brainteasr

[Jules6]

A spider is in a rectangular room measuring 40x10x10metres. The spider is on the 10x10 metre wall, 5metres from the sides and 1 metre above the ground. A fly is on the opposite wall 5 metres from the sides and 1 metre below the ceiling. What is the shortest distance for the spider to walk to the fly (assuming the fly doesnt fly away!!)

[rekstout]

48.7m

[tomd]

I get 50.6m. The shortest route the spider can take is the hypotenuse of a triangle 50m x 8m (50m across the walls and 8m up) This makes root((50^2)+(8^2)) = root(2564) = 50.6

[Shrek2]

I'm not a mathematician, but simply walking down the wall (1m)across the room (40m)and up the wall to the fly (9m) works out at a total of 50m, which is shorter than the root 50.6 posted previously. But 50m is probably not right either!

[DOB]

50m seems about right to me.

[bernardo]

tomd seems to be talking about the hypotenuse of a triangle which is drawn on the (flattened-out) shape of the end walls (where the spider and the fly are at opposite ends) combined with the long wall, involving the spider having to walk 8 metres upwards. If that is the path followed by the spider it would indeed be a right-angle-triangle with vertical 8m and horizontal 50m, i.e the path would be the square root of (50 squared plus 8 squared) i.e. square root of 2564, i.e. 50.6 (ish).

But as Shrek2 says, it is not necessary for the spider to walk along the hypotenuse of a triangle, because the path from the spider to the fly on the flattened-out shape of the end-walls and the floor is a straight path of 50m which does not need to include any sideways element. Therefore the spider could walk 50 metres to get to the fly.

BUT

what nobody has yet mentioned is that the spider can in fact make the journey even shorter by jumping off the wall onto the floor before walking the length of the room. Therefore the journey is reduced to 49 metres; if however the spider is able to include an element of sideways propulsion in his jump off the wall onto the floor, he may land on the floor (for example) 30 cm away from the end wall, in which case the distance he has to walk is only 48.7 metres.

Presumably this scenario (involving a sideways jump onto the floor) is what rekstout had in mind in his first answer.

[Shrek2]

looking at Tomd and Bernardo response, i don't follow how the spider could literally walk through thin air on the hypotenuse of a triangle basis ( although they seem to be able to create webs everywhere!) but based on their replies, should it not be the square root of 1664 = 40.8 (40sq + 8sq)as the room is only 40m long. I was also thinking the spider could spin a web to drop the 1m to the floor, and save his feet a little!! bringing it down to 49m

[bernardo]

To get the triangle and the hypotenuse, you flatten out the floor and walls onto a surface and then draw on that.

[dicky113]

Can it be a rectangular room with pitched walls? e.g if the 10x10 walls are pitched inwards so that the 10m lengths are the 10s in two 6.8.10 right angle triangles then that would leave 40m -8+8 as the ceiling length = 24. 24 plus the 1m and 9m = 34. It could of course have only one 10x10m wall meeting the opposite longer wall in a triangle as seen from the side. It would still be a rectangular room as the floor would be 10x40. In which case the answer is 10m.

[gazzawazza]

Presuming that it's not a trick question, I too came up with 50m.

[bernardo]

I think we can discount dicky113's theory, because the question refers to a "rectangular room measuring 40x10x10metres". I assume "rectangular" means cuboid. In any case, "rest-angular" means "right-angled" which also implies cuboid.

[dicky113]

My theory was heavily sponsored by The Tiger Beer Co. So feel free to discount it :)

[Stav's Back]

Jules, are you gonna tell us the correct answer, and how to get it?

[Jules6]

Hi this is a question from a quiz I'm doing, so don't know the answer yet! THanks to you all for trying to work the answer out, and as soon as i know the answer i'll let u know. :-)

[Stav's Back]

IVE GOT IT! If you flatten the whole room at with the ceiling at the top and the fly wall attached to that, then below the ceiling a wall, then the floor with the spider wall attached to that, then the other wall below that, you can create a triangle with the hypotenuse as the distance the spider walks, and the horizontal distance 42m, and the vertical distance 20m, the the distance for little old spidey is 46.5m. I WIN!!

[Stav's Back]

The calculation is: Root ((42 x 42) + (20 x 20)) = Root (1764 + 400) = Root 2164 =46.5188m

[tomd]

I get Stav's triangle to be 46 x 14, don't understand where you're coming from.

[tomd]

Oh, I see it. I'd attached the spider wall to the one underneath the ceiling, so he had to walk 5m across that. It doesn't make sense that it would be quicker to take a route crossing some of the ceiling, but it works mathematically.

[bernardo]

Wow! But I doubt if the spider is as much of a mathematician as that.

[Jilly]

45 metres

[Ichi]

I agree - 45metres...