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Joined: Jun 2002
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Carpal Tunnel
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Carpal Tunnel
Joined: Jun 2002
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diameter must be irrelevantthanks doc, for giving me one more puzzle that makes me feel inadequate to the task... and is going to bug me all day long!
formerly known as etaoin...
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Pooh-Bah
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Pooh-Bah
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Okay guys, and while you're at it, prove this, will you?:
xn + yn = zn
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Joined: Aug 2001
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Carpal Tunnel
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Carpal Tunnel
Joined: Aug 2001
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Well, this sphere-with-a-hole puzzle is "under-determined". Try it for a six-inch sphere with a hole in it, and applying the principles above you should be able to solve it. (The volume of a sphere is 4/3 pi r-cubed, if you need it) (and note I didn't give the formula for the volume of a cylinder...; based on those blankety-blank principles above, that has to mean it too is irrelevant)
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Joined: Aug 2001
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Carpal Tunnel
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Carpal Tunnel
Joined: Aug 2001
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Okay guys, and while you're at it, prove this, will you?: xn + yn = zn
I know what you mean, but it's already been done...not cleanly, but done. Cf. Andrew Weil.
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Joined: Jun 2002
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Carpal Tunnel
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Carpal Tunnel
Joined: Jun 2002
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you should be able to solve itbut we're going to end up with a formula, right, not an actual number? if that's the case, then I can give up right now... of course, I'll probably give up now anyway...
formerly known as etaoin...
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Joined: Aug 2001
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Carpal Tunnel
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Carpal Tunnel
Joined: Aug 2001
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Well, it'd be something symbolic, like "36 times pi" (because pi always figures into working with circles and spheres), but if you want you could use 3.14 for approximately-pi and multiply it out and get a real number. The trick is to convince yourself that you are doing the right thing by selecting the calculation!
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Joined: Jun 2002
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Carpal Tunnel
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Carpal Tunnel
Joined: Jun 2002
Posts: 7,210 |
how can you figure out how much is left over if you don't know how much you're taking out?
formerly known as etaoin...
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Joined: Aug 2001
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Carpal Tunnel
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Carpal Tunnel
Joined: Aug 2001
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I see I should restate the problem; there was an omission as presented.
"You have a sphere. You drill a hole right through the center and measure the cylinder that the hole forms. The side of that cylinder is six inches long [or three, or two, or whatever]. How much volume is left in the sphere?"
ASSUMING THAT THE PROBLEM HAS A SOLUTION -- not always a valid assumption but we have to make it -- since they've been so adamant about leaving out the other sizes, they mustn't really matter. The volume left behind (volume-of-sphere minus volume-of-hole) must be the same, no matter what the diameter of the sphere is: a larger-diameter hole will require a larger-diameter sphere to make the straight side measure six inches! And if that's so then you can specify whatever numbers makes your life easiest. Conveniently, the smaller the hole, the closer the remaining volume is to the volume of the sphere itself, and with a microscopically-small hole the volume removed is microscopic and negligible, and with a zero-size hole what's left is volume-of-sphere minus volume-of-hole, i.e. volume-of-sphere minus zero, or just plain ol' volume-of-sphere = 4/3 times pi times radius cubed. The diameter of the sphere will be the height of a zero-diameter hole, and the radius half of that, in this case 3 inches.
You have to take it as a leap of faith that the problem is honest and so we can make those assumptions. Or you can invoke a mathematician to prove it. (I'm a cardiologist, Jim, not a mathematician!) But the problem has been around long enough that I suspect it's true as presented.
(It does make it easier to have the problem stated correctly...)
So, eta, you were right the first time. As the problem was originally given, the answer is a formula: 4/3 times pi time the cube of (1/2 the height of the side of the hole).
And if we're reduced to having me as Resident Mathematician, it's a good thing this is a Words board!
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Carpal Tunnel
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Carpal Tunnel
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You're right, wofa. The side (or height) of the cylinder is 6 inches. I left that out by mistake. The problem was given in the Mathematical Games section of Scientific American when Martin Gardner ran it, time back way back. It was stated that you didn't need higher math to solve the problem. Your reasoning about the arbitrariness of the diameter of the sphere was in the canonical answer.
Here's another one from the series of problems (which came about once every eight months or so):
Farmer Brown has ten pigs; seven of them are brown, two are white and one is pink. How many of them can say that they are not the same color as any other of Farmer Brown's pigs?
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Pooh-Bah
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Pooh-Bah
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So ... I guess the answer isn't one - the pink one?
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