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Cylinders in Spheres
41–50 of 60 posts
Re: Cylinders in Spheres
#42It wasn't (initially) clear to me that the cylindrical hole must enter and exit the sphere. With that knowledge the solution seems pretty intuitive.
How can one drill a 6" long hole through a sphere of more than 6 inches in diameter? What I mean is if you drill 6 inches into the earth, you haven't passed through the other side... edit: I think they mean drill a 6 inch hole of maximum width, which of course would just leave a very thin ring of the earth 6 inches tall.
It doesn't say "through the sphere." It says "through the center of the sphere." For example, drill a thin radial hole 6" inches deep through the center of a sphere of radius 5"; you will pass through the center and stop before you reach the far side. The volume remaining then depends on the radius of the drill bit, and the size of the sphere.
It also doesn't even state that the hole must enter the sphere. A large solid sphere that internally contains a six inch hole through its center would qualify too.
This was my objection when I first encountered the problem. Everyone else seemed to understand that the hole must pass through both sides of the sphere, but that's not stated or even implied in the problem.
Re: Cylinders in Spheres
#43The cheat is absolutely brilliant reasoning.
Ehh, sort of. It's logically flawed as phrased, in that the first bit is false: "If the problem is being posed, it must have a constant solution" is false; it could just as well be a niftily-simple symbolic solution too. (I say "as phrased" because as other commenters observe, there are mathematically valid ways to arrive at the result.) This reminds me of one of my favorite joke proofs from when I was in school. (It…
I took it to mean that the problem would not be given in this way unless there was an exact solution. In the stated version, there are no degrees of freedom specified, so there can be no degrees of freedom in the solution. In that way, the proof wouldn't work if the problem stated
"A six inch high cylindrical hole is drilled through the center of a sphere with radius R. How much volume is left in the sphere?"
In that case, the solution might be dependent on R, and we'd have to go through the long version. Without that mention of R, unless the problem is wrongly stated, it must be constant.
So it's still an invalid inductive proof, but it's a lot stronger than just assuming that because there is a question there is a constant answer.
Re: Cylinders in Spheres
#44The cheat is absolutely brilliant reasoning.
It's much more impressive if you can show the answer is independent of the variables and then use the limiting case to find the answer. As another poster wrote, all you have to do is write down the integrals for the volumes and notice that in the difference the R^2 terms will cancel.
Unfortunately, in this case the integrals are trivial to evaluate. I think a much more interesting problem would be one where the same approach I described works, except it's not tractable (or at least not easy) to do the integrals in your head.
Re: Cylinders in Spheres
#45The author never pays off the answer to the initial question- is it a fat cylinder or a skinny one? We know it has height ~1.15R (where R is sphere's radius), but this is not easy to visualize.
If I did my trigonometry right, the edges of the cylinder would be about at the 35.26° latitudes, which seems pretty fat to me.
Re: Cylinders in Spheres
#46Nice, I especially appreciate the "cheat answer" to the Gardner Puzzle at the bottom of the page. I have found that kind of meta-reasoning about questions quite useful, on exams and in games like Trivial Pursuit, for example.
Looking at the other comments, it seems like most people really like the cheat. I admit it's very cute, and you're absolutely right that this sort of thinking can be helpful in artificial situations like exams and games --- I've used it myself. That artificiality is why I don't really like that approach, though. It's a brand of thinking that generally works only on artificial problems, because the key component ("you…
Of course, the key components of the analytic solution (knowing the volume formula for spherical caps, spheres, and cylinders) also don't exist on most problems.
Re: Cylinders in Spheres
#47It reminds me of two other neat problems - 1. Imagine a band stretched taught around the diameter of the earth (which, for the purposes of this question, is a smooth sphere). Now imagine that the band is raised one metre from the ground at every single point along its length. How much longer is it? 2. Imagine perfectly parallel lines painted on the floor, exactly one foot apart, and a rigid needle of length one foot.…
#2 is 3/Pi. It's a good puzzle and it helps to know the answer :)
Maybe it's just me, but that one could use better wording. Because of the "exactly one foot apart" phrase, I interpreted it to mean that there are only two parallel lines, which obviously makes it a poorly defined problem. I probably would have understood it if the phrase was replaced by "at one foot intervals."
Re: Cylinders in Spheres
#48Earlier quoted context omitted.
Likely Kopfball ( http://de.wikipedia.org/wiki/Kopfball_(Show) ), although that is late eighties (and the precursor late seventies)
I found it through some extra searching starting from your link- "Kopf um Kopf" ... Is that the precursor you were thinking of? Here is a video on youtube that shows the awesomeness of this show (don't need to know German to appreciate it- On the start of the show, putting your finger under the device makes it spin in the opposite direction... why? The audience member with the right answer would get a reward.) https:…
The deformed gelatin on the polarizer about half way in was pretty cool.
Re: Cylinders in Spheres
#49Earlier quoted context omitted.
Ehh, sort of. It's logically flawed as phrased, in that the first bit is false: "If the problem is being posed, it must have a constant solution" is false; it could just as well be a niftily-simple symbolic solution too. (I say "as phrased" because as other commenters observe, there are mathematically valid ways to arrive at the result.) This reminds me of one of my favorite joke proofs from when I was in school. (It…
My favorite "cheat" was on an exam in an algorithm-analysis class (lots of discrete math, infinite sums, and proofs). We were asked to prove or find a counterexample to some conjecture or other. Looking at the conjecture, and looking at the clock, I decided the only thing I could complete in time was to find a counterexample. Indeed within a few minutes I found one and wrote it down. I was the last to finish my exam…
In a linear algebra class, we were asked to prove that a theorem is true if and only if X. (I think X may have been that a matrix had a nonzero determinant.) I had no idea how to do it, but I noticed the theorem was trivially true in the case of a 1x1 zero matrix. Since this matrix has a zero determinant, it spoiled the "only if" part. I wrote it as a counter-example, showing that "if AND only-if" was false, and didn't prove the "if." Full credit!
Re: Cylinders in Spheres
#50Regarding the Napkin Ring part of the problem.
Had the author simplified the formula for "Vanswer" rather than plugging in values for h & c, he would have gotten:
Vanswer = (Pi/6) * h^3
From which is it easy to see that the answer in this particular case is 36 * Pi but it also makes clear that the answer does not depend on R.