 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  The Maypole Problem 2 (Posted on 2006-06-12) Continuing on the original problem here, posted by Jer.

During a gale a maypole was broken in such a manner that it struck the ground at a distance twenty feet from the base of the pole.

It was repaired without any gain or loss in its length. Later it broke a second time at a point below the earlier by an amount exactly equal to the difference in length of the two sections of the first break.

How far did it strike this time from the base?

 Submitted by Salil Rating: 2.5000 (2 votes) Solution: (Hide) Let the length of two sections during the first break be X & Y ( top & bottom respectively ) hence X^2 - Y^2 = 400 or ( X + Y ) * ( X - Y ) = 400 during the second the height of the sections are X+( X -Y ) & Y-( X -Y ) respectively let the distance at which it will hit the ground be D Then D^2 = ( X+ ( X-Y ))^2 - ( Y - ( X -Y ))^2 = ( X + Y ) ( 3 *( X-Y )) = 3* ( X+Y) ( X-Y) = 1200 D = SQRT (1200) = 34.64 Subject Author Date Puzzle Solution K Sengupta 2007-05-11 15:09:05 Solution Bractals 2006-06-12 10:33:51 Solution tomarken 2006-06-12 09:30:47 Please log in:

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