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Problems
Contests
National and Regional Contests
China Contests
China Girls Math Olympiad
2021 China Girls Math Olympiad
2021 China Girls Math Olympiad
Part of
China Girls Math Olympiad
Subcontests
(8)
7
1
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CGMO 2021 P7
In an acute triangle
A
B
C
ABC
A
BC
,
A
B
≠
A
C
AB \neq AC
A
B
=
A
C
,
O
O
O
is its circumcenter.
K
K
K
is the reflection of
B
B
B
over
A
C
AC
A
C
and
L
L
L
is the reflection of
C
C
C
over
A
B
AB
A
B
.
X
X
X
is a point within
A
B
C
ABC
A
BC
such that
A
X
⊥
B
C
,
X
K
=
X
L
AX \perp BC, XK=XL
A
X
⊥
BC
,
X
K
=
X
L
. Points
Y
,
Z
Y, Z
Y
,
Z
are on
B
K
‾
,
C
L
‾
\overline{BK}, \overline{CL}
B
K
,
C
L
respectively, satisfying
X
Y
⊥
C
K
,
X
Z
⊥
B
L
XY \perp CK, XZ \perp BL
X
Y
⊥
C
K
,
XZ
⊥
B
L
.Proof that
B
,
C
,
Y
,
O
,
Z
B, C, Y, O, Z
B
,
C
,
Y
,
O
,
Z
lie on a circle.
8
1
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CGMO 2021 P8
Let
m
,
n
m, n
m
,
n
be positive integers, define:
f
(
x
)
=
(
x
−
1
)
(
x
2
−
1
)
⋯
(
x
m
−
1
)
f(x)=(x-1)(x^2-1)\cdots(x^m-1)
f
(
x
)
=
(
x
−
1
)
(
x
2
−
1
)
⋯
(
x
m
−
1
)
,
g
(
x
)
=
(
x
n
+
1
−
1
)
(
x
n
+
2
−
1
)
⋯
(
x
n
+
m
−
1
)
g(x)=(x^{n+1}-1)(x^{n+2}-1)\cdots(x^{n+m}-1)
g
(
x
)
=
(
x
n
+
1
−
1
)
(
x
n
+
2
−
1
)
⋯
(
x
n
+
m
−
1
)
.Show that there exists a polynomial
h
(
x
)
h(x)
h
(
x
)
of degree
m
n
mn
mn
such that
f
(
x
)
h
(
x
)
=
g
(
x
)
f(x)h(x)=g(x)
f
(
x
)
h
(
x
)
=
g
(
x
)
, and its
m
n
+
1
mn+1
mn
+
1
coefficients are all positive integers.
5
1
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CGMO 2021 P5
Proof that if
4
4
4
numbers (not necessarily distinct) are picked from
{
1
,
2
,
⋯
,
20
}
\{1, 2, \cdots, 20\}
{
1
,
2
,
⋯
,
20
}
, one can pick
3
3
3
numbers among them and can label these
3
3
3
as
a
,
b
,
c
a, b, c
a
,
b
,
c
such that
a
x
≡
b
(
m
o
d
c
)
ax \equiv b \;(\bmod\; c)
a
x
≡
b
(
mod
c
)
has integral solutions.
4
1
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CGMO 2021 P4
Call a sequence of positive integers
(
a
n
)
n
≥
1
(a_n)_{n \ge 1}
(
a
n
)
n
≥
1
a "CGMO sequence" if
(
a
n
)
n
≥
1
(a_n)_{n \ge 1}
(
a
n
)
n
≥
1
strictly increases, and for all integers
n
≥
2022
n \ge 2022
n
≥
2022
,
a
n
a_n
a
n
is the smallest integer such that there exists a non-empty subset of
{
a
1
,
a
2
,
⋯
,
a
n
−
1
}
\{a_{1}, a_{2}, \cdots, a_{n-1} \}
{
a
1
,
a
2
,
⋯
,
a
n
−
1
}
A
n
A_n
A
n
where
a
n
⋅
∏
a
∈
A
n
a
a_n \cdot \prod\limits_{a \in A_n} a
a
n
⋅
a
∈
A
n
∏
a
is a perfect square.Proof: there exists
c
1
,
c
2
∈
R
+
c_1, c_2 \in \mathbb{R}^{+}
c
1
,
c
2
∈
R
+
s.t. for any "CGMO sequence"
(
a
n
)
n
≥
1
(a_n)_{n \ge 1}
(
a
n
)
n
≥
1
, there is a positive integer
N
N
N
that satisfies any
n
≥
N
n \ge N
n
≥
N
,
c
1
⋅
n
2
≤
a
n
≤
c
2
⋅
n
2
c_1 \cdot n^2 \le a_n \le c_2 \cdot n^2
c
1
⋅
n
2
≤
a
n
≤
c
2
⋅
n
2
.
3
1
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CGMO 2021 P3
Find the smallest positive integer
n
n
n
, such that one can color every cell of a
n
×
n
n \times n
n
×
n
grid in red, yellow or blue with all the following conditions satisfied: (1) the number of cells colored in each color is the same; (2) if a row contains a red cell, that row must contain a blue cell and cannot contain a yellow cell; (3) if a column contains a blue cell, it must contain a red cell but cannot contain a yellow cell.
2
1
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CGMO 2021 P2
In acute triangle
A
B
C
ABC
A
BC
(
A
B
≠
A
C
AB \neq AC
A
B
=
A
C
),
I
I
I
is its incenter and
J
J
J
is the
A
A
A
-excenter.
X
,
Y
X, Y
X
,
Y
are on minor arcs
A
B
^
\widehat{AB}
A
B
and
A
C
^
\widehat{AC}
A
C
respectively such that
∠
A
X
I
=
∠
A
Y
J
=
9
0
∘
\angle{AXI}=\angle{AYJ}=90^{\circ}
∠
A
X
I
=
∠
A
Y
J
=
9
0
∘
.
K
K
K
is on line
B
C
BC
BC
such that
K
I
=
K
J
KI=KJ
K
I
=
K
J
. Proof that line
A
K
AK
A
K
bisects
X
Y
‾
\overline{XY}
X
Y
.
6
1
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Beautiful inequality on sets
Given a finite set
S
S
S
,
P
(
S
)
P(S)
P
(
S
)
denotes the set of all the subsets of
S
S
S
. For any
f
:
P
(
S
)
→
R
f:P(S)\rightarrow \mathbb{R}
f
:
P
(
S
)
→
R
,prove the following inequality:
∑
A
∈
P
(
S
)
∑
B
∈
P
(
S
)
f
(
A
)
f
(
B
)
2
∣
A
∩
B
∣
≥
0.
\sum_{A\in P(S)}\sum_{B\in P(S)}f(A)f(B)2^{\left| A\cap B \right|}\geq 0.
A
∈
P
(
S
)
∑
B
∈
P
(
S
)
∑
f
(
A
)
f
(
B
)
2
∣
A
∩
B
∣
≥
0.
1
1
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Chinese Girls Mathematical Olympiad 2021, Problem 1
Let
n
∈
N
+
,
n \in \mathbb{N}^+,
n
∈
N
+
,
x
1
,
x
2
,
.
.
.
,
x
n
+
1
,
p
,
q
∈
R
+
x_1,x_2,...,x_{n+1},p,q\in \mathbb{R}^+
x
1
,
x
2
,
...
,
x
n
+
1
,
p
,
q
∈
R
+
,
p
<
q
p<q
p
<
q
and
x
n
+
1
p
>
∑
i
=
1
n
x
i
p
.
x^p_{n+1}>\sum_{i=1}^{n}x^p_{i}.
x
n
+
1
p
>
∑
i
=
1
n
x
i
p
.
Prove that
(
1
)
x
n
+
1
q
>
∑
i
=
1
n
x
i
q
;
(1)x^q_{n+1}>\sum_{i=1}^{n}x^q_{i};
(
1
)
x
n
+
1
q
>
∑
i
=
1
n
x
i
q
;
(
2
)
(
x
n
+
1
p
−
∑
i
=
1
n
x
i
p
)
1
p
<
(
x
n
+
1
q
−
∑
i
=
1
n
x
i
q
)
1
q
.
(2)\left(x^p_{n+1}-\sum_{i=1}^{n}x^p_{i}\right)^{\frac{1}{p}}<\left(x^q_{n+1}-\sum_{i=1}^{n}x^q_{i}\right)^{\frac{1}{q}}.
(
2
)
(
x
n
+
1
p
−
∑
i
=
1
n
x
i
p
)
p
1
<
(
x
n
+
1
q
−
∑
i
=
1
n
x
i
q
)
q
1
.