G'morning David, 

 DW> You've lost me...

Sorry ...

We agreed that the `set algebra' for your problem, given 20cm^2 
circles and a central overlap of 3cm^2 comes up with...

Cover  Segment  Total
Thrice   3        3
Twice    6       18
Once     5       15
       Total =   36

Spurred by your quest for proof that this could not be drawn on a 
flat surface,  I solved the set algebraic formula for the problem, 
using two different scenarios;  

First, I accepted that the 20cm^2 circles must cover 36cms^2 but 
left the 3cm^2 constraint out. This only gets me....

 -> 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

And here the optimum now covers 51cm^2, nothing close to the 36 in 
your problem.  Trying to draw this doesn't succeed - but the 
equilateral triangle now has sides equal to 3cm....

To me,  the original problem was crafted using set logic, 
expecting geometricity to follow - unjustifiably.

What is now unclear to me is why set algebra is so at odds with 
the geometry - I've gone through my texts here without finding a 
mention of such an alarming lack of correspondence at all....

I double-checked by considering a probability problem,  worked out 
the interactions of 3 20% conditions...

All Three -  0.8%
Any Two   -  9.6%
Just One  - 38.4%
Total     - 49.8%

....remembering I failed to do this years ago;  sure enough, 
attempting to set %'s = cm^2 in a true geometric rendering just 
dont work... 

Ask for a model answer, drawn on a flat surface...

:-)
 
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