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Topic: Hard: ROTTEN APPLE (Read 2954 times) |
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Drake
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Hard: ROTTEN APPLE
« on: Aug 8th, 2002, 2:33pm » |
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This one had me a bit confused at first because I assumed there didn't have to be outer holes (ie, the worm could tunnel in to the exact center, make a loop and come back out the first hole). But I guess since a tunnel would technically have to have 2 openings, then this is easy to solve.
The worm could not have gone completely to the center, as a straight line to the centerpoint would take 31mm, and it would thus it could not get back out (assuming no backtracking) because the length of the tunnel is only 61mm. Therefore it must have gone no more than 30.5mm in, thus a hemisphere can be sliced directly through the center, missing the tunnel.
Of course I'd throw the apple away anyways because I don't trust that nothing else crawled in and burrowed through it.. 
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tot
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Re: Hard: ROTTEN APPLE
« Reply #1 on: Aug 10th, 2002, 3:54pm » |
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I agree, throw it away.
The riddle says nothing about width of the tunnel, which must be non-zero in real life, thus part of the tunnel might be visible in the clean part, making it inedible at least for me, unless you can cut it furthter.
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Sergei
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Re: Hard: ROTTEN APPLE
« Reply #2 on: Nov 1st, 2002, 5:24pm » |
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on Aug 8th, 2002, 2:33pm, Drake wrote:
it must have gone no more than 30.5mm in, thus a hemisphere can be sliced directly through the center, missing the tunnel.
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I know it seems pretty obvious, but you skipped the actual proof 
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Patrick Gordon
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Re: Hard: ROTTEN APPLE
« Reply #3 on: Nov 30th, 2002, 3:24am » |
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Let's first assume tunnel's thickness = 0. The tunnel is a curve (C). Call O the center of the sphere, R its radius, A et B the IN and OUT points. The length of the "tunnel" is d < 2R.
Let's now consider the set (E) of points M in space such that MA + MB > d. E is the outside of a revolution ellipsoid (never mind the English; this is how we call it my side of the Atlantic!) whose focuses (again, excuse the English if needed!) are AB. O belongs to this set, since 2R > d.
If the tunnel (C) had a point M belonging to (E), the following would hold:
lenght of (C) = lenght of arc AM + lenght of arc MB >= MA + MB > d,
which contradicts the hypothesis : lenght of (C) = d.
Thus (C) is entirely contained in the ellipsoid (boundary included if need be). From O, which is outside the ellipsoid, one can draw to it a tangent half-cone. Any plane containing O and outside this half-cone therefore partitions space into two portions such that (C) is entirely in it. Such a plane meets the requirement.
If the tunnel has thickness, the 'slack' we have between 2R and d will accomodate for it... up to a point.
Patrick Gordon, senior, Paris France
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Patrick Gordon
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Re: Hard: ROTTEN APPLE
« Reply #4 on: Nov 30th, 2002, 3:56am » |
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Sorry!
In my previous message, towards the end, instead of:
Any plane containing O and outside this half-cone therefore partitions space into two portions such that (C) is entirely in it,
please read:
Any plane containing O and outside this half-cone therefore partitions space into two portions such that (C) is entirely in one of them.
Patrick Gordon
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