MathDB
Problems
Contests
National and Regional Contests
China Contests
National High School Mathematics League
1983 National High School Mathematics League
1983 National High School Mathematics League
Part of
National High School Mathematics League
Subcontests
(11)
11
1
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Inradius of hexahedron and octahedron
For a regular hexahedron and a regular octahedron, all their faces are regular triangles, whose lengths of each side are
a
a
a
. Their inradius are
r
1
,
r
2
r_1,r_2
r
1
,
r
2
.
r
1
r
2
=
m
n
,
gcd
(
m
,
n
)
=
1
\frac{r_1}{r_2}=\frac{m}{n}, \gcd(m,n)=1
r
2
r
1
=
n
m
,
g
cd
(
m
,
n
)
=
1
. Then
m
n
=
mn=
mn
=
________.
10
1
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Count the Number of Triangles
The number of triangles such that lengths of three sides are integers, and the length of the longest side is
11
11
11
is________.
9
1
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Fill In the Blank(EASY!)
In
△
A
B
C
,
sin
A
=
3
5
,
cos
B
=
5
13
\triangle ABC,\sin A=\frac{3}{5},\cos B=\frac{5}{13}
△
A
BC
,
sin
A
=
5
3
,
cos
B
=
13
5
, then
cos
C
=
\cos C=
cos
C
=
________.
8
1
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Geometry Inequality, but...Maybe Incorrect
For any
△
A
B
C
\triangle ABC
△
A
BC
, its girth is
l
l
l
, its circumradius is
R
R
R
, its inscribed radius is
r
r
r
.Which one is true?
(A)
l
>
R
+
r
(B)
l
≤
R
+
r
(C)
l
6
<
R
+
r
<
6
l
(D)
\text{(A)}l>R+r\qquad\text{(B)}l\leq R+r\qquad\text{(C)}\frac{l}{6}<R+r<6l\qquad\text{(D)}
(A)
l
>
R
+
r
(B)
l
≤
R
+
r
(C)
6
l
<
R
+
r
<
6
l
(D)
None above
7
1
Hide problems
The Number of Isosceles Triangles
P
P
P
is a point on the plane which square
A
B
C
D
ABCD
A
BC
D
belongs to, satisfying that
△
P
A
B
,
△
P
B
C
,
△
P
C
D
,
△
P
D
A
\triangle PAB,\triangle PBC,\triangle PCD,\triangle PDA
△
P
A
B
,
△
PBC
,
△
PC
D
,
△
P
D
A
are isosceles triangles. What's the number of such points?
(A)
9
(B)
17
(C)
1
(D)
5
\text{(A)}9\qquad\text{(B)}17\qquad\text{(C)}1\qquad\text{(D)}5
(A)
9
(B)
17
(C)
1
(D)
5
6
1
Hide problems
An inequality
Let
a
,
b
,
c
,
d
,
m
,
n
a,b,c,d,m,n
a
,
b
,
c
,
d
,
m
,
n
be positive real numbers.
P
=
a
b
+
c
d
,
Q
=
m
a
+
n
c
⋅
b
m
+
d
n
P=\sqrt{ab}+\sqrt{cd},Q=\sqrt{ma+nc}\cdot\sqrt{\frac{b}{m}+\frac{d}{n}}
P
=
ab
+
c
d
,
Q
=
ma
+
n
c
⋅
m
b
+
n
d
. Then
(A)
P
≥
Q
(B)
P
≤
Q
(C)
P
<
Q
(D)
\text{(A)}P\geq Q\qquad\text{(B)}P\leq Q\qquad\text{(C)}P<Q\qquad\text{(D)}
(A)
P
≥
Q
(B)
P
≤
Q
(C)
P
<
Q
(D)
Not sure
5
2
Hide problems
Twice Function
f
(
x
)
=
a
x
2
−
c
f(x)=ax^2-c
f
(
x
)
=
a
x
2
−
c
. If
−
4
≤
f
(
1
)
≤
−
1
,
−
z
≤
f
(
2
)
≤
5
-4\leq f(1)\leq -1,-z\leq f(2)\leq 5
−
4
≤
f
(
1
)
≤
−
1
,
−
z
≤
f
(
2
)
≤
5
, then
(A)
7
≤
f
(
3
)
≤
26
(B)
−
4
≤
f
(
3
)
≤
15
(C)
−
1
≤
f
(
3
)
≤
23
(D)
−
28
3
≤
f
(
3
)
≤
35
3
\text{(A)}7\leq f(3)\leq26\qquad\text{(B)}-4\leq f(3)\leq15\qquad\text{(C)}-1\leq f(3)\leq23\qquad\text{(D)}-\frac{28}{3}\leq f(3)\leq\frac{35}{3}
(A)
7
≤
f
(
3
)
≤
26
(B)
−
4
≤
f
(
3
)
≤
15
(C)
−
1
≤
f
(
3
)
≤
23
(D)
−
3
28
≤
f
(
3
)
≤
3
35
Maximum Value of Trigonometric Functions
Function
F
(
x
)
=
∣
cos
2
x
+
2
sin
x
cos
x
−
sin
2
x
+
A
x
+
B
∣
F(x)=|\cos^2x+2\sin x\cos x-\sin^2x+Ax+B|
F
(
x
)
=
∣
cos
2
x
+
2
sin
x
cos
x
−
sin
2
x
+
A
x
+
B
∣
, where
A
,
B
A,B
A
,
B
are two real numbers,
x
∈
[
0
,
3
2
π
]
x\in[0,\frac{3}{2}\pi]
x
∈
[
0
,
2
3
π
]
.
M
M
M
is the maximun value of
F
(
x
)
F(x)
F
(
x
)
. Find the minumum value of
M
M
M
.
4
2
Hide problems
Two Simple Sets
Define two sets:
M
=
{
(
x
,
y
)
∣
y
≥
x
2
}
,
N
=
{
(
x
,
y
)
∣
x
2
+
(
y
−
a
)
2
≤
1
}
M=\{ (x,y)|y\geq x^2\} ,N=\{ (x,y)|x^2+(y-a)^2\leq 1\}
M
=
{(
x
,
y
)
∣
y
≥
x
2
}
,
N
=
{(
x
,
y
)
∣
x
2
+
(
y
−
a
)
2
≤
1
}
. If
M
∪
N
=
N
M\cup N=N
M
∪
N
=
N
, then the range value of
a
a
a
is
(A)
a
≥
1
1
4
(B)
a
=
1
1
4
(C)
a
≥
1
(D)
0
<
a
<
1
\text{(A)}a\geq 1\frac{1}{4}\qquad\text{(B)}a=1\frac{1}{4}\qquad\text{(C)}a\geq 1\qquad\text{(D)}0<a<1
(A)
a
≥
1
4
1
(B)
a
=
1
4
1
(C)
a
≥
1
(D)
0
<
a
<
1
Tetrahedron Problem
In a tetrahedron, lengths of six edges are
2
,
3
,
3
,
4
,
5
,
5
2,3,3,4,5,5
2
,
3
,
3
,
4
,
5
,
5
. Find its largest volume.
3
2
Hide problems
Rational Number Or Not?
In triangle
A
B
C
ABC
A
BC
,
A
B
=
A
C
AB=AC
A
B
=
A
C
,
∣
B
C
∣
|BC|
∣
BC
∣
and
d
(
A
,
B
C
)
d(A,BC)
d
(
A
,
BC
)
are both integers. Then,
sin
A
\sin A
sin
A
and
cos
A
\cos A
cos
A
(A)
\text{(A)}
(A)
one is a rational number while the other is not
(B)
\text{(B)}
(B)
both are rational numbers
(C)
\text{(C)}
(C)
neither is rational number
(D)
\text{(D)}
(D)
not sure
So many midpoints!
In quadrilateral
A
B
C
D
ABCD
A
BC
D
,
S
△
A
B
D
:
S
△
B
C
D
:
S
△
A
B
C
=
3
:
4
:
1
S_{\triangle ABD}:S_{\triangle BCD}:S_{\triangle ABC}=3:4:1
S
△
A
B
D
:
S
△
BC
D
:
S
△
A
BC
=
3
:
4
:
1
.
M
∈
A
C
,
N
∈
C
D
M\in AC,N\in CD
M
∈
A
C
,
N
∈
C
D
, satisfying that
A
M
A
C
=
C
N
C
D
\frac{AM}{AC}=\frac{CN}{CD}
A
C
A
M
=
C
D
CN
. If
B
,
M
,
N
B,M,N
B
,
M
,
N
are collinear, prove that
M
,
N
M,N
M
,
N
are mid points of
A
C
,
C
D
AC,CD
A
C
,
C
D
.
2
2
Hide problems
Estimate
x
=
1
log
1
2
1
3
+
1
log
1
5
1
3
x=\frac{1}{\log_{\frac{1}{2}} \frac{1}{3}}+\frac{1}{\log_{\frac{1}{5}} \frac{1}{3}}
x
=
l
o
g
2
1
3
1
1
+
l
o
g
5
1
3
1
1
, then
(A)
x
∈
(
−
2
,
−
1
)
(B)
x
∈
(
1
,
2
)
(C)
x
∈
(
−
3
,
−
2
)
(D)
x
∈
(
2
,
3
)
\text{(A)}x\in(-2,-1)\qquad\text{(B)}x\in(1,2)\qquad\text{(C)}x\in(-3,-2)\qquad\text{(D)}x\in(2,3)
(A)
x
∈
(
−
2
,
−
1
)
(B)
x
∈
(
1
,
2
)
(C)
x
∈
(
−
3
,
−
2
)
(D)
x
∈
(
2
,
3
)
Interesting Function
Function
f
(
x
)
f(x)
f
(
x
)
is defined on
[
0
,
1
]
[0,1]
[
0
,
1
]
,
f
(
0
)
=
f
(
1
)
f(0)=f(1)
f
(
0
)
=
f
(
1
)
. For any
x
1
,
x
2
∈
[
0
,
1
]
,
∣
f
(
x
1
)
−
f
(
x
2
)
∣
<
∣
x
1
−
x
2
∣
(
x
1
≠
x
2
)
x_1,x_2\in [0,1], |f(x_1)-f(x_2)|<|x_1-x_2|(x_1\neq x_2)
x
1
,
x
2
∈
[
0
,
1
]
,
∣
f
(
x
1
)
−
f
(
x
2
)
∣
<
∣
x
1
−
x
2
∣
(
x
1
=
x
2
)
. Prove that
∣
f
(
x
1
)
−
f
(
x
2
)
∣
<
1
2
|f(x_1)-f(x_2)|<\frac{1}{2}
∣
f
(
x
1
)
−
f
(
x
2
)
∣
<
2
1
.
1
2
Hide problems
Even Numbers
p
,
q
p,q
p
,
q
are nonnegative integers.Given two conditions: A:
p
3
−
q
3
p^3-q^3
p
3
−
q
3
is an even number. B:
p
+
q
p+q
p
+
q
is an even number. Then, which one of the followings are true?
(
A
)
(\text{A})
(
A
)
A is sufficient but unnecessary condition of B.
(
B
)
(\text{B})
(
B
)
A is necessary but insufficient condition of B.
(
C
)
(\text{C})
(
C
)
A is sufficient and necessary condition of B.
(
D
)
(\text{D})
(
D
)
A is insufficient and unnecessary condition of B.
Is This a Problem?
Prove that
arcsin
x
+
arccos
x
=
π
2
\arcsin x+\arccos x=\frac{\pi}{2}
arcsin
x
+
arccos
x
=
2
π
, where
x
∈
[
−
1
,
1
]
x\in[-1,1]
x
∈
[
−
1
,
1
]
.