-> -> Accepting the 20cm^2 circles must cover 36cms^2 only gets you....  
->   
-> -> 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   
->   
-> -> Setting the thrice to 3 cm^2 then has the total coverage rise....  
->   
-> -> Thrice  3             3.0  
-> -> Twice   1.0237947     3.0713841  
-> -> Once   14.952411     44.857233  
-> ->         Totalling... 50.9286171  
->   
-> -> Neither of these optimums fit the specifications given.  
->   
-> -> I'd conclude that the problem is geometrically impossible...  
->    
-> You've lost me...  
  
Okay. I understand now. 
  
But how did you calculate these figures? I assume you must have used a 
computer. But it also looks to me like you *assumed* that the centres 
of the circles are at the vertices of an equilateral triangle, and 
there is nothing in the problem that states that this must be so. So 
all you've proved is that it's impossible for the circles to be in the 
equilateral-triangle arrangement. Some other arrangement may be 
possible. 
  
The other possibility that I've tried is to have two circles exactly 
superimposed, and the third partly overlapping them. It's pretty 
obvious that this leads to the sum of the "Once" and
"Thrice" areas 
equalling 20 cm^2, which isn't going to be any good. 
  
The question is: Can *any* arrangement of the circles exactly fulfill 
the problem's description of the situation? If so, what? And if not, 
can we prove it? 
  
                             dow 
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