MathDB
Problems
Contests
National and Regional Contests
China Contests
National High School Mathematics League
1984 National High School Mathematics League
1984 National High School Mathematics League
Part of
National High School Mathematics League
Subcontests
(10)
10
1
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Solve the Trigonometric Equation
All solutions of
cos
x
=
cos
x
4
\cos x=\cos\frac{x}{4}
cos
x
=
cos
4
x
are________, it has_________different solutions in
(
0
,
24
π
)
(0,24\pi)
(
0
,
24
π
)
.
9
1
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Easy Geometry
A
(
−
1
,
0
)
,
B
(
1
,
0
)
A(-1,0),B(1,0)
A
(
−
1
,
0
)
,
B
(
1
,
0
)
.
D
(
x
,
0
)
D(x,0)
D
(
x
,
0
)
is a point on
A
B
AB
A
B
.
C
D
⊥
A
B
CD\perp AB
C
D
⊥
A
B
, and
C
C
C
is a point on unit circle. When
x
∈
x\in
x
∈
________, segments
A
D
,
B
D
,
C
D
AD,BD,CD
A
D
,
B
D
,
C
D
can be three sides of a acute triangle.
8
1
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A Funcion about 3D Geometry
Lengths of five edges of a tetrahedron are
1
1
1
, while the last one is
x
x
x
. Its volume is
F
(
x
)
F(x)
F
(
x
)
. On its domain of definition, we have
(A)
\text{(A)}
(A)
F
(
x
)
F(x)
F
(
x
)
is an increasing function, it has no maximum value.
(B)
\text{(B)}
(B)
F
(
x
)
F(x)
F
(
x
)
is an increasing function, it has maximum value.
(C)
\text{(C)}
(C)
F
(
x
)
F(x)
F
(
x
)
is not an increasing function, it has no maximum value.
(D)
\text{(D)}
(D)
F
(
x
)
F(x)
F
(
x
)
is an increasing function, it has maximum value.
7
1
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Rotate!!!
A moving point
P
(
x
,
y
)
P(x,y)
P
(
x
,
y
)
rotate anticlockwise around unit circle, who seangular speed is
ω
\omega
ω
. Then how does
Q
(
−
2
x
y
,
y
2
−
x
2
)
Q(-2xy,y^2-x^2)
Q
(
−
2
x
y
,
y
2
−
x
2
)
moves?
(A)
\text{(A)}
(A)
Rotate clockwise around unit circle, who seangular speed is
ω
\omega
ω
.
(B)
\text{(B)}
(B)
Rotate anticlockwise around unit circle, who seangular speed is
ω
\omega
ω
.
(C)
\text{(C)}
(C)
Rotate clockwise around unit circle, who seangular speed is
2
ω
2\omega
2
ω
.
(D)
\text{(D)}
(D)
Rotate anticlockwise around unit circle, who seangular speed is
2
ω
2\omega
2
ω
.
6
1
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Some Equalities about Function
If
F
(
1
−
x
1
+
x
)
=
x
F(\frac{1-x}{1+x})=x
F
(
1
+
x
1
−
x
)
=
x
, then
(A)
F
(
−
2
−
x
)
=
−
2
−
F
(
x
)
(B)
F
(
−
x
)
=
F
(
1
+
x
1
−
x
)
\text{(A)}F(-2-x)=-2-F(x)\qquad\text{(B)}F(-x)=F(\frac{1+x}{1-x})
(A)
F
(
−
2
−
x
)
=
−
2
−
F
(
x
)
(B)
F
(
−
x
)
=
F
(
1
−
x
1
+
x
)
(C)
F
(
1
x
)
=
F
(
x
)
(D)
F
(
F
(
−
x
)
)
=
−
x
\text{(C)}F(\frac{1}{x})=F(x)\qquad\text{(D)}F(F(-x))=-x
(C)
F
(
x
1
)
=
F
(
x
)
(D)
F
(
F
(
−
x
))
=
−
x
5
2
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Odd Funcion or Not?
If
a
>
0
,
a
≠
1
a>0,a\neq1
a
>
0
,
a
=
1
,
F
(
x
)
F(x)
F
(
x
)
is an odd function.
G
(
x
)
=
F
(
x
)
⋅
(
1
a
x
−
1
+
1
2
)
G(x)=F(x)\cdot(\frac{1}{a^x-1}+\frac{1}{2})
G
(
x
)
=
F
(
x
)
⋅
(
a
x
−
1
1
+
2
1
)
, then
G
(
x
)
G(x)
G
(
x
)
is
(A)
\text{(A)}
(A)
odd function
(B)
\text{(B)}
(B)
even function
(C)
\text{(C)}
(C)
not odd or even function
(D)
\text{(D)}
(D)
not sure
Maybe the Easiest Inequality You've Seen!
x
1
,
x
2
,
⋯
,
x
n
x_1,x_2,\cdots,x_n
x
1
,
x
2
,
⋯
,
x
n
are positive real numbers. Prove that
x
1
2
x
2
+
x
2
2
x
3
+
⋯
+
x
n
2
x
1
≥
x
1
+
x
2
+
⋯
x
n
.
\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+\cdots+\frac{x_n^2}{x_1}\geq x_1+x_2+\cdots x_n.
x
2
x
1
2
+
x
3
x
2
2
+
⋯
+
x
1
x
n
2
≥
x
1
+
x
2
+
⋯
x
n
.
4
2
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The Number of Real Roots
The number of real roots of the equation
sin
x
=
lg
x
\sin x=\lg x
sin
x
=
l
g
x
is
(A)
1
(B)
2
(C)
3
(D)
\text{(A)}1\qquad\text{(B)}2\qquad\text{(C)}3\qquad\text{(D)}
(A)
1
(B)
2
(C)
3
(D)
more than
3
3
3
Complete Alternation
Define
a
n
a_n
a
n
: the last digit of
1
2
+
2
2
+
3
2
+
⋯
+
n
2
1^2+2^2+3^2+\cdots+n^2
1
2
+
2
2
+
3
2
+
⋯
+
n
2
. Prove that
0.
a
1
a
2
a
3
⋯
‾
\overline{0.a_1 a_2 a_3 \cdots}
0.
a
1
a
2
a
3
⋯
is a rational number.
3
2
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Polar Coordinates Problem
For any integers
1
≤
n
≤
m
≤
5
1\leq n\leq m\leq5
1
≤
n
≤
m
≤
5
, how many hyperbolas does the equation
ρ
=
1
1
−
C
m
n
cos
θ
\rho=\frac{1}{1-\text{C}_m^n \cos\theta}
ρ
=
1
−
C
m
n
c
o
s
θ
1
represent? Note:
C
m
n
=
m
!
n
!
(
m
−
n
)
!
\text{C}_m^n=\frac{m!}{n!(m-n)!}
C
m
n
=
n
!
(
m
−
n
)!
m
!
.
(A)
15
(B)
10
(C)
7
(D)
6
\text{(A)}15\qquad\text{(B)}10\qquad\text{(C)}7\qquad\text{(D)}6
(A)
15
(B)
10
(C)
7
(D)
6
Largest Area
In
△
A
B
C
\triangle ABC
△
A
BC
,
P
P
P
is a point on
B
C
BC
BC
.
F
∈
A
B
,
E
∈
A
C
,
P
F
/
/
C
A
,
P
E
/
/
B
A
F\in AB,E\in AC,PF//CA,PE//BA
F
∈
A
B
,
E
∈
A
C
,
PF
//
C
A
,
PE
//
B
A
. If
S
△
A
B
C
=
1
S_{\triangle ABC}=1
S
△
A
BC
=
1
, prove that at least one of
S
△
B
P
F
,
S
△
P
C
E
,
S
P
E
A
F
S_{\triangle BPF},S_{\triangle PCE},S_{PEAF}
S
△
BPF
,
S
△
PCE
,
S
PE
A
F
is not less than
4
9
\frac{4}{9}
9
4
.
2
2
Hide problems
Choose the figure
Which figure's shaded part satisfies the inequality
log
x
(
log
x
y
2
)
>
0
\log_x(\log_x y^2)>0
lo
g
x
(
lo
g
x
y
2
)
>
0
? https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNi83LzY2ZWE5OGJmZjlhNzI1NDM5ZjdiNjZmYTcyZTFkMzEzZjUzMzk5LnBuZw==&rn=NGRkLnBuZw==
Skew Lines
a
,
b
a,b
a
,
b
are two skew lines, the angle they form is
θ
\theta
θ
. Length of their common perpendicular
A
A
′
AA'
A
A
′
is
d
d
d
(
A
′
∈
a
,
A
∈
b
)
A'\in a,A\in b)
A
′
∈
a
,
A
∈
b
)
.
E
∈
a
,
F
∈
b
,
∣
A
′
E
∣
=
m
,
∣
A
F
∣
=
n
E\in a,F\in b,|A'E|=m,|AF|=n
E
∈
a
,
F
∈
b
,
∣
A
′
E
∣
=
m
,
∣
A
F
∣
=
n
. Calculate
∣
E
F
∣
|EF|
∣
EF
∣
.
1
2
Hide problems
Complex Plane
On complex plane, what figure does the set
{
Z
‾
2
∣
arg
Z
=
α
,
α
∈
[
0
,
2
π
)
}
\{ \overline{Z}^2|\arg Z=\alpha,\alpha\in[0,2\pi)\}
{
Z
2
∣
ar
g
Z
=
α
,
α
∈
[
0
,
2
π
)}
stands for?
(A)
\text{(A)}
(A)
half-line
arg
Z
=
2
α
\arg Z=2\alpha
ar
g
Z
=
2
α
(B)
\text{(B)}
(B)
half-line
arg
Z
=
−
2
α
\arg Z=-2\alpha
ar
g
Z
=
−
2
α
(C)
\text{(C)}
(C)
half-line
arg
Z
=
α
\arg Z=\alpha
ar
g
Z
=
α
(D)
\text{(D)}
(D)
None above
True Or False?
Judge the following statements. If true, prove it; If false, give a counter-example. (a)
P
,
Q
P,Q
P
,
Q
are two different points on the sime side of line
l
l
l
. There exists two different circles, passing
P
,
Q
P,Q
P
,
Q
, and tangent to line
l
l
l
. (b) If
a
>
0
,
b
>
0
,
a
≠
1
,
b
≠
1
a>0,b>0,a\neq1,b\neq1
a
>
0
,
b
>
0
,
a
=
1
,
b
=
1
, then
log
a
b
+
log
b
a
≥
2
\log_ab+\log_ba\geq2
lo
g
a
b
+
lo
g
b
a
≥
2
. (c)
A
,
B
A,B
A
,
B
are two sets of points on coordinate plane.
C
r
=
{
(
x
,
y
)
∣
x
2
+
y
2
≤
r
2
}
C_r=\{(x,y)|x^2+y^2\leq r^2\}
C
r
=
{(
x
,
y
)
∣
x
2
+
y
2
≤
r
2
}
. For any
r
≥
0
r\geq0
r
≥
0
,
C
r
∪
A
⊂
C
r
∪
B
C_r\cup A\subset C_r\cup B
C
r
∪
A
⊂
C
r
∪
B
, then
A
⊂
B
A\subset B
A
⊂
B
.