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CTK Exchange
itineracy
Member since Jul-29-07
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Jul-29-07, 08:31 AM (EST) |
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"Gardner's Torus cutting puzzle... 13 pieces?"
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In a well-known puzzle, Martin Gardner asked for the maximum number of pieces into which a torus could be sliced by three planes. Based on the formula X(n)=(n^3+3n^2+8n)/6 for n planes, the answer is given as 13. The formula however, assumes the torus to be a solid, where I thought a torus was supposed to be a surface.
Now, when slicing the torus in such a way that it would produce 13 pieces if it were a donut, it actually produces 14 pieces if it were an inner tube (see attached image). My question: is there a way to modify the above formula for solid pieces in order to yield the number of surface pieces, or do I have to go about it in an entirely different manner? Yours truly,
Walter |
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https://www.cut-the-knot.org/htdocs/dcforum/User_files/46ac73787316dae1.gif
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Itineracy
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Dec-23-07, 08:48 AM (EST) |
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2. "RE: Gardner's Torus cutting puzzle... 13 pieces?"
In response to message #1
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Hi Predrag, I totally forgot about me having posted my question here... :-O
So much for the huge delay in answering to your response... Interesting puzzles you put up.
Re: https://www.mathlinks.ro/Forum/viewtopic.php?t=70759
I think the solution would be A3F1
I'll be happy to provide you with the rationale of that, but will not do so on this public forum... If my solution is the correct one, that would spoil the fun somewhat, me reckons... Cheers |
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itineracy
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Nov-02-08, 11:22 PM (EST) |
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3. "RE: Gardner's Torus cutting puzzle... 13 pieces?"
In response to message #1
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I think I found it: 2^(c+1)-2, where c is the number of cuts... Thanks for the input.
And please feel free to comment, should I be mistaken... :P |
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itineracy
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Jan-02-09, 10:10 AM (EST) |
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4. "RE: Gardner's Torus cutting puzzle... 13 pieces?"
In response to message #3
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Hm... 2^(c+1) - 2 doesn't seem to bring home the bacon...
I'm more inclined to (c^3 + 5c) / 3 now... Hard time proving it, though... |
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