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brainteasr
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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!!)
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No best answer has yet been selected by Jules6. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.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.
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.
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
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.
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!!
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