
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 ?

Are you in Calculus?minimizing this with calculus is easy.but if you are in geometry I have that solution also=]..

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).

Say Man, Do Your Own Homework!!!! You Just Trying To Hide This Question As A "Puzzle"!! Good Try Though.

The bridge can be built anywhere as long as it is between the houses.

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 cx. Write and expression for the sum of the hypotenuses of the two triangles:f(x) = sqrt(a^2 + x^2) + sqrt(b^2 + (cx)^2)find f'(x) and set it equal to zero to find the maximaf'(x) = x/(sqrt(a^2 + x^2))  (cx)/(sqrt(b^2 + (cx)^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.

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.

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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