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?
As before, I am assigning unit length planks and letting the circle's radius change.
4 planks Plan 1
Place the first plank with endpoints AB. Let M be the midpoint of AB. Then place the second plank starting at M and ending at C, on the circle. Let O be the center of the circle. Let N be the point such that NO is perpendicular to MC. Then place the third plank starting at N and ending on the circle at D. Let P be the point on ND such that PO is perpendicular to ND. The fourth plank starts at P, passes through O and ends on the circle at E.
AM=MB=1/2. Let the radius be r. Then AO=OE=OD=r and OR=1-r. Let MN=x, MO=y, NO=z, and NR=w. Then NC=1-x and RD=1-w. AMO, MNO, NRO, ORD, and CNO are all right triangles. By the Pythagorean Theorem:
AMO: r^2 = (1/2)^2 + y^2
MNO: y^2 = x^2 + z^2
CNO: r^2 = z^2 + (1-x)^2
ORD: r^2 = (1-r)^2 + (1-w)^2
NRO: z^2 = w^2 + (1-r)^2
The first three equations can be combined to show x=3/8, independent of the radius. Substituting x back into the third equation allows the last three equations to show w=39/128. Then the fourth equation can be solved with r=24305/32768=0.74173
4 planks Plan 2
Place the first two planks like before. Points A,B,M,C,N,O will be the same. The third plank will be laid parallel to the second plank on the opposite side of the circle. Its endpoints are D and E and its midpoint is P. The fourth plank is perpendictular to MC and ED, joins points N and P, and passes through O.
AM=MB=1/2. MN=3/8 and NC=5/8, like in part 1. Let r be the radius. Then AO=CO=EO=r. Let NO=x. Then RO=1-x. Triangles CNO and ERO are both right triangles. By the Pythagorean Theorem:
CNO: x^2 + (5/8)^2 = r^2
ERO: (1-x)^2+ (1/2)^2 = r^2
Setting the expressions for r^2 equal gives x=55/128. Then r=5*sqrt(377)/128=0.75846
4 planks Plan 3
Place the first two planks on the circumfrence of the circle, sharing a common endpoint. The two planks will be AB and BC. Take point D on the circle so that BD is a diameter, passing through the center O. Place the third plank so that its endpoints are on AB and BC and is perpendicular to diameter BD. Let M be the end on AB and let N be the end on BC. Let P be the midpoint of MN, which is also where MN intersects BD. RD is the fourth plank's location.
MR=RN=1/2. Let r be the radius. Then AO=BO=DO=r, RO=1-r, and BR=2r-1. Let AM=y. Then MB=1-y. Triangles BRM and MRO are both right triangles. By the Pythagorean Theorem:
BRM: (2r-1)^2 + (1/2) = (1-y)^2
CRO: (1-r)^2 + (1/2) = x^2
Let angle OBA be theta. Then by the Law of Cosines:
ABO: r^2 = 1^2 + r^2 - 2*1*r*cos(theta)
MBO: x^2 = (1-y)^2 + r^2 - 2*(1-y)*r*cos(theta)
The first cosine equation simplifies to cos(theta) = 1/(2r). Then the second cosine equation becomes x^2 = (1-y)^2 + r^2 - (1-y).
This can be combined with the equation from triangle CRO to form -2r+5/4 = y^2-y. The equation from BRM can be rearranged to form 4r^2-4r+1/4 = y^2-2y. Then y = -4r^2+2r+1
Substituting the expression for y into one of the equations results in a cubic equation for r of 16r^3-16r^2+4r-5/4. This has a positive root at r=0.73313
4 planks Plan 4
Place the first two planks on the circle's circumfrence, but not sharing an endpoint. Let AB be the first plank with midpoint M. Let CD be the second plank with midpoint N. The third plank joins the midpoints M and N. Let the midpoint of MN be P. The fourth plank starts at P, passes through the center O, and ends on the circle at E.
AM=MB=CN=ND=MP=PN=1/2. Let the radius be r. Then AO=EO=r and RO=1-r. Let MO=x. AMO and MRO are both right triangles. By the Pythagorean Theorem:
AMO: (1/2)^2 + x^2 = r^2
MRO: (1-r)^2+ (1/2)^2 = x^2
From these equations, r=3/4=0.75
4 plank Summary: The smallest radius was from plan 3, where I abandoned the idea of always placing endpoints of new planks on the spot of the previous plank closest to the center. The best result was from plan 2, r=5*sqrt(377)/128=0.75846, where the supporting planks were divided 2/1. This best result is equivalent to a plank length 1.3185 for Larry's unit radius circle.