This suggests to me, something OTHER than an hyperbola might be involved???

Cheers

Don

Posted on: 23 February 2002 by ken c

Damn, Ken, I wanted Bam to work on this one for weeks, returning some of our ladder effort.

ooops!!! sorry omer, a bit thick, didnt "read" that one. i am sure you have plenty others??

enjoy

ken

Posted on: 23 February 2002 by ken c

This suggests to me, something OTHER than an hyperbola might be involved???

don, the hyperbola theory was really me clutching at the last straw (is this how you say it -- you know me and my french!!) -- trying to read too much into bam's hints. to be honest, you had done much more analysis on it than i have -- showing that perhaps this is a cul de sac, unless you transform the axes??? ignore last comment...

enjoy

ken

Posted on: 23 February 2002 by ken c

in the attached diagram, the large circle has diameter 10cm. each of the inner circles are identical size and they each touch the large circle and 2 adjacent circles symmetrically as shown (not very well i a'm afraid). non of the circles "cross" each other -- diagram not very good in this respect

find the radius of the largest circle which will fit in the middle.

enjoy

ken

Posted on: 23 February 2002 by bam

Been away a couple of days and have now just caught up. You all seem to be making such excellent progress. Do you want another hint yet? Interestingly, there was a brief aha moment a few pages back when I thought you had started down the right path but that quickly fizzled out and attention returned to going around in various circles.
BAM
[I shall be wandering incognito at tomorrow's show for personal security reasons and will not now, as I had written to Don in an email, be carrying a ladder]

Posted on: 23 February 2002 by ken c

prove that the product of 4 consecutive numbers is always one less than a square number.

so for example: 2 * 3 * 4 * 5 = 120 which is 1 less than 11^2

enjoy

ken

Posted on: 23 February 2002 by ken c

Interestingly, there was a brief aha moment a few pages back when I thought you had started down the right path but that quickly fizzled out and attention returned to going around in various circles.

you teaser you.

what pages exactly!!!

enjoy

ken

Posted on: 23 February 2002 by bam

Ken, is it 0.858cm?

Posted on: 23 February 2002 by ken c

Ken, is it 0.858cm?

oh bam, you are awfully close!!! but no.

this suggests to me you have the right method, but you have made a careless mistake. i will also double check that i havent goofed in stating the problem -- though looks OK to me?

enjoy

ken

Posted on: 23 February 2002 by Paul Ranson

n(n+1)(n+2)(n+3) = m^2 - 1

where n and m are numbers.

so,

n^4 + 6n^3 + 11n^2 + 6n = m^2 - 1

a suspicious symmetry to those coefficients...

(n^2 + 3n + 1)(n^2 + 3n + 1) = m^2

Which I think is sufficient to prove the assertion.

Paul

Posted on: 23 February 2002 by ken c

n^4 + 6n^3 + 11n^2 + 6n = m^2 - 1
a suspicious symmetry to those coefficients...
(n^2 + 3n + 1)(n^2 + 3n + 1) = m^2

Which I think is sufficient to prove the assertion.

that is correct. but you have a way with words. why "suspicious" and "i think"???

enjoy

ken

Posted on: 23 February 2002 by bam

Hmm.
My reasoning is this. Imagine a quadrant of the large circle that contains a small circle.
the radius of the large circle is 5cm
let r be the radius of the small circle
let p be the distance between the centre of the large circle (the right-angled corner of the quadrant) and the centre of the small circle.

For the small circle to touch the circumference of the large circle p+r=5

For the small circle to touch another small circle, or equivalently both sides of the quadrant, psin45=r

Solving gives r=5/(1+sqrt(2)) = 2.071

The distance from the centre of the large circle to the edge of the small circle must be 5-2r = 0.858. This is also the largest radius of a circle in the middle that just touches each of the small circles.

Posted on: 23 February 2002 by Paul Ranson

my head is spinning round? (is that the Yardbirds?)

There's obviously a bunch of symmetry. If we sit a circle of radius 'r' on some axes so that it touches both, then its centre is at (r, r). The distance from the origin to that centre is r*sqrt2. In Ken's problem this circle is bounded by a circle centred on the origin and radius 5, again from symmetry the point the small circle touches the larger must be up the line formed by the origin and the centre of the small circle, which is a radius of the outer circle.

So 5 = r*sqrt2 + r, r = 5/(1 + sqrt2), and Ken's answer is 5 - 2r, which calculates out at about 0.858.

Paul

[This message was edited by Paul Ranson on SUNDAY 24 February 2002 at 01:07.]

Posted on: 23 February 2002 by bam

Paul, I hope two wrongs make a right.

Variant:
What is the radius of the largest centre circle if there are three small circles within the larger one?

Posted on: 23 February 2002 by Paul Ranson

quote:
why "suspicious" and "i think"???

At what point does a 'proof' become obvious? I'm not sure what words are required to round things off, and I hadn't considered negative values of n, although on reflection it's obviously true for all integers.

Paul

Posted on: 23 February 2002 by ken c

apologies. 0.858 is indeed correct. i mis-stated the problem -- meant to say radius of the big circle is 10cm. hence your answer was half what i expected.

two rights means i am wrong...

go to bed ken...

enjoy(?)

ken

[This message was edited by ken c on SUNDAY 24 February 2002 at 01:30.]

Posted on: 23 February 2002 by ken c

paul,

I'm not sure what words are required to round things off...

i think i sort of see what you mean. in the context of your solution, perhaps no words were required?

enjoy

ken

Posted on: 24 February 2002 by ken c

prove that the product of any 3 consecutive integers is a multiple of 6.

enjoy

ken

Posted on: 24 February 2002 by Don Atkinson

Ken,

let the integers be n; (n+1); (n+2). these also are factors of the product n(n+1)(n+2).

two consecutive integers: one must be even and divisible by two hence have a factor of two, or be two.

three consecutive numbers: one must be divisible by three, or be three, hence have a factor of three.

it doesn't matter if the number that is even is also divisible by three, what is important is that at least one factor is 2 and another factor is 3.

Hence at least two factors of the product are 2 & 3 and the product is therefore divisible by its factors ie 2*3 = 6

Not a 'proof' as such but a sort of 'logic'

Any good ? (be careful! if you want 'elegant' you will have to wait for Bam!)

Cheers

Don

Posted on: 24 February 2002 by ken c

Not a 'proof' as such but a sort of 'logic'

Any good ? (be careful! if you want 'elegant' you will have to wait for Bam!)

don, elegant enough for me. well done!!!

i hope this was a useful distraction from the ladder problem.

enjoy

ken

Posted on: 24 February 2002 by ken c

Variant:
What is the radius of the largest centre circle if there are three small circles within the larger one?

bam, interesting variant.

equilateral triangle symmetry. i have a really elegant solution, but it won't fit in this email, so i await solutions from others --don? paul? omer?

enjoy

ken

Posted on: 24 February 2002 by Don Atkinson

bam, interesting variant.

equilateral triangle symmetry. i have a really elegant solution, but it won't fit in this email, so i await solutions from others --don? paul? omer?

Nothing elegant but 0.359 seems to ring out when I tinkle the triangle, assuming, of course the big circle is 10 dia.

Haven't check it yet!

Now, i haven't thought the next variant through at all (believe me!) but if we put five (5) circles inside the big one, do you recon there's a "golden ratio" in there somewhere?

Cheers

Don

Posted on: 24 February 2002 by ken c

this is NOT a puzzle. its just something i read about which is rather interesting, but i am not sure i can "explain" it. number theory has a lot of very interesting/intriguing results.

well, here goes:

take the number 142857 and multiples of it as follows:

142857x1 = (to start you off...)
142857x2 =
142857x3 =
142857x4 =
142857x5 =
142857x6 =

what do you notice??

now arrange the digits of the results of the multiplications (1 row per result) as a 6x6 matrix and compute the row sums and column sums. curious isnt it??

any ideas??

in fact you can continue the pattern beyond 6 and there will be interesting things to observe?

enjoy

ken

[This message was edited by ken c on MONDAY 25 February 2002 at 12:28.]