moje_ime Posted June 13, 2006 Report Share Posted June 13, 2006 Hello! I need a help.Here is the question:There is a bug in a room. The room is 4 meters high, 4 meters wide and 10 meters long.The bug is sitting on the left (4x4) wall, one meter up from the ground, in the middle (position A).The bug must come to the goal (position :confused: which is on the totally right side, in the middle of the right wall (4x4), three meters up from the ground.Can the bug pass the distance less than 14 meters?(Sorry for my bad English). Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted June 13, 2006 Report Share Posted June 13, 2006 The distance between A and B is 10.2m, it can travel this path in under 14meters if that is what you meant :confused: Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted June 13, 2006 Report Share Posted June 13, 2006 I think it means by crawling across walls, floor and ceiling. I also think it is homework. I can see exactly two routes of exactly 14 metres, and I can't see any briefer ones. Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted June 13, 2006 Report Share Posted June 13, 2006 oh so not a flying bug :confused: yeah 2 possible routes at 14 but dont see any less Quote Link to comment Share on other sites More sharing options...
moje_ime Posted June 13, 2006 Author Report Share Posted June 13, 2006 It is not flying bug, just "walking".I have found 13,928 m.Any less? Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted June 13, 2006 Report Share Posted June 13, 2006 Well, I was trying to minimize an alternative possibility but it takes a bit more doing than I hoped for, unless I've bungled the polynomials you would need to find any solutions to: [math]2a \sqrt{4a^2-32a+164} + (4a-16) \sqrt{1+a^2} = 0[/math] Anybody in the mood? Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted June 13, 2006 Report Share Posted June 13, 2006 Gosh LaTeX doesn't seem to be working right. :doh: :doh:, :doh: and :doh: again! Who's idea was it to have slashes and backslashes in sowtware languages!?!?! :hihi: Anyway you appear to have beat me to it, or sort of... did you actually minimize the idea or did you just hit on a value that makes it under 14? Quote Link to comment Share on other sites More sharing options...
Tim_Lou Posted June 13, 2006 Report Share Posted June 13, 2006 edit: it can be done......i just figured it out, i think 13.9284 is the minimum..it seems that the bug DOES cross the edge in the 10 by 4 rectangle. (i thought it shouldn't previously, I was wrong)see picture. there are many ways of "un-folding" the "room". the shortest path in these foldings are shown.[math]a=\sqrt{14^2+1}[/math][math]b=14[/math][math]c=\sqrt{13^2+5^2}=13.9284...[/math][math]d=\sqrt{12^2+8^2}=14.4222...[/math]it can be less than 14 Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted June 13, 2006 Report Share Posted June 13, 2006 I agree Tim, if I haven't bungled things, after a quadrature my equation simplifies to a second degree one: [math]384a^2+128a+256[/math] which has a negative discriminant... no minima! I have found 13,928 m.How did you get that, then? Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted June 13, 2006 Report Share Posted June 13, 2006 No, wait, that's: [math]384a^2+128a-256[/math] which has two solutions, -1 which I bet isn't good for the original equation, and 2/3. I don't have time to try the whole thing though, I'll tell you tomorow. Quote Link to comment Share on other sites More sharing options...
Pyrotex Posted June 13, 2006 Report Share Posted June 13, 2006 oh so not a flying bug :hihi: yeah 2 possible routes at 14 but dont see any lessThere are actually FOUR possible routes. This can be most easily seen by unfolding the room into six flat rectangles, keeping their attachments so that the distance from A to B is unbroken. There are four ways of doing this, according to how you "unfold" the wall with the point B. See diagram. In upper left we have AB = 14.0 mIn upper right we have AB = SQRT( 13^2 + 5^2 ) = SQRT( 194 ) = 13.9 mIn lower left we have AB = SQRT( 12^2 + 8^2 ) = SQRT( 208 ) = 14.4 mIn lower right we have AB = SQRT( 13^2 + 11^2 ) = SQRT( 290) = 17.0 m There is a shorter distance from A to B, by going across TWO of the long walls instead of ONE. On second thought, there are only THREE routes. Note the 4th one below, crosses some "empty space" near B. Ignore that one. Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted June 14, 2006 Report Share Posted June 14, 2006 yeah I see it now.. thats what you get for rushing! But I did mean I see to at length of 14 (also saw a few more than 14) but couldnt see any less. Very tricky problem :shrug: Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted June 14, 2006 Report Share Posted June 14, 2006 Yeah, I only thought of the unfoldings after catching my commuter train! :hihi: The one I had in mind yesterday was the lower left and indeed my calculations got [math]\small 4\sqrt{13}[/math] but instead of unfolding and then just drawing a straight line, I was using a parameter and minimizing. All roads lead to Rome. ;) Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 14, 2006 Report Share Posted June 14, 2006 There are actually FOUR possible routes. This can be most easily seen by unfolding the room into six flat rectangles, keeping their attachments so that the distance from A to B is unbroken. There are four ways of doing this, according to how you "unfold" the wall with the point B. See diagram. In upper left we have AB = 14.0 mIn upper right we have AB = SQRT( 13^2 + 5^2 ) = SQRT( 194 ) = 13.9 mIn lower left we have AB = SQRT( 12^2 + 8^2 ) = SQRT( 208 ) = 14.4 mIn lower right we have AB = SQRT( 13^2 + 11^2 ) = SQRT( 290) = 17.0 m There is a shorter distance from A to B, by going across TWO of the long walls instead of ONE. On second thought, there are only THREE routes. Note the 4th one below, crosses some "empty space" near B. Ignore that one. Simply wonderful; wonderfully simple? Either way, you rock Pyro. Having squashed the bug in the box, may I ask you to employ your method to a bug outside a box? Your unfolding diagrams have a similarity to my prints from rolling an inked box as if unfolding sides in order. Here are photos of some of the prints:http://hypography.com/gallery/showimage.php?i=406&c=3&userid=796http://hypography.com/gallery/showimage.php?i=407&c=3&userid=796I have never satisfied myself as to a solution asking essentially how many ways can the bug walk from one face to another & visit each face only once?The thread discussing this, including a photo of the box & I think my last guestimate of 72 different bug routes (so to speak) is here:http://hypography.com/forums/social-sciences/1606-mathemagical-box-mind-experiment.html?highlight=mathemagical+box Now where's that flyswatter?:shrug: Quote Link to comment Share on other sites More sharing options...
ughaibu Posted June 14, 2006 Report Share Posted June 14, 2006 How long are the bug's legs and how much does it weigh? Is one looking for the distance travelled by it's centre of gravity? Quote Link to comment Share on other sites More sharing options...
Turtle Posted June 14, 2006 Report Share Posted June 14, 2006 How long are the bug's legs and how much does it weigh? Is one looking for the distance travelled by it's centre of gravity?You misunderstood my question as needing a distance solution whereas it requires a number-of-possible-paths solution.:shrug: Quote Link to comment Share on other sites More sharing options...
ughaibu Posted June 14, 2006 Report Share Posted June 14, 2006 Turtle: From post 1: "Can the bug pass the distance less than 14 meters?" Apologies for the confusion. Quote Link to comment Share on other sites More sharing options...
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