A puzzle for you

 BBS: Inland Empire Archive
Date: 02-28-93 (15:23)             Number: 376
From: RICK PEDLEY                  Refer#: NONE
  To: CHARLES GRAHAM                Recvd: NO  
Subj: A puzzle for you               Conf: (2) Quik_Bas
 On 02-25-93 Charles Graham wrote to All...

 CG> 'PUZZLE.BAS - EGA or better required
 CG> '
 CG> 'Assume the Earth's circumference at the Equator is 25,000
 CG> 'miles. You run a piece of string around the Equator so the
 CG> 'ends of the string meet exactly. The string is 25,000 miles
 CG> 'long.
 CG> '
 CG> 'You add 1 foot to the length of the string. If the now
 CG> '25,000-mile-plus-1-foot long string could be suspended evenly
 CG> 'above the Equator, how far off the ground would the string
 CG> 'be?
 CG> '
 CG> 'No tricks. Just math and logic.

No tricks, but a surprising answer. Let's give a few more people
a chance to solve it.

COLOR 14, 0
LOCATE 17, 1
PRINT "Here's another problem for all you analytical geometry experts:"
PRINT "An 80' rope is suspended from the tops of two vertical 50' poles."
PRINT "The lowest point of the curve is suspended 10' above the ground."
PRINT "How far apart are the two poles?"
LINE (45, 46)-(49, 180), 6, BF
LINE (551, 46)-(555, 180), 6, BF

LINE (20, 180)-(580, 183), 2, BF

FOR x! = 0 TO 250
   y! = x! ^ 2 * .001
   PSET (x! + 300, 110 - y!), 15
   PSET (300 - x!, 110 - y!), 15
   PSET (x! + 300, 111 - y!), 15
   PSET (300 - x!, 111 - y!), 15

... OFFLINE 1.50

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