7 ms·

Cylinders in Spheres

118 points by stansmith 12y ago

chton 12y ago
The cheat is absolutely brilliant reasoning.
- jerf 12y ago
  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's so simple I'm sure it has other originations but AFAIK I independently recreated it.) "The problem begins with the phrase 'Show that...' or similar. All previous problems that began with that phrase have been provable. Therefore, by induction, the claim I am being asked to prove must be correct. QED." Unfortunately, this proof line hit a snag about halfway through my graph theory class in which we were assigned a problem out of the book that turned out to ask you to prove a false statement. (Well, a snag above and beyond the fact that that is an invalid inductive proof in general, ahem.) It was a typo and clearly accidental, but it was enough to break my proof forevermore. May you have better luck with it.
  - thedufer 12y ago
    Its flawed as a "proof", but it might contextually justify the answer. That exact reasoning accounts for pretty much all of my limited success at middle school/high school math competitions. You can usually skip long, difficult steps by simply assuming there is a single constant answer (as the rules of the contests usually make clear must be the case).
  - colanderman 12y ago
    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 and handed in my booklet; the prof looked it over and noted I was the only one (out of 5) in the class to get that particular question correct. I explained to him my "reasoning". He said, "yes, that is how you're supposed to do it!"
    - millstone 12y ago
      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!
  - chton 12y ago
    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.
- phkahler 12y ago
  Agreed. Given that it is a puzzle and not a request for proof. Once you realize that the hole and sphere diameter are related and the situation has one degree of freedom not specified, it is perfectly valid to conclude that the answer must be independent of that and choose a case you can work out in your head.
- nilkn 12y ago
  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.
drcode 12y ago
In the early eighties there was a TV show on German network television and I remember them presenting this puzzle, and I figured out the "cheat solution" as a kid. It was this awesome TV game show that consisted entirely of Martin Gardner-style puzzles and other fiendish physics/biology puzzles, and contestants who were all scientists. Does anyone by any chance remember the name of this German TV show? I would really like to look up more information about it! It deserves to be remembered.
- Someone 12y ago
  Likely Kopfball (http://de.wikipedia.org/wiki/Kopfball_(Show) http://de.wikipedia.org/wiki/Kopfball_(Show)), although that is late eighties (and the precursor late seventies)
  - drcode 12y ago
    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://www.youtube.com/watch?v=1ObdE9n3UF4 https://www.youtube.com/watch?v=1ObdE9n3UF4 Take a quick peek at a few other random spots of the show in the video and marvel at the awesome scientific experiments on live TV: I think that was one of the most bad ass shows ever, especially since it didn't have that pejorative "science shows are only for kids" thing happening.
    - Someone 12y ago
      I'm not sure. Wikipedia claims that he precursor also was called Kopfball, and has some overlap in the times when the series ran. I just googled the "kopf" that I remembered in combination with "wissenschaft" (science) and "fernsehen" (television), and that popped up, and I thought "that must be it". I should have been more cautious, though. German TV at the time had quite a few interesting programs about science (in a broad sense) I remember programs where they taught differentiation and integration, live chess (or at least, it seemed like that; if a player takes ten minutes to think about a move, the commenters would explain what the players might be thinking about) by such players as Karpov, Timman, and Anand (http://en.m.wikipedia.org/wiki/Chess_of_the_Grandmasters http://en.m.wikipedia.org/wiki/Chess_of_the_Grandmasters), Hobbythek (http://de.m.wikipedia.org/wiki/Hobbythek http://de.m.wikipedia.org/wiki/Hobbythek) about DIY, (with subjects such as "we build a hot air balloon", "book binding", and "stereo photography"), and that program about personal computing whose name I don't remember.
davtbaum 12y ago
It wasn't (initially) clear to me that the cylindrical hole must enter and exit the sphere. With that knowledge the solution seems pretty intuitive.
- thedufer 12y ago
  Nor that the endcap(s) would not be included in either the 6 inches or the resulting volume. It seems to me that there must be a much better phrasing that brings these points home, although it might require a diagram.
- TheLoneWolfling 12y ago
  I've always heard it described as a circular napkin ring where the depth of the inner hole is an inch.
- fleitz 12y ago
  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.
  - dotrob 12y ago
    In the puzzle statement, "A six inch high cylindrical hole is drilled through the center of a sphere," through is the keyword, rather than into. I made the same initial mistake of misreading through as into.
    - delluminatus 12y ago
      In that case, the statement from the article "We could have a sphere as large as a planet, bore a hole 6" in length through it..." seems inconsistent. Either the cylinder is not 6" long, or it does not go through the sphere.
      - 12y ago
carlob 12y ago
One of those cases where using integrals rather than geometry is much simpler. \pi \int_-3^3 (R^2 - x^2) dx = \pi (6 R^2 - 18) is the volume of the rotational solid without removing the cylinder. While the volume of the cylinder is given by: \pi \int_-3^3 (R^2 - 3^2) dx = \pi 6 (R^2 - 9) As you can see the difference between the two volumes is 36 \pi. You can actually show the solution does not depend on \pi without evaluating the integral, and then compute the 'cheat' case, which would not be a cheat once you've made this observation.
- stephencanon 12y ago
  Nit: the solution doesn’t depend on “R” (as you note, this is clear from looking at the structure of the two integrals); it does depend on \pi.
  - carlob 12y ago
    typo! And I'm out of the edit window.
jbaskette 12y ago
I remember reading and solving this the problem as the end of the article (“A six inch high cylindrical hole is drilled through the center of a sphere. How much volume is left in the sphere?”) as a kid. I did it the hard way using the formula's, but the whole point of the puzzle was what this article called the "cheat" answer. It reduces the solution to utter simplicity by application of some elegant logic. It's not something of a cheat -- it's the whole point of the puzzle. I really was a kid. I was taken in by Gardner's April 1st column that claimed among other things that a proven solution for "Chess" indicated that white would always win and that the opening move was P-KR4 (h4 in modern notation).
- sparky_z 12y ago
  It only wouldn't be a cheat if the question were framed as "Surprisingly, the volume of the ring is constant, regardless of the radii of the sphere and circle. What is that constant volume?" Otherwise, you don't really know your answer is correct. Maybe, they wanted the answer in terms of R1 and R2. And if you are able to trick them into confirming it, then you're essentially using social engineering to leverage somebody else's work (the person who discovered the surprising result) in formulating your answer. There's no way that the original discoverer could have used that method. That's what makes it a cheat. Back in middle school geometry, I came up with my own strategy for proving theorems. "You wouldn't ask me to prove it if it weren't true, therefore it must be true. Q.E.D." It didn't go over well.
  - eru 12y ago
    That's why in Uni they usually asked: prove this statement right or wrong (or undecidable).
throwaway283719 12y ago
It 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. If you throw the needle to the floor at random, what is the probability that it crosses one of the lines? (This one has a nice 'cheat' solution just like the OP article).
- KRuchan 12y ago
  Maybe I am missing something here, but why is 1 a neat problem? You are increasing the radius by 1 meter so the length of the band is now 2pi(r_e + 1) making the increment 2*pi meters. Is the surprisingly low increment the point of the problem?
  - Jun8 12y ago
    Most people when faced with 1 and knowing the huge circumference of the Earth estimate (using Sytem 1 type thinking, http://en.wikipedia.org/wiki/Dual_process_theory#System_1 http://en.wikipedia.org/wiki/Dual_process_theory#System_1) it would be longer by kilometers.
  - JoshTriplett 12y ago
    Yes, exactly; if you don't start out by doing the math, the result can seem unintuitive.
- stansmith 12y ago
  If you read to the bottom of the cylinder article, there's a link to another blog post by the same guy about other Martin Gardner puzzles, and the 'band around the Earth' puzzle is the first one! http://www.datagenetics.com/blog/may12012/index.html http://www.datagenetics.com/blog/may12012/index.html
- eps 12y ago
  #2 is 3/Pi. It's a good puzzle and it helps to know the answer :)
svantana 12y ago
Nice, 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.
- bkcooper 12y ago
  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 wouldn't be asking me this if it didn't have a well-defined answer") doesn't exist on most problems. Proving that it's constant and then using the r -> 0 trick to calculate the constant is much more satisfactory to me.
  - baddox 12y ago
    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.
  - vorg 12y ago
    > that approach is a brand of thinking that generally works only on artificial problems, because the key component ("you wouldn't be asking me this if it didn't have a well-defined answer") doesn't exist on most problems Perhaps changing the footnote hint "no more information is given" to an integral part of the problem spec, and saying "no more information is required", would make the problem less artificial. There would then be a self-referencial component in the problem definition, self-referentiality being fairly common in nature and engineering.
eps 12y ago
There is a simpler puzzle that's similar to Gardner's, but as unintuitive. Take an orange and wrap a string around its diameter. Now extend the string by 1 inch and redistribute it around the orange so that it floats an even distance from it. Do the same with the Earth, i.e. wrap, extend by 1 inch and even out into a circle. The gap betwen the string and the orange/Earth - which one is bigger?
- nemetroid 12y ago
  Spoiler below! The relationship between radius and circumference is linear (C = 2pi * r). When the circumference is increased by 1, the radius increases by 1/2pi. Therefore, the gap has the same size. Let g be the gap. Then C + 1 = 2pi * (r + g) 2pi * r + 1 = 2pi * (r + g) r + 1 / 2pi = r + g 1 / 2pi = g
hammock 12y ago
The 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.
- Chinjut 12y ago
  The cylinder is sqrt(2) (~= 1.41) times as wide as it is high. So... a little fat? (I don't know the healthy baseline for cylinders...)
- dazmax 12y ago
  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.
scotty79 12y ago
I once encountered question that had cheat answer. It was about trapezoid and it seemed that it had some data missing. I was amazed that just assuming that the question had one answer allowed to reduce the problem to trivially calculable one by consistently manipulating the variables that were not given. I was also pretty proud of myself for finding this solution.
deleted 12y ago
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drjoe047 12y ago
Regarding 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.
quarterwave 12y ago
To estimate an answer for the first problem: consider two cylinders with zero volume, the one equatorial (r=R) and the other polar (r=0). Both correspond to zero volume. Hence the maximum volume occurs (hand wave) for a height somewhere between 0 and 2R. Not knowing better, the initial estimate for h is to bisect this interval, to give h_est = R (compare 1 with 2/sqrt(3) ~= 1.15). It would be nice to construct a first order correction term. Any ideas?
e3pi 12y ago
...and its Archimedean inverse(his favorite!): a Wilson(tm)* soccer ball enjoys 2/3 the volume and 2/3 the surface content of its minimal cylindrical official Wilson(tm) shipping clear plastic blister pack. *citation: Tom Hanks in "Castaway"
tempestn 12y ago
I'm just pleased that I finally found a math puzzle on HN that I was able to solve without looking ahead! (Although I called the radius of the cylinder r/a, where r is the radius of the sphere, which ended up making my math a bit messy...) Regardless, deserves an upvote just for the cheat answer at the end. I like that kind of reasoning!
cousin_it 12y ago
I solved the puzzle in my head before getting to the second paragraph in the article. Here's the reasoning: 1) The area of a circle has a fixed ratio to the area of a square inscribed in that circle. 2) Therefore the volume of a cylinder has a fixed ratio to the volume of a square box of the same height, which sits inside that cylinder. 3) Therefore the biggest cylinder corresponds to the biggest box that can fit inside the sphere. 4) That box is obviously a cube, because what else could it be? 5) If a cube is inscribed in a unit sphere centered at the origin, the corners have coordinates ±1/√3, ±1/√3, ±1/√3. 6) Now you can calculate the volume of the cylinder in your head. Do it!
- logicallee 12y ago
  Congrats! But I hope all that is a humblebrag, not a reflection of your standards as a technical interviewer ;-)
  - cousin_it 12y ago
    Yeah, that was definitely a brag, nothing humble about it :-) When I was a bit younger, I was in fact the typical asshole interviewer who would ask lambda calculus questions. Now I mostly stay away from interviewing, because I can emphasize much more with the pressure that candidates feel.
- Chinjut 12y ago
  Step 4 seems a bit glib... :) It would follow straightforwardly if we knew that the largest rectangle in a circle is a square. The easiest way I can think of right now to see this is by Lagrange optimization (which we could also apply just as well directly to the 3d problem): the area of a rectangle of width and height <x, y> has a gradient of <y, x>, which can only be normal to the circle (which has normal vector <x, y>) when x = y. Another way is to think of <x, y> as proportional to <cos(t), sin(t)>; the area is then cos(t) * sin(t), which is proportional to sin(2t), and thus clearly maximized when x = y.

118 points by stansmith 12y ago

chton 12y ago

The cheat is absolutely brilliant reasoning.

jerf 12y ago
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's so simple I'm sure it has other originations but AFAIK I independently recreated it.) "The problem begins with the phrase 'Show that...' or similar. All previous problems that began with that phrase have been provable. Therefore, by induction, the claim I am being asked to prove must be correct. QED." Unfortunately, this proof line hit a snag about halfway through my graph theory class in which we were assigned a problem out of the book that turned out to ask you to prove a false statement. (Well, a snag above and beyond the fact that that is an invalid inductive proof in general, ahem.) It was a typo and clearly accidental, but it was enough to break my proof forevermore. May you have better luck with it.
- thedufer 12y ago
  Its flawed as a "proof", but it might contextually justify the answer. That exact reasoning accounts for pretty much all of my limited success at middle school/high school math competitions. You can usually skip long, difficult steps by simply assuming there is a single constant answer (as the rules of the contests usually make clear must be the case).
- colanderman 12y ago
  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 and handed in my booklet; the prof looked it over and noted I was the only one (out of 5) in the class to get that particular question correct. I explained to him my "reasoning". He said, "yes, that is how you're supposed to do it!"
  - millstone 12y ago
    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!
- chton 12y ago
  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.
phkahler 12y ago
Agreed. Given that it is a puzzle and not a request for proof. Once you realize that the hole and sphere diameter are related and the situation has one degree of freedom not specified, it is perfectly valid to conclude that the answer must be independent of that and choose a case you can work out in your head.
nilkn 12y ago
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.

drcode 12y ago

In the early eighties there was a TV show on German network television and I remember them presenting this puzzle, and I figured out the "cheat solution" as a kid. It was this awesome TV game show that consisted entirely of Martin Gardner-style puzzles and other fiendish physics/biology puzzles, and contestants who were all scientists. Does anyone by any chance remember the name of this German TV show? I would really like to look up more information about it! It deserves to be remembered.

Someone 12y ago
Likely Kopfball (http://de.wikipedia.org/wiki/Kopfball_(Show) http://de.wikipedia.org/wiki/Kopfball_(Show)), although that is late eighties (and the precursor late seventies)
- drcode 12y ago
  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://www.youtube.com/watch?v=1ObdE9n3UF4 https://www.youtube.com/watch?v=1ObdE9n3UF4 Take a quick peek at a few other random spots of the show in the video and marvel at the awesome scientific experiments on live TV: I think that was one of the most bad ass shows ever, especially since it didn't have that pejorative "science shows are only for kids" thing happening.
  - Someone 12y ago
    I'm not sure. Wikipedia claims that he precursor also was called Kopfball, and has some overlap in the times when the series ran. I just googled the "kopf" that I remembered in combination with "wissenschaft" (science) and "fernsehen" (television), and that popped up, and I thought "that must be it". I should have been more cautious, though. German TV at the time had quite a few interesting programs about science (in a broad sense) I remember programs where they taught differentiation and integration, live chess (or at least, it seemed like that; if a player takes ten minutes to think about a move, the commenters would explain what the players might be thinking about) by such players as Karpov, Timman, and Anand (http://en.m.wikipedia.org/wiki/Chess_of_the_Grandmasters http://en.m.wikipedia.org/wiki/Chess_of_the_Grandmasters), Hobbythek (http://de.m.wikipedia.org/wiki/Hobbythek http://de.m.wikipedia.org/wiki/Hobbythek) about DIY, (with subjects such as "we build a hot air balloon", "book binding", and "stereo photography"), and that program about personal computing whose name I don't remember.

davtbaum 12y ago

It wasn't (initially) clear to me that the cylindrical hole must enter and exit the sphere. With that knowledge the solution seems pretty intuitive.

thedufer 12y ago
Nor that the endcap(s) would not be included in either the 6 inches or the resulting volume. It seems to me that there must be a much better phrasing that brings these points home, although it might require a diagram.
TheLoneWolfling 12y ago
I've always heard it described as a circular napkin ring where the depth of the inner hole is an inch.
fleitz 12y ago
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.
- dotrob 12y ago
  In the puzzle statement, "A six inch high cylindrical hole is drilled through the center of a sphere," through is the keyword, rather than into. I made the same initial mistake of misreading through as into.
  - delluminatus 12y ago
    In that case, the statement from the article "We could have a sphere as large as a planet, bore a hole 6" in length through it..." seems inconsistent. Either the cylinder is not 6" long, or it does not go through the sphere.
    - 12y ago

carlob 12y ago

One of those cases where using integrals rather than geometry is much simpler. \pi \int_-3^3 (R^2 - x^2) dx = \pi (6 R^2 - 18) is the volume of the rotational solid without removing the cylinder. While the volume of the cylinder is given by: \pi \int_-3^3 (R^2 - 3^2) dx = \pi 6 (R^2 - 9) As you can see the difference between the two volumes is 36 \pi. You can actually show the solution does not depend on \pi without evaluating the integral, and then compute the 'cheat' case, which would not be a cheat once you've made this observation.

stephencanon 12y ago
Nit: the solution doesn’t depend on “R” (as you note, this is clear from looking at the structure of the two integrals); it does depend on \pi.
- carlob 12y ago
  typo! And I'm out of the edit window.

jbaskette 12y ago

I remember reading and solving this the problem as the end of the article (“A six inch high cylindrical hole is drilled through the center of a sphere. How much volume is left in the sphere?”) as a kid. I did it the hard way using the formula's, but the whole point of the puzzle was what this article called the "cheat" answer. It reduces the solution to utter simplicity by application of some elegant logic. It's not something of a cheat -- it's the whole point of the puzzle. I really was a kid. I was taken in by Gardner's April 1st column that claimed among other things that a proven solution for "Chess" indicated that white would always win and that the opening move was P-KR4 (h4 in modern notation).

sparky_z 12y ago
It only wouldn't be a cheat if the question were framed as "Surprisingly, the volume of the ring is constant, regardless of the radii of the sphere and circle. What is that constant volume?" Otherwise, you don't really know your answer is correct. Maybe, they wanted the answer in terms of R1 and R2. And if you are able to trick them into confirming it, then you're essentially using social engineering to leverage somebody else's work (the person who discovered the surprising result) in formulating your answer. There's no way that the original discoverer could have used that method. That's what makes it a cheat. Back in middle school geometry, I came up with my own strategy for proving theorems. "You wouldn't ask me to prove it if it weren't true, therefore it must be true. Q.E.D." It didn't go over well.
- eru 12y ago
  That's why in Uni they usually asked: prove this statement right or wrong (or undecidable).

throwaway283719 12y ago

It 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. If you throw the needle to the floor at random, what is the probability that it crosses one of the lines? (This one has a nice 'cheat' solution just like the OP article).

KRuchan 12y ago
Maybe I am missing something here, but why is 1 a neat problem? You are increasing the radius by 1 meter so the length of the band is now 2pi(r_e + 1) making the increment 2*pi meters. Is the surprisingly low increment the point of the problem?
- Jun8 12y ago
  Most people when faced with 1 and knowing the huge circumference of the Earth estimate (using Sytem 1 type thinking, http://en.wikipedia.org/wiki/Dual_process_theory#System_1 http://en.wikipedia.org/wiki/Dual_process_theory#System_1) it would be longer by kilometers.
- JoshTriplett 12y ago
  Yes, exactly; if you don't start out by doing the math, the result can seem unintuitive.
stansmith 12y ago
If you read to the bottom of the cylinder article, there's a link to another blog post by the same guy about other Martin Gardner puzzles, and the 'band around the Earth' puzzle is the first one! http://www.datagenetics.com/blog/may12012/index.html http://www.datagenetics.com/blog/may12012/index.html
eps 12y ago
#2 is 3/Pi. It's a good puzzle and it helps to know the answer :)

svantana 12y ago

Nice, 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.

bkcooper 12y ago
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 wouldn't be asking me this if it didn't have a well-defined answer") doesn't exist on most problems. Proving that it's constant and then using the r -> 0 trick to calculate the constant is much more satisfactory to me.
- baddox 12y ago
  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.
- vorg 12y ago
  > that approach is a brand of thinking that generally works only on artificial problems, because the key component ("you wouldn't be asking me this if it didn't have a well-defined answer") doesn't exist on most problems Perhaps changing the footnote hint "no more information is given" to an integral part of the problem spec, and saying "no more information is required", would make the problem less artificial. There would then be a self-referencial component in the problem definition, self-referentiality being fairly common in nature and engineering.

eps 12y ago

There is a simpler puzzle that's similar to Gardner's, but as unintuitive. Take an orange and wrap a string around its diameter. Now extend the string by 1 inch and redistribute it around the orange so that it floats an even distance from it. Do the same with the Earth, i.e. wrap, extend by 1 inch and even out into a circle. The gap betwen the string and the orange/Earth - which one is bigger?

nemetroid 12y ago
Spoiler below! The relationship between radius and circumference is linear (C = 2pi * r). When the circumference is increased by 1, the radius increases by 1/2pi. Therefore, the gap has the same size. Let g be the gap. Then C + 1 = 2pi * (r + g) 2pi * r + 1 = 2pi * (r + g) r + 1 / 2pi = r + g 1 / 2pi = g

hammock 12y ago

The 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.

Chinjut 12y ago
The cylinder is sqrt(2) (~= 1.41) times as wide as it is high. So... a little fat? (I don't know the healthy baseline for cylinders...)
dazmax 12y ago
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.

scotty79 12y ago

I once encountered question that had cheat answer. It was about trapezoid and it seemed that it had some data missing. I was amazed that just assuming that the question had one answer allowed to reduce the problem to trivially calculable one by consistently manipulating the variables that were not given. I was also pretty proud of myself for finding this solution.

deleted 12y ago

[deleted]

drjoe047 12y ago

Regarding 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.

quarterwave 12y ago

To estimate an answer for the first problem: consider two cylinders with zero volume, the one equatorial (r=R) and the other polar (r=0). Both correspond to zero volume. Hence the maximum volume occurs (hand wave) for a height somewhere between 0 and 2R. Not knowing better, the initial estimate for h is to bisect this interval, to give h_est = R (compare 1 with 2/sqrt(3) ~= 1.15). It would be nice to construct a first order correction term. Any ideas?

e3pi 12y ago

...and its Archimedean inverse(his favorite!): a Wilson(tm)* soccer ball enjoys 2/3 the volume and 2/3 the surface content of its minimal cylindrical official Wilson(tm) shipping clear plastic blister pack. *citation: Tom Hanks in "Castaway"

tempestn 12y ago

I'm just pleased that I finally found a math puzzle on HN that I was able to solve without looking ahead! (Although I called the radius of the cylinder r/a, where r is the radius of the sphere, which ended up making my math a bit messy...) Regardless, deserves an upvote just for the cheat answer at the end. I like that kind of reasoning!

cousin_it 12y ago

I solved the puzzle in my head before getting to the second paragraph in the article. Here's the reasoning: 1) The area of a circle has a fixed ratio to the area of a square inscribed in that circle. 2) Therefore the volume of a cylinder has a fixed ratio to the volume of a square box of the same height, which sits inside that cylinder. 3) Therefore the biggest cylinder corresponds to the biggest box that can fit inside the sphere. 4) That box is obviously a cube, because what else could it be? 5) If a cube is inscribed in a unit sphere centered at the origin, the corners have coordinates ±1/√3, ±1/√3, ±1/√3. 6) Now you can calculate the volume of the cylinder in your head. Do it!

logicallee 12y ago
Congrats! But I hope all that is a humblebrag, not a reflection of your standards as a technical interviewer ;-)
- cousin_it 12y ago
  Yeah, that was definitely a brag, nothing humble about it :-) When I was a bit younger, I was in fact the typical asshole interviewer who would ask lambda calculus questions. Now I mostly stay away from interviewing, because I can emphasize much more with the pressure that candidates feel.
Chinjut 12y ago
Step 4 seems a bit glib... :) It would follow straightforwardly if we knew that the largest rectangle in a circle is a square. The easiest way I can think of right now to see this is by Lagrange optimization (which we could also apply just as well directly to the 3d problem): the area of a rectangle of width and height <x, y> has a gradient of <y, x>, which can only be normal to the circle (which has normal vector <x, y>) when x = y. Another way is to think of <x, y> as proportional to <cos(t), sin(t)>; the area is then cos(t) * sin(t), which is proportional to sin(2t), and thus clearly maximized when x = y.