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France Team Selection Test
2005 France Team Selection Test
5
<PAC=2<CPA
<PAC=2<CPA
Source: French TST 2005 pb 5.
May 27, 2005
trigonometry
geometry solved
geometry
Problem Statement
Let
A
B
C
ABC
A
BC
be a triangle such that
B
C
=
A
C
+
1
2
A
B
BC=AC+\frac{1}{2}AB
BC
=
A
C
+
2
1
A
B
. Let
P
P
P
be a point of
A
B
AB
A
B
such that
A
P
=
3
P
B
AP=3PB
A
P
=
3
PB
. Show that
P
A
C
^
=
2
C
P
A
^
.
\widehat{PAC} = 2 \widehat{CPA}.
P
A
C
=
2
CP
A
.
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