Conference thebay::joyoflex

>    Given 2 points A and B situated at equidistance from each other,
>    how can I move A without B noticing it ?
    
What are you talking about Roger? How could points NOT be the same distance
from each other?

How can points notice each other?

Is this a riddle?  If you give us more clues you may get more answers. 

	Steve 

DYO780::BRAIN_BOGGLERS  ??

kp7 etc etc

    re .1
    
    I read the joke on .0 in a book the other day, and thought
    it was extremely funny; but then I'm known to have a strange
    sense of humour at times. Oh well, forget it.
    
    	roger

    	I think the two points have to form an equilateral triangle in
    n-space. Then if you subject them to a gravitational field for a moment
    they bend, even though they are still noticing each other. (Every
    couple in a field has its moment).

    well, in order for you to move A without B noticing you can:
    
    distract B while a lovely new point C wanders past in naughty
    underthings.
    
    wave huge quantities of cash at B or otherwise assault B's senses
    with high pressure sales techniques.
    
    have A tell B to close its eyes and count to 20 million, thus allowing
    it to sneak off peacably and be back prior to B's completing the
    count.
    
    fold the plane on which they exist at the same rate A moves, so
    if A and B can only see on that plane, B will not see A move if
    it is folded in particular directions with respect to A's motion.
    
    Have a nuclear bomb create a superb diversionary tactic.
    
    
    ps:  I love the "every couple in a field has its moment..."
    
    -Jody
    

   Easy.
   
   If we assume that B is always noticing A, then all A has to do is to
   move in a circle around B.  B will never notice that A is moving since
   B is always noticing A and will move to keep noticing A.  
   
   If we assume B is not noticing A, then A can move anywhere, anytime
   and the question becomes moot.
   
   QED.
   
   
   							Terza
   
   
   P.S.-Actually, the first would still be true if B only noticed A some
   of the time.  It would have to find A first to be able to notice if
   A moved or not.  If there were no other point of reference, and none
   was provided in the original question, then B wouldn't be able to
   determine whether A had moved or not.
   
   
   

Re .0

     How about moving A while it's dark.

    If points A and B are on the same piece of paper, you can move the
    paper (which moves point A) but to point B it will always appear
    to be in the same place.  If it is also required that point B remain
    in the same place which A moves, simply rotate the paper about point
    B. 
    
    Everything is relative, right Cousin?

    
    
    Actually, 481.0 may be more related to the JEC process than to
    Geometry, Brain_bogglers, Puzzles, etc..

    :-)
    jc
    p.s. Of course, it may be more related to itself than anything other!

A is a point; hence, subtends no angle.  Therefore A can move anywhere 
along the line A-B without B's noticing it.

Moffatt Kennedy
Greenvile, SC

    	But A is always along the line A-B anyway.... Maybe we need to
    provide Betty with a compass so that she can work out what direction to
    go to find Andy.

    Sorry, I think it's pointless to try and refer to A and B as Andy and
Betty.  Andy and Betty comprise many points, subtending many inherent angles.
The attempt is poignant, but provides too many red-herring points of view.
    Before we get to the point of no return, I think we need to clarify:
                      Does B have the capacity to notice?

>    Given 2 points A and B situated at equidistance from each other,
>    how can I move A without B noticing it ?

OK, so this is getting serious. I'm still having trouble with the idea that it
is mere chance that A and B are the same distance from each other. My Euclidean
space does not allow for A to be nearer to B than B is to A. 

If you have some other dimensions to play with that allow you to bend the rules
then chances are you can move A at will, and B will be so confused it won't
know where to look, and so won't notice that A has gone walkabout. 

I think this is a trick question. Roger, are you there?

	Steve

    Thanks for all the suggestions. It's good to see that there still
    exist people who are stimulated by the mathematics (instead of 
    wasting their time in futile pastimes).

    Now remember that A and B are both inside a circle C with radius r;
    in what direction do you stroke C to make sure he becomes vicious ?
        
    	roger

    In order to turn C into a vicious circle, you must apply negative
    behavior modification techniques in the -x direction.  As the pressure
    increases, the circle will feel stress and will begin to deform
    in the -x direction.  C will begin to come around to your way of
    thinking (figuratively speaking, of course) within a short period
    of time (t).  As it spins, this pressure gradually affects
    all sides, forming the popular skating move known as the death spiral.

    -jody
    

   Re: .14
   
   Circle C with radius r should only be stroked counterclockwise, that
   is, to the left, a sinistra.
   
   
   							Terza
   
   p.s.- But if C becomes vicious, will B finay notice A?

    > Circle C with radius r should only be stroked counterclockwise, that
    > is, to the left, a sinistra.

    But, of course, if you stroke its bottom it should be to the right
    to get the same effect.
    
    Jeff.

    Hey, do you think we can put as many replies here as in the
    word association football topic (396)? (-:
    
    Dan

    G'day from Downunder,
    
    Now if you draw a line from A to B you get A-B. Since a point has
    only position and no size, then A-B=0. ergo nothing changes so B
    doesn't notice whatever A does. Or maybe
    
    A and B are equidistant in time not space and so that would add
    another dimension so to speak....
    
    Howabout if there was a six-foot high brick wall between A & B then A
    could even make rude gestures at B and B would not notice.
    
    Now if A was BEFORE a Mirror and B was its reflection in the mirror
    then the resultant view of A by A would Always be A which would
    make B redundant so he can go home and wouldn't care what happened
    to A. 
    
    Now if A was on top of B (ie occupying the same point) on one plane
    then by lowering A away from a casual observer, there is no way
    they would B noticing A go-away.
    
    
    re.2 if you saw the problem, did you see the answer?
                                    
    
    In the immortal words of Tom Sawyer's enemy, Judge Thatcher's son
    
    I holler 'Nuff' Whats the answer?  How about a clue??
    
    Derek
     

    re .19
    Conceptually, I would agree with your approach.
    Let me just give you a couple of additional hints: 
    
    . Could you specify what is the _longest_ distance 
      from A to B ?
    
    . The circle (C) should never be confused with the point C 
      (still going around in naughty underthings)

        	roger

>    > Circle C with radius r should only be stroked counterclockwise, that
>    > is, to the left, a sinistra.
>
>    But, of course, if you stroke its bottom it should be to the right
>    to get the same effect.

If you stroke it's bottom, that's "a Sinatra." (It helps if you address 
the circle as "honey.")
					>>>==>PStJTT

    B, being a point, can't have stereoscopic vision; so A can gallivant
    directly towards or away from B without fear of detection.
    
    The stroking of the circle is, by definition, tangential.
    
    b

    G'day again,
     Well now....
    
    All this talk of circles and touching and...  enough to take one's
    mind off the loci..
    
    
    Now how about the longest line?  Seems to me if there can be a longest
    line, then the line length must be finite. Now a plane is not finite,
    therefore, arguably, A  and B cannot lie on a plane (or a plain or
    a steppe); so where can A and B reside?
    
    How about on a sphere? Now the longest line from A to B  can be defined
    being of length 'Pi times R' where Pi is that well known Mathematician's
    friend with a value often given as 22/7 and R is the Radius of the said
    sphere.
    
    NOW, if A is 'over the horizon' from B and given that the sphere is not
    transparent, then A cannot ever be seen from B so he can go
    skinny-dipping or whatever takes his fancy without B noticing. 
    
    What d'yer reckon? Am I getting warmer??
    
    Or perhaps (word association here warmer = colder = polar = old problem
    about going S then W then N and getting back to start point...)

    Howabout if A, being over the horizon from B travels around the sphere
    at a constant radius from B ? A will remain 'equidistant from B' regardless
    of how far he travels.

    Is that close?
    
    Derek
    
    
    

    > Now how about the longest line?  Seems to me if there can be a longest
    > line, then the line length must be finite.
    
    How about the largest number? [Eric, please don't type it in this
    file :-)].  Same difference.

    > NOW, if A is 'over the horizon' from B
    
    Unless A is sitting on top of B, it will always be 'over the horizon'!

    Ah yer ah er G'day,
    
>        > NOW, if A is 'over the horizon' from B
    
>    Unless A is sitting on top of B, it will always be 'over the horizon'!

         'Pends how big R is in the sphere and how high B is relative
    to the surface of the sphere.... unless.. and here we go off at another
    tangent... A is along the tangent and is therefore free to move...
    No that would put them back on the same plane.
         
    Suppose B was a point above the sphere, and A ran around the parallel
    of latitude (as it were) coincident with the point of contact of
    the tangent from B to the sphere?  Range of view of
     B ~= SQRT(ht of B in feet)+3 miles on the earth (given its not
    foggy etc).
    
    If A were to sit on B's back, piggy back style, provided A did not
    eat in B's ear, B might be content to carry A around thereby A moving
    without B noticing (A is attached to a large helium balloon with
    lift = A's mass so B can't feel A sitting on his back)
    
    or suppose A and B were along the centre line of a mobius strip
    and A and B start off at a run along the centre (one should always
    tear down the dotted line.......rip..... (these jokes??? are getting
    worse) they could run forever and B would not notice A had moved
    - given that they ran at the same speed and A did not stop at
    McDonalds for a McPuke and chips.
    
    Now the title of the last note was "pretty Close" followed by a
    smiley face.... Is that a hint I ask myself like 'A & B  must be
    pretty close' like coincident.   or was it merely that we're getting
    warmer? I must confess I need the weekend to mull this one over
    some more - and as I must go and get my car out of the menders -
    I'll see you on Monday (net permitting) - in the meantime how about
    a tiny hint (you can send it under plain brown paper cover to me
    via e-mail so's not to spoil it for others.. ;-)
    
    
    Derek

re .25 (Derek)

>        the tangent from B to the sphere?  Range of view of
>     B ~= SQRT(ht of B in feet)+3 miles on the earth (given its not
>    foggy etc).

    This is true to a certain extend, if you know what I mean.

    But what if A happens to move faster than the light ? 
    The theory says there will be a distortion of 1.75; recent research
    however indicate that it is more likely to be 21.43.

    In that case, supposing B is aged 25 (this is just an example, don't 
    take it literally), how many revolutions will it take for him to get back
    to the age of 8 ?

    	roger 

    Re: .25
    
    Aha. I'd supposed B to be fixed to the surface of the sphere.  If
    B is allowed to move, or, heaven forfend, even jump upwards, we
    are into a whole new sphere game.
    
    Re: .26
    
    > But what if A happens to move faster than light ? 
    
    I believe that A would become infinitely massive, and could no longer
    be contained in non-dimensional space.  In the resultant explosion
    as A metamorphoses into a solid object, B would be blown away and
    cease caring whether A was there or not.  In the absence of an
    observer, A would also cease to exist, plummeting back though time.
    I estimate the time it would take to reach the age of 8, as postulated,
    would be 7.4 femtoseconds.
    
    Jeff.

�    I believe that A would become infinitely massive
    
    You believe well. 'A' mass is:
                m0
    m = ----------------
            ___________
           /       v�
        \ /  1 - -----
         V         c�
    
    v is the speed of A
    c is the speed of light 
    m0 is naturally the mass of A when v=0.
    
    Oeuf Corse, no wonder how much energy it would take ( E = mc� ).
    
    Pierre

    Re. .27, .28
    
    Since m0 (initial mass of A at rest) is zero (A is a zero-dimensional
    point), its mass will still be zero no matter how close to "c" (the
    speed of light, not to be confused with "C", still wearing naughty
    underthings) A may venture. Similarly, it would require no energy
    at all to reach the speed of light. A is probably therefore already
    moving at light speed, and B has merely overlooked the fact.
    
    Andy

    Re .29:
    
    A zero-mass item going at light speed is a photon.  Therefore, B
    can't see A.  If B did, A would be absorbed (hence, destroyed).
    
    Steve Kallis, Jr.

    >A zero-mass item going at light speed is a photon.  Therefore, B
    >can't see A.  If B did, A would be absorbed (hence, destroyed).
     
    Not necessarily -- A could be a neutrino, which also has a rest mass 
    of zero.  And neutrinos aren't easily absorbed.
    
    					andrew

    	The original question presupposes that Betty is aware of the
    continued existance of Andy. Now photons and neutrinos are only
    detectable when they are absorbed like falling into blancmange, or
    bounced like Eeyore. Hence Andy cannot be a photon or neutrino.
    
    	The fact that Andy also has zero dimensions excludes most other
    physical explanations for this awareness, and I think only leaves
    telepathy, which means that Andy (and Betty) are sentient beings, which
    justifies the personalisation.
    
    	My personal theory is that Andy (and maybe Betty too) is one of the
    angels that danced on the top of a pinhead. Perhaps his real name is
    Abdiel.

    It has not been proved that neutrinos have zero rest mass ...

    	Will anyone who finds a stationary neutrino please put it in an
    inter-office envelope and send it to me. When I have enough I will
    weigh them and post the result here.

    re:.34
    
    There's a whole bunch of them heading your way right now.
    
    It's up to you to catch them. :-)
    
    --- jerry

    re: .34
    
    Did you try looking in a neutrino station?
    
    --bonnie

    [The following just showed up on the usenet, and this seemed an
    appropriate place to put it.]
    
    <From: [email protected] (andy frederick)>
    >
    >This next one is from my calculus teacher in my freshman year.  He had a 
    >horrible habit of explaining the wrong way too much and oversimplifying the
    >correct way so that we were always confused.  When I heard this I just had
    >to write it down.  It sums up trigonometry beautifully.  Unfortunately, he
    >said it with a straight face:
    > 
    >  "Whatever works with sine will usually work with cosine --
    >   just replace all the things with co-things."
 
    -b

    
    
    And give me an answer do,
    
    I'm half crazy......
    
    about this problem. I've exhausted my brain and my friends' brains
    too.
    
    Derek - even a concrete hint would do!
    

    The author of this "riddle" (and the longest path) is Jean Tardieu;
    Derek, in .38, would like a *concrete* hint, so here's another from
    same:
      Given a wall, what's happening behind it?
    
    John.

I'm examining the wall.  It's made of brick, but I don't think I'll
stick around.  See ya.