Posted August 21, 2007 Maybe this has already been posted. A friend asked me this a while back, and I answered her in less than a minute. She said I was a genius. But I said there were two ways to arrive at the answer, and I simply chose the easier way. A 6-inch [long] hole is drilled through [the center of] a sphere. What is the volume of the remaining portion of the sphere? The hard way involves calculus. The easy way uses logic. 0 Share this post Link to post Share on other sites

0 Posted February 8, 2008 Is it? Did Jkyle1980 post the right approach in drilling?? I think not!! According to his picture, the diameter of the hole = diameter of sphere - 6 inch!! Now that cannot ge right according to the answer and to the posts on this thread! The 6 inchs in his picture cannot be seen from this angle, they represent the HEIGHT of the remaining part of earth once you put the earth bracelet flat on a plane!! Be nice. He got the picture. You got the dimensions. 0 Share this post Link to post Share on other sites

0 Posted February 8, 2008 Be nice. He got the picture. You got the dimensions. No disrespect intended at all!! It's just that if Jkyle1980 uses this picture to make his calculations and reasonning, he will never get the 36pi answer... And it's not at all because he's "slow" as he claims (and he's not), it's simply because according to his picture, the Radius is very relevant to the calculation of the remaining volume and the remaining bracelet of earth has a volume much bigger than 36pi!! The 6 inchs are not reduced from the diameter of the drill, they are not the THICKNESS of the remaining crest, they are the HEIGHT of the remaining bracelet if it's laid on its side on a plane!! If what I'm saying is wrong, I may be misunderstanding something very important... Please take another look at the picture and tell me if it portraits the Original Puzzle. 0 Share this post Link to post Share on other sites

0 Posted February 8, 2008 The 6 inches is along the axis of the hole - the height of the cylindrical part of what's removed. All as you say. And JK "got the idea" of how a 6" hole could be drilled thru the earth. He labeled it wrong, but he got the idea. 0 Share this post Link to post Share on other sites

0 Posted February 8, 2008 None taken, roolstar. Actually, between the discussion over the last picture I finally get what 6" inch hole is being talked about. Thanks for the lengthy, yet helpful, explanation. Where can I mail the check for my tuition? 0 Share this post Link to post Share on other sites

0 Posted February 8, 2008 None taken, roolstar. Actually, between the discussion over the last picture I finally get what 6" inch hole is being talked about. Thanks for the lengthy, yet helpful, explanation. Where can I mail the check for my tuition? Sorry for the detour there guys... I apparently didn't get that the picture portraits the misunderstood version that JK had at first... To tell you the truth, I thought I was the one who didn't get it at first!!! I even made some calculations about the "HEIGHT" of the cylinder in JK's picture and ended up with around 2Km!!! ANyway, I promise I will post a new puzzle very soon to regain your respect 0 Share this post Link to post Share on other sites

0 Posted February 9, 2008 (edited) Eek, I don't understand the question. I know there is somethng wrong with my comprehension of the question, because then I would need the radius of the sphere. So right now, I will do some post digging and look for what i don't get. I'm so confused about what information I have exactly )= Edited February 9, 2008 by Eugenia 0 Share this post Link to post Share on other sites

0 Posted February 10, 2008 Eek, I don't understand the question. I know there is somethng wrong with my comprehension of the question, because then I would need the radius of the sphere. So right now, I will do some post digging and look for what i don't get. I'm so confused about what information I have exactly )= It's a wonderful result, really. The answer is the same for any radius. Except the radius must be at least 3" -- so that a 6" hole can be made. Read through the thread to get more information. 0 Share this post Link to post Share on other sites

0 Posted February 21, 2008 So I have been reading this forum for almost a year and it took this problem to get me to register. i read all fo the responses and I STILL don't get it. Can someone draw a correct diagram showing hwo this might work for various R lengths or at least pointing to the different parts of the cylinder/hole and sphere being referred to in the problem? I haven't had any calculus at all, but I want to at least understand the problem well enough to comprehend the logic answer. Thanks! 0 Share this post Link to post Share on other sites

0 Posted February 22, 2008 sunburntpixY, Thanks for the inquiry to my puzzle. I don't have a diagram that i can post. But this may be helpful... It will be helpful to visualize the following ideas if you draw a circle on a piece of paper and think of it as a cross section of a sphere. The center of the circle is the center of the sphere. Now draw parallel lines -- equal distances from, and either side of, the center of the circle. Make sure the lines go all the way through the circle. The lines represent a cross section of a hole drilled through the sphere. The distance between the lines is the diameter of the drill. The area between the lines is a cross section of the hole. The area outside the lines but inside the circle is a cross section of the portion of the sphere that remains after the hole is drilled. And, most importantly to understand, the length of the lines between the points that they intersect the circle, from end to end, is the length of the hole. Note that the length of the hole is less than the diameter of the circle [sphere]. Repeat. The length of the hole is not the same thing as the diameter of the sphere. Finally, note that if you had drawn the lines farther apart -- go ahead and draw another pair of lines that are farther apart -- the length of the hole decreases. OK - now think through these statements. [1] For there to be anything left of the sphere, the diameter of the drill better be less than the diameter of the sphere. [2] For the hole to be 6 inches long [length is measured in the direction the drill traveled while it was making the hole] the sphere had to have been at least 6 inches in diameter. You can't drill a 6-inch hole through a 4-inch-diameter sphere. But you can drill a 6-inch hole through a 7-inch-diameter sphere. [3] In all cases, we are talking about drilling a hole completely through the sphere. You can look through from beginning of the hole to its end, and see all the way through the sphere. You could string the sphere onto a rope, so long as the rope is thinner than the drill, of course. And, in all cases, what we mean by the length of the hole is the distance between the beginning and the end of the hole. You have to see this length as being different from the diameter of the original sphere. [3] If the diameter of the drill is negligibly small, a negligible amount of the sphere is removed, and the remaining volume is essentially the original volume: [4pi/3]r^3 and the length of the hole is essentially the diameter of the sphere. Only in this case is the length of the hole the same as the diameter of the sphere. Think of a pearl, with the tiniest of holes, string on the slenderest of threads. In the case of a vanishingly small [diameter] drill, the length of the hole approaches the diameter of the sphere, and the remaining volume approaches the initial volume. [4] If the diameter of the drill is negligibly less than the diameter of the sphere, the remaining portion of the sphere is a tiny band of material at the "equator" of the sphere - think of the drill entering the sphere at its "North Pole" and exiting the sphere at its "South Pole", and the length of the hole - the height of the remaining volume - is now much less than the diameter of the original sphere. In the case of a vanishingly small difference between the sphere's diameter and the drill's diameter, the remaining volume approaches zero - the entire sphere has been drilled away - and the length of the hole [obviously] approaches zero as well. This would be the case if you drew your two lines tangent to your circle instead of intersecting the circle. If you really understand this point, you can see how a 6-inch hole could be drilled [somewhat disastrously to be sure] through the Earth! Remember, the length of the hole is defined as the height of the remaining portion of the sphere, not as the length of the drill. Or think of it this way: if you crawled inside the hole and painted the inside surface of the hole, the painted surface would be a circular cylinder. The length of that cylinder is the length of the hole. Now if you can visualize all of that, you understand what "length of the hole" and "remaining volume" mean. The puzzle is cryptically worded. Intentionally. It's meant to make the solver think through all of these things so that the question is even understood. That's part of the challenge of the puzzle. If you can visualize the "length of the hole" then the question at least is understandable: If you drill a 6-inch hole through a sphere, what is the volume of the remaining portion of the sphere? And, if you've read this thread, you've heard it asserted that in every case - for every sphere whose diameter is at least 6 inches - the remaining volume is the same. A person with calculus skills can compute that volume and see that the answer can be expressed in terms of the length of the hole - only. An amazing result! But another person - one who does not like tedious calculations - simply takes note of the fact that an answer has been requested even though the original diameter of the sphere is not given. The only way that can be a reasonable request is if the answer does not depend on the original diameter. The person who reaches that conclusion smiles, rubs his/her hands together, and computes the answer for the simplest case: a 6-inch-diameter sphere. Remember, we said that's the case where the hole has zero diameter and thus removes zero volume, so that the remaining volume is the original volume: For a 6-inch diameter sphere, V = [4pi/3]r^3 = 36pi cubic inches. And that's the correct answer. Hope that makes this thread easier to follow. You're not alone in asking about what it means. - bn 0 Share this post Link to post Share on other sites

0 Posted February 22, 2008 Does this help? I think this is exactly what Bonanova was describing... 0 Share this post Link to post Share on other sites

0 Posted February 24, 2008 Maybe this has already been posted. A friend asked me this a while back, and I answered her in less than a minute. She said I was a genius. But I said there were two ways to arrive at the answer, and I simply chose the easier way. A 6-inch hole is drilled through a sphere. What is the volume of the remaining portion of the sphere? The hard way involves calculus. The easy way uses logic. Hey, how bought you post the answer to the easy way? its been driving me nuts for about a week 0 Share this post Link to post Share on other sites

0 Posted February 24, 2008 Hey, how bought you post the answer to the easy way? its been driving me nuts for about a week The easy way is to suppose the answer is the same for any sphere [with diameter not less than 6 inches], and calculate the answer for a 6-inch diameter sphere. The answer is 36pi cubic inches - exactly the volume of the sphere. The hard way, and why the easy way works, can be found earlier in this thread. 0 Share this post Link to post Share on other sites

0 Posted February 24, 2008 This is a hell of a puzzle. The problem is most people didnot understand the problem!!! 0 Share this post Link to post Share on other sites

0 Posted February 24, 2008 (edited) So we can visualize an infinitely elastic sphere of 6 inch in diameter. If we make a cylinder with variable diameter go through it (the cylinder can get wider and wider), and increase its diameter to any value we want: the sphere will stretch around the cylinder and will remain 6 inchs tall! The volume of the sphere will not change but the sphere will become more & more flattened around the cylinder... WOW Edited February 24, 2008 by roolstar 0 Share this post Link to post Share on other sites

0 Posted February 25, 2008 This may help those that are not quite seeing how the hole is being drilled. Imagine instead that you are drilling a hole into the top of a mountain. You have a drill that is 4 ft. wide. When you begin, the drill is wider than the very top of the mountain so all dirt is removed. If you stopped at this point, you wouldn't have a hole on top, you'd just have a mountain with a flat top. You have to keep drilling until the diameter of the cross-section of the mountain at the point your drill has reached is greater than 4 ft. (diameter of the drill) before you begin actually making a hole. See the picture below. It is not to scale, and I made it in PowerPoint when I should have been working so cut me some slack. Now imagine that you were drilling through a sphere instead. Same idea of the hole not starting until there is matter left on the outside of the drill and stopping as soon as the drill busted through the bottom. This is the easiest way for me to see it. 0 Share this post Link to post Share on other sites

0 Posted March 2, 2008 Maybe this has already been posted. A friend asked me this a while back, and I answered her in less than a minute. She said I was a genius. But I said there were two ways to arrive at the answer, and I simply chose the easier way. A 6-inch hole is drilled through a sphere. What is the volume of the remaining portion of the sphere? The hard way involves calculus. The easy way uses logic. Thinking of it differently, the cylinder expands to the sphere, right? So, at any time a hole is drilled through the sphere, the remaining portion always measures 6 inches in height. Therefore, a 6-inch sphere cannot be drilled and must be the "ideal" volume. Therefore the volume would be 36*pi, right? 0 Share this post Link to post Share on other sites

0 Posted March 2, 2008 Thinking of it differently, the cylinder expands to the sphere, right? So, at any time a hole is drilled through the sphere, the remaining portion always measures 6 inches in height. Therefore, a 6-inch sphere cannot be drilled and must be the "ideal" volume. Therefore the volume would be 36*pi, right? Bingo! 0 Share this post Link to post Share on other sites

0 Posted March 2, 2008 you have to know some type of measurment besides a 6in hole 0 Share this post Link to post Share on other sites

0 Posted March 3, 2008 you have to know some type of measurment besides a 6in hole It certainly seems so at first glance. But are you certain? 0 Share this post Link to post Share on other sites

0 Posted March 3, 2008 (edited) What I think is that drilling a hole doesnot reduce the Volume of Sphere. Yes it will increase the density as Surface area will be increased due to drilling of hole., but the volume will remain same.Am I right? Edited March 3, 2008 by Knight 0 Share this post Link to post Share on other sites

0 Posted March 3, 2008 What I think is that drilling a hole does not reduce the [1] Volume of Sphere. Yes it will increase the [2] density as [3] Surface area will be increased due to drilling of hole, but the volume will remain same. Am I right? [1] The volume of a sphere can be reduced only by reducing its radius, as it remains a sphere. In the puzzle, the drilled sphere is no longer a sphere - it's like a bead that can be put on a string. The drill removes some volume of material from the sphere. The question asks what is the volume of the remaining portion of the sphere. [2] Mass removed is proportional to the volume removed, so density remains the same. Density has nothing to do with the puzzle, which is purely geometric. [3] Surface area does not necessarily increase. It may actually decrease if the sphere is originally large enough. 0 Share this post Link to post Share on other sites

0 Posted March 3, 2008 A 6-inch hole is drilled through a sphere. What is the volume of the remaining portion of the sphere? 1 - Are we to assume the axis is the point of entry and exit? If not size of sphere is not obtainable 2 - diameter of hole is not included, can not deduce the REMAINING volume. Could you not use a liquid to deduce the volume to avoid calculus????? I should get the girl!! 0 Share this post Link to post Share on other sites

0 Posted March 3, 2008 1 - Are we to assume the axis is the point of entry and exit? If not size of sphere is not obtainable 2 - diameter of hole is not included, can not deduce the REMAINING volume. [1] Axis of drill passes through center of sphere. And out the other side, of course [2] Yes you can. 0 Share this post Link to post Share on other sites

0 Posted March 3, 2008 Yes you can. yes if I want to give a formula and not give a figure, I will have to go for displacement/eureka. Will you accept that as a "method/answer" Not gonna give the lady up without a fight are you 0 Share this post Link to post Share on other sites

0 Posted March 3, 2008 yes if I want to give a formula and not give a figure, I will have to go for displacement/eureka. [1] Will you accept that as a "method/answer" [2] Not gonna give the lady up without a fight are you [1] Eureka is always good; displacement seems irrelevant. The method is up to you; the OP just asks for the answer. [2] If you saw the lady, you might become interested. It's entirely up to her; have at it, and good luck. 0 Share this post Link to post Share on other sites

Posted

Maybe this has already been posted. A friend asked me this a while back, and I answered her in less than a minute.

She said I was a genius. But I said there were two ways to arrive at the answer, and I simply chose the easier way.

A 6-inch [long] hole is drilled through [the center of] a sphere.

What is the volume of the remaining portion of the sphere?

The hard way involves calculus. The easy way uses logic.

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