There is a room with a height of 2.5 meters, the ground has an area of 4 X 2 meters. On one narrow side of the room there sits a fly centrally 20 centimeters below the ceiling, on the opposite wall there is the spider 60 centimeters above the floor (also centrally). Fly: "I am sure you are hungry and you want to eat me to have dinner!" Spider: "Yeah sure, I am very hungry and I can run 2 meters in one minute." The fly calculates a few seconds and says: "I am tired, I need a nap. I am gonna sleep now here exactly 3 minutes and then I fly away!!!" Will the spider have dinner, or has the fly calculated enough time to escape? Btw: The spider has to walk any time. No jumps, no drop, no tricks. The fly remains at its place 3 minutes sleeping.

nice question, yen.. the spider has to go up 1.9 metres. then he has to cross the ceiling. 4 metres. then he has to go down 0.6 metres. the hungry predator has to cover a total distance of 6.5 metres. in 3 minutes he will cover only 6 metres. so he will still be 0.5 metres away when the fly takes to the wings, if the fly set her alarm clock correctly.. no dinner for the spider.. from a more philosophical point of view, darwin could have told you that spiders would be extinct by now, if they did not do their calculations correctly... and they are not...

Lol, flies can't speak either. The creation of the room is the easiest thing. A room is a cuboid. The ground is 4 x 2 meters. This means 2 opposite walls are 2 meters the other 2 opposite are 4 meters. The room is 2.5 meters high. The fly and the spider are sitting opposite centrally on the narrow sides, means to their left and right are 1 meter till the corners. The fly sits 20 centimeters below the ceiling, the spider 60 centimeters above the floor. A good start.....and to prove the philosophical point of view you just need to be as good in maths as the spider, right? Sometimes 1.38 seconds can change your entire life.

no thanks button here.. wonder how many seconds the spider needed to work it out..if he takes as long as me he will still get no dinner.. i loved history, back then.. 1066, the invasions, the life of harald hardrada..

Nobody? I have to admit it is not easy, it took me a lot of time to figure it. OK, one more hint. How would one create a model of the room by using a paper, scissors, a pencil and a ruler? The manufacturer of cardboard boxes do that every time. Boxes are made of a special cut, the 6 surfaces of the cuboid can be arranged different ways. One gets one piece of paper, with a special shape and the delineated lines for the edges ('the flat room'). When cut out, the delineated lines need to be fold to become the edges (and finally glued), in that way that one gets a cuboid.

Cool, a drawing of it. The spider needs to find a way which is shorter than 6 meters, yes. The spider sits 60 cm above the floor on the opposite wall of the fly means 1.90 are left to the ceiling. (The room's height is 2.50 meters). So if the spider travels up 170 cm he is 20 cm below the ceiling, but still on the same side, exactly at the same height as the fly, but on the opposite wall of the room. The spider must be on a surface all the time and cannot travel through the air. See the conditions at first post "The spider has to walk any time" The spider and the fly sit centrally, means to the corners it are 1 meter each time, the drawing is perfect.

If he walks then he will never reach in time, it will be 190 cm up and 400 cm across and 20 cm down, so the fly lives. Being a spider you would think he would go 170 cm up and then just web it across.

Ever saw a spider going up 170 cm and then building a 4 meters spiderweb, going across it and 20 cm down the opposite wall and that altogether within 3 minutes? To walk <6 meters a particular way within 3 minutes seems realizable for a spider, though.

very toughtfull, that drawing, R29k, i used it to show the path that i thought would lead to the spiders dinner. untill i thought it over.. but i will still leave him hungry, i think..[below] the spider crosses to the long wall, climbing slightly in the process. he then crosses the long wall, climbing untill he is slightly below the fly. he crosses half the short wall, climbing untill he reaches the spot where the fly was 3 minutes ago. the devil of it is, that the climbing makes his path on the short walls slightly longer than 1 metre, and his path on the long wall slightly over 4 metres. so he still has over 6 metres to cover. the mathematicians amongst us, if any, could use pythagoras`theorem to calculate how much longer. but no matter, it is still over 6 metres, so he will be late for dinner.. and i cannot think of shorter route, that would prove this old world is as wicked as it really is..

Yes, nodnar. This is spiteful. The one who has created that riddle had been very clever and the spider reaches the fly by a very narrow margin only. I had been only able to find a solution by tinkering a model of the room using paper, a pencil, scissors and a ruler. Wow, nice little spider, lol. It doesn't matter, since when applying maths one assumes the spider to be a point. The bigger, the less distance. I let you all think about..........it had been fun to solve it. Tinker a folded box, or at least create its shape 'a flat room', lol.

@ R29k, my turn for posting a drawing. [it aint no masterpiece, for sure, how m$ ruined mspaint in w7 btw, had not used it yet..] yen provided the solution. if i had been more attentive at school all those 47 years ago, i could poblably prove that the spider arrives with 1.38 seconds to spare to have dinner...