MathDB
Problems
Contests
National and Regional Contests
China Contests
China Team Selection Test
1989 China Team Selection Test
1989 China Team Selection Test
Part of
China Team Selection Test
Subcontests
(4)
4
2
Hide problems
P_1 Q_1 R_1 is congruent to triangle P_2 Q_2 R_2
Given triangle
A
B
C
ABC
A
BC
, squares
A
B
E
F
,
B
C
G
H
,
C
A
I
J
ABEF, BCGH, CAIJ
A
BEF
,
BCG
H
,
C
A
I
J
are constructed externally on side
A
B
,
B
C
,
C
A
AB, BC, CA
A
B
,
BC
,
C
A
, respectively. Let
A
H
∩
B
J
=
P
1
AH \cap BJ = P_1
A
H
∩
B
J
=
P
1
,
B
J
∩
C
F
=
Q
1
BJ \cap CF = Q_1
B
J
∩
CF
=
Q
1
,
C
F
∩
A
H
=
R
1
CF \cap AH = R_1
CF
∩
A
H
=
R
1
,
A
G
∩
C
E
=
P
2
AG \cap CE = P_2
A
G
∩
CE
=
P
2
,
B
I
∩
A
G
=
Q
2
BI \cap AG = Q_2
B
I
∩
A
G
=
Q
2
,
C
E
∩
B
I
=
R
2
CE \cap BI = R_2
CE
∩
B
I
=
R
2
. Prove that triangle
P
1
Q
1
R
1
P_1 Q_1 R_1
P
1
Q
1
R
1
is congruent to triangle
P
2
Q
2
R
2
P_2 Q_2 R_2
P
2
Q
2
R
2
.
Sum of partitions
∀
n
∈
N
\forall n \in \mathbb{N}
∀
n
∈
N
,
P
(
n
)
P(n)
P
(
n
)
denotes the number of the partition of
n
n
n
as the sum of positive integers (disregarding the order of the parts), e.g. since
4
=
1
+
1
+
1
+
1
=
1
+
1
+
2
=
1
+
3
=
2
+
2
=
4
4 = 1+1+1+1 = 1+1+2 = 1+3 = 2+2 = 4
4
=
1
+
1
+
1
+
1
=
1
+
1
+
2
=
1
+
3
=
2
+
2
=
4
, so
P
(
4
)
=
5
P(4)=5
P
(
4
)
=
5
. "Dispersion" of a partition denotes the number of different parts in that partitation. And denote
q
(
n
)
q(n)
q
(
n
)
is the sum of all the dispersions, e.g.
q
(
4
)
=
1
+
2
+
2
+
1
+
1
=
7
q(4)=1+2+2+1+1=7
q
(
4
)
=
1
+
2
+
2
+
1
+
1
=
7
.
n
≥
1
n \geq 1
n
≥
1
. Prove that (1)
q
(
n
)
=
1
+
∑
i
=
1
n
−
1
P
(
i
)
.
q(n) = 1 + \sum^{n-1}_{i=1} P(i).
q
(
n
)
=
1
+
∑
i
=
1
n
−
1
P
(
i
)
.
(2)
1
+
∑
i
=
1
n
−
1
P
(
i
)
≤
2
⋅
n
⋅
P
(
n
)
1 + \sum^{n-1}_{i=1} P(i) \leq \sqrt{2} \cdot n \cdot P(n)
1
+
∑
i
=
1
n
−
1
P
(
i
)
≤
2
⋅
n
⋅
P
(
n
)
.
3
2
Hide problems
non-zero roots in the unitary circumference
Find the greatest
n
n
n
such that
(
z
+
1
)
n
=
z
n
+
1
(z+1)^n = z^n + 1
(
z
+
1
)
n
=
z
n
+
1
has all its non-zero roots in the unitary circumference, e.g.
(
α
+
1
)
n
=
α
n
+
1
,
α
≠
0
(\alpha+1)^n = \alpha^n + 1, \alpha \neq 0
(
α
+
1
)
n
=
α
n
+
1
,
α
=
0
implies
∣
α
∣
=
1.
|\alpha| = 1.
∣
α
∣
=
1.
1989 equal circles are arbitrarily placed on the table
1989
1989
1989
equal circles are arbitrarily placed on the table without overlap. What is the least number of colors are needed such that all the circles can be painted with any two tangential circles colored differently.
2
2
Hide problems
There no terms with 3^x*5^y
Let
v
0
=
0
,
v
1
=
1
v_0 = 0, v_1 = 1
v
0
=
0
,
v
1
=
1
and
v
n
+
1
=
8
⋅
v
n
−
v
n
−
1
,
v_{n+1} = 8 \cdot v_n - v_{n-1},
v
n
+
1
=
8
⋅
v
n
−
v
n
−
1
,
n
=
1
,
2
,
.
.
.
n = 1,2, ...
n
=
1
,
2
,
...
. Prove that in the sequence
{
v
n
}
\{v_n\}
{
v
n
}
there aren't terms of the form
3
α
⋅
5
β
3^{\alpha} \cdot 5^{\beta}
3
α
⋅
5
β
with
α
,
β
∈
N
.
\alpha, \beta \in \mathbb{N}.
α
,
β
∈
N
.
find the range of angle BAC
A
D
AD
A
D
is the altitude on side
B
C
BC
BC
of triangle
A
B
C
ABC
A
BC
. If
B
C
+
A
D
−
A
B
−
A
C
=
0
BC+AD-AB-AC = 0
BC
+
A
D
−
A
B
−
A
C
=
0
, find the range of
∠
B
A
C
\angle BAC
∠
B
A
C
. Alternative formulation. Let
A
D
AD
A
D
be the altitude of triangle
A
B
C
ABC
A
BC
to the side
B
C
BC
BC
. If
B
C
+
A
D
=
A
B
+
A
C
BC+AD=AB+AC
BC
+
A
D
=
A
B
+
A
C
, then find the range of
∠
A
\angle{A}
∠
A
.
1
2
Hide problems
maximum value of the coincident triangle area
A triangle of sides
3
2
,
5
2
,
2
\frac{3}{2}, \frac{\sqrt{5}}{2}, \sqrt{2}
2
3
,
2
5
,
2
is folded along a variable line perpendicular to the side of
3
2
.
\frac{3}{2}.
2
3
.
Find the maximum value of the coincident area.
Does there exist a function f^{1989}(n) = 2n
Let
N
=
{
1
,
2
,
…
}
.
\mathbb{N} = \{1,2, \ldots\}.
N
=
{
1
,
2
,
…
}
.
Does there exists a function
f
:
N
↦
N
f: \mathbb{N} \mapsto \mathbb{N}
f
:
N
↦
N
such that
∀
n
∈
N
,
\forall n \in \mathbb{N},
∀
n
∈
N
,
f
1989
(
n
)
=
2
⋅
n
f^{1989}(n) = 2 \cdot n
f
1989
(
n
)
=
2
⋅
n
?