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Rectangle pegs in a round hole (Posted on 2019-04-28) Difficulty: 4 of 5
What is the maximum fractional coverage of the area of a circle by three, non-overlapping rectangles, all of which lie internal to the circle? (The rectangles can be of different sizes)

No Solution Yet Submitted by Danish Ahmed Khan    
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Confirming Comment 3 of 3 |
As a check, I ran my approach out to the bitter end. First I differentiated my area function and set it to zero. However the result was so horrific to solve, I instead opted to just test the function to see if it peaked at sqrt(2-2/sqrt(5)) as found by B. Smith. It did (see below). So his approach (extending the rotational symmetry of a circle to simplify the problem) worked beautifully. 

        program yyyo

        implicit real*8 (a-z)

        integer ib

        x2b=sqrt(2.d0-2.d0/sqrt(5.d0))

        xa=2.d0*(sqrt(5.d0)-1.d0)

        do ib=52571,52576

        b=ib/100000.

        term1 =  b*(1-b**2)**0.5d0

        term2a = ( (b+ sqrt(b**2+8.d0) )/4.d0 )**2.d0

        term2 =   sqrt(1.d0 - term2a)

        term3 = (b+  sqrt(b**2.d0+8.d0) )/4.d0 -b

        a = 4.d0*(term1+term2*term3)

        print*,x2b,2*b,xa,a

        enddo

 sqrt(2-2/sqrt(5))   trial height      2 (sqrt(5)-1)         resulting area  

 1.05146222423  1.05141997337  2.47213595499  2.47213595356   1.05146222423  1.05144000053  2.47213595499  2.47213595460   1.05146222423  1.05146002769  2.47213595499  2.47213595499   1.05146222423  1.05148005485  2.47213595499  2.47213595474   1.05146222423  1.05149996280  2.47213595499  2.47213595385       

Comparisons of trial heights and resulting areas. The peak occurs on the 3d line.  


Edited on April 29, 2019, 4:19 pm
  Posted by Steven Lord on 2019-04-29 14:54:14

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