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?

I decided to let the planks have a constant length of 1 unit and let the size of the circle scale. These answers can be converted to plank lengths Larry is looking for by taking the reciprocal of the radii I give.

1 plank

Obvious, the plank equals the diameter.

Radius = 0.5

2 planks

Place one plank as chord AB. Let the midpoint of AB be M. The second plank is segment MC, with C being on the circle. The center of the circle, O, is on MC.

Let r be the length of the radius. Then AO=BO=CO=r, MO=1-r, and AM=MB=1/2. Then by the Pythagorean Theorem, r^2 = (1/2)^2 + (1-r)^2. That makes r=5/8=0.625

3 planks Plan 1

Place one plank as chord AB. Let the midpoint of AB be M. Place the second plank starting at M and ending on the circle at C. Let O be the center of the circle. Let N be the point on plank MC such than NO is perpendicular to MC. The third plank starts at N, passes through O, and ends on the circle at D.

Let r be the radius, let x be the length of MN, let y be the length of MO. Then NO=1-r and NC=1-x. Triangles AMO, MNO, CNO are all right triangles. By the Pythagorean Theorem:

AMO: (1/2)^2 + y^2 = r^2

MNO: x^2 + (1-r)^2 = y^2

CNO: (1-r)^2 + (1-x)^2 = r^2

The first two equations can be combined and simplified to (1-x)^2 = 2r-1

The third equation can be simplified to x^2 = 2r - 5/4

Solving the resulting equations for r gives r=89/128=0.6953125

3 planks Plan 2

Place the first two planks on opposite sides of the circle parallel to each other. Join their midpoints with the third plank. The four endpoints of the first two planks are the corners of a square with an edge of 1. Then the circle has a radius of sqrt(2)/2 = 0.7071

3 planks Summary: The second plan, which divides the supporing planks on either side of the final plank is the better plan with a radius of 0.7071

Final Summary: When expressed as short planks spanning Larry's unit circle:

1 plank has length 2

2 planks have length 8/5 = 1.6

3 planks have best length sqrt(2) = 1.4142