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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.
The 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…
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.…
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…
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.
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.…
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 2 pi (r_e + 1) making the increment 2*pi meters. Is the surprisingly low increment the point of the problem?
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.…
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 2 pi (r_e + 1) making the increment 2*pi meters. Is the surprisingly low increment the point of the problem?
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.…
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.
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…
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.
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.