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International Contests
Iranian Geometry Olympiad
2014 Iran Geometry Olympiad (senior)
2014 Iran Geometry Olympiad (senior)
Part of
Iranian Geometry Olympiad
Subcontests
(4)
3:
1
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Iranian geometry olympiad 2014(4)
Let
A
B
C
ABC
A
BC
be an acute triangle.A circle with diameter
B
C
BC
BC
meets
A
B
AB
A
B
and
A
C
AC
A
C
at
E
E
E
and
F
F
F
,respectively.
M
M
M
is midpoint of
B
C
BC
BC
and
P
P
P
is point of intersection
A
M
AM
A
M
with
E
F
EF
EF
.
X
X
X
is an arbitary point on arc
E
F
EF
EF
and
Y
Y
Y
is the second intersection of
X
P
XP
XP
with a circle with diameter
B
C
BC
BC
.Prove that
∡
X
A
Y
=
∡
X
Y
M
\measuredangle XAY=\measuredangle XYM
∡
X
A
Y
=
∡
X
Y
M
. Author:Ali zo'alam , Iran
4:
1
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Iranian geometry olympiad 2014(3)
A tangent line to circumcircle of acute triangle
A
B
C
ABC
A
BC
(
A
C
>
A
B
AC>AB
A
C
>
A
B
) at
A
A
A
intersects with the extension of
B
C
BC
BC
at
P
P
P
.
O
O
O
is the circumcenter of triangle
A
B
C
ABC
A
BC
.Point
X
X
X
lying on
O
P
OP
OP
such that
∡
A
X
P
=
9
0
∘
\measuredangle AXP=90^\circ
∡
A
XP
=
9
0
∘
.Points
E
E
E
and
F
F
F
lying on
A
B
AB
A
B
and
A
C
AC
A
C
,respectively,and they are in one side of line
O
P
OP
OP
such that
∡
E
X
P
=
∡
A
C
X
\measuredangle EXP=\measuredangle ACX
∡
EXP
=
∡
A
CX
and
∡
F
X
O
=
∡
A
B
X
\measuredangle FXO=\measuredangle ABX
∡
FXO
=
∡
A
BX
.
K
K
K
,
L
L
L
are points of intersection
E
F
EF
EF
with circumcircle of triangle
A
B
C
ABC
A
BC
.prove that
O
P
OP
OP
is tangent to circumcircle of triangle
K
L
X
KLX
K
L
X
.Author:Mehdi E'tesami Fard , Iran
5:
1
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Iranian geometry olympiad 2014(2)
Two points
P
P
P
and
Q
Q
Q
lying on side
B
C
BC
BC
of triangle
A
B
C
ABC
A
BC
and their distance from the midpoint of
B
C
BC
BC
are equal.The perpendiculars from
P
P
P
and
Q
Q
Q
to
B
C
BC
BC
intersect
A
C
AC
A
C
and
A
B
AB
A
B
at
E
E
E
and
F
F
F
,respectively.
M
M
M
is point of intersection
P
F
PF
PF
and
E
Q
EQ
EQ
.If
H
1
H_1
H
1
and
H
2
H_2
H
2
be the orthocenters of triangles
B
F
P
BFP
BFP
and
C
E
Q
CEQ
CEQ
, respectively, prove that
A
M
⊥
H
1
H
2
AM\perp H_1H_2
A
M
⊥
H
1
H
2
.Author:Mehdi E'tesami Fard , Iran
2:
1
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Iranian geometry olympiad 2014
In the Quadrilateral
A
B
C
D
ABCD
A
BC
D
we have
∡
B
=
∡
D
=
6
0
∘
\measuredangle B=\measuredangle D = 60^\circ
∡
B
=
∡
D
=
6
0
∘
.
M
M
M
is midpoint of side
A
D
AD
A
D
.The line through
M
M
M
parallel to
C
D
CD
C
D
meets
B
C
BC
BC
at
P
P
P
.Point
X
X
X
lying on
C
D
CD
C
D
such that
B
X
=
M
X
BX=MX
BX
=
MX
.Prove that
A
B
=
B
P
AB=BP
A
B
=
BP
if and only if
∡
M
X
B
=
6
0
∘
\measuredangle MXB=60^\circ
∡
MXB
=
6
0
∘
.Author: Davoud Vakili, Iran