MathDB
Problems
Contests
National and Regional Contests
China Contests
China Team Selection Test
1992 China Team Selection Test
1992 China Team Selection Test
Part of
China Team Selection Test
Subcontests
(3)
3
2
Hide problems
p | (x^2_0 - x_0 + 3)
For any prime
p
p
p
, prove that there exists integer
x
0
x_0
x
0
such that
p
∣
(
x
0
2
−
x
0
+
3
)
p | (x^2_0 - x_0 + 3)
p
∣
(
x
0
2
−
x
0
+
3
)
⇔
\Leftrightarrow
⇔
there exists integer
y
0
y_0
y
0
such that
p
∣
(
y
0
2
−
y
0
+
25
)
.
p | (y^2_0 - y_0 + 25).
p
∣
(
y
0
2
−
y
0
+
25
)
.
China TST 1992 inequality
For any
n
,
T
≥
2
,
n
,
T
∈
N
n,T \geq 2, n, T \in \mathbb{N}
n
,
T
≥
2
,
n
,
T
∈
N
, find all
a
∈
N
a \in \mathbb{N}
a
∈
N
such that
∀
a
i
>
0
,
i
=
1
,
2
,
…
,
n
\forall a_i > 0, i = 1, 2, \ldots, n
∀
a
i
>
0
,
i
=
1
,
2
,
…
,
n
, we have
∑
k
=
1
n
a
⋅
k
+
a
2
4
S
k
<
T
2
⋅
∑
k
=
1
n
1
a
k
,
\sum^n_{k=1} \frac{a \cdot k + \frac{a^2}{4}}{S_k} < T^2 \cdot \sum^n_{k=1} \frac{1}{a_k},
k
=
1
∑
n
S
k
a
⋅
k
+
4
a
2
<
T
2
⋅
k
=
1
∑
n
a
k
1
,
where
S
k
=
∑
i
=
1
k
a
i
.
S_k = \sum^k_{i=1} a_i.
S
k
=
∑
i
=
1
k
a
i
.
2
2
Hide problems
sum equality implies product-sum inequality
Let
n
≥
2
,
n
∈
N
,
n \geq 2, n \in \mathbb{N},
n
≥
2
,
n
∈
N
,
find the least positive real number
λ
\lambda
λ
such that for arbitrary
a
i
∈
R
a_i \in \mathbb{R}
a
i
∈
R
with
i
=
1
,
2
,
…
,
n
i = 1, 2, \ldots, n
i
=
1
,
2
,
…
,
n
and
b
i
∈
[
0
,
1
2
]
b_i \in \left[0, \frac{1}{2}\right]
b
i
∈
[
0
,
2
1
]
with
i
=
1
,
2
,
…
,
n
i = 1, 2, \ldots, n
i
=
1
,
2
,
…
,
n
, the following holds:
∑
i
=
1
n
a
i
=
∑
i
=
1
n
b
i
=
1
⇒
∏
i
=
1
n
a
i
≤
λ
∑
i
=
1
n
a
i
b
i
.
\sum^n_{i=1} a_i = \sum^n_{i=1} b_i = 1 \Rightarrow \prod^n_{i=1} a_i \leq \lambda \sum^n_{i=1} a_i b_i.
i
=
1
∑
n
a
i
=
i
=
1
∑
n
b
i
=
1
⇒
i
=
1
∏
n
a
i
≤
λ
i
=
1
∑
n
a
i
b
i
.
(3n + 1) * (3n + 1) table
A
(
3
n
+
1
)
×
(
3
n
+
1
)
(3n + 1) \times (3n + 1)
(
3
n
+
1
)
×
(
3
n
+
1
)
table
(
n
∈
N
)
(n \in \mathbb{N})
(
n
∈
N
)
is given. Prove that deleting any one of its squares yields a shape cuttable into pieces of the following form and its rotations: ''L" shape formed by cutting one square from a
2
×
2
2 \times 2
2
×
2
squares.
1
2
Hide problems
16 students took part in a competition
16 students took part in a competition. All problems were multiple choice style. Each problem had four choices. It was said that any two students had at most one answer in common, find the maximum number of problems.
ABC is congruent to triangle A'B'C'
A triangle
A
B
C
ABC
A
BC
is given in the plane with
A
B
=
7
,
AB = \sqrt{7},
A
B
=
7
,
B
C
=
13
BC = \sqrt{13}
BC
=
13
and
C
A
=
19
,
CA = \sqrt{19},
C
A
=
19
,
circles are drawn with centers at
A
,
B
A,B
A
,
B
and
C
C
C
and radii
1
3
,
\frac{1}{3},
3
1
,
2
3
\frac{2}{3}
3
2
and
1
,
1,
1
,
respectively. Prove that there are points
A
′
,
B
′
,
C
′
A',B',C'
A
′
,
B
′
,
C
′
on these three circles respectively such that triangle
A
B
C
ABC
A
BC
is congruent to triangle
A
′
B
′
C
′
.
A'B'C'.
A
′
B
′
C
′
.