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Cover the Circumference (Posted on 2010-05-25) Difficulty: 4 of 5
Start with a unit circle and draw four line segments each of length L. The first line is a chord: it begins and ends on the circle. Subsequent lines start from the midpoint of the previous line and end on the circle. If the fourth line ends where the first line begins, what is L?



Note that this sketch is an approximation.

No Solution Yet Submitted by Larry    
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Partly analytic (spoiler) | Comment 3 of 4 |
Let r1......r5 denote the radii to the points A, H, I, J, K(mid-point of JA).

Using the cosine rule in triangles HCO and ICO and then eliminating angle C

gives:   r32 = (2 + 2r22 - L2)/4


The same recurrence relation will connect any adjacent pair of radii, so,

starting with r1 = 1 gives:

r12 = 1,   r22 = 1 - L2/4,   r32 = 1 - 3L2/8,   r42 = 1 - 7L2/16,   r52 = 1 - 15L2/32

Let Q1, Q2, ... Q5 denote the angles between successive radii r1,....r5,r1 ,

cosQn = (rn2 + rn+12 - L2/4)/(2rnrn+1) , which gives:

            cosQ1 = sqrt(1 - L2/4)

            cosQ2 = (1 - 7L2/16)/sqrt[(1 - L2/4)(1 - 3L2/8)]

            cosQ3 = (1 - 17L2/32)/sqrt[(1 - 3L2/8)(1 - 7L2/16)]

            cosQ4 = (1 - 37L2/64)/sqrt[(1 - 7L2/16)(1 - 15L2/32)]

            cosQ5 = (1 - 23L2/64)/sqrt(1 - 15L2/32)

Now all we need to do (!) is solve the equation Q1 + Q2 + Q3 + Q4 + Q5 = 2*pi

Sadly, I can’t find an exact solution for L. But my Maple software finds

L = 1.38247107.., which agrees with Charlie’s previous posting.


There are some interesting patterns in the above cosines, and I notice that the product of all five cosines contains no square roots, so there may be something still to discover.

  Posted by Harry on 2010-05-26 22:21:28
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