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        riddles >> hard >> Hard- rotten apple
        
(Message started by: Your name here on Jul 26^th, 2002, 1:03pm)

Title: Hard- rotten apple
Post by Your name here on Jul 26^th, 2002, 1:03pm

"An apple is in the shape of a ball of radius 31 mm. A worm gets into the apple and digs a tunnel of total length 61 mm, and then leaves the apple. (The tunnel need not be a straight line.) Prove that one can cut the apple with a straight slice through the center so that one of the two halves is not rotten."

If the radius is 31 mm then the diameter is 62 mm. In three dimensions, this means a straight line from the skin of the apple through the exact centre out the other side would measure 62 mm. Since the wormhole is shorter than this (61 mm) it must lie on one side of some line which runs through the centre. Extend the line into a plane and it will slice the apple in two, with the wormhole on one side.

This appears self-evident to me but I don't know how to mathematically express it. I'd like to see the mathematical proof.

Title: Re: Hard- rotten apple
Post by Pietro Kreitlon Carolino on Aug 25^th, 2002, 1:34am

For the following proof, it is helpful to draw a picture! ;)

Let A be the point where the worm enters the apple, B the point where it leaves said apple, C the center of the spherical aforementioned apple and r the path described by the worm. Let s be a straight line bisecting angle ACB and P the plane perpendicular to s through C. I say that r does not intersect P.

For let Q be the plane containing AC and BC (and therefore s), d the diameter of the apple contained in Q (which is also in the intersection of P with Q), and suppose that r does in fact intersect P at a point X. Naturally, the length of r must be greater equal to AX + XB, because "the straight line is the least distance between two points". Now, let us suppose the curve r projected orthogonally onto Q, and in particular X projected onto X'. Since the length of the projection must be less than or equal to the length of the original curve, we must have length(r) <= AX'+ X'B, where X' is a point on the diameter d orthogonal to s.

Now we are left with a simple problem of minimization, namely of the sum AX'+ X'B with respect to X'. Instead of attributing variables to the distances and performing ugly differentiations on expressions involving square roots, let us suppose that B' is the mirror image of B with respect to d, i.e. BB' is perpendicular to d and is bisected by it. Then AX' + X'B equals AX' + X'B', and the least value this expression can take is the length of the straight line AB' (which means X' must be the very point C, the center of the sphere). Clearly, then, the sum AX' + X'B cannot be less than AC + BC = d, the diameter of the apple. We have then proved that:

d <= AX' + X'B <= AX + XB <= length(r).

However, the hypothesis is that length(r) < d; therefore the point X cannot exist, i.e. the curve r does not intersect P, which is what we wished to prove. Q.E.D. :)

Title: Re: Hard- rotten apple
Post by Patrick Gordon on Dec 2^nd, 2002, 1:39pm

Dear Pietro,

I found your early solution interesting. I too had first considered orthogonal projections. But I then found out that it is just as easy to reason directly in 3D, as shown in my proof.

PS Are you an overseas vistor like myself? Acaso eres argentino?

Title: Re: Hard- rotten apple
Post by Pietro K.C. on Dec 6^th, 2002, 12:40pm

No, I'm not argentinian, I'm brazilian! :) Close.

I refrain from writing more because this is a public thread. If you would register, then we could exchange private messages, or e-mails.

But thanks for the compliments. I find the "replace point B by symmetrical point B' " especially neat. I think it is due to Fermat, when showing that light, in order to minimize the distance traveled between a point, a mirror, and another point, will reflect off the mirror making the incidence and reflection angles equal.

Title: Re: Hard- rotten apple
Post by mattian on Jul 13^th, 2004, 9:50am

How fat is the worm?

Title: Re: Hard- rotten apple
Post by Flo Niebling on Mar 11^th, 2005, 6:57am

I think that this riddle is quite badly phrased. The worm could dig a tunnel to the core, eat the core, waste some more millimeters (making the tunnel 61mm long) and crawl through the tunnel through which he came back to the outside.

Title: Re: Hard- rotten apple
Post by Icarus on Mar 11^th, 2005, 1:00pm

No: Its only a tunnel if it connects two places on the outside. Otherwise, it is just a hole.

Title: Re: Hard- rotten apple
Post by Leonid Broukhis on Mar 11^th, 2005, 5:03pm

on 03/11/05 at 13:00:20, Icarus wrote:

No: Its only a tunnel if it connects two places on the outside. Otherwise, it is just a hole.

Fine. Is a Y-shaped hole a tunnel, if two ends of the Y are at the surface?

Title: Re: Hard- rotten apple
Post by Sjoerd Job Postmus on May 24^th, 2005, 8:57am

I've thought about it, and I think I could prove it, not 'mathematically', but it's proveable.

We have a circle with diameter 62.

*draws circle*.

On the circumference of the circle, we will randomly put two dots. Now, consider the worm making the 'cut'. See it as two straight lines, connected with the two dots on the circumference.

Using this (http://www.maa.org/mathland/9_2_fig1.gif) method to draw an ellipse, you can clearly see that if the worm would dig in the most efficient way, he could only touch the circumference of the second ellipse. If he dug in-efficiently, he would not even touch the circumference.

Thus, it is proven that the worm can not reach the center. On the other hand, it is also proven, that the worm can only reach one side of the circle. Just look at the picture we just drawed... And you will see that the worm is limited to one side.

Now, if we knew all this, and had to cut the apple, how would we do it?

We'd place the apple so that the two entry/exit points were exactly 31mm from the floor, so that they were on the exact side of the 'ball'. Then, we'd cut the apple parrallel to the imaginative line between the entry/exit points, and we'd have one nice side of the apple. :)

Title: Re: Hard- rotten apple
Post by Deedlit on May 24^th, 2005, 11:00am

on 05/24/05 at 08:57:37, Sjoerd Job Postmus wrote:

I've thought about it, and I think I could prove it, not 'mathematically', but it's proveable.

I'm not sure what it means to be provable, but not mathematically. I think it's better to say that you have a heuristic argument, or that the proof is not quite rigorous, or that such-and-such is left unjustified.

Of course, different people will accept different things as being justified without an explicit proof. This problem is one of those times people will probably disagree.

Your proof looks reasonable, except that you made it two-dimensional rather than three-dimensional. For the original problem, you can just say the same thing except with a sphere and an ellipsoid. As you seem to be aware, a couple of things are asserted without proof: First, that the set of paths of a fixed length stay entirely within a particular ellipsoid; and secondly, that the ellipsoid is tangent to the given plane at the center of the sphere. This is visually obvious, so I imagine some people will be willing to accept these without proof, and some won't. Probably, the standard is considerably lower in a puzzle forum than in, say, a mathematical journal!

Fortunately, in this case Pietro gives an elementary and elegant argument why we have to stay above the plane, so I'd say that should definitely be the preferred proof.