MathDB
Problems
Contests
National and Regional Contests
China Contests
National High School Mathematics League
1989 National High School Mathematics League
1989 National High School Mathematics League
Part of
National High School Mathematics League
Subcontests
(15)
15
1
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Inequality or Not?
For any positive integer
n
n
n
,
a
n
>
0
a_n>0
a
n
>
0
, and
∑
j
=
1
n
a
j
3
=
(
∑
j
=
1
n
a
j
)
2
\sum_{j=1}^{n}a_j^3=\left(\sum_{j=1}^{n}a_j\right)^2
∑
j
=
1
n
a
j
3
=
(
∑
j
=
1
n
a
j
)
2
. Prove that
a
n
=
n
a_n=n
a
n
=
n
14
1
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3D Geometry Problem
In regular triangular pyramid
S
−
A
B
C
S-ABC
S
−
A
BC
, hieght
S
O
=
3
SO=3
SO
=
3
, length of sides of bottom surface is
6
6
6
. Projection of
A
A
A
on plane
S
B
C
SBC
SBC
is
O
′
O'
O
′
.
P
∈
A
O
′
,
A
P
P
O
′
=
8
P\in AO',\frac{AP}{PO'}=8
P
∈
A
O
′
,
P
O
′
A
P
=
8
. Draw a plane parallel to plane
A
B
C
ABC
A
BC
and passes
P
P
P
. Find the area of the cross section.
13
1
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Inequality
a
1
,
a
2
,
⋯
,
a
n
a_1,a_2,\cdots,a_n
a
1
,
a
2
,
⋯
,
a
n
are positive numbers, satisfying that
a
1
a
2
⋯
a
n
=
1
a_1a_2\cdots a_n=1
a
1
a
2
⋯
a
n
=
1
. Prove that
(
2
+
a
1
)
(
2
+
a
2
)
⋯
(
2
+
a
n
)
≥
3
n
(2+a_1)(2+a_2)\cdots(2+a_n)\geq3^n
(
2
+
a
1
)
(
2
+
a
2
)
⋯
(
2
+
a
n
)
≥
3
n
12
1
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The Minumum Value
s
,
t
∈
R
s,t\in\mathbb{R}
s
,
t
∈
R
, then the minumum value of
(
s
+
5
−
3
∣
cos
t
∣
)
2
+
(
s
−
2
∣
sin
t
∣
)
2
(s+5-3|\cos t|)^2+(s-2|\sin t|)^2
(
s
+
5
−
3∣
cos
t
∣
)
2
+
(
s
−
2∣
sin
t
∣
)
2
is________.
11
1
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Ways of taking out numbers
From
1
,
2
,
⋯
,
14
1,2,\cdots,14
1
,
2
,
⋯
,
14
, take out three numbers
a
1
<
a
2
<
a
3
a_1<a_2<a_3
a
1
<
a
2
<
a
3
, satisfying that
a
2
−
a
1
≥
3
,
a
3
−
a
2
≥
3
a_2-a_1\geq3,a_3-a_2\geq3
a
2
−
a
1
≥
3
,
a
3
−
a
2
≥
3
. Then the number of different ways of taking out numbers is________.
10
1
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Gaussian Function
A positive number, if its fractional part, integeral part, and itself are geometric series, then the number is________.
9
1
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Three Functions
Functions
f
0
(
x
)
=
∣
x
∣
,
f
1
(
x
)
=
∣
f
0
(
x
)
−
1
∣
,
f
2
(
x
)
=
∣
f
1
(
x
)
−
2
∣
f_0(x)=|x|,f_1(x)=|f_0(x)-1|,f_2(x)=|f_1(x)-2|
f
0
(
x
)
=
∣
x
∣
,
f
1
(
x
)
=
∣
f
0
(
x
)
−
1∣
,
f
2
(
x
)
=
∣
f
1
(
x
)
−
2∣
. Area of the closed part between the figure of
f
2
(
x
)
f_2(x)
f
2
(
x
)
and
x
x
x
-axis is________.
8
1
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Coordinate
Line
l
:
2
x
+
y
=
10
l:2x+y=10
l
:
2
x
+
y
=
10
, line
l
′
l'
l
′
passes
(
−
10
,
0
)
(-10,0)
(
−
10
,
0
)
, and
l
′
⊥
l
l'\perp l
l
′
⊥
l
, then the coordinate of the intersection of
l
l
l
and
l
′
l'
l
′
is________.
7
1
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Solve the Inequality
If
log
a
2
<
1
\log_{a}\sqrt2<1
lo
g
a
2
<
1
, then the range value of
a
a
a
is________.
6
1
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Two Sets Again
Set
M
=
{
u
∣
u
=
12
m
+
8
n
+
4
l
,
m
,
n
,
l
∈
Z
}
,
N
=
{
u
∣
u
=
20
p
+
16
q
+
12
x
,
p
,
q
,
x
∈
Z
}
M=\{u|u=12m+8n+4l,m,n,l\in\mathbb{Z}\},N=\{u|u=20p+16q+12x,p,q,x\in\mathbb{Z}\}
M
=
{
u
∣
u
=
12
m
+
8
n
+
4
l
,
m
,
n
,
l
∈
Z
}
,
N
=
{
u
∣
u
=
20
p
+
16
q
+
12
x
,
p
,
q
,
x
∈
Z
}
. Then
(A)
M
=
N
(B)
M
⊄
N
,
N
⊄
M
(C)
M
⊂
N
(D)
N
⊂
M
\text{(A)}M=N\qquad\text{(B)}M\not\subset N,N\not\subset M\qquad\text{(C)}M\subset N\qquad\text{(D)}N\subset M
(A)
M
=
N
(B)
M
⊂
N
,
N
⊂
M
(C)
M
⊂
N
(D)
N
⊂
M
5
1
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Two Sets
If
M
=
{
z
∈
C
∣
z
=
t
1
+
t
+
i
1
+
t
t
,
t
∈
R
,
t
≠
0
,
t
≠
−
1
}
M=\{z\in\mathbb{C}|z=\frac{t}{1+t}+\text{i}\frac{1+t}{t},t\in\mathbb{R},t\neq0,t\neq-1\}
M
=
{
z
∈
C
∣
z
=
1
+
t
t
+
i
t
1
+
t
,
t
∈
R
,
t
=
0
,
t
=
−
1
}
,
N
=
{
z
∈
C
∣
z
=
2
[
cos
(
arcsin
t
)
+
i
cos
(
arccos
t
)
]
,
t
∈
R
,
∣
t
∣
≤
1
}
N=\{z\in\mathbb{C}|z=\sqrt2[\cos(\arcsin t)+\text{i}\cos(\arccos t)],t\in\mathbb{R},|t|\leq1\}
N
=
{
z
∈
C
∣
z
=
2
[
cos
(
arcsin
t
)
+
i
cos
(
arccos
t
)]
,
t
∈
R
,
∣
t
∣
≤
1
}
, then
∣
M
∩
N
∣
|M\cap N|
∣
M
∩
N
∣
is
(A)
0
(B)
1
(C)
2
(D)
4
\text{(A)}0\qquad\text{(B)}1\qquad\text{(C)}2\qquad\text{(D)}4
(A)
0
(B)
1
(C)
2
(D)
4
4
1
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Count the Number of Triangles
Three points of a triangle are among 8 vertex of a cube. So the number of such acute triangles is
(A)
0
(B)
6
(C)
8
(D)
24
\text{(A)}0\qquad\text{(B)}6\qquad\text{(C)}8\qquad\text{(D)}24
(A)
0
(B)
6
(C)
8
(D)
24
3
2
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Figure of Function
For any function
f
(
x
)
f(x)
f
(
x
)
, in the same rectangular coordinates, figures of function
y
=
f
(
x
−
1
)
y=f(x-1)
y
=
f
(
x
−
1
)
and
y
=
f
(
−
x
+
1
)
y=f(-x+1)
y
=
f
(
−
x
+
1
)
(A)
\text{(A)}
(A)
are symmetrical about
x
x
x
-axis
(B)
\text{(B)}
(B)
are symmetrical about line
x
=
1
x=1
x
=
1
(C)
\text{(C)}
(C)
are symmetrical about line
x
=
−
1
x=-1
x
=
−
1
(D)
\text{(D)}
(D)
are symmetrical about
y
y
y
-axis
Square Table
Given
n
×
n
n\times n
n
×
n
square table, fill in each square with
−
1
-1
−
1
or
1
1
1
. For
n
n
n
squares that any two of them are neither in the same line nor in the same column, we call the product of the numbers in the squares basic term. Prove that she sum of all basic terms is always a multiple of
4
4
4
.
2
2
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Anti-trigonometric Function
Range of function
f
(
x
)
=
arctan
x
+
1
2
arcsin
x
f(x)=\arctan x+\frac{1}{2}\arcsin x
f
(
x
)
=
arctan
x
+
2
1
arcsin
x
is
(A)
(
−
π
,
π
)
(B)
[
−
3
4
π
,
3
4
π
]
(C)
(
−
3
4
π
,
3
4
π
)
(D)
[
−
1
2
π
,
1
2
π
]
\text{(A)}(-\pi,\pi)\qquad\text{(B)}[-\frac{3}{4}\pi,\frac{3}{4}\pi]\qquad\text{(C)}(-\frac{3}{4}\pi,\frac{3}{4}\pi)\qquad\text{(D)}[-\frac{1}{2}\pi,\frac{1}{2}\pi]
(A)
(
−
π
,
π
)
(B)
[
−
4
3
π
,
4
3
π
]
(C)
(
−
4
3
π
,
4
3
π
)
(D)
[
−
2
1
π
,
2
1
π
]
Inequality
x
i
∈
R
(
i
=
1
,
2
,
⋯
,
n
;
n
≥
2
)
x_i\in\mathbb{R}(i=1,2,\cdots,n;n\geq2)
x
i
∈
R
(
i
=
1
,
2
,
⋯
,
n
;
n
≥
2
)
, satisfying that
∑
i
=
1
n
∣
x
i
∣
=
1
,
∑
i
=
1
n
x
i
=
0
\sum_{i=1}^n|x_i|=1,\sum_{i=1}^nx_i=0
∑
i
=
1
n
∣
x
i
∣
=
1
,
∑
i
=
1
n
x
i
=
0
. Prove that
∣
∑
i
=
1
n
x
i
i
∣
≤
1
2
−
1
2
n
|\sum_{i=1}^n\frac{x_i}{i}|\leq\frac{1}{2}-\frac{1}{2n}
∣
∑
i
=
1
n
i
x
i
∣
≤
2
1
−
2
n
1
1
2
Hide problems
Complex Plane
On complex plane, if
A
,
B
A,B
A
,
B
are two angles of acute triangle
A
B
C
ABC
A
BC
, then the point
z
=
(
cos
B
−
sin
A
)
+
i
(
sin
B
−
cos
A
)
z=(\cos B-\sin A)+\text{i}(\sin B-\cos A)
z
=
(
cos
B
−
sin
A
)
+
i
(
sin
B
−
cos
A
)
corresponding to is in
(A)
\text{(A)}
(A)
Quadrant I
(B)
\text{(B)}
(B)
Quadrant II
(C)
\text{(C)}
(C)
Quadrant III
(D)
\text{(D)}
(D)
Quadrant IV
Geometry Problem
In
△
A
B
C
\triangle ABC
△
A
BC
,
A
B
>
A
C
AB>AC
A
B
>
A
C
, bisector of outer angle
∠
A
\angle A
∠
A
intersects circumcircle of
△
A
B
C
\triangle ABC
△
A
BC
at
E
E
E
. Projection of
E
E
E
on
A
B
AB
A
B
is
F
F
F
. Prove that
2
A
F
=
A
B
−
A
C
2AF=AB-AC
2
A
F
=
A
B
−
A
C
.