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  1. #31
    CC International Master Cat's Avatar
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    Bugger! Well done KB, it's been about 25yrs since I did my statistics and I guess my rustiness shows in my inelegant solution.

  2. #32
    Reader in Slood Dynamics Rincewind's Avatar
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    Quote Originally Posted by Kevin Bonham
    Feel free to post more, except I'm allergic to trig, geometry and calculus.
    OK geometry it is!

    Actually, I won't try composing any more problems but I do have a couple of favorites that like minded individuals might enjoy. The first one is a geometry one. It's cute and not very difficult.

    I'll post it tonight. (Perhaps I should start a new thread).
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  3. #33
    Illuminati Bill Gletsos's Avatar
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    Firstly let me congratulate all of you on excellent submissions. =D> =D> =D>

    Before someone asks why I didin't bother entering there were two reasons.

    Firstly I was/am busy doing ratings related stuff and secondly and more importantly I had seen this problem previously in a slightly modified form and therefore did not think it was fair to put in a response.

    With regards to Matt's solution let me say I have seen this reasoning applied in similar cases, although to be honest I had forgotten about it until seeing Matt's response. #-o

    One feature it has is its sheer simplicity. 8)
    I commend him on it. Well done Matt. =D>

    One interesting aspect is the following.

    From the figures it can be seen that
    ((2^18 )-1)/((2^19)-1) = (2^-1) - (1/(2^20 - 2^1))

    To the less mathematically inclined it is far from obvious that this is true. :-s :?

  4. #34
    Reader in Slood Dynamics Rincewind's Avatar
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    Quote Originally Posted by Bill Gletsos
    From the figures it can be seen that
    ((2^18 )-1)/((2^19)-1) = (2^-1) - (1/(2^20 - 2^1))
    That's pretty much covered in my derivation of the elegant solution. It goes

    P = (2^(n-2) - 1)/(2^(n-1) - 1)

    multiply by 2/2

    P = (2^(n-1) - 2)/(2^n - 2)

    Separting the numerator into 2^(n-1) -1 and -1

    P = (2^(n-1) - 1)/(2(2^(n-1) - 1)) - 1/(2^n - 2)

    Simplifying the first term you get

    P = 1/2 - 1/(2^n - 2)

    QED

    But you're right - it is not immediately apparent. And it is pleasing that two different lines of reasoning can reach mathematical equivalents which are not obvious.

    Actually I prefer the first form for calculating the probabilities but I favoured the second form in my solution as it more clearly demonstrates the monotonical increasing and bounded at 0.5 behaviour of the solution for n marbles.

    BTW as a further postscript had Dr Evil ensured the existence of marbles of both colours in the bag by a different method then different probabilities might result. EG say he looks after 19 marbles have been put in and if they are all the same colour then he adds a marble of the opposite colour, else he add one randomly. This is a similar problem and a similar answer (different colour is favoured) but with a slightly different probability.

    Just goes to show, you can't assume nothing.
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  5. #35
    Illuminati Bill Gletsos's Avatar
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    Quote Originally Posted by Barry Cox
    That's pretty much covered in my derivation of the elegant solution.
    But you hadn't published your elegant solution.

  6. #36
    Reader in Slood Dynamics Rincewind's Avatar
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    Then what's this?

    Quote Originally Posted by Barry Cox
    OK. It is the hour of reckoning. For my worked solution look at the following link. If you have any questions let me know here or by PM, whatever is your preference.

    Worked Solution
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  7. #37
    Illuminati Bill Gletsos's Avatar
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    Quote Originally Posted by Barry Cox
    Then what's this?

    Quote Originally Posted by Barry Cox
    OK. It is the hour of reckoning. For my worked solution look at the following link. If you have any questions let me know here or by PM, whatever is your preference.

    Worked Solution
    #-o #-o

  8. #38
    Reader in Slood Dynamics Rincewind's Avatar
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    Recreational Mathematics

    Q4. The Three Circles Puzzle

    Three conguent circles of radius r, colinear and centres are 2r apart (ie each are touching the adjacent circle at only one point).

    The point A lies on the line passing through the centres of the circles and the first cirlce, but not the middle circle. The point D is tangential to the third circle. B and C are the points where the line AD passes through the middle circle.

    Find the length of the chord BC in terms of r.

    The following diagram should help clear up any confusion...



    Send solution to me via PM before nexts Wednesday night, 21-Jan at 24:00 GMT+11.
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  9. #39
    Reader in Slood Dynamics Rincewind's Avatar
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    Not even one solution received yet. You guys are allergic to geometry, aren't you?

    Have a crack at it, you might be (pleasantly) surprised.

    If anyone wants a hint, send me a PM.
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  10. #40
    CC Grandmaster arosar's Avatar
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    Hey Bazza...I've been meaning to ask you man. No need to answer if you don't want to, but I think you already told us anyways. You a teacher or something. I atually thought you were in IT.

    AR

  11. #41
    Reader in Slood Dynamics Rincewind's Avatar
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    Quote Originally Posted by arosar
    Hey Bazza...I've been meaning to ask you man. No need to answer if you don't want to, but I think you already told us anyways. You a teacher or something. I atually thought you were in IT.
    Yes, I'm in IT, not a teacher.

    Puzzles are a hobby though. Cryptic crosswords, maths puzzles, logicical puzzles, etc, doesn't matter much.


    BTW I only got one response to the geometry puzzle. You know who you were. It wasn't right.

    So here are some hints.


    The answer is a rational coefficient of r.

    It can be reasoned using Euclidean geometry and Pythagoras' theorem.

    Key facts.

    A line originating from a circle which is perpendicular to a chord of that circle, bisects the chord.

    The same is true as the chord becomes diminishingly small (i.e. the perpendicular to a tangent passes through the centre).
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  12. #42
    CC International Master Cat's Avatar
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    When's the new deadline BJC?

  13. #43
    Reader in Slood Dynamics Rincewind's Avatar
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    When the Noachian deluge of entries abates.

    The next puzzle will be pure number theory so the sooner I get the entries the better.
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

  14. #44
    Account Permanently Banned PHAT's Avatar
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    Quote Originally Posted by Barry Cox
    When the Noachian deluge of entries abates.

    The next puzzle will be pure number theory so the sooner I get the entries the better.
    Is this over?

  15. #45
    Reader in Slood Dynamics Rincewind's Avatar
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    Quote Originally Posted by Matthew Sweeney
    Is this over?
    No. Still no entries since the massive hints I dropped. Perhaps geometry is even less popular than I thought.

    I'll post the solution when I find the text for the next puzzle which is much more difficult but related to number theory so might be more popular.

    BTW Welcome back. You missed the start of the Club Rapid.
    So einfach wie möglich, aber nicht einfacher - Albert Einstein

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