I'll give this one a try....these are always interesting......Each circle has an area of pi....so the middle area must be pi/2
See the following pic....
Let circle A be centered at (0, 0)
We need to first solve this.....
Area of sector ABECA - area of triangle ABC = pi/4 so we have
(1/2)[Θ - sin(Θ)] = pi/4.......with a lttle help from WolframAlpha...Θ = about 2.30988 rads = about 132.3463751816785913° ...and (1/2) of this = 66.17318759083929565°
So the point B is given by :
[ cos(66.17318759083929565°), sin (66.17318759083929565°) ] = (0.403973421087, 0.914770722671)
And by symmetry, circle B is centered at twice the distance from A to G = point D =
( 2*.403973421087, 0) = ( 0.807946842174 , 0 )
So the distance between the circles' centers is just 2AG = AD = about .808
Proof.....area between segment BC and arc BEC of circle A =
Which is about pi/4.......and by symmetry....this is the same area between segment BC and arc BFC of circle
B.......so 2(pi/4) = pi/2
I'll give this one a try....these are always interesting......Each circle has an area of pi....so the middle area must be pi/2
See the following pic....
Let circle A be centered at (0, 0)
We need to first solve this.....
Area of sector ABECA - area of triangle ABC = pi/4 so we have
(1/2)[Θ - sin(Θ)] = pi/4.......with a lttle help from WolframAlpha...Θ = about 2.30988 rads = about 132.3463751816785913° ...and (1/2) of this = 66.17318759083929565°
So the point B is given by :
[ cos(66.17318759083929565°), sin (66.17318759083929565°) ] = (0.403973421087, 0.914770722671)
And by symmetry, circle B is centered at twice the distance from A to G = point D =
( 2*.403973421087, 0) = ( 0.807946842174 , 0 )
So the distance between the circles' centers is just 2AG = AD = about .808
Proof.....area between segment BC and arc BEC of circle A =
Which is about pi/4.......and by symmetry....this is the same area between segment BC and arc BFC of circle
B.......so 2(pi/4) = pi/2