Ready for a challenge?

Question

A room is 10 meters long, 4 meters wide, and 4 meters high. A spider is at the middle of an end wall, 1/3 meter from the floor. A fly is at the middle of the other end wall, 1/3 meter from the ceiling, too frightened to move. The spider crawls to the fly. What is the shortest distance? (14 meters is not the correct answer.)

Answer 1

The width of the room is irrelivant.

They are sqrt ( 10 ^2 + (4 - 2/3)^2) = sqrt( 100+100/9) = sqrt(1000/9) = 10*sqrt(10)/3 = 10.54 m approximately apart.

The horizontal distance is clearly 10. The vertical distance is the difference of their heights (4-1/3) for the fly and 1/3 for the spider.

Then using pythagoreans theorem you get the above

EDIT he crawled :P next time ill read... ill come back to this later

To be frank i think the answer is 14. I have conjectures that take too much to say online... It is clear the path must be symmetric.

Note 1 if the path is on the bottom the minimum is 14. The direct path. Conjecture omitted

If he goes across the side then he must go up to a point (0,b) from (2,1/3) b>1/3 then across from (0,b) to (10,4-b) then from (0,4-b) to (2,4-1/3).

This is necessarily greater than 14 for all b (the min is about 14.39)

His other option is to go up to (0,b) from (2,1/3) then across from (0,b) to (5,4) then from (5,4) to (0,b) (first horizontal wall then cieling) then from (0,4-b) to (2,4-1/3)

The min on this is about 15.8

Answer 2

He never goes to the floor or ceiling. If x is the shortest distance, then

x^2 = (4 - 2/3)^2 + 14^2 = 100/9 + 196 = 207 1/9

x = 14.39 m

Oh -- if he had a ready-made spider web so he didn't have to crawl along the walls, the shortest distance x is given by

x^2 = (4 - 2/3)^2 + 10^2 = 100/9 + 100 = 111 1/9
x = 10.54 m

but only if the spider web is a taut line.

Answer 3

At what point horizontally is the spider on the wall, cause you have the vertically down it's 1/3 of a meter from the floor but the spider can be at different points on the wall and still be 1/3 of a meter from the floor, the same goes for the fly. And define middle, do you mean exact middle or just middle, cause middle can be a very broad phrase. ask your question correctly and precisely or don't ask it at all.

Answer 4

What you described make a right angled triangle.

Since the spider is at 1/3 meters=0.33 meters above the floor
the fly is 1/3 m=0.33 meters below the ceiling.

Now you can get two answer because you didnt mention that fly and spider are on the 10 meter long wall or 4 meter wide wall.

If it is 4 meter long wall, then u get the answer as by solving right angled triangle. to find hypotenuse in it

√[ (3.334)² + (10)² ] = 10.54 meters <==pathagorus theorm)

If it is on 10 meter long wall, then u will get.

√[ (3.334)² + (4)² ] = 5.20 meters

----------------------------------------------------------

let me know if i did it wrong.

Answer 5

The shortest distance would be to the fly swatter to get the fly and then the spider, in that order!

Answer 6

I am assuming you meant the spider was on one of the square walls and the fly was on one of the rectangle walls.

Answer = 7.781745 (Approximately)

Answer 7

Which end? If it is the short end, it is 8 meters I think (but my math is not that good).

Answer 8

can u re-check your question? 14m seems to be the correct answer if the end walls are 4X4 and the fly/spider are in the centre (2m from edge).

Answer 9

I think the answer is 7.54 ft my work was a little complicated so I won't write it, seeing as there is a huge chance that I'm wrong.

Answer 10

8 meters ????