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        riddles >> hard >> Hard:  ROTTEN APPLE
        
(Message started by: Drake on Aug 8^th, 2002, 2:33pm)

Title: Hard: ROTTEN APPLE
Post by Drake on Aug 8^th, 2002, 2:33pm

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.. ;)

Title: Re: Hard: ROTTEN APPLE
Post by tot on Aug 10^th, 2002, 3:54pm

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.

Title: Re: Hard: ROTTEN APPLE
Post by Sergei on Nov 1^st, 2002, 5:24pm

on 08/08/02 at 14:33:52, 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.

I know it seems pretty obvious, but you skipped the actual proof :)

Title: Re: Hard:  ROTTEN APPLE
Post by Patrick Gordon on Nov 30^th, 2002, 3:24am

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

Title: Re: Hard: ROTTEN APPLE
Post by Patrick Gordon on Nov 30^th, 2002, 3:56am

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

Title: Re: Hard: ROTTEN APPLE
Post by Pietro K.C. on Nov 30^th, 2002, 8:45am

I like your proof a lot! :)

If you would like to see a different one, perhaps a bit more complicated, check out

http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1027806985

This is from before I became a member!