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Topic: Fellow Travellers (Read 1841 times) |
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ThudnBlunder
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The dewdrop slides into the shining Sea
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Fellow Travellers
« on: Jan 26th, 2009, 9:45pm » |
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There are four straight roads on a flat plain, none of which are parallel. Nor do any three pass through the same point. Along each road plods a traveller at a constant speed, but not necessarily at the same speed as the others. It is known that i) traveller 1 met travellers 2, 3, and 4 ii) traveller 2 also met travellers 3 and 4 Prove that travellers 3 and 4 also met.
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« Last Edit: Jan 26th, 2009, 9:45pm by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
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Re: Fellow Travellers
« Reply #2 on: Jan 27th, 2009, 12:22pm » |
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on Jan 27th, 2009, 8:23am, Grimbal wrote: Fellow Travellers seeks a proof that H will surely meet E if velocities remain constant, whereas Ghost Ships seeks to ensure, by varying velocity if necessary, that they will not meet. Are these problems equivalent then? Four ghostly galleons – call them E, F, G and H, – sail on a ghostly sea so foggy that visibility is nearly zero. Each pursues its course steadily, changing neither its speed nor heading. G collides with H amidships; but since they are ghostly galleons they pass through each other with no damage nor change in course. As they part, H’s captain hears G’s say “Damnation! That’s our third collision this night!” A little while later, F runs into H amidships with the same effect (none) and H’s captain hears the same outburst from F’s. What can H’s captain do to avoid a third collision and yet reach his original destination, whatever it may be, and why will doing that succeed?
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Grimbal
wu::riddles Moderator Uberpuzzler
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Re: Fellow Travellers
« Reply #3 on: Jan 27th, 2009, 1:58pm » |
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Sorry, you are right. It is not quite the same. Travelers is some kind of prelude to the ghost ships, i.e. prove that there is a problem before seeing what can be done about it. Here is a sketch of a proof: Considering the problem in 3D, the 3rd dimension being time. There is a unique plane that contains the trajectory of travelers 1 and 2. If travelers 3 and 4 met 1 and 2, they must evolve in the same plane. And therefore, since their paths are not parallel, they must meet at some place and some time.
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