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Moldova Team Selection Test
2005 Moldova Team Selection Test
1
alpha+beta+gamma=1
alpha+beta+gamma=1
Source: Moldova TST 2005
April 9, 2005
analytic geometry
geometry solved
geometry
Problem Statement
In triangle
A
B
C
ABC
A
BC
,
M
∈
(
B
C
)
M\in(BC)
M
∈
(
BC
)
,
B
M
B
C
=
α
\frac{BM}{BC}=\alpha
BC
BM
=
α
,
N
∈
(
C
A
)
N\in(CA)
N
∈
(
C
A
)
,
C
N
C
A
=
β
\frac{CN}{CA}=\beta
C
A
CN
=
β
,
P
∈
(
A
B
)
P\in(AB)
P
∈
(
A
B
)
,
A
P
A
B
=
γ
\frac{AP}{AB}=\gamma
A
B
A
P
=
γ
. Let
A
M
∩
B
N
=
{
D
}
AM\cap BN=\{D\}
A
M
∩
BN
=
{
D
}
,
B
N
∩
C
P
=
{
E
}
BN\cap CP=\{E\}
BN
∩
CP
=
{
E
}
,
C
P
∩
A
M
=
{
F
}
CP\cap AM=\{F\}
CP
∩
A
M
=
{
F
}
. Prove that
S
D
E
F
=
S
B
M
D
+
S
C
N
E
+
S
A
P
F
S_{DEF}=S_{BMD}+S_{CNE}+S_{APF}
S
D
EF
=
S
BM
D
+
S
CNE
+
S
A
PF
iff
α
+
β
+
γ
=
1
\alpha+\beta+\gamma=1
α
+
β
+
γ
=
1
.
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