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National and Regional Contests
Iran Contests
Iran MO (2nd Round)
1999 Iran MO (2nd round)
1999 Iran MO (2nd round)
Part of
Iran MO (2nd Round)
Subcontests
(3)
3
2
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100*100 garden - Iran NMO 1999 (Second Round) Problem3
We have a
100
×
100
100\times100
100
×
100
garden and we’ve plant
10000
10000
10000
trees in the
1
×
1
1\times1
1
×
1
squares (exactly one in each.). Find the maximum number of trees that we can cut such that on the segment between each two cut trees, there exists at least one uncut tree.
A_1,...,A_n points - Iran NMO 1999 (Second Round) Problem6
Let
A
1
,
A
2
,
⋯
,
A
n
A_1,A_2,\cdots,A_n
A
1
,
A
2
,
⋯
,
A
n
be
n
n
n
distinct points on the plane (
n
>
1
n>1
n
>
1
). We consider all the segments
A
i
A
j
A_iA_j
A
i
A
j
where
i
<
j
≤
n
i<j\leq{n}
i
<
j
≤
n
and color the midpoints of them. What's the minimum number of colored points? (In fact, if
k
k
k
colored points coincide, we count them
1
1
1
.)
2
2
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Making triangles - Iran NMO 1999 (Second Round) Problem2
A
B
C
ABC
A
BC
is a triangle with
∠
B
>
4
5
∘
\angle{B}>45^{\circ}
∠
B
>
4
5
∘
,
∠
C
>
4
5
∘
\angle{C}>45^{\circ}
∠
C
>
4
5
∘
. We draw the isosceles triangles
C
A
M
,
B
A
N
CAM,BAN
C
A
M
,
B
A
N
on the sides
A
C
,
A
B
AC,AB
A
C
,
A
B
and outside the triangle, respectively, such that
∠
C
A
M
=
∠
B
A
N
=
9
0
∘
\angle{CAM}=\angle{BAN}=90^{\circ}
∠
C
A
M
=
∠
B
A
N
=
9
0
∘
. And we draw isosceles triangle
B
P
C
BPC
BPC
on the side
B
C
BC
BC
and inside the triangle such that
∠
B
P
C
=
9
0
∘
\angle{BPC}=90^{\circ}
∠
BPC
=
9
0
∘
. Prove that
Δ
M
P
N
\Delta{MPN}
Δ
MPN
is an isosceles triangle, too, and
∠
M
P
N
=
9
0
∘
\angle{MPN}=90^{\circ}
∠
MPN
=
9
0
∘
.
S_PQR/S_A'B'C' - Iran NMO 1999 (Second Round) Problem5
Let
A
B
C
ABC
A
BC
be a triangle and points
P
,
Q
,
R
P,Q,R
P
,
Q
,
R
be on the sides
A
B
,
B
C
,
A
C
AB,BC,AC
A
B
,
BC
,
A
C
, respectively. Now, let
A
′
,
B
′
,
C
′
A',B',C'
A
′
,
B
′
,
C
′
be on the segments
P
R
,
Q
P
,
R
Q
PR,QP,RQ
PR
,
QP
,
RQ
in a way that
A
B
∣
∣
A
′
B
′
AB||A'B'
A
B
∣∣
A
′
B
′
,
B
C
∣
∣
B
′
C
′
BC||B'C'
BC
∣∣
B
′
C
′
and
A
C
∣
∣
A
′
C
′
AC||A'C'
A
C
∣∣
A
′
C
′
. Prove that:
A
B
A
′
B
′
=
S
P
Q
R
S
A
′
B
′
C
′
.
\frac{AB}{A'B'}=\frac{S_{PQR}}{S_{A'B'C'}}.
A
′
B
′
A
B
=
S
A
′
B
′
C
′
S
PQR
.
Where
S
X
Y
Z
S_{XYZ}
S
X
Y
Z
is the surface of the triangle
X
Y
Z
XYZ
X
Y
Z
.
1
2
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Powers of 2 - Iran NMO 1999 (Second Round) Problem1
Does there exist a positive integer that is a power of
2
2
2
and we get another power of
2
2
2
by swapping its digits? Justify your answer.
m= a_1/1+... - Iran NMO 1999 (Second Round) Problem4
Find all positive integers
m
m
m
such that there exist positive integers
a
1
,
a
2
,
…
,
a
1378
a_1,a_2,\ldots,a_{1378}
a
1
,
a
2
,
…
,
a
1378
such that:
m
=
∑
k
=
1
1378
k
a
k
.
m=\sum_{k=1}^{1378}{\frac{k}{a_k}}.
m
=
k
=
1
∑
1378
a
k
k
.