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Contests
National and Regional Contests
Iran Contests
Iran MO (2nd Round)
1991 Iran MO (2nd round)
1991 Iran MO (2nd round)
Part of
Iran MO (2nd Round)
Subcontests
(3)
3
2
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The old Cauchy function [Iran Second Round 1991]
Let
f
:
R
→
R
f : \mathbb R \to \mathbb R
f
:
R
→
R
be a function such that
f
(
1
)
=
1
f(1)=1
f
(
1
)
=
1
and
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
f(x+y)=f(x)+f(y)
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
And for all
x
∈
R
/
{
0
}
x \in \mathbb R / \{0\}
x
∈
R
/
{
0
}
we have
f
(
1
x
)
=
1
f
(
x
)
.
f\left( \frac 1x \right) = \frac{1}{f(x)}.
f
(
x
1
)
=
f
(
x
)
1
.
Find all such functions
f
.
f.
f
.
Three group of mathematicians [Iran Second Round 1991]
Three groups
A
,
B
A, B
A
,
B
and
C
C
C
of mathematicians from different countries have invited to a ceremony. We have formed meetings such that three mathematicians participate in every meeting and there is exactly one mathematician from each group in every meeting. Also every two mathematicians have participated in exactly one meeting with each other.(a) Prove that if this is possible, then number of mathematicians of the groups is equal.(b) Prove that if there exist
3
3
3
mathematicians in each group, then that work is possible.(c) Prove that if number mathematicians of the groups be equal, then that work is possible.
2
2
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Tetragonal ABCD [Iran Second Round 1991]
Let
A
B
C
D
ABCD
A
BC
D
be a tetragonal.(a) If the plane
(
P
)
(P)
(
P
)
cuts
A
B
C
D
,
ABCD,
A
BC
D
,
find the necessary and sufficient condition such that the area formed from the intersection of the plane
(
P
)
(P)
(
P
)
and the tetragonal be a parallelogram. Prove that the problem has three solutions in this case.(b) Consider one of the solutions of (a). Find the situation of the plane
(
P
)
(P)
(
P
)
for which the parallelogram has maximum area.(c) Find a plane
(
P
)
(P)
(
P
)
for which the parallelogram be a lozenge and then find the length side of his lozenge in terms of the length of the edges of
A
B
C
D
.
ABCD.
A
BC
D
.
Inequalituy with the incenter [Iran Second Round 1991]
Triangle
A
B
C
ABC
A
BC
is inscribed in circle
C
.
C.
C
.
The bisectors of the angles
A
,
B
A,B
A
,
B
and
C
C
C
meet the circle
C
C
C
again at the points
A
′
,
B
′
,
C
′
A', B', C'
A
′
,
B
′
,
C
′
. Let
I
I
I
be the incenter of
A
B
C
,
ABC,
A
BC
,
prove that
I
A
′
I
A
+
I
B
′
I
B
+
I
C
′
I
C
≥
3
\frac{IA'}{IA} + \frac{IB'}{IB}+\frac{IC'}{IC} \geq 3
I
A
I
A
′
+
I
B
I
B
′
+
I
C
I
C
′
≥
3
,
I
A
′
+
I
B
′
+
I
C
′
≥
I
A
+
I
B
+
I
C
, IA'+IB'+IC' \geq IA+IB+IC
,
I
A
′
+
I
B
′
+
I
C
′
≥
I
A
+
I
B
+
I
C
1
2
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On the equation x+x^2=y+y^2+y^3 [Iran Second Round 1991]
Prove that the equation
x
+
x
2
=
y
+
y
2
+
y
3
x+x^2=y+y^2+y^3
x
+
x
2
=
y
+
y
2
+
y
3
do not have any solutions in positive integers.
Six points with rational coordinates[Iran Second Round 1991]
Prove that there exist at least six points with rational coordinates on the curve of the equation
y
3
=
x
3
+
x
+
137
0
1370
y^3=x^3+x+1370^{1370}
y
3
=
x
3
+
x
+
137
0
1370