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Contests
National and Regional Contests
China Contests
China National Olympiad
1990 China National Olympiad
1990 China National Olympiad
Part of
China National Olympiad
Subcontests
(6)
6
1
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China Mathematical Olympiad 1990 problem6
A convex
n
n
n
-gon and its
n
−
3
n-3
n
−
3
diagonals which have no common point inside the polygon form a subdivision graph. Show that if and only if
3
∣
n
3|n
3∣
n
, there exists a subdivision graph that can be drawn in one closed stroke. (i.e. start from a certain vertex, get through every edges and diagonals exactly one time, finally back to the starting vertex.)
5
1
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China mathematical olympiad 1990 problem5
Given a finite set
X
X
X
, let
f
f
f
be a rule such that
f
f
f
maps every even-element-subset
E
E
E
of
X
X
X
(i.e.
E
⊆
X
E \subseteq X
E
⊆
X
,
∣
E
∣
|E|
∣
E
∣
is even) into a real number
f
(
E
)
f(E)
f
(
E
)
. Suppose that
f
f
f
satisfies the following conditions: (I) there exists an even-element-subset
D
D
D
of
X
X
X
such that
f
(
D
)
>
1990
f(D)>1990
f
(
D
)
>
1990
; (II) for any two disjoint even-element-subsets
A
,
B
A,B
A
,
B
of
X
X
X
, equation
f
(
A
∪
B
)
=
f
(
A
)
+
f
(
B
)
−
1990
f(A\cup B)=f(A)+f(B)-1990
f
(
A
∪
B
)
=
f
(
A
)
+
f
(
B
)
−
1990
holds. Prove that there exist two subsets
P
,
Q
P,Q
P
,
Q
of
X
X
X
satisfying: (1)
P
∩
Q
=
∅
P\cap Q=\emptyset
P
∩
Q
=
∅
,
P
∪
Q
=
X
P\cup Q=X
P
∪
Q
=
X
; (2) for any non-even-element-subset
S
S
S
of
P
P
P
(i.e.
S
⊆
P
S\subseteq P
S
⊆
P
,
∣
S
∣
|S|
∣
S
∣
is odd), we have
f
(
S
)
>
1990
f(S)>1990
f
(
S
)
>
1990
; (3) for any even-element-subset
T
T
T
of
Q
Q
Q
, we have
f
(
T
)
≤
1990
f(T)\le 1990
f
(
T
)
≤
1990
.
4
1
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China Mathematical Olympiad 1990 problem4
Given a positive integer number
a
a
a
and two real numbers
A
A
A
and
B
B
B
, find a necessary and sufficient condition on
A
A
A
and
B
B
B
for the following system of equations to have integer solution:
{
x
2
+
y
2
+
z
2
=
(
B
a
)
2
x
2
(
A
x
2
+
B
y
2
)
+
y
2
(
A
y
2
+
B
z
2
)
+
z
2
(
A
z
2
+
B
x
2
)
=
1
4
(
2
A
+
B
)
(
B
a
)
4
\left\{\begin{array}{cc} x^2+y^2+z^2=(Ba)^2\\ x^2(Ax^2+By^2)+y^2(Ay^2+Bz^2)+z^2(Az^2+Bx^2)=\dfrac{1}{4}(2A+B)(Ba)^4\end{array}\right.
{
x
2
+
y
2
+
z
2
=
(
B
a
)
2
x
2
(
A
x
2
+
B
y
2
)
+
y
2
(
A
y
2
+
B
z
2
)
+
z
2
(
A
z
2
+
B
x
2
)
=
4
1
(
2
A
+
B
)
(
B
a
)
4
3
1
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China Mathematical Olympiad 1990 problem3
A function
f
(
x
)
f(x)
f
(
x
)
defined for
x
≥
0
x\ge 0
x
≥
0
satisfies the following conditions: i. for
x
,
y
≥
0
x,y\ge 0
x
,
y
≥
0
,
f
(
x
)
f
(
y
)
≤
x
2
f
(
y
/
2
)
+
y
2
f
(
x
/
2
)
f(x)f(y)\le x^2f(y/2)+y^2f(x/2)
f
(
x
)
f
(
y
)
≤
x
2
f
(
y
/2
)
+
y
2
f
(
x
/2
)
; ii. there exists a constant
M
M
M
(
M
>
0
M>0
M
>
0
), such that
∣
f
(
x
)
∣
≤
M
|f(x)|\le M
∣
f
(
x
)
∣
≤
M
when
0
≤
x
≤
1
0\le x\le 1
0
≤
x
≤
1
. Prove that
f
(
x
)
≤
x
2
f(x)\le x^2
f
(
x
)
≤
x
2
.
2
1
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China Mathematical Olympiad 1990 problem2
Let
x
x
x
be a natural number. We call
{
x
0
,
x
1
,
…
,
x
l
}
\{x_0,x_1,\dots ,x_l\}
{
x
0
,
x
1
,
…
,
x
l
}
a factor link of
x
x
x
if the sequence
{
x
0
,
x
1
,
…
,
x
l
}
\{x_0,x_1,\dots ,x_l\}
{
x
0
,
x
1
,
…
,
x
l
}
satisfies the following conditions: (1)
x
0
=
1
,
x
l
=
x
x_0=1, x_l=x
x
0
=
1
,
x
l
=
x
; (2)
x
i
−
1
<
x
i
,
x
i
−
1
∣
x
i
,
i
=
1
,
2
,
…
,
l
x_{i-1}<x_i, x_{i-1}|x_i, i=1,2,\dots,l
x
i
−
1
<
x
i
,
x
i
−
1
∣
x
i
,
i
=
1
,
2
,
…
,
l
. Meanwhile, we define
l
l
l
as the length of the factor link
{
x
0
,
x
1
,
…
,
x
l
}
\{x_0,x_1,\dots ,x_l\}
{
x
0
,
x
1
,
…
,
x
l
}
. Denote by
L
(
x
)
L(x)
L
(
x
)
and
R
(
x
)
R(x)
R
(
x
)
the length and the number of the longest factor link of
x
x
x
respectively. For
x
=
5
k
×
3
1
m
×
199
0
n
x=5^k\times 31^m\times 1990^n
x
=
5
k
×
3
1
m
×
199
0
n
, where
k
,
m
,
n
k,m,n
k
,
m
,
n
are natural numbers, find the value of
L
(
x
)
L(x)
L
(
x
)
and
R
(
x
)
R(x)
R
(
x
)
.
1
1
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China Mathematical Olympiad 1990 problem1
Given a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
, side
A
B
AB
A
B
is not parallel to side
C
D
CD
C
D
. The circle
O
1
O_1
O
1
passing through
A
A
A
and
B
B
B
is tangent to side
C
D
CD
C
D
at
P
P
P
. The circle
O
2
O_2
O
2
passing through
C
C
C
and
D
D
D
is tangent to side
A
B
AB
A
B
at
Q
Q
Q
. Circle
O
1
O_1
O
1
and circle
O
2
O_2
O
2
meet at
E
E
E
and
F
F
F
. Prove that
E
F
EF
EF
bisects segment
P
Q
PQ
PQ
if and only if
B
C
∥
A
D
BC\parallel AD
BC
∥
A
D
.