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### The Spider and the Fly

```
Date: 12/23/1999 at 02:52:24
From: keith
Subject: Algebra 1 honors

Hi:

I have not tried this question myself because I don't know how to
start the problem. Here it is:

A spider and a fly are on opposite walls of a rectangular room. The
spider asks the fly if it can come over and "visit" the fly. The fly
believes that the shortest distance, walking on one of the room's
surfaces, is 42 feet. (That is the distance if the spider walks
straight up the wall, over the ceiling, and straight down the opposite
wall.) So the fly agrees to the visit, provided the spider can find a
path that is shorter than 42 feet. Use the "unfolded room" and the
Distance Formula to explain the fly's fatal miscalculation.

Any help given would be greatly appreciated.
```

```
Date: 12/23/1999 at 09:41:24
From: Doctor Rob
Subject: Re: Algebra 1 honors

Thanks for writing to Ask Dr. Math, Keith.

Whether the fly lives or dies depends on the location of the spider
and the fly on their respective walls, which you have not told in the
above description. If both are centered horizontally between the side
walls, the fly is D feet below the ceiling, the spider is d feet below
the ceiling, the distance between their walls is L, the width of the
room is W, and its height is H, then you know that L + D + d = 42, or
L = 42 - d - D. If you unfold the room this way,

H
o---------o
|         |
|      W/2|
|         |
W|     F---o
|       D |
|         |
|         |
o-------------o---------o-------------o---------o
|      W      |    H    |      W      |    H    |
|             |         |             |         |
|             |         |             |         |
|             |         |             |         |
|             |         |             |         |
|L   Floor    |L        |L  Ceiling   |L        |L
|             |         |             |         |
|             |         |             |         |
|             |         |             |         |
|             |         |             |         |
|      W      |    H    |      W      |    H    |
o-------------o---------o-------------o---------o
|             |
|             |
|H            |H
|      S      |
|     d|  W/2 |
o------o------o
W

and draw a diagonal line from S to F, this might be shorter than
42 feet. The horizontal distance from S to F in this diagram is
W/2 + (H-D), and the vertical distance is (H-d) + L + W/2. You figure
out using the distance formula (or the Pythagorean Theorem) if this is
possible.

If, however, you unfold the room this way,

W
o-------------o
|             |
|     Fly     |
|H     F      |H
|      |D     |
|  W/2 |      |
o-------------o---------o-------------o---------o
|      W      |    H    |      W      |    H    |
|             |         |             |         |
|             |         |             |         |
|             |         |             |         |
|             |         |             |         |
|L   Floor    |L        |L  Ceiling   |L        |L
|             |         |             |         |
|             |         |             |         |
|             |         |             |         |
|             |         |             |         |
|      W      |    H    |      W      |    H    |
o-------------o---------o-------------o---------o
|             |
|             |
|H  Spider    |H
|      S      |
|     d|  W/2 |
o------o------o
W

and draw a diagonal from S to F, this also might be shorter than
42 feet. The horizontal distance is now W/2 + H + W/2, and the
vertical distance is (H-d) + L + D. Again, you should figure out using
the distance formula (or the Pythagorean Theorem) whether or not this
is possible.

HINT: Replace L by 42 - d - D first.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Polyhedra
High School Triangles and Other Polygons
Middle School Geometry
Middle School Polyhedra
Middle School Triangles and Other Polygons

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