 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?

 See The Solution Submitted by Salil Rating: 2.5000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 1 of 3

Let H be the height of the pole.
Let x be the length of the upper part of the pole that breaks off in the first case.
Let D be the distance from the base in the second case.

That gives us:

20^2 + (H-x)^2 = x^2 for the first case, and

D^2 + (2H - 3x)^2 = (3x - H)^2 for the second case.

Simplifying these equations we get:

400 + H^2 - 2xH = 0 for the first case, and

D^2 + 3H^2 - 6xH = 0 for the second case.

Multiplying the first equation by three and setting them equal to each other gives:

D^2 = 1200

D = 34.64

So the pole strikes 34.64 feet from the base the second time.

 Posted by tomarken on 2006-06-12 09:30:47 Please log in:

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