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China Contests
South East Mathematical Olympiad
2021 South East Mathematical Olympiad
2021 South East Mathematical Olympiad
Part of
South East Mathematical Olympiad
Subcontests
(8)
6
2
Hide problems
China South East Mathematical Olympiad 2021 Grade10 P6
Let
A
B
C
D
ABCD
A
BC
D
be a cyclic quadrilateral. Let
E
E
E
be a point on side
B
C
,
BC,
BC
,
F
F
F
be a point on side
A
E
,
AE,
A
E
,
G
G
G
be a point on the exterior angle bisector of
∠
B
C
D
,
\angle BCD,
∠
BC
D
,
such that
E
G
=
F
G
,
EG=FG,
EG
=
FG
,
∠
E
A
G
=
1
2
∠
B
A
D
.
\angle EAG=\dfrac12\angle BAD.
∠
E
A
G
=
2
1
∠
B
A
D
.
Prove that
A
B
⋅
A
F
=
A
D
⋅
A
E
.
AB\cdot AF=AD\cdot AE.
A
B
⋅
A
F
=
A
D
⋅
A
E
.
China South East Mathematical Olympiad 2021 Grade11 P6
Let
A
B
C
D
ABCD
A
BC
D
be a cyclic quadrilateral. The internal angle bisector of
∠
B
A
D
\angle BAD
∠
B
A
D
and line
B
C
BC
BC
intersect at
E
.
E.
E
.
M
M
M
is the midpoint of segment
A
E
.
AE.
A
E
.
The exterior angle bisector of
∠
B
C
D
\angle BCD
∠
BC
D
and line
A
D
AD
A
D
intersect at
F
.
F.
F
.
The lines
M
F
MF
MF
and
A
B
AB
A
B
intersect at
G
.
G.
G
.
Prove that if
A
B
=
2
A
D
,
AB=2AD,
A
B
=
2
A
D
,
then
M
F
=
2
M
G
.
MF=2MG.
MF
=
2
MG
.
8
2
Hide problems
China South East Mathematical Olympiad 2021 Grade10 P8
Determine all the pairs of positive integers
(
a
,
b
)
,
(a,b),
(
a
,
b
)
,
such that
14
φ
2
(
a
)
−
φ
(
a
b
)
+
22
φ
2
(
b
)
=
a
2
+
b
2
,
14\varphi^2(a)-\varphi(ab)+22\varphi^2(b)=a^2+b^2,
14
φ
2
(
a
)
−
φ
(
ab
)
+
22
φ
2
(
b
)
=
a
2
+
b
2
,
where
φ
(
n
)
\varphi(n)
φ
(
n
)
is Euler's totient function.
China South East Mathematical Olympiad 2021 Grade11 P8
A sequence
{
z
n
}
\{z_n\}
{
z
n
}
satisfies that for any positive integer
i
,
i,
i
,
z
i
∈
{
0
,
1
,
⋯
,
9
}
z_i\in\{0,1,\cdots,9\}
z
i
∈
{
0
,
1
,
⋯
,
9
}
and
z
i
≡
i
−
1
(
m
o
d
10
)
.
z_i\equiv i-1 \pmod {10}.
z
i
≡
i
−
1
(
mod
10
)
.
Suppose there is
2021
2021
2021
non-negative reals
x
1
,
x
2
,
⋯
,
x
2021
x_1,x_2,\cdots,x_{2021}
x
1
,
x
2
,
⋯
,
x
2021
such that for
k
=
1
,
2
,
⋯
,
2021
,
k=1,2,\cdots,2021,
k
=
1
,
2
,
⋯
,
2021
,
∑
i
=
1
k
x
i
≥
∑
i
=
1
k
z
i
,
∑
i
=
1
k
x
i
≤
∑
i
=
1
k
z
i
+
∑
j
=
1
10
10
−
j
50
z
k
+
j
.
\sum_{i=1}^kx_i\geq\sum_{i=1}^kz_i,\sum_{i=1}^kx_i\leq\sum_{i=1}^kz_i+\sum_{j=1}^{10}\dfrac{10-j}{50}z_{k+j}.
i
=
1
∑
k
x
i
≥
i
=
1
∑
k
z
i
,
i
=
1
∑
k
x
i
≤
i
=
1
∑
k
z
i
+
j
=
1
∑
10
50
10
−
j
z
k
+
j
.
Determine the least possible value of
∑
i
=
1
2021
x
i
2
.
\sum_{i=1}^{2021}x_i^2.
∑
i
=
1
2021
x
i
2
.
7
2
Hide problems
China South East Mathematical Olympiad 2021 Grade10 P7
Let
a
,
b
,
c
a,b,c
a
,
b
,
c
be pairwise distinct positive real, Prove that
a
b
+
b
c
+
c
a
(
a
+
b
)
(
b
+
c
)
(
c
+
a
)
<
1
7
(
1
∣
a
−
b
∣
+
1
∣
b
−
c
∣
+
1
∣
c
−
a
∣
)
.
\dfrac{ab+bc+ca}{(a+b)(b+c)(c+a)}<\dfrac17(\dfrac{1}{|a-b|}+\dfrac{1}{|b-c|}+\dfrac{1}{|c-a|}).
(
a
+
b
)
(
b
+
c
)
(
c
+
a
)
ab
+
b
c
+
c
a
<
7
1
(
∣
a
−
b
∣
1
+
∣
b
−
c
∣
1
+
∣
c
−
a
∣
1
)
.
China South East Mathematical Olympiad 2021 Grade11 P7
Determine all the pairs of positive odd integers
(
a
,
b
)
,
(a,b),
(
a
,
b
)
,
such that
a
,
b
>
1
a,b>1
a
,
b
>
1
and
7
φ
2
(
a
)
−
φ
(
a
b
)
+
11
φ
2
(
b
)
=
2
(
a
2
+
b
2
)
,
7\varphi^2(a)-\varphi(ab)+11\varphi^2(b)=2(a^2+b^2),
7
φ
2
(
a
)
−
φ
(
ab
)
+
11
φ
2
(
b
)
=
2
(
a
2
+
b
2
)
,
where
φ
(
n
)
\varphi(n)
φ
(
n
)
is Euler's totient function.
5
2
Hide problems
Number of a sequence
To commemorate the
43
r
d
43rd
43
r
d
anniversary of the restoration of mathematics competitions, a mathematics enthusiast arranges the first
2021
2021
2021
integers
1
,
2
,
…
,
2021
1,2,\dots,2021
1
,
2
,
…
,
2021
into a sequence
{
a
n
}
\{a_n\}
{
a
n
}
in a certain order, so that the sum of any consecutive
43
43
43
items in the sequence is a multiple of
43
43
43
. (1) If the sequence of numbers is connected end to end into a circle, prove that the sum of any consecutive
43
43
43
items on the circle is also a multiple of
43
43
43
; (2) Determine the number of sequences
{
a
n
}
\{a_n\}
{
a
n
}
that meets the conditions of the question.
China South East Mathematical Olympiad 2021 Grade11 P5
Let
A
=
{
a
1
,
a
2
,
⋯
,
a
n
,
b
1
,
b
2
,
⋯
,
b
n
}
A=\{a_1,a_2,\cdots,a_n,b_1,b_2,\cdots,b_n\}
A
=
{
a
1
,
a
2
,
⋯
,
a
n
,
b
1
,
b
2
,
⋯
,
b
n
}
be a set with
2
n
2n
2
n
distinct elements, and
B
i
⊆
A
B_i\subseteq A
B
i
⊆
A
for any
i
=
1
,
2
,
⋯
,
m
.
i=1,2,\cdots,m.
i
=
1
,
2
,
⋯
,
m
.
If
⋃
i
=
1
m
B
i
=
A
,
\bigcup_{i=1}^m B_i=A,
⋃
i
=
1
m
B
i
=
A
,
we say that the ordered
m
−
m-
m
−
tuple
(
B
1
,
B
2
,
⋯
,
B
m
)
(B_1,B_2,\cdots,B_m)
(
B
1
,
B
2
,
⋯
,
B
m
)
is an ordered
m
−
m-
m
−
covering of
A
.
A.
A
.
If
(
B
1
,
B
2
,
⋯
,
B
m
)
(B_1,B_2,\cdots,B_m)
(
B
1
,
B
2
,
⋯
,
B
m
)
is an ordered
m
−
m-
m
−
covering of
A
,
A,
A
,
and for any
i
=
1
,
2
,
⋯
,
m
i=1,2,\cdots,m
i
=
1
,
2
,
⋯
,
m
and any
j
=
1
,
2
,
⋯
,
n
,
j=1,2,\cdots,n,
j
=
1
,
2
,
⋯
,
n
,
{
a
j
,
b
j
}
\{a_j,b_j\}
{
a
j
,
b
j
}
is not a subset of
B
i
,
B_i,
B
i
,
then we say that ordered
m
−
m-
m
−
tuple
(
B
1
,
B
2
,
⋯
,
B
m
)
(B_1,B_2,\cdots,B_m)
(
B
1
,
B
2
,
⋯
,
B
m
)
is an ordered
m
−
m-
m
−
covering of
A
A
A
without pairs. Define
a
(
m
,
n
)
a(m,n)
a
(
m
,
n
)
as the number of the ordered
m
−
m-
m
−
coverings of
A
,
A,
A
,
and
b
(
m
,
n
)
b(m,n)
b
(
m
,
n
)
as the number of the ordered
m
−
m-
m
−
coverings of
A
A
A
without pairs.
(
1
)
(1)
(
1
)
Calculate
a
(
m
,
n
)
a(m,n)
a
(
m
,
n
)
and
b
(
m
,
n
)
.
b(m,n).
b
(
m
,
n
)
.
(
2
)
(2)
(
2
)
Let
m
≥
2
,
m\geq2,
m
≥
2
,
and there is at least one positive integer
n
,
n,
n
,
such that
a
(
m
,
n
)
b
(
m
,
n
)
≤
2021
,
\dfrac{a(m,n)}{b(m,n)}\leq2021,
b
(
m
,
n
)
a
(
m
,
n
)
≤
2021
,
Determine the greatest possible values of
m
.
m.
m
.
1
1
Hide problems
China South East Mathematical Olympiad 2021 Grade10 P1
A sequence
{
a
n
}
\{a_n\}
{
a
n
}
is defined recursively by
a
1
=
1
2
,
a_1=\frac{1}{2},
a
1
=
2
1
,
and for
n
≥
2
,
n\ge 2,
n
≥
2
,
0
<
a
n
≤
a
n
−
1
0<a_n\leq a_{n-1}
0
<
a
n
≤
a
n
−
1
and
a
n
2
(
a
n
−
1
+
1
)
+
a
n
−
1
2
(
a
n
+
1
)
−
2
a
n
a
n
−
1
(
a
n
a
n
−
1
+
a
n
+
1
)
=
0.
a_n^2(a_{n-1}+1)+a_{n-1}^2(a_n+1)-2a_na_{n-1}(a_na_{n-1}+a_n+1)=0.
a
n
2
(
a
n
−
1
+
1
)
+
a
n
−
1
2
(
a
n
+
1
)
−
2
a
n
a
n
−
1
(
a
n
a
n
−
1
+
a
n
+
1
)
=
0.
(
1
)
(1)
(
1
)
Determine the general formula of the sequence
{
a
n
}
;
\{a_n\};
{
a
n
}
;
(
2
)
(2)
(
2
)
Let
S
n
=
a
1
+
⋯
+
a
n
.
S_n=a_1+\cdots+a_n.
S
n
=
a
1
+
⋯
+
a
n
.
Prove that for
n
≥
1
,
n\ge 1,
n
≥
1
,
ln
(
n
2
+
1
)
<
S
n
<
ln
(
n
+
1
)
.
\ln\left(\frac{n}{2}+1\right)<S_n<\ln(n+1).
ln
(
2
n
+
1
)
<
S
n
<
ln
(
n
+
1
)
.
2
2
Hide problems
Centers in a isosceles triangle
In
△
A
B
C
\triangle ABC
△
A
BC
,
A
B
=
A
C
>
B
C
AB=AC>BC
A
B
=
A
C
>
BC
, point
O
,
H
O,H
O
,
H
are the circumcenter and orthocenter of
△
A
B
C
\triangle ABC
△
A
BC
respectively,
G
G
G
is the midpoint of segment
A
H
AH
A
H
,
B
E
BE
BE
is the altitude on
A
C
AC
A
C
. Prove that if
O
E
∥
B
C
OE\parallel BC
OE
∥
BC
, then
H
H
H
is the incenter of
△
G
B
C
\triangle GBC
△
GBC
.
China South East Mathematical Olympiad 2021 Grade11 P2
Let
p
≥
5
p\geq 5
p
≥
5
be a prime number, and set
M
=
{
1
,
2
,
⋯
,
p
−
1
}
.
M=\{1,2,\cdots,p-1\}.
M
=
{
1
,
2
,
⋯
,
p
−
1
}
.
Define
T
=
{
(
n
,
x
n
)
:
p
∣
n
x
n
−
1
and
n
,
x
n
∈
M
}
.
T=\{(n,x_n):p|nx_n-1\ \textup{and}\ n,x_n\in M\}.
T
=
{(
n
,
x
n
)
:
p
∣
n
x
n
−
1
and
n
,
x
n
∈
M
}
.
If
∑
(
n
,
x
n
)
∈
T
n
[
n
x
n
p
]
≡
k
(
m
o
d
p
)
,
\sum_{(n,x_n)\in T}n\left[\dfrac{nx_n}{p}\right]\equiv k \pmod {p},
∑
(
n
,
x
n
)
∈
T
n
[
p
n
x
n
]
≡
k
(
mod
p
)
,
with
0
≤
k
≤
p
−
1
,
0\leq k\leq p-1,
0
≤
k
≤
p
−
1
,
where
[
α
]
\left[\alpha\right]
[
α
]
denotes the largest integer that does not exceed
α
,
\alpha,
α
,
determine the value of
k
.
k.
k
.
4
2
Hide problems
2021China South East Mathematical Olympiad Grade10 P4
Suppose there are
n
≥
5
n\geq{5}
n
≥
5
different points arbitrarily arranged on a circle, the labels are
1
,
2
,
…
1, 2,\dots
1
,
2
,
…
, and
n
n
n
, and the permutation is
S
S
S
. For a permutation , a “descending chain” refers to several consecutive points on the circle , and its labels is a clockwise descending sequence (the length of sequence is at least
2
2
2
), and the descending chain cannot be extended to longer .The point with the largest label in the chain is called the "starting point of descent", and the other points in the chain are called the “non-starting point of descent” . For example: there are two descending chains
5
,
2
5, 2
5
,
2
and
4
,
1
4, 1
4
,
1
in
5
,
2
,
4
,
1
,
3
5, 2, 4, 1, 3
5
,
2
,
4
,
1
,
3
arranged in a clockwise direction, and
5
5
5
and
4
4
4
are their starting points of descent respectively, and
2
,
1
2, 1
2
,
1
is the non-starting point of descent . Consider the following operations: in the first round, find all descending chains in the permutation
S
S
S
, delete all non-starting points of descent , and then repeat the first round of operations for the arrangement of the remaining points, until no more descending chains can be found. Let
G
(
S
)
G(S)
G
(
S
)
be the number of all descending chains that permutation
S
S
S
has appeared in the operations,
A
(
S
)
A(S)
A
(
S
)
be the average value of
G
(
S
)
G(S)
G
(
S
)
of all possible n-point permutations
S
S
S
. (1) Find
A
(
5
)
A(5)
A
(
5
)
. (2)For
n
≥
6
n\ge{6}
n
≥
6
, prove that
83
120
n
−
1
2
≤
A
(
S
)
≤
101
120
n
−
1
2
.
\frac{83}{120}n-\frac{1}{2} \le A(S) \le \frac{101}{120}n-\frac{1}{2}.
120
83
n
−
2
1
≤
A
(
S
)
≤
120
101
n
−
2
1
.
China South East Mathematical Olympiad 2021 Grade11 P4
For positive integer
k
,
k,
k
,
we say that it is a Taurus integer if we can delete one element from the set
M
k
=
{
1
,
2
,
⋯
,
k
}
,
M_k=\{1,2,\cdots,k\},
M
k
=
{
1
,
2
,
⋯
,
k
}
,
such that the sum of remaining
k
−
1
k-1
k
−
1
elements is a positive perfect square. For example,
7
7
7
is a Taurus integer, because if we delete
3
3
3
from
M
7
=
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
,
M_7=\{1,2,3,4,5,6,7\},
M
7
=
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
,
the sum of remaining
6
6
6
elements is
25
,
25,
25
,
which is a positive perfect square.
(
1
)
(1)
(
1
)
Determine whether
2021
2021
2021
is a Taurus integer.
(
2
)
(2)
(
2
)
For positive integer
n
,
n,
n
,
determine the number of Taurus integers in
{
1
,
2
,
⋯
,
n
}
.
\{1,2,\cdots,n\}.
{
1
,
2
,
⋯
,
n
}
.
3
2
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China South East Mathematical Olympiad 2021 Grade10 Q3
Let
p
p
p
be an odd prime and
{
u
i
}
i
≥
0
\{u_i\}_{i\ge 0}
{
u
i
}
i
≥
0
be an integer sequence. Let
v
n
=
∑
i
=
0
n
C
n
i
p
i
u
i
v_n=\sum_{i=0}^{n} C_{n}^{i} p^iu_i
v
n
=
∑
i
=
0
n
C
n
i
p
i
u
i
where
C
n
i
C_n^i
C
n
i
denotes the binomial coefficients. If
v
n
=
0
v_n=0
v
n
=
0
holds for infinitely many
n
n
n
, prove that it holds for every positive integer
n
n
n
.
China South East Mathematical Olympiad 2021 Grade11 Q3
Let
a
,
b
,
c
≥
0
a,b,c\geq 0
a
,
b
,
c
≥
0
and
a
2
+
b
2
+
c
2
≤
1.
a^2+b^2+c^2\leq 1.
a
2
+
b
2
+
c
2
≤
1.
Prove that
a
a
2
+
b
c
+
1
+
b
b
2
+
c
a
+
1
+
c
c
2
+
a
b
+
1
+
3
a
b
c
<
3
\frac{a}{a^2+bc+1}+\frac{b}{b^2+ca+1}+\frac{c}{c^2+ab+1}+3abc<\sqrt 3
a
2
+
b
c
+
1
a
+
b
2
+
c
a
+
1
b
+
c
2
+
ab
+
1
c
+
3
ab
c
<
3