In the figure shown above, 3 unit circles are squeezed into the smallest square possible.
If the ratio of the area of the yellow region to the area of the white region can be expressed as
where are all positive integers with , find .
Hint: The first step is to find the square side length.
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R i g h t v e r t i c a l s i d e o f t h e s q u a r e = r a d i u s o f O 1 + p r o j e c t i o n o f c e n t e r d i s t a n c e O 1 O 2 , 1 5 o t o v e r t i c a l + r a d i u s o f O 2 . W h i t e a r e a = ( a r e a o f e q u i l a t e r a l t r i a n g l e o f c e n t e r s O 1 O 2 O 3 ) − ( s u m o f a r e a s o f ‘ 3 c o l o r e d s e c t o r s o f t h e 3 c i r c l e s ) T h e R a t i o = w h i t e a r e a g r e e n a r e a = w h i t e a r e a ( a r e a o f t h e s q u a r e ) − ( W h i t e a r e a ) − ( s u m o f a r e a s o f t h e t h r e e c i r c l e s . ) = 3 − π / 2 ( 6 + 2 ∗ 2 + 2 6 + 3 ) − ( 3 − π / 2 ) − ( 3 ∗ π / 2 ) = 1 2 − π 1 2 + 3 2 + 9 6 + − π − 6 ∗ π = d − π a + c + b − 5 ∗ π ∴ c d a b = 1 2 ∗ 3 2 1 2 ∗ 9 6 = 3 .