MathDB
Problems
Contests
National and Regional Contests
China Contests
National High School Mathematics League
2003 National High School Mathematics League
2003 National High School Mathematics League
Part of
National High School Mathematics League
Subcontests
(15)
15
1
Hide problems
Paper Folding
A circle
O
O
O
with radius of
R
R
R
is drawn on a piece of paper.
A
A
A
is a fixed point inside circle
O
O
O
, and
O
A
=
a
OA=a
O
A
=
a
. Fold the paper, so that a point
A
′
A'
A
′
on the circle is coincident with
A
A
A
. For all such foldings, a kink mark is remained. Find the set of points on a certain kink mark.
14
1
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Complex Number
A
,
B
,
C
A,B,C
A
,
B
,
C
are points that three complex numbers
z
0
=
a
i
,
z
1
=
1
2
+
b
i
,
z
2
=
1
+
c
i
(
a
,
b
,
c
∈
R
)
z_0=a\text{i},z_1=\frac{1}{2}+b\text{i},z_2=1+c\text{i}(a,b,c\in\mathbb{R})
z
0
=
a
i
,
z
1
=
2
1
+
b
i
,
z
2
=
1
+
c
i
(
a
,
b
,
c
∈
R
)
refer to on complex plane (not collinear). Prove that curve
Z
=
Z
0
cos
4
t
+
2
Z
1
cos
2
t
sin
2
t
+
Z
2
sin
4
t
(
t
∈
R
)
Z=Z_0\cos^4t+2Z_1\cos^2t\sin^2t+Z_2\sin^4t(t\in\mathbb{R})
Z
=
Z
0
cos
4
t
+
2
Z
1
cos
2
t
sin
2
t
+
Z
2
sin
4
t
(
t
∈
R
)
has only one common point with the perpendicular bisector of
A
C
AC
A
C
, and find the point.
13
1
Hide problems
Inequality
Prove that
2
1
+
x
+
2
x
−
3
+
15
−
3
x
<
2
19
2\sqrt{1+x}+\sqrt{2x-3}+\sqrt{15-3x}<2\sqrt{19}
2
1
+
x
+
2
x
−
3
+
15
−
3
x
<
2
19
, where
3
2
≤
x
≤
5
\frac{3}{2}\leq x\leq5
2
3
≤
x
≤
5
.
12
1
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Limitation
M
n
=
{
0.
a
1
a
2
⋯
a
n
‾
∣
a
i
∈
0
,
1
,
i
=
1
,
2
,
⋯
,
n
,
a
n
=
1
}
M_n=\{\overline{0.a_1a_2\cdots a_n}|a_i\in{0,1},i=1,2,\cdots,n,a_n=1\}
M
n
=
{
0.
a
1
a
2
⋯
a
n
∣
a
i
∈
0
,
1
,
i
=
1
,
2
,
⋯
,
n
,
a
n
=
1
}
.
T
n
=
∣
M
n
∣
,
S
n
=
∑
x
∈
M
n
x
T_n=|M_n|,S_n=\sum_{x\in M_n}x
T
n
=
∣
M
n
∣
,
S
n
=
∑
x
∈
M
n
x
, then
lim
n
→
∞
S
n
T
n
=
\lim_{n\to\infty}\frac{S_n}{T_n}=
lim
n
→
∞
T
n
S
n
=
________.
11
1
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Circular Column and Spheres
Eight spheres with radius of
1
1
1
are put into a circular column. There are two floors, and each sphere is tangent to adjacent four spheres, one of the bottom surfaces, and the flank. Then the height of the circular column is________.
10
1
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Logarithm
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
are positive integers, and
log
a
b
=
3
2
,
log
c
d
=
5
4
\log_{a}b=\frac{3}{2},\log_{c}d=\frac{5}{4}
lo
g
a
b
=
2
3
,
lo
g
c
d
=
4
5
. If
a
−
c
=
9
a-c=9
a
−
c
=
9
, then
b
−
d
=
b-d=
b
−
d
=
________.
9
1
Hide problems
Two Sets
Two sets
A
=
{
x
∈
R
∣
x
2
−
4
x
+
3
<
0
}
,
B
=
{
x
∈
R
∣
2
1
−
x
+
a
≤
0
,
x
2
−
2
(
a
+
7
)
x
+
5
≤
0
}
A=\{x\in\mathbb{R}|x^2-4x+3<0\},B=\{x\in\mathbb{R}|2^{1-x}+a\leq0,x^2-2(a+7)x+5\leq0\}
A
=
{
x
∈
R
∣
x
2
−
4
x
+
3
<
0
}
,
B
=
{
x
∈
R
∣
2
1
−
x
+
a
≤
0
,
x
2
−
2
(
a
+
7
)
x
+
5
≤
0
}
. If
A
⊆
B
A\subseteq B
A
⊆
B
, then the range value of real number
a
a
a
is________.
8
1
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Ellipse Problem
F
1
,
F
2
F_1,F_2
F
1
,
F
2
are two focal points of ellipse
x
2
9
+
y
2
4
=
1
\frac{x^2}{9}+\frac{y^2}{4}=1
9
x
2
+
4
y
2
=
1
,
P
P
P
is a point on the ellipse, and
∣
P
F
1
∣
:
∣
P
F
2
∣
=
2
:
1
|PF_1|:|PF_2|=2:1
∣
P
F
1
∣
:
∣
P
F
2
∣
=
2
:
1
, then the area of
△
P
F
1
F
2
\triangle PF_1F_2
△
P
F
1
F
2
is________.
7
1
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Solve the Inequality
The solution set for inequality
∣
x
∣
3
−
2
x
2
−
4
∣
x
∣
+
3
<
0
|x|^3-2x^2-4|x|+3<0
∣
x
∣
3
−
2
x
2
−
4∣
x
∣
+
3
<
0
is________.
6
1
Hide problems
Tetrahedron
In tetrahedron
A
B
C
D
ABCD
A
BC
D
,
A
B
=
1
,
C
D
=
3
AB=1,CD=3
A
B
=
1
,
C
D
=
3
, the distance between
A
B
AB
A
B
and
C
D
CD
C
D
is
2
2
2
, the intersection angle between
A
B
AB
A
B
and
C
D
CD
C
D
is
π
3
\frac{\pi}{3}
3
π
, then the volume of tetrahedron
A
B
C
D
ABCD
A
BC
D
is
(A)
3
2
(B)
1
2
(C)
1
3
(D)
3
3
\text{(A)}\frac{\sqrt3}{2}\qquad\text{(B)}\frac{1}{2}\qquad\text{(C)}\frac{1}{3}\qquad\text{(D)}\frac{\sqrt3}{3}
(A)
2
3
(B)
2
1
(C)
3
1
(D)
3
3
5
1
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The Minumum Value
If
x
,
y
∈
(
−
2
,
2
)
,
x
y
=
−
1
x,y\in(-2,2),xy=-1
x
,
y
∈
(
−
2
,
2
)
,
x
y
=
−
1
, then the minumum value of
u
=
4
4
−
x
2
+
9
9
−
y
2
u=\frac{4}{4-x^2}+\frac{9}{9-y^2}
u
=
4
−
x
2
4
+
9
−
y
2
9
is
(A)
8
5
(B)
24
11
(C)
12
7
(D)
12
5
\text{(A)}\frac{8}{5}\qquad\text{(B)}\frac{24}{11}\qquad\text{(C)}\frac{12}{7}\qquad\text{(D)}\frac{12}{5}\qquad
(A)
5
8
(B)
11
24
(C)
7
12
(D)
5
12
4
1
Hide problems
Trigonometry
If
x
∈
[
−
5
π
12
,
−
π
3
]
x\in\left[-\frac{5\pi}{12},-\frac{\pi}{3}\right]
x
∈
[
−
12
5
π
,
−
3
π
]
, then the maximum value of
y
=
tan
(
x
+
2
π
3
)
−
tan
(
x
+
π
6
)
+
cos
(
x
+
π
6
)
y=\tan\left(x+\frac{2\pi}{3}\right)-\tan\left(x+\frac{\pi}{6}\right)+\cos\left(x+\frac{\pi}{6}\right)
y
=
tan
(
x
+
3
2
π
)
−
tan
(
x
+
6
π
)
+
cos
(
x
+
6
π
)
is
(A)
12
5
2
(B)
11
6
2
(C)
11
6
3
(D)
12
5
3
\text{(A)}\frac{12}{5}\sqrt2\qquad\text{(B)}\frac{11}{6}\sqrt2\qquad\text{(C)}\frac{11}{6}\sqrt3\qquad\text{(D)}\frac{12}{5}\sqrt3
(A)
5
12
2
(B)
6
11
2
(C)
6
11
3
(D)
5
12
3
3
2
Hide problems
Parabola
Line passes the focal point
F
F
F
of parabola
y
2
=
8
(
x
+
2
)
y^2=8(x+2)
y
2
=
8
(
x
+
2
)
with bank angle of
6
0
∘
60^{\circ}
6
0
∘
intersects the parabola at
A
,
B
A,B
A
,
B
. Perpendicular bisector of
A
B
AB
A
B
intersects
x
x
x
-axis at
P
P
P
, then the length of
P
F
PF
PF
is
(A)
16
3
(B)
8
3
(C)
16
3
3
(D)
8
3
\text{(A)}\frac{16}{3}\qquad\text{(B)}\frac{8}{3}\qquad\text{(C)}\frac{16}{3}\sqrt3\qquad\text{(D)}8\sqrt3
(A)
3
16
(B)
3
8
(C)
3
16
3
(D)
8
3
Graph Theory
A space figure is consisted of
n
n
n
vertexes and
l
l
l
lines connecting these vertices, where
n
=
q
2
+
q
+
1
,
l
≥
1
2
q
(
q
+
1
)
2
+
1
,
q
≥
2
,
q
∈
Z
+
n=q^2+q+1, l\geq\frac{1}{2}q(q+1)^2+1,q\geq2,q\in\mathbb{Z}_+
n
=
q
2
+
q
+
1
,
l
≥
2
1
q
(
q
+
1
)
2
+
1
,
q
≥
2
,
q
∈
Z
+
. The figure satisfies: every four vertices are not coplane, every vertex is connected by at least one line, and there is a vertex connected by at least
q
+
2
q+2
q
+
2
lines. Prove that there exists a space quadrilateral in the figure. Note: a space quadrilateral is figure with four vertices
A
,
B
,
C
,
D
A, B, C, D
A
,
B
,
C
,
D
and four lines
A
B
,
B
C
,
C
D
,
D
A
AB, BC, CD, DA
A
B
,
BC
,
C
D
,
D
A
.
2
2
Hide problems
Choose the Figure
If
a
,
b
∈
R
,
a
b
≠
0
a,b\in\mathbb{R},ab\neq 0
a
,
b
∈
R
,
ab
=
0
, the the possible figure of
a
x
−
y
+
b
=
0
ax-y+b=0
a
x
−
y
+
b
=
0
and
b
x
2
+
a
y
2
=
a
b
bx^2+ay^2=ab
b
x
2
+
a
y
2
=
ab
is https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYy80Lzc3NGNjZWNiN2ZjYzIxMTJlYWE5NDlmZmQ0ZjE1NzgwNmNhM2JiLnBuZw==&rn=MTI0MjQ1ODUyMTI0MjQyNTI0MjUyNS5wbmc=
Number Theory
Let the lengths of three sides of a triangle be
l
,
m
,
n
(
l
>
m
>
n
)
l, m, n(l>m>n)
l
,
m
,
n
(
l
>
m
>
n
)
. If
{
3
l
1
0
4
}
=
{
3
m
1
0
4
}
=
{
3
n
1
0
4
}
\left\{\frac{3^l}{10^4}\right\}=\left\{\frac{3^m}{10^4}\right\}=\left\{\frac{3^n}{10^4}\right\}
{
1
0
4
3
l
}
=
{
1
0
4
3
m
}
=
{
1
0
4
3
n
}
, find the minimum value of the perimeter of the triangle. Note:
{
x
}
=
x
−
[
x
]
\{x\}=x-[x]
{
x
}
=
x
−
[
x
]
and
[
x
]
[x]
[
x
]
denotes the integral part of number
x
x
x
.
1
2
Hide problems
Delete the Profect Squares
Delete all perfect squares in
1
,
2
,
3
,
⋯
1,2,3,\cdots
1
,
2
,
3
,
⋯
, then the 2003rd number is
(A)
2046
(B)
2047
(C)
2048
(D)
2049
\text{(A)}2046\qquad\text{(B)}2047\qquad\text{(C)}2048\qquad\text{(D)}2049
(A)
2046
(B)
2047
(C)
2048
(D)
2049
Geometry
Draw two tangents to the circle from point
P
P
P
outside a circle, touching the circle at
A
A
A
and
B
B
B
, then draw a secant line passes
P
P
P
, intersecting the circle at points
C
C
C
and
D
D
D
(
C
C
C
is between
P
P
P
and
D
D
D
).
Q
Q
Q
is a point on the chord
C
D
CD
C
D
such that
∠
D
A
Q
=
∠
P
B
C
\angle DAQ=\angle PBC
∠
D
A
Q
=
∠
PBC
. Prove that
∠
D
B
Q
=
∠
P
A
C
\angle DBQ=\angle PAC
∠
D
BQ
=
∠
P
A
C
.