perplexus.info :: Geometry : The Maypole Problem 2

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

Comments: ( You must be logged in to post comments.)

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:

Login:
Password:
Remember me:

Sign up! \| Forgot password

Search:

Search body:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info