Question 237420
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I have to assume that the 250 foot distance is the distance along the ground from the base of the cathedral tower to the base of the church tower.  Otherwise, this problem gets ludicrously complex.


So if the pigeons leave at the same time and fly at the same speed, they must fly the same distance.  That means that there is a point on the ground along that 250 foot distance that is equidistant from a 200 foot tower and a 150 foot tower.


Let's call the distance from the base of the cathedral tower to the point where the pigeons meet *[tex \Large x].  Now we have a right triangle with a leg that is 200 feet and a leg that is *[tex \Large x] feet.  That means that the hypotenuse squared is *[tex \Large x^2\ +\ 200^2]


On the other side of the picture, the distance from the meeting point to the church tower must be *[tex \Large 250\ -\ x].  Again we have a right triangle with legs 150 and *[tex \Large 250\ -\ x], so the hypotenuse squared must be *[tex \Large (250\ -\ x)^2\ +\ 150^2].


These two hypotenuses are the flight paths of the two pigeons that we decided earlier must be equal in length.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 200^2\ =\ (250\ -\ x)^2\ +\ 150^2]


Just solve for *[tex \Large x] which we very cleverly defined as the distance the problem asks you to discover.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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