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#13 by Bruno Berselli at Thu Apr 22 11:54:46 EDT 2021
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#12 by Peter Luschny at Thu Apr 22 11:44:21 EDT 2021
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#11 by Amiram Eldar at Tue Apr 20 13:41:11 EDT 2021
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#10 by Amiram Eldar at Tue Apr 20 13:41:08 EDT 2021
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| MATHEMATICA
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RealDigits[(Sqrt[20] - 3*ArcCos[2/3])/(4*Pi) - 1/8, 10, 100][[1]] (* Amiram Eldar, Apr 20 2021 *)
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#9 by Wolfdieter Lang at Tue Apr 20 13:35:10 EDT 2021
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#8 by Wolfdieter Lang at Tue Apr 20 13:35:03 EDT 2021
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The radii of the inner and outer Soddy circles, normalized to the radius r of one of the two large circle rcircles are s_i = S_i/r = -3/2 + phi = A176055 - 2 = 0.1180339887... and s_o = S_o/r = 1/2 + phi = A176055 = 2 + s_i = 2.1180339887... Here phi = A001622 (golden ratio).
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#7 by Wesley Ivan Hurt at Tue Apr 20 10:39:39 EDT 2021
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#6 by Wesley Ivan Hurt at Tue Apr 20 10:39:08 EDT 2021
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| COMMENTS
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This constant gives the ratio of the area between three touching circles, one with half of the radius of the two others, and the area of one of the large circular diskdisks.
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#5 by Michel Marcus at Tue Apr 20 06:23:04 EDT 2021
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#4 by Michel Marcus at Tue Apr 20 06:22:22 EDT 2021
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| COMMENTS
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The isosceles triangle with the centers of the circles as corners has two angles alpha = arctan(sqrt(5)/2) = A228496 (about 48,.2 degrees).
The ratio of the perimeter of the boundary of this circular cuspodial triangle and the perimeter of the large circle is alpha/(2*Pi) + 1/4 = 0.3838602364...
The radii of the inner and outer Soddy circles, normalized to the radius of the large circle r are s_i = S_i/r = -3/2 + phi = = A176055 - 2 = 0.1180339887... and s_o = S_o/r = 1/2 + phi = A176055 = 2 + s_i = 2.1180339887... Here phi = A001622 (golden ratio).
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| FORMULA
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a = Equals A/(Pi*r^2) = (sqrt(5)/Pi - 3*arctan(sqrt(5)/2)/(2*Pi) - 1/4)/2, where A is the area between three mutually touching circular disks with radii r, r, and r/2 (in some length unit).
a = Equals sqrt(5)/(2*Pi) - 3*A228496/(4*Pi) - 1/8.
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| EXAMPLE
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a = 0.03009091710766602117945599124597761...
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