-> Well, I've succumbed and determined the radius of one circle to be  
-> 2.5231325 odd cms before calculating the angle of a chord whose  
-> area was 4.5cms^2  -  half of the overlap of 9cms^2 needed between  
-> any two squares  -  at 126.84910 degrees or so. 
 
-> Exhuming ancient instruments (originally my grandfather's), I drew  
-> me a 20cm^2 circle with radius 2.52cms,  constructed two radii at  
-> 126.85 degrees and drew in the chord.  It looked good... 
 
-> Next, I used my compasses to locate the centre of the second  
-> circle from the bases of the chord,  and drew that in.  Still  
-> looked good - about 9cms in the overlap,  and about 11cms on each  
-> side. 
 
-> My equilateral now had sides of about 2.3cms,  and the 3rd circle  
-> fitted physically. 
 
-> However, writing in the logical cm^2's of each segment shows  
-> clearly a very bad fit with the problem as set. 
 
-> The outer single-coverage segments of `5cms' are obviously much  
-> larger than the double-coverage `6's - which are individually  
-> smaller than the central `3' of treble coverage. 
  
I wrote a computer program. First, it calculated the angle that would 
be subtended at the centre of any circle by an area of overlap with an 
adjoining circle that had an area of 9 cm^2, each circle having an area 
of 20 cm^2. I calculated that the angle is the solution (in radians) of 
the equation: 
  
(9/20) * pi - A + sin(A) = 0 
  
A is the angle, of course. 
  
The thing found the angle iteratively. It came to about 2.1 radians, 
but the program calculated it to 15 significant digits. 
  
Then it "imagined" the situation in which three circles, with their 
centres at the vertices of an equilateral triangle, all overlapped to 
this extent. It calculated the area of overlap between any pair of 
circles, which correctly came to 9 cm^2 (with an absolutely teeny 
round-off error). Then it calculated the area where all three circles 
overlapped. It came to about 5.7 cm^2, i.e. *far* more than the 3 cm^3 
specified in the question. 
  
So I conclude that it is impossible for three equal circles, with their 
centres at the vertices of an equilateral triangle, to overlap so that, 
simultaneously, the total area enclosed is 36 cm^3 and the area of 
"triple overlap" is 3 cm^3. 
  
I also thought about having two circles exactly on top of each other, 
and the third partly overlapping. This *almost* works. If the total 
area is 36 cm^3, then the triple-overlap area is 4 cm^3. Close, but no 
cigar. 
  
But I suspect this is the best possible arrangement. All others produce 
a triple-overlap area that is greater than 4 cm^2. 
  
But how to prove it... 
  
                           dow 
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