randals@sri-unix (07/28/82)
A puzzle for you to play with:
A hole six inches long is drilled clear through the center of a
solid sphere. What is the volume of the material remaining?
After getting your answer (and I haven't left anything out),
try to give a physical explanation and/or interpretation of
the suprising result.
Randal L. Schwartz
Tektronix Microcomputer Development Products
Beaverton, Oregon, USA
UUCP: ...!{ucbvax or decvax}!teklabs!tekmdp!randals (ignore return address)
CSNET: teklabs!tekmdp!randals @ tektronix
ARPA: teklabs!tekmdp!randals.tektronix @ udel-relayjarvis (08/03/82)
#R:azure:-108100:pur-ee:6600004:000:261
pur-ee!jarvis Aug 2 23:39:00 1982
Since the hole must go through the center the radius is three
inches. Since the length of a hole through a sphere can be
defined only for an infinitely thin hole the volume is unchanged.
V=4*pi*3sup(3) [I hope I remembered the equation correctly]barry (08/06/82)
Regarding the six-inch hole, I saw an article that seemed to indicate that
the only possible configuration is an infinitely thin hole through a three
inch radius sphere. This is incorrect, as any sphere of radius greater than
three inches can have a six inch long hole bored into it, with the radius of
the hole being equal to the square root of the difference of the radius of
the circle squared and three inches squared. The significance of the
degenerative solution is that the original question indicated that all in
information was given, and that the sphere radius was not, making the
degenerative solution the general one. If the problem is worked out long
hand, with the radius of the circle being left as a variable, the volume of
the sphere remaining after the boring will not contain the variable, and the
volume, in fact will be that of a three inch radius sphere - 36 pi cubic inches.
(For the above, replace radius of the circle with radius of the sphere - I
was thinking in terms of the cross-section.)
Another puzzle in a similar vein follows. The thinkers on the net should be
able to figure this one out MUCH faster than those who try to program it.
Two trains are travelling on a track at fifty kilometres per hour, one
travelling east, and the other west. At the point when they are one
hundred kilometres apart, a very quick bee (travelling at two hundred
kilometres per hour) leaves the front of the east bound train and heads
straight for the other. As soon as it gets to the other train, it turns
around (in zero time) and heads back. The bee continues to shuttle back
and forth between the trains until the two collide, whereapon the bee is
crushed in between. The problem is to figure out how far the bee has
travelled.
Barry Lay
University of Toronto Computing Services
...!utzoo!utcsstat!barry