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Posers and Puzzles

Posers and Puzzles

  1. 07 Nov '02 22:44
    There is a piece of string that just fits snugly around the equator of
    the Earth (ignore any mountains that get in the way - apologies to
    any goat herders living in the Andes).

    The string is then cut at a single point and lifted upon one foot high
    posts that are placed around the entire equator and (for the benefit
    of any pedants reading) are a negligible distance apart.

    The question is: How much extra string must added to join up the two
    ends of the original piece of string (to 6 significant figures please)?

    Hint: The circumference of the Earth is roughly 129,101,050 feet.
  2. Donation Acolyte
    Now With Added BA
    07 Nov '02 22:54
  3. 08 Nov '02 00:17
    Okay...that was a bit easy...

    However, I do like this little puzzle...the way it arouses an initial
    reaction of 'hmm, must be loads' (caused by thinking with the bit of
    the brain that is linked to most foot-in-mouth incidents)...but then
    the logical chunk of the brain kicks in (with a little 'Eureka!' perhaps?),
    to remind us of all those sweaty afternoons spent in maths class
    trying to peek down Miss Warwick's top...

    Where was I? Oh yeah, here's an extension to the problem then...

    Suppose the string is replaced by a big taut elastic band (the posts
    are still in place). You've been volunteered for the lucky job of
    removing the posts. Assuming the elastic band remains taut, what is
    the minimum number of posts that can be left standing so that the
    elastic band doesn't touch the ground? What is the distance between
    the posts?

    Assume the circumference of the Earth really is 129,101,050 ft.
  4. Donation Acolyte
    Now With Added BA
    08 Nov '02 15:08
    I don't like the numbers, so I'll use letters instead (a common ploy among mathmoes). Let r
    be the radius of the Earth in feet, ie 129,101,050/2pi. Pick two consecutive posts that are
    just close enough together for the band not to touch the ground in between (I'm assuming the
    surface at the equator is perfectly smooth, otherwise the answer involves putting the posts on
    top of oceanic ridges/hills/mountains!). Then the band forms a tangent to the sphere at the
    midpoint of the Equator. Now draw a line down through each post. The lines will make an
    angle t, say, at the centre of the Earth. You now have a triangle with altitude r and two sides
    r+1, and you can show t is 2 arccos(r/(r+1)) (not very large!). I'll take t in radians; round t
    down to t' so 2pi is an integer multiple of t', then the distance between posts (along the
    Earth's surface) is rt', and the minimum number of posts is 2pi/t'. (I think it's in the region of
    20000 posts, but I may have pressed the wrong buttons on my calculator.)
  5. 13 Nov '02 02:20
  6. Donation Acolyte
    Now With Added BA
    14 Nov '02 19:25