Spider - fly math problem?

Question

A rectangular room measures 30 feet in length and 12 feet in height, and the ends are 12 feet in width. A spider, rests at a point one foot down from the ceiling at the middle of one end. A fly (food of food to the former) is located one foot up from the floor at the middle of the other end.
The OLD problem, as above, published in various math challenge books, and I see repeated here, as it probably will be again and again years from now, is to find the minimum distance between the spider and the fly, accomplished by "unfolding" the room and computing the pythagorean distance.
What will would probably be MUCH more of a challenge is to determine if the spider CONTINUES on the same direction, (an after dinner walk? :-) ), straight with respect to the "unfolding" of the walls, until (and if?) it EVER returns to its starting position, and if so, how far does the spider travel. Is there an analytic way to do this (preferred), or as an initial answer one could write a computr program?

I think it is a question where "the fun begins". Symmetry is not a reason to suppose that the spider return trip will be the same. This would ONLY be true IF the spider was allowed to backtrack the original trip which he is not. This is sort of a 3-d analog of a problem to predict where an ideal bounce pool ball will be after N cushions given an initial position and angle

Answer 1

Hmmmmm........ It's pretty obvious that if you put the room in a 3-space with {0.0.0} in the 'center' (such that the 'ends' are at x=±15, the 'walls' are at y=±6 and the 'floor' and 'ceiling' are at z=±6) if the spider is at coordinates (x,y,z) and the fly at coordinates (-x,-y,-z) then 'any' path (minimal length or not) that takes the spider through the flys location will (by symetry) also get him back to his starting place after he's travelled the same distance.

But if the fly is *not* at (-x,-y,-z) then the fun begins *big* time


Doug

Answer 2

If you draw this problem out as a 3 dimensional sketch of the room you will see that the problem resolves into a trapeziod with verticals of 11 ft., and 1 foot, a horizontal base of 30 ft, and a sloping top of unknown dimension. Since the vertical legs are at right angles to the base, calculation of the sloping side does not depend on the 30 foot by 1 foot chunk at the bottom, and we may discard that.

The sloping side is the distance from the spider to the fly as the crow flies, or as a bullet would fly. This may be calculated as:

The square root of (10 squared + 30 squared) =

the sq root of ( 100 + 900 ) =

the sq root of 1000 = 31.6 ft.

If the spider had to walk over to the fly, he would traverse down 11 feet, over 30 feet and up 1 foot, or 42 feet total.

Answer 3

Question: When the spider, continuing after eating the fly, encounters the floor, what is its behavior? Does it "bounce" off the floor and continue back up the same wall, or does it travel across the floor, and if so, at what angle?

The rules for traveling to the fly are fixed by fact the distance traveled is to be minimized, but once the spider is past the fly, those rules no longer apply.

Answer 4

If I understand the question correctly -
If the creatures are on opposite walls, then this is my calculation.

Shortest distance x:
x = √(10² + 12²)
x = √(244)
x = 15∙62049925
x ≈ 15∙62 ft.

To walk straight around the walls - distance s:
s = 1 + 12 +11
s = 24 ft.

I'm sure there are computer programs out there that will solve the problem.