The Spider and the FlyDate: 12/23/1999 at 02:52:24 From: keith Subject: Algebra 1 honors Hi: I have not tried this question myself because I don't know how to start the problem. Here it is: A spider and a fly are on opposite walls of a rectangular room. The spider asks the fly if it can come over and "visit" the fly. The fly believes that the shortest distance, walking on one of the room's surfaces, is 42 feet. (That is the distance if the spider walks straight up the wall, over the ceiling, and straight down the opposite wall.) So the fly agrees to the visit, provided the spider can find a path that is shorter than 42 feet. Use the "unfolded room" and the Distance Formula to explain the fly's fatal miscalculation. Any help given would be greatly appreciated. Date: 12/23/1999 at 09:41:24 From: Doctor Rob Subject: Re: Algebra 1 honors Thanks for writing to Ask Dr. Math, Keith. Whether the fly lives or dies depends on the location of the spider and the fly on their respective walls, which you have not told in the above description. If both are centered horizontally between the side walls, the fly is D feet below the ceiling, the spider is d feet below the ceiling, the distance between their walls is L, the width of the room is W, and its height is H, then you know that L + D + d = 42, or L = 42 - d - D. If you unfold the room this way, H o---------o | | | W/2| | | W| F---o | D | | | | | o-------------o---------o-------------o---------o | W | H | W | H | | | | | | | | | | | | | | | | | | | | | |L Floor |L |L Ceiling |L |L | | | | | | | | | | | | | | | | | | | | | W | H | W | H | o-------------o---------o-------------o---------o | | | | |H |H | S | | d| W/2 | o------o------o W and draw a diagonal line from S to F, this might be shorter than 42 feet. The horizontal distance from S to F in this diagram is W/2 + (H-D), and the vertical distance is (H-d) + L + W/2. You figure out using the distance formula (or the Pythagorean Theorem) if this is possible. If, however, you unfold the room this way, W o-------------o | | | Fly | |H F |H | |D | | W/2 | | o-------------o---------o-------------o---------o | W | H | W | H | | | | | | | | | | | | | | | | | | | | | |L Floor |L |L Ceiling |L |L | | | | | | | | | | | | | | | | | | | | | W | H | W | H | o-------------o---------o-------------o---------o | | | | |H Spider |H | S | | d| W/2 | o------o------o W and draw a diagonal from S to F, this also might be shorter than 42 feet. The horizontal distance is now W/2 + H + W/2, and the vertical distance is (H-d) + L + D. Again, you should figure out using the distance formula (or the Pythagorean Theorem) whether or not this is possible. HINT: Replace L by 42 - d - D first. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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