->  DW> "Three circles are drawn on a flat surface. Each circle
encloses an 
->  DW> area of 20 cm^2. They overlap so that the total area that is enclosed 
->  DW> by one or more circles is 36 cm^2. The area that is enclosed by all 
->  DW> three circles is 3 cm^2. What is the total area that is enclosed by 
->  DW> exactly two circles?" 
 
->  DW> The more I play with it, the more interesting it gets. Any thoughts 
->  DW> about it, from anyone, would be welcome. 
 
-> Mmmm - the use of `circles' and `cm^2' is misleading for this  
-> problem can be the province of combinatories, set theory or  
-> Boolean algebra.... 
  
Diagrams that look like the one in my stepson's math homework are used 
in various branches of math. However, the question he had to answer was 
a simple geometrical one, with no underlying significance. It dealt 
only with overlapping circles and enclosed areas. 
  
-> Ergo,  the `total area enclosed by exactly two circles' 
-> is 18 (cm^2),  giving you the following distribution of  
-> coverages... 
 
-> Amount Covered By... 
-> 3 circles =  3 cm^2 
-> 2 circles = 18 cm^2 
-> 1 circle  = 15 cm^2 
-> ~~~~~~~~~~~~~~~~~~~ 
-> Total     = 36 cm^2 
 
-> This all fits the logic of the problem;  I haven't tried to see if  
-> one can actually draw a true-to-scale Venn diagramme with real  
-> circles of it.... 
  
I agree that the answer is 18 cm^2, *if* there is an answer. However, 
the question is: Is it possible to overlap three circles, each with an 
area of 20 cm^2, in such a way that the total enclosed area is 36 cm^2 
and the area enclosed by all three circles is only 3 cm^2? 
  
I have proved that the symmetrical case, in which the centres of the 
circles are at the vertices of an equilateral triangle, is NOT 
possible. Another simple case, in which two circles are exactly 
superimposed and the third circle partly overlaps them, also fails. I 
have developed a strong suspicion that, if the circles overlap 
sufficiently to reduce the total area from 60 to 36 cm^2, then the area 
covered by all three will always be *more than* 3 cm^2. However, I have 
not yet been able to prove this in the general case. 
  
Any ideas about this? 
  
                         dow 
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