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Contests
National and Regional Contests
China Contests
China National Olympiad
1989 China National Olympiad
1989 China National Olympiad
Part of
China National Olympiad
Subcontests
(6)
3
1
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China Mathematical Olympiad 1989 problem3
Let
S
S
S
be the unit circle in the complex plane (i.e. the set of all complex numbers with their moduli equal to
1
1
1
). We define function
f
:
S
→
S
f:S\rightarrow S
f
:
S
→
S
as follow:
∀
z
∈
S
\forall z\in S
∀
z
∈
S
,
f
(
1
)
(
z
)
=
f
(
z
)
,
f
(
2
)
(
z
)
=
f
(
f
(
z
)
)
,
…
,
f^{(1)}(z)=f(z), f^{(2)}(z)=f(f(z)), \dots,
f
(
1
)
(
z
)
=
f
(
z
)
,
f
(
2
)
(
z
)
=
f
(
f
(
z
))
,
…
,
f
(
k
)
(
z
)
=
f
(
f
(
k
−
1
)
(
z
)
)
(
k
>
1
,
k
∈
N
)
,
…
f^{(k)}(z)=f(f^{(k-1)}(z)) (k>1,k\in \mathbb{N}), \dots
f
(
k
)
(
z
)
=
f
(
f
(
k
−
1
)
(
z
))
(
k
>
1
,
k
∈
N
)
,
…
We call
c
c
c
an
n
n
n
-period-point of
f
f
f
if
c
c
c
(
c
∈
S
c\in S
c
∈
S
) and
n
n
n
(
n
∈
N
n\in\mathbb{N}
n
∈
N
) satisfy:
f
(
1
)
(
c
)
≠
c
,
f
(
2
)
(
c
)
≠
c
,
f
(
3
)
(
c
)
≠
c
,
…
,
f
(
n
−
1
)
(
c
)
≠
c
,
f
(
n
)
(
c
)
=
c
f^{(1)}(c) \not=c, f^{(2)}(c) \not=c, f^{(3)}(c) \not=c, \dots, f^{(n-1)}(c) \not=c, f^{(n)}(c)=c
f
(
1
)
(
c
)
=
c
,
f
(
2
)
(
c
)
=
c
,
f
(
3
)
(
c
)
=
c
,
…
,
f
(
n
−
1
)
(
c
)
=
c
,
f
(
n
)
(
c
)
=
c
. Suppose that
f
(
z
)
=
z
m
f(z)=z^m
f
(
z
)
=
z
m
(
z
∈
S
;
m
>
1
,
m
∈
N
z\in S; m>1, m\in \mathbb{N}
z
∈
S
;
m
>
1
,
m
∈
N
), find the number of
1989
1989
1989
-period-point of
f
f
f
.
6
1
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China Mathematical Olympiad 1989 problem6
Find all functions
f
:
(
1
,
+
∞
)
→
(
1
,
+
∞
)
f:(1,+\infty) \rightarrow (1,+\infty)
f
:
(
1
,
+
∞
)
→
(
1
,
+
∞
)
that satisfy the following condition: for arbitrary
x
,
y
>
1
x,y>1
x
,
y
>
1
and
u
,
v
>
0
u,v>0
u
,
v
>
0
, inequality
f
(
x
u
y
v
)
≤
f
(
x
)
1
4
u
f
(
y
)
1
4
v
f(x^uy^v)\le f(x)^{\dfrac{1}{4u}}f(y)^{\dfrac{1}{4v}}
f
(
x
u
y
v
)
≤
f
(
x
)
4
u
1
f
(
y
)
4
v
1
holds.
5
1
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China Mathematical Olympiad 1989 problem5
Given
1989
1989
1989
points in the space, any three of them are not collinear. We divide these points into
30
30
30
groups such that the numbers of points in these groups are different from each other. Consider those triangles whose vertices are points belong to three different groups among the
30
30
30
. Determine the numbers of points of each group such that the number of such triangles attains a maximum.
4
1
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China Mathematical Olympiad 1989 problem4
Given a triangle
A
B
C
ABC
A
BC
, points
D
,
E
,
F
D,E,F
D
,
E
,
F
lie on sides
B
C
,
C
A
,
A
B
BC,CA,AB
BC
,
C
A
,
A
B
respectively. Moreover, the radii of incircles of
△
A
E
F
,
△
B
F
D
,
△
C
D
E
\triangle AEF, \triangle BFD, \triangle CDE
△
A
EF
,
△
BF
D
,
△
C
D
E
are equal to
r
r
r
. Denote by
r
0
r_0
r
0
and
R
R
R
the radii of incircles of
△
D
E
F
\triangle DEF
△
D
EF
and
△
A
B
C
\triangle ABC
△
A
BC
respectively. Prove that
r
+
r
0
=
R
r+r_0=R
r
+
r
0
=
R
.
2
1
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China Mathematical Olympiad 1989 problem2
Let
x
1
,
x
2
,
…
,
x
n
x_1, x_2, \dots ,x_n
x
1
,
x
2
,
…
,
x
n
(
n
≥
2
n\ge 2
n
≥
2
) be positive real numbers satisfying
∑
i
=
1
n
x
i
=
1
\sum^{n}_{i=1}x_i=1
∑
i
=
1
n
x
i
=
1
. Prove that:
∑
i
=
1
n
x
i
1
−
x
i
≥
∑
i
=
1
n
x
i
n
−
1
.
\sum^{n}_{i=1}\dfrac{x_i}{\sqrt{1-x_i}}\ge \dfrac{\sum_{i=1}^{n}\sqrt{x_i}}{\sqrt{n-1}}.
i
=
1
∑
n
1
−
x
i
x
i
≥
n
−
1
∑
i
=
1
n
x
i
.
1
1
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China Mathematical Olympiad 1989 problem1
We are given two point sets
A
A
A
and
B
B
B
which are both composed of finite disjoint arcs on the unit circle. Moreover, the length of each arc in
B
B
B
is equal to
π
m
\dfrac{\pi}{m}
m
π
(
m
∈
N
m \in \mathbb{N}
m
∈
N
). We denote by
A
j
A^j
A
j
the set obtained by a counterclockwise rotation of
A
A
A
about the center of the unit circle for
j
π
m
\dfrac{j\pi}{m}
m
jπ
(
j
=
1
,
2
,
3
,
…
j=1,2,3,\dots
j
=
1
,
2
,
3
,
…
). Show that there exists a natural number
k
k
k
such that
l
(
A
k
∩
B
)
≥
1
2
π
l
(
A
)
l
(
B
)
l(A^k\cap B)\ge \dfrac{1}{2\pi}l(A)l(B)
l
(
A
k
∩
B
)
≥
2
π
1
l
(
A
)
l
(
B
)
.(Here
l
(
X
)
l(X)
l
(
X
)
denotes the sum of lengths of all disjoint arcs in the point set
X
X
X
)