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Contests
National and Regional Contests
Iran Contests
Iran MO (2nd Round)
1998 Iran MO (2nd round)
1998 Iran MO (2nd round)
Part of
Iran MO (2nd Round)
Subcontests
(3)
1
2
Hide problems
a_1<a_2<...<a_n - Iran NMO 1998 (Second Round) Problem1
If
a
1
<
a
2
<
⋯
<
a
n
a_1<a_2<\cdots<a_n
a
1
<
a
2
<
⋯
<
a
n
be real numbers, prove that:
a
1
a
2
4
+
a
2
a
3
4
+
⋯
+
a
n
−
1
a
n
4
+
a
n
a
1
4
≥
a
2
a
1
4
+
a
3
a
2
4
+
⋯
+
a
n
a
n
−
1
4
+
a
1
a
n
4
.
a_1a_2^4+a_2a_3^4+\cdots+a_{n-1}a_n^4+a_na_1^4\geq a_2a_1^4+a_3a_2^4+\cdots+a_na_{n-1}^4+a_1a_n^4.
a
1
a
2
4
+
a
2
a
3
4
+
⋯
+
a
n
−
1
a
n
4
+
a
n
a
1
4
≥
a
2
a
1
4
+
a
3
a
2
4
+
⋯
+
a
n
a
n
−
1
4
+
a
1
a
n
4
.
n = d_1^2+... - Iran NMO 1998 (Second Round) Problem4
Let the positive integer
n
n
n
have at least for positive divisors and
0
<
d
1
<
d
2
<
d
3
<
d
4
0<d_1<d_2<d_3<d_4
0
<
d
1
<
d
2
<
d
3
<
d
4
be its least positive divisors. Find all positive integers
n
n
n
such that:
n
=
d
1
2
+
d
2
2
+
d
3
2
+
d
4
2
.
n=d_1^2+d_2^2+d_3^2+d_4^2.
n
=
d
1
2
+
d
2
2
+
d
3
2
+
d
4
2
.
2
2
Hide problems
Value of angle BAC - Iran NMO 1998 (Second Round) Problem2
Let
A
B
C
ABC
A
BC
be a triangle.
I
I
I
is the incenter of
Δ
A
B
C
\Delta ABC
Δ
A
BC
and
D
D
D
is the meet point of
A
I
AI
A
I
and the circumcircle of
Δ
A
B
C
\Delta ABC
Δ
A
BC
. Let
E
,
F
E,F
E
,
F
be on
B
D
,
C
D
BD,CD
B
D
,
C
D
, respectively such that
I
E
,
I
F
IE,IF
I
E
,
I
F
are perpendicular to
B
D
,
C
D
BD,CD
B
D
,
C
D
, respectively. If
I
E
+
I
F
=
A
D
2
IE+IF=\frac{AD}{2}
I
E
+
I
F
=
2
A
D
, find the value of
∠
B
A
C
\angle BAC
∠
B
A
C
.
AQ+CQ=BP - Iran NMO 1998 (Second Round) Problem5
Let
A
B
C
ABC
A
BC
be a triangle and
A
B
<
A
C
<
B
C
AB<AC<BC
A
B
<
A
C
<
BC
. Let
D
,
E
D,E
D
,
E
be points on the side
B
C
BC
BC
and the line
A
B
AB
A
B
, respectively (
A
A
A
is between
B
,
E
B,E
B
,
E
) such that
B
D
=
B
E
=
A
C
BD=BE=AC
B
D
=
BE
=
A
C
. The circumcircle of
Δ
B
E
D
\Delta BED
Δ
BE
D
meets the side
A
C
AC
A
C
at
P
P
P
and
B
P
BP
BP
meets the circumcircle of
Δ
A
B
C
\Delta ABC
Δ
A
BC
at
Q
Q
Q
. Prove that:
A
Q
+
C
Q
=
B
P
.
AQ+CQ=BP.
A
Q
+
CQ
=
BP
.
3
2
Hide problems
Good n-numbers - Iran NMO 1998 (Second Round) Problem3
Let
n
n
n
be a positive integer. We call
(
a
1
,
a
2
,
⋯
,
a
n
)
(a_1,a_2,\cdots,a_n)
(
a
1
,
a
2
,
⋯
,
a
n
)
a good
n
−
n-
n
−
tuple if
∑
i
=
1
n
a
i
=
2
n
\sum_{i=1}^{n}{a_i}=2n
∑
i
=
1
n
a
i
=
2
n
and there doesn't exist a set of
a
i
a_i
a
i
s such that the sum of them is equal to
n
n
n
. Find all good
n
−
n-
n
−
tuple. (For instance,
(
1
,
1
,
4
)
(1,1,4)
(
1
,
1
,
4
)
is a good
3
−
3-
3
−
tuple, but
(
1
,
2
,
1
,
2
,
4
)
(1,2,1,2,4)
(
1
,
2
,
1
,
2
,
4
)
is not a good
5
−
5-
5
−
tuple.)
f(A,B) - Iran NMO 1998 (Second Round) Problem6
If
A
=
(
a
1
,
⋯
,
a
n
)
A=(a_1,\cdots,a_n)
A
=
(
a
1
,
⋯
,
a
n
)
,
B
=
(
b
1
,
⋯
,
b
n
)
B=(b_1,\cdots,b_n)
B
=
(
b
1
,
⋯
,
b
n
)
be
2
2
2
n
−
n-
n
−
tuple that
a
i
,
b
i
=
0
o
r
1
a_i,b_i=0 \ or \ 1
a
i
,
b
i
=
0
or
1
for
i
=
1
,
2
,
⋯
,
n
i=1,2,\cdots,n
i
=
1
,
2
,
⋯
,
n
, we define
f
(
A
,
B
)
f(A,B)
f
(
A
,
B
)
the number of
1
≤
i
≤
n
1\leq i\leq n
1
≤
i
≤
n
that
a
i
≠
b
i
a_i\ne b_i
a
i
=
b
i
. For instance, if
A
=
(
0
,
1
,
1
)
A=(0,1,1)
A
=
(
0
,
1
,
1
)
,
B
=
(
1
,
1
,
0
)
B=(1,1,0)
B
=
(
1
,
1
,
0
)
, then
f
(
A
,
B
)
=
2
f(A,B)=2
f
(
A
,
B
)
=
2
. Now, let
A
=
(
a
1
,
⋯
,
a
n
)
A=(a_1,\cdots,a_n)
A
=
(
a
1
,
⋯
,
a
n
)
,
B
=
(
b
1
,
⋯
,
b
n
)
B=(b_1,\cdots,b_n)
B
=
(
b
1
,
⋯
,
b
n
)
,
C
=
(
c
1
,
⋯
,
c
n
)
C=(c_1,\cdots,c_n)
C
=
(
c
1
,
⋯
,
c
n
)
be 3
n
−
n-
n
−
tuple, such that for
i
=
1
,
2
,
⋯
,
n
i=1,2,\cdots,n
i
=
1
,
2
,
⋯
,
n
,
a
i
,
b
i
,
c
i
=
0
o
r
1
a_i,b_i,c_i=0 \ or \ 1
a
i
,
b
i
,
c
i
=
0
or
1
and
f
(
A
,
B
)
=
f
(
A
,
C
)
=
f
(
B
,
C
)
=
d
f(A,B)=f(A,C)=f(B,C)=d
f
(
A
,
B
)
=
f
(
A
,
C
)
=
f
(
B
,
C
)
=
d
.
a
)
a)
a
)
Prove that
d
d
d
is even.
b
)
b)
b
)
Prove that there exists a
n
−
n-
n
−
tuple
D
=
(
d
1
,
⋯
,
d
n
)
D=(d_1,\cdots,d_n)
D
=
(
d
1
,
⋯
,
d
n
)
that
d
i
=
0
o
r
1
d_i=0 \ or \ 1
d
i
=
0
or
1
for
i
=
1
,
2
,
⋯
,
n
i=1,2,\cdots,n
i
=
1
,
2
,
⋯
,
n
, such that
f
(
A
,
D
)
=
f
(
B
,
D
)
=
f
(
C
,
D
)
=
d
2
f(A,D)=f(B,D)=f(C,D)=\frac{d}{2}
f
(
A
,
D
)
=
f
(
B
,
D
)
=
f
(
C
,
D
)
=
2
d
.