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Problems
Contests
National and Regional Contests
China Contests
(China) National High School Mathematics League
2022 China Second Round
2022 China Second Round
Part of
(China) National High School Mathematics League
Subcontests
(4)
1
1
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China Second Round MO Test 2 P1
In a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
,
∠
A
B
C
=
∠
A
D
C
=
9
0
∘
\angle ABC = \angle ADC = 90^\circ
∠
A
BC
=
∠
A
D
C
=
9
0
∘
. A point
P
P
P
is chosen from the diagonal
B
D
BD
B
D
such that
∠
A
P
B
=
2
∠
C
P
D
\angle APB = 2\angle CPD
∠
A
PB
=
2∠
CP
D
, points
X
X
X
,
Y
Y
Y
is chosen from the segment
A
P
AP
A
P
such that
∠
A
X
B
=
2
∠
A
D
B
\angle AXB = 2\angle ADB
∠
A
XB
=
2∠
A
D
B
,
∠
A
Y
D
=
2
∠
A
B
D
\angle AYD = 2\angle ABD
∠
A
Y
D
=
2∠
A
B
D
. Prove that:
B
D
=
2
X
Y
BD = 2XY
B
D
=
2
X
Y
.
4
1
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China Second Round Olympiad 2022 Test 2 P4
Find the smallest positive integer
k
k
k
with the following property: if each cell of a
100
×
100
100\times 100
100
×
100
grid is dyed with one color and the number of cells of each color is not more than
104
104
104
, then there is a
k
×
1
k\times1
k
×
1
or
1
×
k
1\times k
1
×
k
rectangle that contains cells of at least three different colors.
2
1
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China Second Round Olympiad 2022 Test 2 P2
Integer
n
n
n
has
k
k
k
different prime factors. Prove that
σ
(
n
)
∣
(
2
n
−
k
)
!
\sigma (n) \mid (2n-k)!
σ
(
n
)
∣
(
2
n
−
k
)!
3
1
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China Second Round Olympiad 2022 Test 2 Q 3
Let
a
1
,
a
2
,
⋯
,
a
100
a_1,a_2,\cdots ,a_{100}
a
1
,
a
2
,
⋯
,
a
100
be non-negative integers such that
(
1
)
(1)
(
1
)
There are positive integers
k
≤
100
k\leq 100
k
≤
100
such that
a
1
≤
a
2
≤
⋯
≤
a
k
a_1\leq a_2\leq \cdots\leq a_{k}
a
1
≤
a
2
≤
⋯
≤
a
k
and
a
i
=
0
a_i=0
a
i
=
0
(
i
>
k
)
;
(i>k);
(
i
>
k
)
;
(
2
)
(2)
(
2
)
a
1
+
a
2
+
a
3
+
⋯
+
a
100
=
100
;
a_1+a_2+a_3+\cdots +a_{100}=100;
a
1
+
a
2
+
a
3
+
⋯
+
a
100
=
100
;
(
3
)
(3)
(
3
)
a
1
+
2
a
2
+
3
a
3
+
⋯
+
100
a
100
=
2022.
a_1+2a_2+3a_3+\cdots +100a_{100}=2022.
a
1
+
2
a
2
+
3
a
3
+
⋯
+
100
a
100
=
2022.
Find the minimum of
a
1
+
2
2
a
2
+
3
2
a
3
+
⋯
+
10
0
2
a
100
.
a_1+2^2a_2+3^2a_3+\cdots +100^2a_{100}.
a
1
+
2
2
a
2
+
3
2
a
3
+
⋯
+
10
0
2
a
100
.