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%I #9 Aug 27 2016 05:43:10
%S 9,2,5,8,3,7,0,5,7,6,1,3,8,8,6,0,0,8,3,7,6,8,9,1,2,1,2,3,3,5,4,1,1,0,
%T 9,6,8,9,2,7,8,2,9,6,1,1,6,8,3,0,0,9,2,2,1,9,6,0,1,6,8,1,5,8,4,2,3,4,
%U 9,0,9,3,1,9,5,3,2,2,9,7,9,2,1,5,1,9,2,7,2,3,5,3,1,8,4,6,6,8,9,5,7,4
%N Decimal expansion of the area cut out by a rotating regular pentagon of width 1 inside a unit square.
%C The part of the square not cut out is made of four congruent regions, one at each corner. Superficially they look like isosceles right triangles, but their "hypotenuses" curve very slightly outward. The region cut out by the pentagon is therefore concave.
%H Jeremy Tan, <a href="https://math.stackexchange.com/questions/1886807/regular-pentagon-in-a-square/1892221#1892221">Regular pentagon in a square</a> (with derivation of constant)
%F A = 1-(sqrt(5)-2)*Pi/10 = 0.92583705761388600837689121233541...
%t First@ RealDigits@ N[1 - (Sqrt@ 5 - 2)/10 Pi, 120] (* _Michael De Vlieger_, Aug 15 2016 *)
%o (PARI) 1-(sqrt(5)-2)*Pi/10
%Y Cf. A066666 (rotating Reuleaux triangle in a square).
%K nonn,cons
%O 0,1
%A _Jeremy Tan_, Aug 14 2016
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