MathDB
Problems
Contests
National and Regional Contests
China Contests
(China) National High School Mathematics League
2019 China Second Round Olympiad
2019 China Second Round Olympiad
Part of
(China) National High School Mathematics League
Subcontests
(4)
4
2
Hide problems
Two element subsets of a graph
Let
V
V
V
be a set of
2019
2019
2019
points in space where any of the four points are not on the same plane, and
E
E
E
be the set of edges connected between them. Find the smallest positive integer
n
n
n
satisfying the following condition: if
E
E
E
has at least
n
n
n
elements, then there exists
908
908
908
two-element subsets of
E
E
E
such that [*]The two edges in each subset share a common vertice, [*]Any of the two subsets do not intersect.
Coloring problem in 2019 China Second Round(B)
Each side of a convex
2019
2019
2019
-gon polygon is dyed with red, yellow and blue, and there are exactly
673
673
673
sides of each kind of color. Prove that there exists at least one way to draw
2016
2016
2016
diagonals to divide the convex
2019
2019
2019
-gon polygon into
2017
2017
2017
triangles, such that any two of the
2016
2016
2016
diagonals don't have intersection inside the
2019
2019
2019
-gon polygon,and for any triangle in all the
2017
2017
2017
triangles, the colors of the three sides of the triangle are all the same, either totally different.
3
2
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Tricky Sequence With NT Properties
Let
m
m
m
be an integer where
∣
m
∣
≥
2
|m|\ge 2
∣
m
∣
≥
2
. Let
a
1
,
a
2
,
⋯
a_1,a_2,\cdots
a
1
,
a
2
,
⋯
be a sequence of integers such that
a
1
,
a
2
a_1,a_2
a
1
,
a
2
are not both zero, and for any positive integer
n
n
n
,
a
n
+
2
=
a
n
+
1
−
m
a
n
a_{n+2}=a_{n+1}-ma_n
a
n
+
2
=
a
n
+
1
−
m
a
n
.Prove that if positive integers
r
>
s
≥
2
r>s\ge 2
r
>
s
≥
2
satisfy
a
r
=
a
s
=
a
1
a_r=a_s=a_1
a
r
=
a
s
=
a
1
, then
r
−
s
≥
∣
m
∣
r-s\ge |m|
r
−
s
≥
∣
m
∣
.
Geometry in 2019 China Second Round(B)
Point
A
,
B
,
C
,
D
,
E
A,B,C,D,E
A
,
B
,
C
,
D
,
E
lie on a line in this order, such that
B
C
=
C
D
=
A
B
⋅
D
E
,
BC=CD=\sqrt{AB\cdot DE},
BC
=
C
D
=
A
B
⋅
D
E
,
P
P
P
doesn't lie on the line, and satisfys that
P
B
=
P
D
.
PB=PD.
PB
=
P
D
.
Point
K
,
L
K,L
K
,
L
lie on the segment
P
B
,
P
D
,
PB,PD,
PB
,
P
D
,
respectively, such that
K
C
KC
K
C
bisects
∠
B
K
E
,
\angle BKE,
∠
B
K
E
,
and
L
C
LC
L
C
bisects
∠
A
L
D
.
\angle ALD.
∠
A
L
D
.
Prove that
A
,
K
,
L
,
E
A,K,L,E
A
,
K
,
L
,
E
are concyclic.
2
2
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Weird Inequality with 2019 variables
Let
a
1
,
a
2
,
⋯
,
a
n
a_1,a_2,\cdots,a_n
a
1
,
a
2
,
⋯
,
a
n
be integers such that
1
=
a
1
≤
a
2
≤
⋯
≤
a
2019
=
99
1=a_1\le a_2\le \cdots\le a_{2019}=99
1
=
a
1
≤
a
2
≤
⋯
≤
a
2019
=
99
. Find the minimum
f
0
f_0
f
0
of the expression
f
=
(
a
1
2
+
a
2
2
+
⋯
+
a
2019
2
)
−
(
a
1
a
3
+
a
2
a
4
+
⋯
+
a
2017
a
2019
)
,
f=(a_1^2+a_2^2+\cdots+a_{2019}^2)-(a_1a_3+a_2a_4+\cdots+a_{2017}a_{2019}),
f
=
(
a
1
2
+
a
2
2
+
⋯
+
a
2019
2
)
−
(
a
1
a
3
+
a
2
a
4
+
⋯
+
a
2017
a
2019
)
,
and determine the number of sequences
(
a
1
,
a
2
,
⋯
,
a
n
)
(a_1,a_2,\cdots,a_n)
(
a
1
,
a
2
,
⋯
,
a
n
)
such that
f
=
f
0
f=f_0
f
=
f
0
.
A number theory in China Second Round(B)
Find all the positive integers
n
n
n
such that:
(
1
)
(1)
(
1
)
n
n
n
has at least
4
4
4
positive divisors.
(
2
)
(2)
(
2
)
if all positive divisors of
n
n
n
are
d
1
,
d
2
,
⋯
,
d
k
,
d_1,d_2,\cdots ,d_k,
d
1
,
d
2
,
⋯
,
d
k
,
then
d
2
−
d
1
,
d
3
−
d
2
,
⋯
,
d
k
−
d
k
−
1
d_2-d_1,d_3-d_2,\cdots ,d_k-d_{k-1}
d
2
−
d
1
,
d
3
−
d
2
,
⋯
,
d
k
−
d
k
−
1
form a geometric sequence.
1
2
Hide problems
Surprisingly Easy Geo
In acute triangle
△
A
B
C
\triangle ABC
△
A
BC
,
M
M
M
is the midpoint of segment
B
C
BC
BC
. Point
P
P
P
lies in the interior of
△
A
B
C
\triangle ABC
△
A
BC
such that
A
P
AP
A
P
bisects
∠
B
A
C
\angle BAC
∠
B
A
C
. Line
M
P
MP
MP
intersects the circumcircles of
△
A
B
P
,
△
A
C
P
\triangle ABP,\triangle ACP
△
A
BP
,
△
A
CP
at
D
,
E
D,E
D
,
E
respectively. Prove that if
D
E
=
M
P
DE=MP
D
E
=
MP
, then
B
C
=
2
B
P
BC=2BP
BC
=
2
BP
.
Inequality with 100 variables.
Suppose that
a
1
,
a
2
,
⋯
,
a
100
∈
R
+
a_1,a_2,\cdots,a_{100}\in\mathbb{R}^+
a
1
,
a
2
,
⋯
,
a
100
∈
R
+
such that
a
i
≥
a
101
−
i
(
i
=
1
,
2
,
⋯
,
50
)
.
a_i\geq a_{101-i}\,(i=1,2,\cdots,50).
a
i
≥
a
101
−
i
(
i
=
1
,
2
,
⋯
,
50
)
.
Let
x
k
=
k
a
k
+
1
a
1
+
a
2
+
⋯
+
a
k
(
k
=
1
,
2
,
⋯
,
99
)
.
x_k=\frac{ka_{k+1}}{a_1+a_2+\cdots+a_k}\,(k=1,2,\cdots,99).
x
k
=
a
1
+
a
2
+
⋯
+
a
k
k
a
k
+
1
(
k
=
1
,
2
,
⋯
,
99
)
.
Prove that
x
1
x
2
2
⋯
x
99
99
≤
1.
x_1x_2^2\cdots x_{99}^{99}\leq 1.
x
1
x
2
2
⋯
x
99
99
≤
1.