RE: [math-fun] How we "do mathematics"
I wonder how many of you took a look at Eric Angelini's two lovely problems which I reproduce here. The first is a wonderful example of a Gardnerian Aha problem. Writing things down would be of no use here, -so much for my theory on the essentiality of pen and paper. The lovely and slightly surprising answer "came to me" a couple of hours after reading the e mail, sitting on a bench in the botanical garden of the town of Lund, Sweeden. Later while trying to get to sleep I worked out the general case where Mr. A brings a liters and Ms. B brings b liters where 2b>a>b. As for the second problem which I just finished, I defy ANYBODY to solve it without doing at least some writing. Remember the answer has to be a number.The paper in front of me contains four drawings, circles, coordinate axes, line segments with various labels, followed by a mess of symbols like x, y ,R, a, pi, integral signs, square root signs. I don't plan to have it framed. David .
A (very) old one I like :
Two Death Valley marathonians friends complete their run together and seek shadow in their tent nearby the finish line.
The first one had prepared a fresh 5-liter tank of water in a cooling box and his friend a similar 3-liter one.
They were about to start drinking when a journalist enters the tent desperately asking for water.
OK, the water is divided into three equal parts -- and drunk.
The thankful journalist insists to pay his share and leaves 8 dollars to be split between the two friends.
How?
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At 07:29 AM 8/6/2007, you wrote:
it's the kind of thing that throws people because it sounds like there's not enough information.
Old Boniface he took his cheer, Then he bored a hole through a solid sphere, Clear through the center, straight and strong, And the hole was just six inches long.
Now tell me, when the end was gained, What volume in the sphere remained? Sounds like I haven't told enough, But I have, and the answer isn't tough!
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There also is a meta-theoretic answer to this puzzle. Assume the puzzle can be solved. Then it must be solvable with a hole of any diameter, even zero. But if you drill a hole of zero diameter that is six inches long, you leave behind the volume of a six inch diameter sphere.
http://www.faqs.org/faqs/puzzles/archive/geometry/part1/
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On Aug 9, 2007, at 6:17 AM, David Gale wrote:
I wonder how many of you took a look at Eric Angelini's two lovely problems which I reproduce here. ... As for the second problem which I just finished, I defy ANYBODY to solve it without doing at least some writing. Remember the answer has to be a number.The paper in front of me contains four drawings, circles, coordinate axes, line segments with various labels, followed by a mess of symbols like x, y ,R, a, pi, integral signs, square root signs. I don't plan to have it framed.
Old Boniface he took his cheer, Then he bored a hole through a solid sphere, Clear through the center, straight and strong, And the hole was just six inches long.
Now tell me, when the end was gained, What volume in the sphere remained? Sounds like I haven't told enough, But I have, and the answer isn't tough!
SPOILER below white space (in head solution that's easy if you're familiar with this kind of thing).
This actually has an "instant" solution you can see once you get the right mental image. I knew to look for it because it's related to the famous observation of Archimedes that the volume of a ball equals the volume of the smallest solid cylinder containing it, minus the volume of the double cone from the center of the cylinder to its two ends. I won't go through that, but for the current problem: Imagine cutting the bead by half-planes that are tangent to the cylinder and go off to one side (let's say counterclockwise, as seen from the end of the bead closer to you). Any planar cut of a sphere is a circle. Make a bunch of these cuts, closely-spaced, to cut the ball into wedges between two half-disks---slightly fudged only because part of one side of the wedge wraps slightly around the cylinder. After shaving off a tiny bit where one side of the wedge wraps around a small angle of the cylinder, these wedges can be slid together so they form a nearly perfect 6" diameter ball. I think "the volume of a ball whose diameter is the length of the hole" is a better answer than a number. I'd love to see a mechanical version of this! There are a bunch of other similar dissection constructions. Here's one: if you take a closed curve c in space (actually, in any dimension) whose curvature is always greater than some constant J, then the volume of the neighborhood of radius r about c only depends on r and on the length of c, provided r < 1/J. Can you propose a flexible model made of hard, incompressible pieces for a necklace of this sort? Bill
On Thursday 09 August 2007, Bill Thurston wrote: [for the problem about a sphere with a cylindrical hole of given length]
SPOILER below white space (in head solution that's easy if you're familiar with this kind of thing).
[and some more spoiler space ...] ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... If you don't mind making use of the fact that you know the problem as posed has an answer, it's even easier: you weren't told how wide the cylinder is, so take it to be infinitely thin. Then what you have *is* a spherical ball whose diameter equals the length of the cylinder. -- g
participants (3)
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Bill Thurston -
David Gale -
Gareth McCaughan