Géométrie I - La chasse aux angles

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Olympiades Suisses de Mathématiques

<90◦
90◦180◦
4
180◦
g AB C
∠(g, CA) + γ+∠(CB, g) = 180◦.
∠(g, CA)α∠(CB, g) = β
g AB α +β+γ= 180◦
n n −2
n(n−2) ·180◦

AB
C
αβ
γg
αβ
ε
AB
a
a
C
D
4ABC AB BC D
AB BD =BC  =∠ACD
γ−α= 30◦
∠CDB 4ADC
A C
∠CDB =α+.
4BCD
∠CDB =∠BCD =γ−.
α+=∠CDB =γ−
⇔α+ 2=γ
⇔2=γ−α= 30◦
⇔= 15◦
α, γ, 
4ABC ∠ACB = 90◦P
P B AB P B =CB P C
wα∠CAB

ADB
C
α
γ-ε
ε
α+ε
a
a
AB
C
aa
P
w
α
P P1
AB C P2
AB
γ= 90◦α β
β= 90◦−α
∠CBP1=α.
4BP1C
∠BP1C=∠P1CB.
α+ 2 ·∠P1CB = 180◦
⇔∠P1CB = 90◦−α
2.
Q wαCB ∠CAQ =α
2
∠AQC = 90◦−α
2=∠P1CB.
∠AQC ∠P1CB wαP1C
P2∠P2BC = 180◦−α
∠BCP2=α
2.
R wαCP24QCR
∠QCR +∠RQC =∠BCP2+ (90◦−∠CAQ) = α
2+ 90◦−α 2= 90◦.
∠CRQ = 90◦wα⊥∠CP2

½α α
90˚-½α
AB
P
C
Q
½α
90˚-½α w
α
1
90˚-½α
½α
½α
w
α
α
P
2
B
C
Q
R
A
∠CAB =α
∠AQC = 90◦−α
2α
O A B AOB
d
AB
d
AB ∠AOB ≤180◦∠AOB ≥180◦
O
ABC ∠BAC =α
d
BC
A2α
4ABC
4ABC O
ABC α
α1=∠OAB α2=∠CAO O
4OAB 4OCA
∠ABO =α1,∠OCA =α2.
AO D
∠DOB 4OAB ∠COD 4OCA