MTH 495 – Blog Post 6 – Spider Geometry?!

The following problem comes from the NRICH website at http://nrich.maths.org/2365/2365. Here is the problem:

The Spider and the Fly

Stage: 4

A spider is sitting in the middle of one of the smallest walls in my living room and a fly is resting by the side of the window on the opposite wall, 1.5m above the ground and 0.5m from the adjacent wall.

The room is 5m long, 4m wide and 2.5m high.

What is the shortest distance the spider would have to crawl to catch the fly?

If the fly walks down the wall, is there a point at which the spider would be better changing its route?

I am attaching my work below. However, to start I was fairly certain that the spider was going to have to take an angled route to the fly over then entire distance traveled. This way, the spider would be covering the horizontal and vertical distances at the same time. Even so, I wanted to look at a few different cases to see if I might be missing something.

Here are the cases I came up with, where the vertical distance was covered diagonally over various portions of the horizontal distance:

1. 1^st wall
2. 1^st & 2^nd walls
3. 2^nd & 3^rd walls
4. 1^st & 3^rd walls
5. 1^st, 2^nd, & 3^rd walls
6. 2^nd wall
7. 3^rd wall
8. Horizontal distance 1^st, then vertical

Okay, as I was saying, I was fairly certain that the 5^th case listed was the best option. Also, I figured that there were other possible cases that needed to be considered but I figured that this was at least a starting point.

To figure out the distances, I essentially used the net containing the three walls in question, unfolded it, and use the Pythagorean Theorem to calculate the diagonal distances.

I started with case 8 (Worst case scenario! So I had something to compare the shorter distances to), and I found that the distances for cases 8, 5, 7, 3, 2, & 1, in the order I worked on them, was:

8 – 7.75m

5 – 7.50416551m (This is almost 0.25m shorter than the worst case.)

7 – 7.559m

3 – 7.505678886m

2 – 7.504462863m

1 – 7.515564437m

I stopped looking at cases here because I was fairly certain that I had found the shortest distance. Then I thought of another case. What if the spider had a thread of silk that went direct from where it was standing, to the fly’s location? This was tricky, and I am still not 100% certain I got the right distance. I don’t like that I can’t see it clearly in my head, but here is how I see it.

First, imagine there is a string that goes directly from the wall the spider is on, straight across, as in parallel to the floor, to the wall the fly is on. That distance would be 5m but the spider would still have to be travel 0.5m horizontally and 0.25m vertically, a total of 5.75m. Well, this is already much shorter, but could it be shorter if the thread was attached to the fly?

To get this distance, I used the Pythagorean Theorem to get the diagonal distance on the fly wall from the point straight across from the spider to the location of the fly. This gave me a distance of 0.5590169944m. Now, if you picture our right triangle as being slightly rotated, so that the base is still the line from the spider's position to the point on the fly’s wall, directly across from the spider, then the side of the triangle that gives us the height would be that diagonal lime we just calculated. The hypotenuse of this triangle then, would be our silk thread. To get the distance of this line we have:

Sqrt((0.5590169944)² + 5²) = 5.031152949m

Now, I maintain that this is the shortest possible distance from the spider’s current location, to the fly’s current location, without creating a worm hole or something like this.

However, there are other cases, that are far more likely, that could be shorter than the one using the three walls. These are using the ceiling or using the floor. Since the fly is closer to the ceiling than the floor, and the spider could get squashed on the floor, and the ceiling is far less likely to have huge obstacles on it, like a couch for example, I decided to go with the ceiling.

This gives us:

Sqrt((1.5)² + (7.25²) = 7.403546447m

Which is the shortest distance of the more likely cases/scenarios. Not only that, for the reasons already discussed and many more, it is also likely the safest path for the spider to take. Even so, I personally like the string theory the best.

Hmmm . . . I wonder if it was a spider we, instead of just a straight, lone, and magically gravity resistant string, if we could use graph theory to find the shortest distance, assuming the spider had to walk along the treads and not along the top of the web, and what that might do to our distance. Would the spider give up on the web and just take the wall after all?

Hey, to get a better idea of my approach, check out the notes below. If you see something I missed or anything else worth talking about, let me know.

Thank you for reading,

Jerry