Géométrie - Mathematiques pour les PLP.

r O0C0=r(C)r
OCO0=r(O)
ArrCA
C0=r(C)A=r(A)
r A
π2πOA O0
B A (OO0)A
OO0BC C0
CC0A B
(AB)CC0
geo10011.fig 29/11/04
C0
M
B
C
O0
M0
O
A
B0
MCB M0=r(M)B0=r(B)
r
(AMB)(AM0B0) = r((AMB))
−→
BA, −−→
BM−−→
B0A, −−−→ B0M0
−→
BA, −−→
BM0−−→
B0A, −−−→ B0M0C0
mes −→
BA, −−→
BM0= mes −−→
B0A, −−−→
B0M0(mod π)
((BA),(BM 0)) ((B0A),(B0M0))
mes −−→ BM0,−→
BA+ mes −−→
B0A, −−−→ B0M0= 0 (mod π)
−→
BA, −−→
BM=−−→
B0A, −−−→
B0M0
mes −−→
BM0,−→
BA+ mes −→
BA, −−→
BM= 0 (mod π)
mes −−→
BM0,−−→
BM= 0 (mod π)
−−→
BM0,−−→ BMM B M0
M
A B

• CACBCs(BC)(BC)
s(AC)(AC)
• HAH(BC)
HA=s(H)H=s(HA)
HAH
(BC)C(ABC)
H=s(HA)HAsCA
Cs
HCB
C H CACB
geo10012.fig 29/11/04
D
FC
C
A
CB
FBF
O
BOB
H
CAC
HA
OA
FA
OAOBCACB
OAOBO s(BC)s(AC)
r=s(AC)◦s(BC)s(BC )(BC)
s(AC)(AC)
s(BC)CA=s(BC)(C)⇐⇒ C =s(BC)(CA)
s(AC)(s(BC)(CA)) = s(AC)(C)s(AC)(s(BC)(CA)) = CB
s(AC)(s(BC)(FA)) = s(AC )(FB)s(AC)(s(BC)(FA)) = F
CAFArCBFB
rs(BC)s(AC)
r(BC)(AC)C
r θr= 2 ((BC),(AC))

A B C
(ABC)C
•(ABC)C
CA
D
FC
O
FBF
B C
F0FA
H C F 0F
C F FBF0FAC
FBC FA
•(ABC)C
(BC)(AC)
θrr0π
r C θr
C
CACB=r(CA)C H
H FAFB=r(FA)
(ABC)
A B
(ABC)C H FAFB=r(FA)
A B C (ABC)C
H
H FBFC
FC(HFB)
HFAFBFCD
b
C
(ABC)

DHFAFBFC
D(ABC)
F
DFAFBFC
(BC) (AC)(AB)P Q R
25/11/04 25/11/04
D
FCR
CA
CB
FBQ
OF
OB
B
H
CAC
OA
FA
P
P(BC)s(BC)F FA
P F FA(BC)
[F, FA]
P FAP
D
P FAPD
FD
M∈ P ⇐⇒ d(M, F ) = d(M, D)
(BC)
[F, FA]P
PQRP
FD(BC) (AC)
(AB)