... using the formula B( 1/2+ α, 1/2+ α)=4−αB( 1/2, 1/2+ α), weget (2.11), and the lemma follows.Let Sn= S(1/n, 1/2+ 1/n, a) with a =(2s/ sin(2π/n)) 1/2 . Then (2.11) with α =1/n, β1= 1/2+ 1/n givesm(Sn; ... ≤12παklog4αkαkB( 1/2, 1/2+ αk)(σktan παk) 1/2 (4.4)=12παklogπ 1/2 4αkΓ( 1/2+ αk)(σktan παk) 1/2 Γ(αk).Taking into account the proportionality property (4.2) and (4.4), we ... for every k =1, ,n, Tkis an isoscelestriangle having area A/n and the angle 2π/n at the base vertex ak0. Therefore, for every k =1, ,n, Skis an isosceles infinite triangle having the angle...