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Problems
Contests
National and Regional Contests
China Contests
National High School Mathematics League
1992 National High School Mathematics League
1992 National High School Mathematics League
Part of
National High School Mathematics League
Subcontests
(15)
15
1
Hide problems
Mathematical Induction
n
n
n
is a natural number,
f
n
(
x
)
=
x
n
+
1
−
x
−
n
−
1
x
−
x
−
1
(
x
≠
0
,
±
1
)
f_n(x)=\frac{x^{n+1}-x^{-n-1}}{x-x^{-1}}(x\neq0,\pm1)
f
n
(
x
)
=
x
−
x
−
1
x
n
+
1
−
x
−
n
−
1
(
x
=
0
,
±
1
)
, let
y
=
x
+
1
x
y=x+\frac{1}{x}
y
=
x
+
x
1
. (a) Prove that
f
n
+
1
(
x
)
=
y
f
n
(
x
)
−
f
n
−
1
(
x
)
f_{n+1}(x)=yf_n(x)-f_{n-1}(x)
f
n
+
1
(
x
)
=
y
f
n
(
x
)
−
f
n
−
1
(
x
)
(b) Prove with mathematical induction:
f
n
(
x
)
=
{
y
n
−
C
n
−
1
1
y
n
−
2
+
⋯
+
(
−
1
)
i
C
n
−
i
i
y
n
−
2
i
+
⋯
+
(
−
1
)
n
2
(
i
=
1
,
2
,
⋯
,
n
2
,
n
is even
)
y
n
−
C
n
−
1
1
y
n
−
2
+
⋯
+
(
−
1
)
i
C
n
−
i
i
y
n
−
2
i
+
⋯
+
(
−
1
)
n
−
1
2
C
n
+
1
2
n
−
1
2
y
(
i
=
1
,
2
,
⋯
,
n
−
1
2
,
n
is odd
)
f_n(x)=\begin{cases} y^n-\text{C}_{n-1}^{1}y^{n-2}+\cdots+(-1)^i\text{C}_{n-i}^{i}y^{n-2i}+\cdots+(-1)^{\frac{n}{2}}(i=1,2,\cdots,\frac{n}{2},n\text{ is even})\\ y^n-\text{C}_{n-1}^{1}y^{n-2}+\cdots+(-1)^i\text{C}_{n-i}^{i}y^{n-2i}+\cdots+(-1)^{\frac{n-1}{2}}\text{C}_{\frac{n+1}{2}}^{\frac{n-1}{2}}y(i=1,2,\cdots,\frac{n-1}{2},n\text{ is odd}) \end{cases}
f
n
(
x
)
=
{
y
n
−
C
n
−
1
1
y
n
−
2
+
⋯
+
(
−
1
)
i
C
n
−
i
i
y
n
−
2
i
+
⋯
+
(
−
1
)
2
n
(
i
=
1
,
2
,
⋯
,
2
n
,
n
is even
)
y
n
−
C
n
−
1
1
y
n
−
2
+
⋯
+
(
−
1
)
i
C
n
−
i
i
y
n
−
2
i
+
⋯
+
(
−
1
)
2
n
−
1
C
2
n
+
1
2
n
−
1
y
(
i
=
1
,
2
,
⋯
,
2
n
−
1
,
n
is odd
)
.
14
1
Hide problems
3D Geometry Problem
l
,
m
l,m
l
,
m
are skew lines. Three points
A
,
B
,
C
A,B,C
A
,
B
,
C
on line
l
l
l
satisfy that
A
B
=
B
C
AB=BC
A
B
=
BC
. Projection of
A
,
B
,
C
A,B,C
A
,
B
,
C
on
m
m
m
are
D
,
E
,
F
D,E,F
D
,
E
,
F
. If
∣
A
D
∣
=
15
,
∣
B
E
∣
=
7
2
∣
C
F
∣
=
10
|AD|=\sqrt{15},|BE|=\frac{7}{2}|CF|=\sqrt{10}
∣
A
D
∣
=
15
,
∣
BE
∣
=
2
7
∣
CF
∣
=
10
, find the distance between
l
l
l
and
m
m
m
.
13
1
Hide problems
A Very Old Problem
Prove that
16
<
∑
i
=
1
80
1
i
<
17
16<\sum_{i=1}^{80}\frac{1}{\sqrt{i}}<17
16
<
∑
i
=
1
80
i
1
<
17
.
12
1
Hide problems
The Maximum Value
The maximum value of function
f
(
x
)
=
x
4
−
3
x
2
−
6
x
+
13
−
x
4
−
x
2
+
1
f(x)=\sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}
f
(
x
)
=
x
4
−
3
x
2
−
6
x
+
13
−
x
4
−
x
2
+
1
is________.
11
1
Hide problems
Recurrence Sequence
For real numbers
a
1
,
a
2
,
⋯
,
a
100
a_1,a_2,\cdots,a_{100}
a
1
,
a
2
,
⋯
,
a
100
,
a
1
=
a
2
=
1
,
a
3
=
2
a_1=a_2=1,a_3=2
a
1
=
a
2
=
1
,
a
3
=
2
. For any positive integer
n
n
n
,
a
n
a
n
+
1
a
n
+
2
≠
1
,
a
n
a
n
+
1
a
n
+
2
a
n
+
3
=
a
n
+
a
n
+
1
+
a
n
+
2
+
a
n
+
3
a_na_{n+1}a_{n+2}\neq1,a_na_{n+1}a_{n+2}a_{n+3}=a_n+a_{n+1}+a_{n+2}+a_{n+3}
a
n
a
n
+
1
a
n
+
2
=
1
,
a
n
a
n
+
1
a
n
+
2
a
n
+
3
=
a
n
+
a
n
+
1
+
a
n
+
2
+
a
n
+
3
, then
a
1
+
a
2
+
⋯
+
a
100
=
a_1+a_2+\cdots+a_{100}=
a
1
+
a
2
+
⋯
+
a
100
=
________.
10
1
Hide problems
Complex Problem
z
1
,
z
2
z_1,z_2
z
1
,
z
2
are complex numbers.
∣
z
1
∣
=
3
,
∣
z
2
∣
=
5
,
∣
z
1
+
z
2
∣
=
7
|z_1|=3,|z_2|=5,|z_1+z_2|=7
∣
z
1
∣
=
3
,
∣
z
2
∣
=
5
,
∣
z
1
+
z
2
∣
=
7
, then
arg
(
z
2
z
1
)
3
=
\arg(\frac{z_2}{z_1})^3=
ar
g
(
z
1
z
2
)
3
=
________.
9
1
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Skew Lines
From eight edges and eight diagonal of surfaces of a cube, choose
k
k
k
lines. If any two lines of them are skew lines, then the maximum value of
k
k
k
is________.
8
1
Hide problems
Trigonometric Equation
Then number of solutions to
cos
7
x
=
cos
5
x
\cos7x=\cos5x
cos
7
x
=
cos
5
x
in
[
0
,
π
]
[0,\pi]
[
0
,
π
]
is________.
7
1
Hide problems
Arithmetic Sequence and Geometric Series
For real numbers
x
,
y
,
z
x,y,z
x
,
y
,
z
,
3
x
,
4
y
,
5
z
3x,4y,5z
3
x
,
4
y
,
5
z
are geometric series,
1
x
,
1
y
,
1
z
\frac{1}{x},\frac{1}{y},\frac{1}{z}
x
1
,
y
1
,
z
1
are arithmetic sequence. Then
x
z
+
z
x
=
\frac{x}{z}+\frac{z}{x}=
z
x
+
x
z
=
________.
6
1
Hide problems
Function Problem
f
(
x
)
f(x)
f
(
x
)
is a function defined on
R
\mathbb{R}
R
, satisfying:
f
(
10
+
x
)
=
f
(
10
−
x
)
,
f
(
20
−
x
)
=
−
f
(
20
+
x
)
f(10+x)=f(10-x),f(20-x)=-f(20+x)
f
(
10
+
x
)
=
f
(
10
−
x
)
,
f
(
20
−
x
)
=
−
f
(
20
+
x
)
. Then,
f
(
x
)
f(x)
f
(
x
)
is
(A)
\text{(A)}
(A)
even function, but not periodic function
(B)
\text{(B)}
(B)
even function, and periodic function
(C)
\text{(C)}
(C)
odd function, but not periodic function
(D)
\text{(D)}
(D)
odd function, and periodic function
5
1
Hide problems
Complex Plane
Points on complex plane that complex numbers
z
1
,
z
2
z_1,z_2
z
1
,
z
2
corresponding to are
A
,
B
A,B
A
,
B
, and
∣
z
1
∣
=
4
,
4
z
1
2
−
2
z
1
z
2
+
z
2
2
=
0
|z_1|=4,4z_1^2-2z_1z_2+z_2^2=0
∣
z
1
∣
=
4
,
4
z
1
2
−
2
z
1
z
2
+
z
2
2
=
0
.
O
O
O
is original point, then the area of
△
O
A
B
\triangle OAB
△
O
A
B
is
(A)
8
3
(B)
4
3
(C)
6
3
(D)
12
3
\text{(A)}8\sqrt3\qquad\text{(B)}4\sqrt3\qquad\text{(C)}6\sqrt3\qquad\text{(D)}12\sqrt3
(A)
8
3
(B)
4
3
(C)
6
3
(D)
12
3
4
1
Hide problems
Triangle Problem
In
△
A
B
C
\triangle ABC
△
A
BC
,
b
≠
1
b\neq1
b
=
1
. If
C
A
\frac{C}{A}
A
C
and
sin
B
sin
A
\frac{\sin B}{\sin A}
s
i
n
A
s
i
n
B
are solutions to equation
log
b
x
=
log
b
(
4
x
−
4
)
\log_{\sqrt{b}}x=\log_{b}(4x-4)
lo
g
b
x
=
lo
g
b
(
4
x
−
4
)
, then
△
A
B
C
\triangle ABC
△
A
BC
(A)
\text{(A)}
(A)
is an isosceles triangle, but not right-angled triangle
(B)
\text{(B)}
(B)
is a right-angled triangle, but not isosceles triangle
(C)
\text{(C)}
(C)
is an isosceles right-angled triangle
(D)
\text{(D)}
(D)
is neither a right-angled triangle nor an isosceles triangle
3
2
Hide problems
Tetrahedron Problem
Areas of four surfaces of a tetrahedron are
S
1
,
S
2
,
S
3
,
S
4
S_1,S_2,S_3,S_4
S
1
,
S
2
,
S
3
,
S
4
. And the largest one of them is
S
S
S
.
λ
=
S
1
+
S
2
+
S
3
+
S
4
S
\lambda=\frac{S_1+S_2+S_3+S_4}{S}
λ
=
S
S
1
+
S
2
+
S
3
+
S
4
, then
λ
\lambda
λ
always satisfies
(A)
2
<
λ
≤
4
(B)
3
<
λ
<
4
(C)
2.5
<
λ
≤
4.5
(D)
3.5
<
λ
<
5.5
\text{(A)}2<\lambda\leq4\qquad\text{(B)}3<\lambda<4\qquad\text{(C)}2.5<\lambda\leq4.5\qquad\text{(D)}3.5<\lambda<5.5
(A)
2
<
λ
≤
4
(B)
3
<
λ
<
4
(C)
2.5
<
λ
≤
4.5
(D)
3.5
<
λ
<
5.5
Area Smaller, Smaller...
In coordinate system, there are six points
P
i
(
x
i
,
y
i
)
(
i
=
1
,
2
,
⋯
,
6
)
P_i(x_i,y_i)(i=1,2,\cdots,6)
P
i
(
x
i
,
y
i
)
(
i
=
1
,
2
,
⋯
,
6
)
, satisfying: (1)
x
i
,
y
i
∈
{
−
2
,
−
1
,
0
,
1
,
2
}
x_i,y_i\in\{-2,-1,0,1,2\}
x
i
,
y
i
∈
{
−
2
,
−
1
,
0
,
1
,
2
}
. (2) For any three points, they are not collinear. Prove that there exists a triangle
△
P
i
P
j
P
k
(
1
≤
i
<
j
<
k
≤
6
)
\triangle P_iP_jP_k(1\leq i<j<k\leq6)
△
P
i
P
j
P
k
(
1
≤
i
<
j
<
k
≤
6
)
, its area is not larger than
2
2
2
.
2
2
Hide problems
Part of Unit Circle
The equation of unit circle in Quadrant I, III, IV (
(
−
1
,
0
)
,
(
1
,
0
)
,
(
0
,
−
1
)
,
(
0
,
1
)
(-1,0),(1,0),(0,-1),(0,1)
(
−
1
,
0
)
,
(
1
,
0
)
,
(
0
,
−
1
)
,
(
0
,
1
)
included) is
(A)
(
x
+
1
−
y
2
)
(
y
+
1
−
x
2
)
=
0
\text{(A)}(x+\sqrt{1-y^2})(y+\sqrt{1-x^2})=0
(A)
(
x
+
1
−
y
2
)
(
y
+
1
−
x
2
)
=
0
(B)
(
x
−
1
−
y
2
)
(
y
−
1
−
x
2
)
=
0
\text{(B)}(x-\sqrt{1-y^2})(y-\sqrt{1-x^2})=0
(B)
(
x
−
1
−
y
2
)
(
y
−
1
−
x
2
)
=
0
(C)
(
x
+
1
−
y
2
)
(
y
−
1
−
x
2
)
=
0
\text{(C)}(x+\sqrt{1-y^2})(y-\sqrt{1-x^2})=0
(C)
(
x
+
1
−
y
2
)
(
y
−
1
−
x
2
)
=
0
(D)
(
x
−
1
−
y
2
)
(
y
+
1
−
x
2
)
=
0
\text{(D)}(x-\sqrt{1-y^2})(y+\sqrt{1-x^2})=0
(D)
(
x
−
1
−
y
2
)
(
y
+
1
−
x
2
)
=
0
Capacity
Define set
S
n
=
{
1
,
2
,
⋯
,
n
}
S_n=\{1,2,\cdots,n\}
S
n
=
{
1
,
2
,
⋯
,
n
}
.
X
X
X
is a subset of
S
n
S_n
S
n
. We call sum of all numbers in
X
X
X
capacity (capacity of empty set is
0
0
0
). If capacity of
X
X
X
is odd/even, then we call it odd/even subset. (a) Prove that the number of odd subsets and even subsets of
S
n
S_n
S
n
are the same. (b) Prove that the sum of capacity of all odd subsets and even subsets are the same when
n
≥
3
n\geq3
n
≥
3
. (c) Calculate the sum of capacity of all odd subsets when
n
≥
3
n\geq3
n
≥
3
.
1
2
Hide problems
Parabola Problem
For any positive integer
n
n
n
,
A
n
A_n
A
n
and
B
n
B_n
B
n
are intersection of parabola
y
=
(
n
2
+
n
)
x
2
−
(
2
n
+
1
)
x
+
1
y=(n^2+n)x^2-(2n+1)x+1
y
=
(
n
2
+
n
)
x
2
−
(
2
n
+
1
)
x
+
1
and
x
x
x
-axis. Then, the value of
∣
A
1
B
1
∣
+
∣
A
2
B
2
∣
+
⋯
+
∣
A
1992
B
1992
∣
|A_1B_1|+|A_2B_2|+\cdots+|A_{1992}B_{1992}|
∣
A
1
B
1
∣
+
∣
A
2
B
2
∣
+
⋯
+
∣
A
1992
B
1992
∣
is
(A)
1991
1992
(B)
1992
1993
(C)
1991
1993
(D)
1993
1992
\text{(A)}\frac{1991}{1992}\qquad\text{(B)}\frac{1992}{1993}\qquad\text{(C)}\frac{1991}{1993}\qquad\text{(D)}\frac{1993}{1992}
(A)
1992
1991
(B)
1993
1992
(C)
1993
1991
(D)
1992
1993
Geometry~
A
1
A
2
A
3
A
4
A_1A_2A_3A_4
A
1
A
2
A
3
A
4
is cyclic quadrilateral of
⊙
O
\odot O
⊙
O
.
H
1
,
H
2
,
H
3
,
H
4
H_1,H_2,H_3,H_4
H
1
,
H
2
,
H
3
,
H
4
are orthocentres of
△
A
2
A
3
A
4
,
△
A
3
A
4
A
1
,
△
A
4
A
1
A
2
,
△
A
1
A
2
A
3
\triangle A_2A_3A_4,\triangle A_3A_4A_1,\triangle A_4A_1A_2,\triangle A_1A_2A_3
△
A
2
A
3
A
4
,
△
A
3
A
4
A
1
,
△
A
4
A
1
A
2
,
△
A
1
A
2
A
3
. Prove that
H
1
,
H
2
,
H
3
,
H
4
H_1,H_2,H_3,H_4
H
1
,
H
2
,
H
3
,
H
4
are concyclic, and determine its center.