Le cercle, une gure optimale Kikoo <3 Bieber 17.05.12

∀n∈N∗, n ≥3
h ρ b
ρ h b
AT=b·h
2
sin π
n=
b
2
ρ⇒b 2= sin π n·ρ⇒b= 2 sin π n·ρ
cos π
n=h ρ⇒h=ρcos π n
AT=2 sin π
ncos π nρ2
2=sin 2π
nρ2
2
An−gone =YAT=n· AT=n
2sin 2π nρ2
∗∗∗

Pn−gone =Yb= 2nsin π
n·ρ
∞
∞
Pn−gone
An−gone
An−gone ≤p2
4ntan π
n<p2
4π
p= 2nsin π
n·ρ p2= 4n2sin2π n·ρ2
p2
4ntan π
n=4n2sin2π
n·ρ2
4ntan π
n=4n2sin2π
n·ρ2
4nsin(π
n)
cos(π
n)
=nsin π
ncos π nρ2=n 2sin 2π nρ2=An−gone
∀θ∈Rtan(θ)≈θ
∗∗∗
An−gone =n
2sin 2π nρ2
dAn
dn=sin 2π
n
2−πcos 2π
n
n=nsin 2π
n−2πcos 2π n
2n

nsin 2π
n−2πcos 2π n>0nsin 2π n>2πcos 2π n
ntan 2π
n>2π , ∀n > 4
An−gone = 1
Pn−gone = 2rntan π
n
An−gone = 1 ⇒ρ2=1
n
2sin 2π n=2
n·1
sin 2π
n⇒ρ=1
qnsin π
ncos π n
Pn−gone = 4 ·n
2·2 sin π
ncos π n
2 cos π
n·ρ2
ρ=2An
cos π
nρ=2qnsin π
ncos π n
cos π
n= 2√nssin π
n
cos π
n= 2rntan π
n
∗∗∗
N∀n > 4
lim
n→∞ An= lim n→∞
n
2·sin 2π nρ2= lim
n→∞ n
22π nρ2=πρ2
n→ ∞
lim
n→∞ 2nsin π
nρ= lim
n→∞ 2n·π
n·ρ= 2πρ
n→ ∞