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Caucasus Mathematical Olympiad
2019 Caucasus Mathematical Olympiad
5
Different relation in a triangle
Different relation in a triangle
Source: IV Caucasus Mathematic Olympiad
April 7, 2019
trigonometry
Law of Sines
geometry
Problem Statement
Given a triangle
A
B
C
ABC
A
BC
with
B
C
=
a
BC=a
BC
=
a
,
C
A
=
b
CA=b
C
A
=
b
,
A
B
=
c
AB=c
A
B
=
c
,
∠
B
A
C
=
α
\angle BAC = \alpha
∠
B
A
C
=
α
,
∠
C
B
A
=
β
\angle CBA = \beta
∠
CB
A
=
β
,
∠
A
C
B
=
γ
\angle ACB = \gamma
∠
A
CB
=
γ
. Prove that
a
sin
(
β
−
γ
)
+
b
sin
(
γ
−
α
)
+
c
sin
(
α
−
β
)
=
0.
a \sin(\beta-\gamma) + b \sin(\gamma-\alpha) +c\sin(\alpha-\beta) = 0.
a
sin
(
β
−
γ
)
+
b
sin
(
γ
−
α
)
+
c
sin
(
α
−
β
)
=
0.
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