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 Cover the Center (Posted on 2010-06-04)
Can you cover the center of a unit circle, such a Indi's circular pit?

You have a number (N) of identical planks of length L (0 < L <= 2). The planks have small width and negligible thickness, but very high strength and rigidity. No plank extends outside the circle. Both endpoints of each plank, with the exception of the N-th plank, must rest on either the circle or another plank. Neither weaving of planks nor cantilever designs are allowed. Covering the center means that the N-th plank crosses over the center of the circle.

For a given N, L is the minimum length plank necessary.
Obviously if N=1, then L=2.

What is the smallest L when N=2? N=3? N=4?

For larger N, is there an optimum algorithm to minimize L?

Can you determine a general relationship between N and L?

 No Solution Yet Submitted by Larry No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Four Plank Analysis Comment 6 of 6 |
(In reply to Four Plank Analysis by Brian Smith)

I have not yet computed the length, but I believe the minimal plank length for N=4 can be found by finding a chord of the circle where a concentric circle within the circle is tangent to the midpoint of that chord to which itself can have a chord of the same length parallel to the first chord to which a line segment through the center of the circle from the midpoint of the inner chord to a point to the outer circle is equal in length to both chords.
The line segment and inner chord would represent two of the planks with the end points of the plank of the inner chord connecting to the middle of the two other planks which are tangent to the virtual inner circle and its chord.

The solution may be similar to one of your solutions in your four plank analysis, but my brain feels scrambled right now so I have not reviewed the four plank analyzations you presented.

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Using a spreadsheet, the Pythagorean Theorem, and a range of chord lengths from 1 to 2, I found that the plank length would be approximately 1.43143. This value is higher than yours. I am not sure I understood your method(s).

Edited on June 7, 2010, 4:21 am
 Posted by Dej Mar on 2010-06-06 15:49:09

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