Minimum Distance to Cross Rectangular Room

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In summary, a fly in a rectangular room measuring 30 feet in length, 12 feet in height, and 12 feet in width needs to walk from a point one foot down from the ceiling at the middle of one end to a point one foot up from the floor at the middle of the other end. After considering various paths, it is determined that the shortest path for the fly to walk is 40 feet, crossing 5 different surfaces (the two short walls, one long wall, the ceiling, and the floor).
  • #1
ehrenfest
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Homework Statement


A rectangular room measures 30 feet in length and 12 feet in height, and the ends are 12 feet in width. A fly rests at a point one foot down from the ceiling at the middle of one end and wants to get to a point one foot up from the floor at the middle of other end. Can it get there by walking less than 40 feet.

Homework Equations


The Attempt at a Solution



I get that it needs sqrt(1658) > sqrt(1600) feet to get there. What do other people get?
 
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  • #2
Could you please show the method?
 
  • #3
Basically I unfolded the room and found a straight line along the walls from the starting point to the destination. I get that the bug walks along four different walls on this shortest path. Is it wrong?
 
  • #4
Shooting star said:
Could you please show the method?

Pretend the room is a cardboard box and think of various ways to cut it up so it lies flat and you can draw a straight line between the two points. The shortest one I can find is sqrt(1658) as well.
 
  • #5
I get that the bug walks along three surfaces, the one where it is, the one where it ends up, and the wall in between these two. But I get sqrt(42^2+10^2), which is even higher.

Tell me about your four walls.
 
  • #6
There's a shorter path crossing four surfaces. Angle up to the ceiling, go along the ceiling for a while, cross on to the side wall and then move on to the end wall. Get those scissors out.
 
  • #7
As far as I can tell, the answer to the original question is no. But I'm waiting for a better solution.

Incited :mad: by Dick's snippy post (well, he was talking about scissors :wink:), I have destroyed a few rooms and ultimately found the shortest path as 40 ft. This comes from sqrt[(1+30+1)^2 + (12+12)^2] = 40. I'm sure you guys can visualize the way I've opened up the room, without giving any thought to the fact I'm scared of shooting stars falling from the sky.

What about < 40 ft?
 
  • #8
Oooops. The way I was thinking was sort of ceiling obsessive. Go to a side wall, traverse the wall, then back through the end wall. sqrt(10^2+36^2). It IS less than 40! Thanks for pushing me into rethinking this. I'm still not sure I see the 40 route.
 
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  • #9
Alternative solution;

Yes, by flying there instead =D
 
  • #10
Hmmmm the only way I see is by flying. The shortest path I get is 42 exactly.
 
  • #11
MaWM said:
Hmmmm the only way I see is by flying. The shortest path I get is 42 exactly.

Read the question. The fly has to walk. If it can fly the shortest distance is sqrt(10^2+30^2). Which is definitely not 42.
 
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  • #12
Dick said:
Read the question. The fly has to walk. If it can fly the shortest distance is sqrt(10^2+30^2). Which is definitely not 42.

I read it. It doesn't say the fly has to walk. And I don't see how you can get sqrt(10^2+30^2).
 
  • #13
I was actually joking when I last posted in this thread >.< Even if it didn't explicitly state, I assumed it had to walk :(
 
  • #14
Well Gib, I don't see why they would bother making it a fly instead of.. say... a spider. Seems like they are inviting ambiguity. Or.. maybe they did it on purpose and flying is the correct answer.
 
  • #15
Far as I can see.. there are four paths the fly could take (if walking): If it crosses the ceiling or floor, the distance is 1+30+11=42. If it crosses one of the two side walls, the distance is sqrt(42^2 + 10^2) which is even longer.
 
  • #16
Dick said:
Oooops. The way I was thinking was sort of ceiling obsessive. Go to a side wall, traverse the wall, then back through the end wall. sqrt(10^2+36^2). It IS less than 40! Thanks for pushing me into rethinking this. I'm still not sure I see the 40 route.

I'm very bad at making picture files, and I don't have a scanner, so I can't show you the path right now. I'll try -- maybe ask someone to make a bmp or whatever file.

But could you show us your path? Or describe in detail?
 
  • #17
you nearly got it! you should have gone a little further.

you should unfold to two dimension in such a way that you draw the small sidewalls up-left and down-right, and (from up to down) the ceiling, one big sidewall and the ground in the middle.

Now draw a straight line. it's 40 ft.

You could also try take 6 walls, but as you can see that would only be longer. I think 40 ft. is the minimum.
 
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  • #18
oh wait someone said this already >_> never mind then
 
  • #19
Dick said:
Oooops. The way I was thinking was sort of ceiling obsessive. Go to a side wall, traverse the wall, then back through the end wall. sqrt(10^2+36^2). It IS less than 40! Thanks for pushing me into rethinking this. I'm still not sure I see the 40 route.

The 40 ft route may be possible to see even without the diagram because of the way I've written it: sqrt[(1+30+1)^2 + (12+12)^2].

But where are you gettting the 10 from? Please confirm once more.
 
  • #20
MaWM said:
Well Gib, I don't see why they would bother making it a fly instead of.. say... a spider. Seems like they are inviting ambiguity. Or.. maybe they did it on purpose and flying is the correct answer.

In the book where I found the problem (problem solving through problems) the fly has a broken wing and does explicitly have to walk. I should have written that.
 
  • #21
Shooting star said:
The 40 ft route may be possible to see even without the diagram because of the way I've written it: sqrt[(1+30+1)^2 + (12+12)^2].

But where are you gettting the 10 from? Please confirm once more.

I see it now. Its really sqrt[(1+30+1)^2 + (6+12+6)^2]. 5 different surfaces are used (the two short walls, one long wall, the ceiling, and the floor). I think that is the shortest (also the wording of the actual problem implies that it is).
 
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  • #22
If the fly has to walk, then it has to cover the distance through the walls. Had it been able to fly, the answer would have been sqrt(30^2+10^2) ~ 31.6227766016838. The reason for this is, the vertical height is 12, and the fly sites 1 foot below the ceiling (11 feet). It wants to go the 1 feet above the ground, i.e. the total vertical distance difference is thus 10.

Now, whether the fly gets to the top or the bottom of the starting wall, the least distance to the other wall is the straight distance (30), i.e. the perpendicular distance.

It has to travel another 1 to get up to the ceiling, and 10 more to get down to 1 feet after reaching the other side.

Regards,
Sleek.
 
  • #23
@ehrenfest

I think i get it too. Maybe 40 is the shortest. I'll look into it when I get a bit more time :).

Regards,
Sleek.
 
  • #24
Shooting star said:
The 40 ft route may be possible to see even without the diagram because of the way I've written it: sqrt[(1+30+1)^2 + (12+12)^2].

But where are you gettting the 10 from? Please confirm once more.

I got those numbers by temporarily losing the ability to add correctly. But I see your sqrt[(1+30+1)^2 + (12+12)^2] now. On my diagram that route looked longer than the sqrt(37^2+17^2) one. Apparently I should try harder to draw to scale in a problem like this.
 
  • #25
So, what shall we decide on the answer to be? The sooner the better, otherwise this thread will ramble on and the original interesting question will be forgotten.
 
  • #26
40 ft. is minimum, I'm quite sure. If you go over 5 walls, the straight line makes 40 ft. If you go over 5 other walls, then the distance becomes greater or equal.

If you go over 6 or more walls the minimum distance is way more, and for all lesser number of walls we tried all reasonable configurations of straight lines when the room is unfolded, which all end up more than 40 ft.
 
  • #27
> Can it get there by walking less than 40 feet?

That was the original Q in the first post. The answer is no.

We'll wait for Dick to confirm.
 
  • #28
Shooting star said:
> Can it get there by walking less than 40 feet?

That was the original Q in the first post. The answer is no.

We'll wait for Dick to confirm.

Why? I don't even think 'we' have to decide. That's up to ehrenfest. But I think jacobrhcp summed it up. We've looked at basically all paths hitting less than or equal to 5 surfaces and 5 is least. Clearly 6 or more is considerably longer.
 
  • #29
Dick said:
Why? I don't even think 'we' have to decide. That's up to ehrenfest. But I think jacobrhcp summed it up. We've looked at basically all paths hitting less than or equal to 5 surfaces and 5 is least. Clearly 6 or more is considerably longer.

No, not because of we or ehrenfest. I thought you may just come up with a new idea.
 
  • #30
Shooting star said:
No, not because of we or ehrenfest. I thought you may just come up with a new idea.

Nope, I can't see any way to squeeze out a shorter path.
 

1. What is the minimum distance to cross a rectangular room?

The minimum distance to cross a rectangular room is the shortest possible distance between two opposite corners of the room. This can be calculated using the Pythagorean theorem, where the distance is equal to the square root of the sum of the squares of the length and width of the room.

2. How is the minimum distance affected by the size of the room?

The minimum distance is directly proportional to the size of the room. As the length and width of the room increase, the minimum distance also increases. This is because the longer the distance between two opposite corners, the larger the hypotenuse will be in the Pythagorean theorem calculation.

3. Does the shape of the room affect the minimum distance?

Yes, the shape of the room can affect the minimum distance. For a rectangular room, the minimum distance is the shortest distance between two opposite corners. However, for other shapes such as a square or a circle, the minimum distance may be different.

4. Can the minimum distance be calculated for rooms with irregular shapes?

Yes, the minimum distance can still be calculated for rooms with irregular shapes. However, the calculation may be more complex and may require breaking down the shape into smaller, simpler shapes in order to apply the Pythagorean theorem.

5. How can the minimum distance be used in real life applications?

The concept of minimum distance is used in various real life applications such as urban planning, architecture, and navigation. It helps in determining the most efficient and shortest routes for transportation and can also be used in optimizing the layout of buildings and structures.

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