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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.)
 Five and Six Planks | Comment 5 of 6 |

For these cases, I am only detailing the best ones.  Again, I am assigning unit length planks and letting the circle's radius change.

Five 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 on MC such that NO is perpendicular to MC.  Planks three and four form an identical structure rotated 180 degrees from the first two planks.  The fifth plank will span the center between the two structures, and O will be its midpoint.

Since the structure is symmetrical, only one half needs to be examined.  AM=MB=NO=1/2  Let the radius be r.  Then AO=CO=r.  Let MO=x and MN=y.  Then CN=1-y.  MNO, CNO, and AMO are all right triangles.  By the Pythagorean Theorem:
AMO: r^2 = (1/2)^2 + x^2
MNO: x^2 = y^2 + (1/2)^2
CNO: r^2 = (1/2)^2 + (1-y)^2
The equations for AMO and MNO can be combined to form r^2 = 1/2 + y^2.  This with the equation from CNO gives y=3/8 and r=sqrt(41)/8=0.80039

Five planks Plan 2
Place three planks on the circumfrence: AB, CD, and EF.  Let the midpoint of AB be M and the midpoint of CD be N.  The fourth plank joins M and N.  Let its midpoint be P and let the midpoint of EF be Q.  The fifth plank is perpendictular to MN and EF, joins P and Q, and crosses the center O.

AM=MB=CN=ND=EQ=QF=MP=PN=1/2  Let the radius be r.  Then AO=DO=EO=r.  Let MO=NO=QO=x.  Then PO=1-x.  EQO and MPO are both right triangles.  By the Pythagorean Theorem:
EQO: r^2 = x^2 + (1/2)^2
MPO: (1-x)^2 + (1/2)^2 = x^2
The last equation gives x=5/8.  Then r=sqrt(41)/8=0.80039

Six Planks Plan 1
Place three planks on the circumfrence: AB, CD, and EF.  Let the midpoint of AB be M and the midpoint of CD be N.  The fourth plank joins M and N.  Let P be the midpoint of EF.  Then the fifth plank goes from P to a point on the circumfrence G.  Let Q be the midpoint of MN and R be the midpoint of GP.  The last plank joins Q and R, contains the center O, and is perpendicular to MN and GP.

AM=MB=CN=ND=EP=PF=MQ=QN=1/2  Let r be the radius.  Then AO=DO=FO=GO=r.  PR=3/8 and RG=5/8, similar to previous constructions.  Let OR=y and ON=z.  Then OQ=1-y.  GRO, NQO, and DNO are all right triangles.  By the Pythagorean Theorem:
GRO: r^2 = y^2 + (5/8)^2
NQO: z^2 = (1-y)^2 + (1/2)^2
DNO: r^2 = z^2 + (1/2)^2
The last two equations can be combined to form r^2 = (1-y)^2 + 1/2.  Then setting the two expressions for r^2 equal yields y=71/128.  Then r=sqrt(11441)/128=0.83565.

Six planks Plan 2
Place three planks on the circumfrence: AB, CD, and EF.  Let the midpoint of AB be M, the midpoint of CD be N, and the midpoint of EF be P.  The fourth plank goes from M to the circumfrence point G, between B and C.  Let Q be the point on MG closest to the center.  Then the fifth plank connects Q and N.  Let R be the point on QN closest to the center.  Then the sixth plank joins P and R, is perpendicular to QN and EF, and passes through the center O.

AM=MB=CN=ND=EP=PF=1/2  MQ=3/8 and QG=5/8, just like in previous constructions. Let r be the radius. Then AO=DO=EO=r.  Let PO=x.  Then MO=NO=x and RO=1-x.  Let QO=y and QR=z.  Then RN=1-z.  EPO, NRO, QRO, and MQO are all right triangles.  By the Pythagorean Theorem:
EPO: x^2 + (1/2)^2 = r^2
NRO: (1-z)^2 + (1-x)^2 = x^2
QRO: z^2 + (1-x)^2 = y^2
MQO: y^2 + (3/8)^2 = x^2
The second equation simplifies to x = z^2/2 + 73/128.  The third and fourth equations combine to make z^2 + (1-x)^2 = x^2 - (3/8)^2.  These last two equations give z=55/128.  Back substituting results in r=0.83011

Summary:  Interestingly, the best two plans for five planks end up with the same r=0.80039.  The best plan for six planks with r=0.83565 had the more even distribution of planks on either side.  This follows the idea behind the best plan for four planks with the planks forming a balanced branching structure going outward from the center.  For Larry's constant radius of 1, five planks have a length of 1.24939 and six planks have a length of 1.19667.

 Posted by Brian Smith on 2010-06-05 17:38:43

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