MathDB
Problems
Contests
National and Regional Contests
China Contests
China Team Selection Test
2024 China Team Selection Test
2024 China Team Selection Test
Part of
China Team Selection Test
Subcontests
(24)
24
1
Hide problems
Last one
Let
N
=
1
0
2024
N=10^{2024}
N
=
1
0
2024
.
S
S
S
is a square in the Cartesian plane with side length
N
N
N
and the sides parallel to the coordinate axes. Inside there are
N
N
N
points
P
1
P_1
P
1
,
P
2
P_2
P
2
,
…
\dots
…
,
P
N
P_N
P
N
all of which have different
x
x
x
coordinates, and the absolute value of the slope of any connected line between these points is at most
1
1
1
. Prove that there exists a line
l
l
l
such that at least
2024
2024
2024
of these points is at most distance
1
1
1
away from
l
l
l
.
23
1
Hide problems
Complex polynomial
P
(
z
)
=
a
n
z
n
+
⋯
+
a
1
z
+
z
0
P(z)=a_nz^n+\dots+a_1z+z_0
P
(
z
)
=
a
n
z
n
+
⋯
+
a
1
z
+
z
0
, with
a
n
≠
0
a_n\neq 0
a
n
=
0
is a polynomial with complex coefficients, such that when
∣
z
∣
=
1
|z|=1
∣
z
∣
=
1
,
∣
P
(
z
)
∣
≤
1
|P(z)|\leq 1
∣
P
(
z
)
∣
≤
1
. Prove that for any
0
≤
k
≤
n
−
1
0\leq k\leq n-1
0
≤
k
≤
n
−
1
,
∣
a
k
∣
≤
1
−
∣
a
n
∣
2
|a_k|\leq 1-|a_n|^2
∣
a
k
∣
≤
1
−
∣
a
n
∣
2
.Proposed by Jingjun Han
22
1
Hide problems
XEY is fixed
A
B
C
ABC
A
BC
is an isosceles triangle, with
A
B
=
A
C
AB=AC
A
B
=
A
C
.
D
D
D
is a moving point such that
A
D
∥
B
C
AD\parallel BC
A
D
∥
BC
,
B
D
>
C
D
BD>CD
B
D
>
C
D
. Moving point
E
E
E
is on the arc of
B
C
BC
BC
in circumcircle of
A
B
C
ABC
A
BC
not containing
A
A
A
, such that
E
B
<
E
C
EB<EC
EB
<
EC
. Ray
B
C
BC
BC
contains point
F
F
F
with
∠
A
D
E
=
∠
D
F
E
\angle ADE=\angle DFE
∠
A
D
E
=
∠
D
FE
. If ray
F
D
FD
F
D
intersects ray
B
A
BA
B
A
at
X
X
X
, and intersects ray
C
A
CA
C
A
at
Y
Y
Y
, prove that
∠
X
E
Y
\angle XEY
∠
XE
Y
is a fixed angle.
21
1
Hide problems
Inequality with Strange Condition
Let integer
n
≥
3
,
n\ge 3,
n
≥
3
,
(
n
2
)
\tbinom n2
(
2
n
)
nonnegative real numbers
a
i
,
j
a_{i,j}
a
i
,
j
satisfy
a
i
,
j
+
a
j
,
k
≤
a
i
,
k
a_{i,j}+a_{j,k}\le a_{i,k}
a
i
,
j
+
a
j
,
k
≤
a
i
,
k
holds for all
1
≤
i
<
j
<
k
≤
n
1\le i <j<k\le n
1
≤
i
<
j
<
k
≤
n
. Proof
⌊
n
2
4
⌋
∑
1
≤
i
<
j
≤
n
a
i
,
j
4
≥
(
∑
1
≤
i
<
j
≤
n
a
i
,
j
2
)
2
.
\left\lfloor\frac{n^2}4\right\rfloor\sum_{1\le i<j\le n}a_{i,j}^4\ge \left(\sum_{1\le i<j\le n}a_{i,j}^2\right)^2.
⌊
4
n
2
⌋
1
≤
i
<
j
≤
n
∑
a
i
,
j
4
≥
(
1
≤
i
<
j
≤
n
∑
a
i
,
j
2
)
2
.
Proposed by Jingjun Han
20
1
Hide problems
20240327 is good
A positive integer is a good number, if its base
10
10
10
representation can be split into at least
5
5
5
sections, each section with a non-zero digit, and after interpreting each section as a positive integer (omitting leading zero digits), they can be split into two groups, such that each group can be reordered to form a geometric sequence (if a group has
1
1
1
or
2
2
2
numbers, it is also a geometric sequence), for example
20240327
20240327
20240327
is a good number, since after splitting it as
2
∣
02
∣
403
∣
2
∣
7
2|02|403|2|7
2∣02∣403∣2∣7
,
2
∣
02
∣
2
2|02|2
2∣02∣2
and
403
∣
7
403|7
403∣7
form two groups of geometric sequences.If
a
>
1
a>1
a
>
1
,
m
>
2
m>2
m
>
2
,
p
=
1
+
a
+
a
2
+
⋯
+
a
m
p=1+a+a^2+\dots+a^m
p
=
1
+
a
+
a
2
+
⋯
+
a
m
is a prime, prove that
1
0
p
−
1
−
1
p
\frac{10^{p-1}-1}{p}
p
1
0
p
−
1
−
1
is a good number.
19
1
Hide problems
Colorful trapezoid
n
n
n
is a positive integer. An equilateral triangle of side length
3
n
3n
3
n
is split into
9
n
2
9n^2
9
n
2
unit equilateral triangles, each colored one of red, yellow, blue, such that each color appears
3
n
2
3n^2
3
n
2
times. We call a trapezoid formed by three unit equilateral triangles as a "standard trapezoid". If a "standard trapezoid" contains all three colors, we call it a "colorful trapezoid". Find the maximum possible number of "colorful trapezoids".
16
1
Hide problems
NT with condition that is expected to be almost always true by PNT
m
>
1
m>1
m
>
1
is an integer such that
[
2
m
−
m
+
1
,
2
m
]
[2m-\sqrt{m}+1, 2m]
[
2
m
−
m
+
1
,
2
m
]
contains a prime. Prove that for any pairwise distinct positive integers
a
1
a_1
a
1
,
a
2
a_2
a
2
,
…
\dots
…
,
a
m
a_m
a
m
, there is always
1
≤
i
,
j
≤
m
1\leq i,j\leq m
1
≤
i
,
j
≤
m
such that
a
i
(
a
i
,
a
j
)
≥
m
\frac{a_i}{(a_i, a_j)}\geq m
(
a
i
,
a
j
)
a
i
≥
m
.
17
1
Hide problems
Geometry with two cases
A
B
C
D
E
ABCDE
A
BC
D
E
is a convex pentagon with
B
D
=
C
D
=
A
C
BD=CD=AC
B
D
=
C
D
=
A
C
, and
B
B
B
,
C
C
C
,
D
D
D
,
E
E
E
are concyclic. If
∠
B
A
C
+
∠
A
E
D
=
18
0
∘
\angle BAC+\angle AED=180^{\circ}
∠
B
A
C
+
∠
A
E
D
=
18
0
∘
and
∠
D
C
A
=
∠
B
D
E
\angle DCA=\angle BDE
∠
D
C
A
=
∠
B
D
E
, prove that
A
B
=
D
E
AB=DE
A
B
=
D
E
or
A
B
=
2
A
E
AB=2AE
A
B
=
2
A
E
.
18
1
Hide problems
Finally an Inequality
Let
m
,
n
∈
Z
≥
0
,
m,n\in\mathbb Z_{\ge 0},
m
,
n
∈
Z
≥
0
,
a
0
,
a
1
,
…
,
a
m
,
b
0
,
b
1
,
…
,
b
n
∈
R
≥
0
a_0,a_1,\ldots ,a_m,b_0,b_1,\ldots ,b_n\in\mathbb R_{\ge 0}
a
0
,
a
1
,
…
,
a
m
,
b
0
,
b
1
,
…
,
b
n
∈
R
≥
0
For any integer
0
≤
k
≤
m
+
n
,
0\le k\le m+n,
0
≤
k
≤
m
+
n
,
define
c
k
:
=
max
i
+
j
=
k
a
i
b
j
.
c_k:=\max_{i+j=k}a_ib_j.
c
k
:=
max
i
+
j
=
k
a
i
b
j
.
Proof
1
m
+
n
+
1
∑
k
=
0
m
+
n
c
k
≥
1
(
m
+
1
)
(
n
+
1
)
∑
i
=
0
m
a
i
∑
j
=
0
n
b
j
.
\frac 1{m+n+1}\sum_{k=0}^{m+n}c_k\ge\frac 1{(m+1)(n+1)}\sum_{i=0}^{m}a_i\sum_{j=0}^{n}b_j.
m
+
n
+
1
1
k
=
0
∑
m
+
n
c
k
≥
(
m
+
1
)
(
n
+
1
)
1
i
=
0
∑
m
a
i
j
=
0
∑
n
b
j
.
14
1
Hide problems
n-good sets
For a positive integer
n
n
n
and a subset
S
S
S
of
{
1
,
2
,
…
,
n
}
\{1, 2, \dots, n\}
{
1
,
2
,
…
,
n
}
, let
S
S
S
be "
n
n
n
-good" if and only if for any
x
x
x
,
y
∈
S
y\in S
y
∈
S
(allowed to be same), if
x
+
y
≤
n
x+y\leq n
x
+
y
≤
n
, then
x
+
y
∈
S
x+y\in S
x
+
y
∈
S
. Let
r
n
r_n
r
n
be the smallest real number such that for any positive integer
m
≤
n
m\leq n
m
≤
n
, there is always a
m
m
m
-element "
n
n
n
-good" set, so that the sum of its elements is not more than
m
⋅
r
n
m\cdot r_n
m
⋅
r
n
. Prove that there exists a real number
α
\alpha
α
such that for any positive integer
n
n
n
,
∣
r
n
−
α
n
∣
≤
2024.
|r_n-\alpha n|\leq 2024.
∣
r
n
−
α
n
∣
≤
2024.
15
1
Hide problems
Fractional part of exponent of solution to polynomial
n
>
1
n>1
n
>
1
is an integer. Let real number
x
>
1
x>1
x
>
1
satisfy
x
101
−
n
x
100
+
n
x
−
1
=
0.
x^{101}-nx^{100}+nx-1=0.
x
101
−
n
x
100
+
n
x
−
1
=
0.
Prove that for any real
0
<
a
<
b
<
1
0<a<b<1
0
<
a
<
b
<
1
, there exists a positive integer
m
m
m
so that
a
<
{
x
m
}
<
b
.
a<\{x^m\}<b.
a
<
{
x
m
}
<
b
.
Proposed by Chenjie Yu
13
1
Hide problems
Catalan identity
For a natural number
n
n
n
, let
C
n
=
1
n
+
1
(
2
n
n
)
=
(
2
n
)
!
n
!
(
n
+
1
)
!
C_n=\frac{1}{n+1}\binom{2n}{n}=\frac{(2n)!}{n!(n+1)!}
C
n
=
n
+
1
1
(
n
2
n
)
=
n
!
(
n
+
1
)!
(
2
n
)!
be the
n
n
n
-th Catalan number. Prove that for any natural number
m
m
m
,
∑
i
+
j
+
k
=
m
C
i
+
j
C
j
+
k
C
k
+
i
=
3
2
m
+
3
C
2
m
+
1
.
\sum_{i+j+k=m} C_{i+j}C_{j+k}C_{k+i}=\frac{3}{2m+3}C_{2m+1}.
i
+
j
+
k
=
m
∑
C
i
+
j
C
j
+
k
C
k
+
i
=
2
m
+
3
3
C
2
m
+
1
.
Proposed by Bin Wang
11
1
Hide problems
Writing numbers on a blackboard
There is number
1
1
1
on the blackboard initially. The first step is to erase
1
1
1
and write two nonnegative reals whose sum is
1
1
1
. Call the smaller number of the two
L
2
L_2
L
2
. For integer
k
≥
2
k \ge 2
k
≥
2
, the
k
{k}
k
the step is to erase a number on the blackboard arbitrarily and write two nonnegative reals whose sum is the number erased just now. Call the smallest number of the
k
+
1
k+1
k
+
1
on the blackboard
L
k
+
1
L_{k+1}
L
k
+
1
. Find the maximum of
L
2
+
L
3
+
⋯
+
L
2024
L_2+L_3+\cdots+L_{2024}
L
2
+
L
3
+
⋯
+
L
2024
.
10
1
Hide problems
difference of roots
Let
M
M
M
be a positive integer.
f
(
x
)
:
=
x
3
+
a
x
2
+
b
x
+
c
∈
Z
[
x
]
f(x):=x^3+ax^2+bx+c\in\mathbb Z[x]
f
(
x
)
:=
x
3
+
a
x
2
+
b
x
+
c
∈
Z
[
x
]
satisfy
∣
a
∣
,
∣
b
∣
,
∣
c
∣
≤
M
.
|a|,|b|,|c|\le M.
∣
a
∣
,
∣
b
∣
,
∣
c
∣
≤
M
.
x
1
,
x
2
x_1,x_2
x
1
,
x
2
are different roots of
f
.
f.
f
.
Prove that
∣
x
1
−
x
2
∣
>
1
M
2
+
3
M
+
1
.
|x_1-x_2|>\frac 1{M^2+3M+1}.
∣
x
1
−
x
2
∣
>
M
2
+
3
M
+
1
1
.
12
1
Hide problems
Difficult NT
Given positive odd number
m
m
m
and integer
a
.
{a}.
a
.
Proof: For any real number
c
,
c,
c
,
#
{
x
∈
Z
∩
[
c
,
c
+
m
]
∣
x
2
≡
a
(
m
o
d
m
)
}
≤
2
+
log
2
m
.
\#\left\{x\in\mathbb Z\cap [c,c+\sqrt m]\mid x^2\equiv a\pmod m\right\}\le 2+\log_2m.
#
{
x
∈
Z
∩
[
c
,
c
+
m
]
∣
x
2
≡
a
(
mod
m
)
}
≤
2
+
lo
g
2
m
.
Proposed by Yinghua Ai
9
1
Hide problems
coloring of the positive integers
Color the positive integers by four colors
c
1
,
c
2
,
c
3
,
c
4
c_1,c_2,c_3,c_4
c
1
,
c
2
,
c
3
,
c
4
. (1)Prove that there exists a positive integer
n
n
n
and
i
,
j
∈
{
1
,
2
,
3
,
4
}
i,j\in\{1,2,3,4\}
i
,
j
∈
{
1
,
2
,
3
,
4
}
,such that among all the positive divisors of
n
n
n
, the number of divisors with color
c
i
c_i
c
i
is at least greater than the number of divisors with color
c
j
c_j
c
j
by
3
3
3
. (2)Prove that for any positive integer
A
A
A
,there exists a positive integer
n
n
n
and
i
,
j
∈
{
1
,
2
,
3
,
4
}
i,j\in\{1,2,3,4\}
i
,
j
∈
{
1
,
2
,
3
,
4
}
,such that among all the positive divisors of
n
n
n
, the number of divisors with color
c
i
c_i
c
i
is at least greater than the number of divisors with color
c
j
c_j
c
j
by
A
A
A
.
7
1
Hide problems
sum of reciprocals
For coprime positive integers
a
,
b
a,b
a
,
b
,denote
(
a
−
1
m
o
d
b
)
(a^{-1}\bmod{b})
(
a
−
1
mod
b
)
by the only integer
0
≤
m
<
b
0\leq m<b
0
≤
m
<
b
such that
a
m
≡
1
(
m
o
d
b
)
am\equiv 1\pmod{b}
am
≡
1
(
mod
b
)
(1)Prove that for pairwise coprime integers
a
,
b
,
c
a,b,c
a
,
b
,
c
,
1
<
a
<
b
<
c
1<a<b<c
1
<
a
<
b
<
c
,we have
(
a
−
1
m
o
d
b
)
+
(
b
−
1
m
o
d
c
)
+
(
c
−
1
m
o
d
a
)
>
a
.
(a^{-1}\bmod{b})+(b^{-1}\bmod{c})+(c^{-1}\bmod{a})>\sqrt a.
(
a
−
1
mod
b
)
+
(
b
−
1
mod
c
)
+
(
c
−
1
mod
a
)
>
a
.
(2)Prove that for any positive integer
M
M
M
,there exists pairwise coprime integers
a
,
b
,
c
a,b,c
a
,
b
,
c
,
M
<
a
<
b
<
c
M<a<b<c
M
<
a
<
b
<
c
such that
(
a
−
1
m
o
d
b
)
+
(
b
−
1
m
o
d
c
)
+
(
c
−
1
m
o
d
a
)
<
100
a
.
(a^{-1}\bmod{b})+(b^{-1}\bmod{c})+(c^{-1}\bmod{a})< 100\sqrt a.
(
a
−
1
mod
b
)
+
(
b
−
1
mod
c
)
+
(
c
−
1
mod
a
)
<
100
a
.
8
1
Hide problems
Golden ratio
In
△
A
B
C
\triangle {ABC}
△
A
BC
, tangents of the circumcircle
⊙
O
\odot {O}
⊙
O
at
B
,
C
B, C
B
,
C
and at
A
,
B
A, B
A
,
B
intersects at
X
,
Y
X, Y
X
,
Y
respectively.
A
X
AX
A
X
cuts
B
C
BC
BC
at
D
{D}
D
and
C
Y
CY
C
Y
cuts
A
B
AB
A
B
at
F
{F}
F
. Ray
D
F
DF
D
F
cuts arc
A
B
AB
A
B
of the circumcircle at
P
{P}
P
.
Q
,
R
Q, R
Q
,
R
are on segments
A
B
,
A
C
AB, AC
A
B
,
A
C
such that
P
,
Q
,
R
P, Q, R
P
,
Q
,
R
are collinear and
Q
R
∥
B
O
QR \parallel BO
QR
∥
BO
. If
P
Q
2
=
P
R
⋅
Q
R
PQ^2=PR \cdot QR
P
Q
2
=
PR
⋅
QR
, find
∠
A
C
B
\angle ACB
∠
A
CB
.
6
1
Hide problems
Combinatorial Geometry
Let
m
,
n
>
2
m,n>2
m
,
n
>
2
be integers. A regular
n
{n}
n
-sided polygon region
T
\mathcal T
T
on a plane contains a regular
m
{m}
m
-sided polygon region with a side length of
1
{}{}{}1
1
. Prove that any regular
m
{m}
m
-sided polygon region
S
\mathcal S
S
on the plane with side length
cos
π
/
[
m
,
n
]
\cos{\pi}/[m,n]
cos
π
/
[
m
,
n
]
can be translated inside
T
.
\mathcal T.
T
.
In other words, there exists a vector
α
⃗
,
\vec\alpha,
α
,
such that for each point in
S
,
\mathcal S,
S
,
after translating the vector
α
⃗
\vec\alpha
α
at that point, it fall into
T
.
\mathcal T.
T
.
Note: The polygonal area includes both the interior and boundaries.
4
1
Hide problems
Standard NT
Let
n
n
n
be a positive square free integer,
S
S
S
is a subset of
[
n
]
:
=
{
1
,
2
,
…
,
n
}
[n]:=\{1,2,\ldots ,n\}
[
n
]
:=
{
1
,
2
,
…
,
n
}
such that
∣
S
∣
≥
n
/
2.
|S|\ge n/2.
∣
S
∣
≥
n
/2.
Prove that there exists three elements
a
,
b
,
c
∈
S
a,b,c\in S
a
,
b
,
c
∈
S
(can be same), satisfy
a
b
≡
c
(
m
o
d
n
)
.
ab\equiv c\pmod n.
ab
≡
c
(
mod
n
)
.
5
1
Hide problems
FE in TST
Find all functions
f
:
N
+
→
N
+
,
f:\mathbb N_+\to \mathbb N_+,
f
:
N
+
→
N
+
,
such that for all positive integer
a
,
b
,
a,b,
a
,
b
,
∑
k
=
0
2
b
f
(
a
+
k
)
=
(
2
b
+
1
)
f
(
f
(
a
)
+
b
)
.
\sum_{k=0}^{2b}f(a+k)=(2b+1)f(f(a)+b).
k
=
0
∑
2
b
f
(
a
+
k
)
=
(
2
b
+
1
)
f
(
f
(
a
)
+
b
)
.
1
1
Hide problems
Polyhedron Colouring
It is known that each vertex of the convex polyhedron
P
P
P
belongs to three different faces, and each vertex of
P
P
P
can be dyed black and white, so that the two endpoints of each edge of
P
P
P
are different colors. Proof: The interior of each edge of
P
P
P
can be dyed red, yellow, and blue, so that the colors of the three edges connected to each vertex are different, and each face contains two colors of edges.
3
1
Hide problems
Nightmare NT comes again
Given positive integer
M
.
M.
M
.
For any
n
∈
N
+
,
n\in\mathbb N_+,
n
∈
N
+
,
let
h
(
n
)
h(n)
h
(
n
)
be the number of elements in
[
n
]
[n]
[
n
]
that are coprime to
M
.
M.
M
.
Define
β
:
=
h
(
M
)
M
.
\beta :=\frac {h(M)}M.
β
:=
M
h
(
M
)
.
Proof: there are at least
M
3
\frac M3
3
M
elements
n
n
n
in
[
M
]
,
[M],
[
M
]
,
satisfy
∣
h
(
n
)
−
β
n
∣
≤
β
⋅
2
ω
(
M
)
−
3
+
1.
\left| h(n)-\beta n\right|\le\sqrt{\beta\cdot 2^{\omega(M)-3}}+1.
∣
h
(
n
)
−
β
n
∣
≤
β
⋅
2
ω
(
M
)
−
3
+
1.
Here
[
n
]
:
=
{
1
,
2
,
…
,
n
}
[n]:=\{1,2,\ldots ,n\}
[
n
]
:=
{
1
,
2
,
…
,
n
}
for all positive integer
n
.
n.
n
.
Proposed by Bin Wang
2
1
Hide problems
Kiepert Hyperbola in Chinese TST
In acute triangle
△
A
B
C
\triangle {ABC}
△
A
BC
,
∠
A
>
∠
B
>
∠
C
\angle A > \angle B > \angle C
∠
A
>
∠
B
>
∠
C
.
△
A
C
1
B
\triangle {AC_1B}
△
A
C
1
B
and
△
C
B
1
A
\triangle {CB_1A}
△
C
B
1
A
are isosceles triangles such that
△
A
C
1
B
∼
+
△
C
B
1
A
\triangle {AC_1B} \stackrel{+}{\sim} \triangle {CB_1A}
△
A
C
1
B
∼
+
△
C
B
1
A
. Let lines
B
B
1
,
C
C
1
BB_1, CC_1
B
B
1
,
C
C
1
intersects at
T
{T}
T
. Prove that if all points mentioned above are distinct,
∠
A
T
C
\angle ATC
∠
A
TC
isn't a right angle.