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        riddles >> easy >> Extension of Spider and Fly Problem
        
(Message started by: Sir Col on Dec 10^th, 2004, 4:31pm)

Title: Extension of Spider and Fly Problem
Post by Sir Col on Dec 10^th, 2004, 4:31pm

I am sure that this classic problem is on the website in some guise, but this is an extension.

Original problem...

A spider sits in one corner of a room, S, measuring 4m by 3m by 2m. A fly is caught in a web in the opposite corner, F.

The original question asks for the shortest route from S to F.

By "opening out" the room to produce a net of the cuboid, the first time we see this, we are surprised to note that the shortest route takes an unexepected path. The two candidates being SF₁= [sqrt]41 and SF₂=[sqrt]45; SF₁ being the shortest in this example.

Extension...

Given any sized cuboid room, is the shortest route always found by travelling to some point on the longer base wall?

Title: Re: Extension of Spider and Fly Problem
Post by JocK on Dec 12^th, 2004, 8:34am

on 12/10/04 at 16:31:55, Sir Col wrote:

Given any sized cuboid room, is the shortest route always found by travelling to some point on the longer base wall?

:: [hide]Yes, for the shortest route to the opposite corner the spider has to avoid crawling across the two walls (or floor and ceiling) with smallest area.

Intuitively, this is not surprising as crawling like this, the 'kink' in the double-straight-line path is minimal (i.e. it is the closest approximation to the straight-line path the spider can make)[/hide] ::

Title: Re: Extension of Spider and Fly Problem
Post by Sir Col on Dec 12^th, 2004, 10:17am

True, but if the floor measured a by b, with a>b and SF₁ representing the segment of the "shortest" path that meets the edge of length a, then is SF₁ always less than SF₂?

Title: Re: Extension of Spider and Fly Problem
Post by JocK on Dec 12^th, 2004, 10:43am

on 12/12/04 at 10:17:36, Sir Col wrote:

True, but if the floor measured a by b, with a>b and SF₁ representing the segment of the "shortest" path that meets the edge of length a, then is SF₁ always less than SF₂?

Not sure if I understand you correctly. If a > b then SF₁ is less than SF₂. But if the third dimension c (the height of the room) exceeds a, neither is the shortest path. Again, the shortest path is to avoid the walls of smallest area (in this case the floor and ceiling each of area a x b).

But I think I read you incorrectly and you have something different in mind?

Title: Re: Extension of Spider and Fly Problem
Post by Barukh on Dec 12^th, 2004, 10:44am

[hide]What you are actually asking: Is it true that (b+c)² + a² < (a+c)² + b² whenever a > b?

The answer is yes.[/hide]

Title: Re: Extension of Spider and Fly Problem
Post by Sir Col on Dec 12^th, 2004, 11:00am

That is exactly what I am asking, but I wanted to avoid giving too much away; deriving the inequality you presented goes half way to solving the problem.

Of course you're right, Barukh, the answer is, yes. But perhaps we can leave the proof for someone else? Obviously you're free to answer it, but it's only an "easy" problem and is probably not going to challenge you very much.

Title: Re: Extension of Spider and Fly Problem
Post by Barukh on Dec 12^th, 2004, 11:17pm

on 12/12/04 at 11:00:30, Sir Col wrote:

Of course you're right, Barukh, the answer is, yes. But perhaps we can leave the proof for someone else? Obviously you're free to answer it, but it's only an "easy" problem and is probably not going to challenge you very much.

Sir Col, I apologize, I misinterpreted your question.

My last post is hidden now (better later than never).