So we get the exact radii and ratios to confirm our suspicion of the ratio being the square root of 2. This suggests that if we take the extensions to more circles the ratios would have interesting ratios. Try it and see!!!

Input := 


solA = Solve[{Area3[R2] == 1/3 Pi (2 R)^2, 

				Area1[R1] == 1/3 Pi (2 R)^2},

						{R1, R2}] 

General::spell1: 

   Possible spelling error: new symbol name "solA"

     is similar to existing symbol "sol".



;[o]

General::spell1: 

   Possible spelling error: new symbol name "solA"

     is similar to existing symbol "sol".

Output =


               2             -2 R

{{R1 -> -(Sqrt[-] R), R2 -> -------}, 

               3            Sqrt[3]

                2              2 R

  {R1 -> -(Sqrt[-] R), R2 -> -------}, 

                3            Sqrt[3]

              2            -2 R

  {R1 -> Sqrt[-] R, R2 -> -------}, 

              3           Sqrt[3]

              2             2 R

  {R1 -> Sqrt[-] R, R2 -> -------}}

              3           Sqrt[3]

Input := 


R2/R1/.solA[[4]]

Output =


Sqrt[2]