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A nice puzzle?

This is a discussion on A nice puzzle? within the Puzzles forums, part of the Freshers Zone category; Suppose there is a river of width w. And two houses A and B on the opposite sides of the ...


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Puzzles A puzzle is a problem or enigma that challenges ingenuity. Puzzles are often contrived as a form of entertainment, but they can also stem from serious mathematical or logistical problems, All Puzzle related Information and q&a

 
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  #1  
Old 06-05-2007, 07:47 PM
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Default A nice puzzle?

Suppose there is a river of width w. And two houses A and B on the opposite sides of the river. They are not exactly opposite to the river but at a distance c along the river.Now, A is at a distance a from river and B is at a distance b from the river. (If you can't understand wat c is, take it this way .. if A walks a distance a towards the river and then turns 90 degrees and walks c along the river, then turns back to the original direction and walks a distance b, he will reach B).Now, A and B plan to build a bridge on the river. Bridge has to built perpendicular to the river. Where should the bridge be built so that the walking distance from A to B is minimum ?
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Old 06-05-2007, 10:11 PM
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Are you in Calculus?minimizing this with calculus is easy.but if you are in geometry I have that solution also=]..
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Old 06-06-2007, 12:03 AM
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the distance is still the same no matter how you build the bridge.(if i understand your question correctly, after you walk c distance, you have to cross the river, and walk b distance toward B to reach B).
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Old 06-06-2007, 01:22 AM
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Say Man, Do Your Own Homework!!!! You Just Trying To Hide This Question As A "Puzzle"!! Good Try Though.
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Old 06-06-2007, 02:04 AM
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The bridge can be built anywhere as long as it is between the houses.
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Old 06-06-2007, 02:06 AM
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You need to minimize the sum of the two hypotenuses formed by the diagonal walk to the bridge. Clearly the bridge must lie between the two houses. Say the distance along the river from A that the bridge should lie is x, then the distance on the other side will be c-x. Write and expression for the sum of the hypotenuses of the two triangles:f(x) = sqrt(a^2 + x^2) + sqrt(b^2 + (c-x)^2)find f'(x) and set it equal to zero to find the maximaf'(x) = x/(sqrt(a^2 + x^2)) - (c-x)/(sqrt(b^2 + (c-x)^2)Now set it equal to zero and solve for x. One or more of the solutions will be maximum distances and the other will be the minimum distance.I would like to see some proof of Dr D's conjecture that the distance is minimized when the angles are identical. That's quite a leap of faith.
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Old 06-06-2007, 02:47 AM
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LOL, are A and B transformers? First they're houses and the next they could walk and build bridges. This question isn't making sense to me at all.
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Old 06-06-2007, 05:55 AM
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forget the width of the river. the time to cross the bridge is constant wherever the bridge is. Now draw a line between A and B. That is the shortest path.
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