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Problems
Contests
National and Regional Contests
China Contests
(China) National High School Mathematics League
2021 China Second Round A2
2021 China Second Round A2
Part of
(China) National High School Mathematics League
Subcontests
(4)
1
1
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AT bisects BC
As shown in the figure, in the acute angle
△
A
B
C
\vartriangle ABC
△
A
BC
,
A
B
>
A
C
AB > AC
A
B
>
A
C
,
M
M
M
is the midpoint of the minor arc
B
C
BC
BC
of the circumcircle
Ω
\Omega
Ω
of
△
A
B
C
\vartriangle ABC
△
A
BC
.
K
K
K
is the intersection point of the bisector of the exterior angle
∠
B
A
C
\angle BAC
∠
B
A
C
and the extension line of
B
C
BC
BC
. From point
A
A
A
draw a line perpendicular on
B
C
BC
BC
and take a point
D
D
D
(different from
A
A
A
) on that line , such that
D
M
=
A
M
DM = AM
D
M
=
A
M
. Let the circumscribed circle of
△
A
D
K
\vartriangle ADK
△
A
DK
intersect the circle
Ω
\Omega
Ω
at point
A
A
A
and at another point
T
T
T
. Prove that
A
T
AT
A
T
bisects line segment
B
C
BC
BC
. https://cdn.artofproblemsolving.com/attachments/1/3/6fde30405101620828d63ae31b8c0ffcec972f.png
4
1
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number theory problem
The positive integer formed after writing
k
k
k
consecutive positive integers from smallest to largest is called a
k
−
continuous
k-\text{continuous}
k
−
continuous
number. For example
99100101
99100101
99100101
is a
3
−
continuous
3-\text{continuous}
3
−
continuous
number. Prove that: for
∀
N
\forall N
∀
N
,
k
∈
Z
+
k\in\mathbb Z^+
k
∈
Z
+
, there must be a
k
−
continuous
k-\text{continuous}
k
−
continuous
number that can be divisible by
N
N
N
.
2
1
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combinatorics problem
Find the maximum value of
M
M
M
, if you choose
10
10
10
different real numbers randomly in
[
1
,
M
]
[1,M]
[
1
,
M
]
, there must be
3
3
3
numbers
a
<
b
<
c
a<b<c
a
<
b
<
c
, satisfy
a
x
2
+
b
x
+
c
=
0
ax^2+bx+c=0
a
x
2
+
b
x
+
c
=
0
has no real root.
3
1
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inequality in 2021 China Second Round A2
Given
n
≥
2
n\geq 2
n
≥
2
,
a
1
a_1
a
1
,
a
2
a_2
a
2
,
⋯
\cdots
⋯
,
a
n
∈
R
a_n\in\mathbb {R}
a
n
∈
R
satisfy
a
1
⩾
a
2
⩾
⋯
⩾
a
n
⩾
0
,
a
1
+
a
2
+
⋯
+
a
n
=
n
.
a_1\geqslant a_2\geqslant \cdots \geqslant a_n\geqslant 0,a_1+a_2+\cdots +a_n=n.
a
1
⩾
a
2
⩾
⋯
⩾
a
n
⩾
0
,
a
1
+
a
2
+
⋯
+
a
n
=
n
.
Find the minimum value of
a
1
+
a
1
a
2
+
⋯
+
a
1
a
2
⋯
a
n
a_1+a_1a_2+\cdots +a_1a_2\cdots a_n
a
1
+
a
1
a
2
+
⋯
+
a
1
a
2
⋯
a
n
.