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National and Regional Contests
Iran Contests
Iran MO (2nd Round)
1988 Iran MO (2nd round)
1988 Iran MO (2nd round)
Part of
Iran MO (2nd Round)
Subcontests
(3)
3
2
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Function f(f(m)+f(n))=m+n - [Iran Second Round 1988]
Let
f
:
N
→
N
f : \mathbb N \to \mathbb N
f
:
N
→
N
be a function satisfying f(f(m)+f(n))=m+n, \forall m,n \in \mathbb N. Prove that
f
(
x
)
=
x
f(x)=x
f
(
x
)
=
x
for all
x
∈
N
x \in \mathbb N
x
∈
N
.
All of the numbers are equal - [Iran Second Round 1988]
Let
n
n
n
be a positive integer.
136
9
n
1369^n
136
9
n
positive rational numbers are given with this property: if we remove one of the numbers, then we can divide remain numbers into
1368
1368
1368
sets with equal number of elements such that the product of the numbers of the sets be equal. Prove that all of the numbers are equal.
2
2
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I, J and O are collinear - [Iran Second Round 1988]
In a cyclic quadrilateral
A
B
C
D
ABCD
A
BC
D
, let
I
,
J
I,J
I
,
J
be the midpoints of diagonals
A
C
,
B
D
AC, BD
A
C
,
B
D
respectively and let
O
O
O
be the center of the circle inscribed in
A
B
C
D
.
ABCD.
A
BC
D
.
Prove that
I
,
J
I, J
I
,
J
and
O
O
O
are collinear.
Inequality on altitudes - [Iran Second Round 1988]
In tetrahedron
A
B
C
D
ABCD
A
BC
D
let
h
a
,
h
b
,
h
c
h_a, h_b, h_c
h
a
,
h
b
,
h
c
and
h
d
h_d
h
d
be the lengths of the altitudes from each vertex to the opposite side of that vertex. Prove that
1
h
a
<
1
h
b
+
1
h
c
+
1
h
d
.
\frac{1}{h_a} <\frac{1}{h_b}+\frac{1}{h_c}+\frac{1}{h_d}.
h
a
1
<
h
b
1
+
h
c
1
+
h
d
1
.
1
2
Hide problems
The sum of terms is less than 1 - [Iran Second Round 1988]
Let
{
a
n
}
n
=
1
∞
\{a_n \}_{n=1}^{\infty}
{
a
n
}
n
=
1
∞
be a sequence such that
a
1
=
1
2
a_1=\frac 12
a
1
=
2
1
and
a
n
=
(
2
n
−
3
2
n
)
a
n
−
1
∀
n
≥
2.
a_n=\biggl( \frac{2n-3}{2n} \biggr) a_{n-1} \qquad \forall n \geq 2.
a
n
=
(
2
n
2
n
−
3
)
a
n
−
1
∀
n
≥
2.
Prove that for every positive integer
n
,
n,
n
,
we have
∑
k
=
1
n
a
k
<
1.
\sum_{k=1}^n a_k <1.
∑
k
=
1
n
a
k
<
1.
The fraction has a limit - [Iran Second Round 1988]
(a) Prove that for all positive integers
m
,
n
m,n
m
,
n
we have
∑
k
=
1
n
k
(
k
+
1
)
(
k
+
2
)
⋯
(
k
+
m
−
1
)
=
n
(
n
+
1
)
(
n
+
2
)
⋯
(
n
+
m
)
m
+
1
\sum_{k=1}^n k(k+1)(k+2)\cdots (k+m-1)=\frac{n(n+1)(n+2) \cdots (n+m)}{m+1}
k
=
1
∑
n
k
(
k
+
1
)
(
k
+
2
)
⋯
(
k
+
m
−
1
)
=
m
+
1
n
(
n
+
1
)
(
n
+
2
)
⋯
(
n
+
m
)
(b) Let
P
(
x
)
P(x)
P
(
x
)
be a polynomial with rational coefficients and degree
m
.
m.
m
.
If
n
n
n
tends to infinity, then prove that
∑
k
=
1
n
P
(
k
)
n
m
+
1
\frac{\sum_{k=1}^n P(k)}{n^{m+1}}
n
m
+
1
∑
k
=
1
n
P
(
k
)
Has a limit.