->  -> Cover  Segment Cm^2  Total Cm^2 
 
->  -> Thrice  1.1572999     1.1572999 
->  -> Twice   7.2284667    21.6854002 
->  -> Once    4.3857666    13.1572999 
->  ->         Totalling... 36.0000000 
 
-> Here the optimum area covered by all three circles is 1.16cm^2,  
-> NOT 3 as set.  Worst - when I draw it, the equilateral triangle  
-> has sides equal to the radii,  and the centre section is obviously  
-> close to 5cm^2  - NOTHING like 1,16.... 
 
-> Next, setting the thrice to 3 cm^2 and leaving the total area  
-> unconstrained then has that total coverage rise.... 
 
->  -> Thrice  3             3.0 
->  -> Twice   1.0237947     3.0713841 
->  -> Once   14.952411     44.857233 
->  ->         Totalling... 50.9286171 
  
The situation in which two circles are exactly superimposed and the 
third partly overlaps them leads to solutions that are *almost* 
correct: 
  
Thrice: 4 
Twice: 16 
Once:  16 
Total: 36 
  
Or: 
  
Thrice: 3 
Twice: 17 
Once:  17 
Total: 37 
  
In the first case, the "Thrice" number is just 1 too large, and in the 
second, the Total is 1 too large. 
  
I have a feeling that this is the best that can be done. But can we 
prove that? 
  
                             dow 
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