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Contests
National and Regional Contests
Iran Contests
Iran Team Selection Test
2019 Iran Team Selection Test
2019 Iran Team Selection Test
Part of
Iran Team Selection Test
Subcontests
(6)
6
3
Hide problems
Iran TST
{
a
n
}
n
≥
0
\{a_{n}\}_{n\geq 0}
{
a
n
}
n
≥
0
and
{
b
n
}
n
≥
0
\{b_{n}\}_{n\geq 0}
{
b
n
}
n
≥
0
are two sequences of positive integers that
a
i
,
b
i
∈
{
0
,
1
,
2
,
⋯
,
9
}
a_{i},b_{i}\in \{0,1,2,\cdots,9\}
a
i
,
b
i
∈
{
0
,
1
,
2
,
⋯
,
9
}
. There is an integer number
M
M
M
such that
a
n
,
b
n
≠
0
a_{n},b_{n}\neq 0
a
n
,
b
n
=
0
for all
n
≥
M
n\geq M
n
≥
M
and for each
n
≥
0
n\geq 0
n
≥
0
(
a
n
⋯
a
1
a
0
‾
)
2
+
999
∣
(
b
n
⋯
b
1
b
0
‾
)
2
+
999
(\overline{a_{n}\cdots a_{1}a_{0}})^{2}+999 \mid(\overline{b_{n}\cdots b_{1}b_{0}})^{2}+999
(
a
n
⋯
a
1
a
0
)
2
+
999
∣
(
b
n
⋯
b
1
b
0
)
2
+
999
prove that
a
n
=
b
n
a_{n}=b_{n}
a
n
=
b
n
for
n
≥
0
n\geq 0
n
≥
0
.\\ (Note that
(
x
n
x
n
−
1
…
x
0
‾
)
=
1
0
n
×
x
n
+
⋯
+
10
×
x
1
+
x
0
(\overline{x_nx_{n-1}\dots x_0}) = 10^n\times x_n + \dots + 10\times x_1 + x_0
(
x
n
x
n
−
1
…
x
0
)
=
1
0
n
×
x
n
+
⋯
+
10
×
x
1
+
x
0
.)Proposed by Yahya Motevassel
Iran inequality
x
,
y
x,y
x
,
y
and
z
z
z
are real numbers such that
x
+
y
+
z
=
x
y
+
y
z
+
z
x
x+y+z=xy+yz+zx
x
+
y
+
z
=
x
y
+
yz
+
z
x
. Prove that
x
x
4
+
x
2
+
1
+
y
y
4
+
y
2
+
1
+
z
z
4
+
z
2
+
1
≥
−
1
3
.
\frac{x}{\sqrt{x^4+x^2+1}}+\frac{y}{\sqrt{y^4+y^2+1}}+\frac{z}{\sqrt{z^4+z^2+1}}\geq \frac{-1}{\sqrt{3}}.
x
4
+
x
2
+
1
x
+
y
4
+
y
2
+
1
y
+
z
4
+
z
2
+
1
z
≥
3
−
1
.
Proposed by Navid Safaei
Iran TST
For any positive integer
n
n
n
, define the subset
S
n
S_n
S
n
of natural numbers as follow
S
n
=
{
x
2
+
n
y
2
:
x
,
y
∈
Z
}
.
S_n = \left\{x^2+ny^2 : x,y \in \mathbb{Z} \right\}.
S
n
=
{
x
2
+
n
y
2
:
x
,
y
∈
Z
}
.
Find all positive integers
n
n
n
such that there exists an element of
S
n
S_n
S
n
which doesn't belong to any of the sets
S
1
,
S
2
,
…
,
S
n
−
1
S_1, S_2,\dots,S_{n-1}
S
1
,
S
2
,
…
,
S
n
−
1
.Proposed by Yahya Motevassel
1
3
Hide problems
Iran TST
A table consisting of
5
5
5
columns and
32
32
32
rows, which are filled with zero and one numbers, are "varied", if no two lines are filled in the same way.\\ On the exterior of a cylinder, a table with
32
32
32
rows and
16
16
16
columns is constructed. Is it possible to fill the numbers cells of the table with numbers zero and one, such that any five consecutive columns, table
32
×
5
32\times5
32
×
5
created by these columns, is a varied one?Proposed by Morteza Saghafian
Iranian TST 2019, second exam, day 1, problem 1
S
S
S
is a subset of Natural numbers which has infinite members.
S
’
=
{
x
y
+
y
x
:
x
,
y
∈
S
,
x
≠
y
}
S’=\left\{x^y+y^x: \, x,y\in S, \, x\neq y\right\}
S
’
=
{
x
y
+
y
x
:
x
,
y
∈
S
,
x
=
y
}
Prove the set of prime divisors of
S
’
S’
S
’
has also infinite members Proposed by Yahya Motevassel
Iran polynomial
Find all polynomials
P
(
x
,
y
)
P(x,y)
P
(
x
,
y
)
with real coefficients such that for all real numbers
x
,
y
x,y
x
,
y
and
z
z
z
:
P
(
x
,
2
y
z
)
+
P
(
y
,
2
z
x
)
+
P
(
z
,
2
x
y
)
=
P
(
x
+
y
+
z
,
x
y
+
y
z
+
z
x
)
.
P(x,2yz)+P(y,2zx)+P(z,2xy)=P(x+y+z,xy+yz+zx).
P
(
x
,
2
yz
)
+
P
(
y
,
2
z
x
)
+
P
(
z
,
2
x
y
)
=
P
(
x
+
y
+
z
,
x
y
+
yz
+
z
x
)
.
Proposed by Sina Saleh
4
3
Hide problems
Iran TST
Consider triangle
A
B
C
ABC
A
BC
with orthocenter
H
H
H
. Let points
M
M
M
and
N
N
N
be the midpoints of segments
B
C
BC
BC
and
A
H
AH
A
H
. Point
D
D
D
lies on line
M
H
MH
M
H
so that
A
D
∥
B
C
AD\parallel BC
A
D
∥
BC
and point
K
K
K
lies on line
A
H
AH
A
H
so that
D
N
M
K
DNMK
D
NM
K
is cyclic. Points
E
E
E
and
F
F
F
lie on lines
A
C
AC
A
C
and
A
B
AB
A
B
such that
∠
E
H
M
=
∠
C
\angle EHM=\angle C
∠
E
H
M
=
∠
C
and
∠
F
H
M
=
∠
B
\angle FHM=\angle B
∠
F
H
M
=
∠
B
. Prove that points
D
,
E
,
F
D,E,F
D
,
E
,
F
and
K
K
K
lie on a circle. Proposed by Alireza Dadgarnia
Iran TST
Let
1
<
t
<
2
1<t<2
1
<
t
<
2
be a real number. Prove that for all sufficiently large positive integers like
d
d
d
, there is a monic polynomial
P
(
x
)
P(x)
P
(
x
)
of degree
d
d
d
, such that all of its coefficients are either
+
1
+1
+
1
or
−
1
-1
−
1
and
∣
P
(
t
)
−
2019
∣
<
1.
\left|P(t)-2019\right| <1.
∣
P
(
t
)
−
2019
∣
<
1.
Proposed by Navid Safaei
Iran geometry
Given an acute-angled triangle
A
B
C
ABC
A
BC
with orthocenter
H
H
H
. Reflection of nine-point circle about
A
H
AH
A
H
intersects circumcircle at points
X
X
X
and
Y
Y
Y
. Prove that
A
H
AH
A
H
is the external bisector of
∠
X
H
Y
\angle XHY
∠
X
H
Y
. Proposed by Mohammad Javad Shabani
5
3
Hide problems
Beautiful combinatorial geometry (Iran TST)
Let
P
P
P
be a simple polygon completely in
C
C
C
, a circle with radius
1
1
1
, such that
P
P
P
does not pass through the center of
C
C
C
. The perimeter of
P
P
P
is
36
36
36
. Prove that there is a radius of
C
C
C
that intersects
P
P
P
at least
6
6
6
times, or there is a circle which is concentric with
C
C
C
and have at least
6
6
6
common points with
P
P
P
.Proposed by Seyed Reza Hosseini
Iran TST
A sub-graph of a complete graph with
n
n
n
vertices is chosen such that the number of its edges is a multiple of
3
3
3
and degree of each vertex is an even number. Prove that we can assign a weight to each triangle of the graph such that for each edge of the chosen sub-graph, the sum of the weight of the triangles that contain that edge equals one, and for each edge that is not in the sub-graph, this sum equals zero.Proposed by Morteza Saghafian
Iran TST
Find all functions
f
:
R
→
R
f:\mathbb{R}\rightarrow \mathbb{R}
f
:
R
→
R
such that for all
x
,
y
∈
R
x,y\in \mathbb{R}
x
,
y
∈
R
:
f
(
f
(
x
)
2
−
y
2
)
2
+
f
(
2
x
y
)
2
=
f
(
x
2
+
y
2
)
2
f\left(f(x)^2-y^2\right)^2+f(2xy)^2=f\left(x^2+y^2\right)^2
f
(
f
(
x
)
2
−
y
2
)
2
+
f
(
2
x
y
)
2
=
f
(
x
2
+
y
2
)
2
Proposed by Ali Behrouz - Mojtaba Zare Bidaki
3
3
Hide problems
Iran TST
Numbers
m
m
m
and
n
n
n
are given positive integers. There are
m
n
mn
mn
people in a party, standing in the shape of an
m
×
n
m\times n
m
×
n
grid. Some of these people are police officers and the rest are the guests. Some of the guests may be criminals. The goal is to determine whether there is a criminal between the guests or not.\\ Two people are considered
adjacent
if they have a common side. Any police officer can see their adjacent people and for every one of them, know that they're criminal or not. On the other hand, any criminal will threaten exactly one of their adjacent people (which is likely an officer!) to murder. A threatened officer will be too scared, that they deny the existence of any criminal between their adjacent people.\\ Find the least possible number of officers such that they can take position in the party, in a way that the goal is achievable. (Note that the number of criminals is unknown and it is possible to have zero criminals.)Proposed by Abolfazl Asadi
Iran TST
Point
P
P
P
lies inside of parallelogram
A
B
C
D
ABCD
A
BC
D
. Perpendicular lines to
P
A
,
P
B
,
P
C
PA,PB,PC
P
A
,
PB
,
PC
and
P
D
PD
P
D
through
A
,
B
,
C
A,B,C
A
,
B
,
C
and
D
D
D
construct convex quadrilateral
X
Y
Z
T
XYZT
X
Y
ZT
. Prove that
S
X
Y
Z
T
≥
2
S
A
B
C
D
S_{XYZT}\geq 2S_{ABCD}
S
X
Y
ZT
≥
2
S
A
BC
D
.Proposed by Siamak Ahmadpour
Iran geometry
In triangle
A
B
C
ABC
A
BC
,
M
,
N
M,N
M
,
N
and
P
P
P
are midpoints of sides
B
C
,
C
A
BC,CA
BC
,
C
A
and
A
B
AB
A
B
. Point
K
K
K
lies on segment
N
P
NP
NP
so that
A
K
AK
A
K
bisects
∠
B
K
C
\angle BKC
∠
B
K
C
. Lines
M
N
,
B
K
MN,BK
MN
,
B
K
intersects at
E
E
E
and lines
M
P
,
C
K
MP,CK
MP
,
C
K
intersects at
F
F
F
. Suppose that
H
H
H
be the foot of perpendicular line from
A
A
A
to
B
C
BC
BC
and
L
L
L
the second intersection of circumcircle of triangles
A
K
H
,
H
E
F
AKH, HEF
A
KH
,
H
EF
. Prove that
M
K
,
E
F
MK,EF
M
K
,
EF
and
H
L
HL
H
L
are concurrent.Proposed by Alireza Dadgarnia
2
3
Hide problems
Iranian TST 2019, first exam, day1, problem 2
a
,
a
1
,
a
2
,
…
,
a
n
a, a_1,a_2,\dots ,a_n
a
,
a
1
,
a
2
,
…
,
a
n
are natural numbers. We know that for any natural number
k
k
k
which
a
k
+
1
ak+1
ak
+
1
is square, at least one of
a
1
k
+
1
,
…
,
a
n
k
+
1
a_1k+1,\dots ,a_n k+1
a
1
k
+
1
,
…
,
a
n
k
+
1
is also square. Prove
a
a
a
is one of
a
1
,
…
,
a
n
a_1,\dots ,a_n
a
1
,
…
,
a
n
Proposed by Mohsen Jamali
Iranian TST 2019, second exam, day 1, problem 2
In a triangle
A
B
C
ABC
A
BC
,
∠
A
\angle A
∠
A
is
6
0
∘
60^\circ
6
0
∘
. On sides
A
B
AB
A
B
and
A
C
AC
A
C
we make two equilateral triangles (outside the triangle
A
B
C
ABC
A
BC
)
A
B
K
ABK
A
B
K
and
A
C
L
ACL
A
C
L
.
C
K
CK
C
K
and
A
B
AB
A
B
intersect at
S
S
S
,
A
C
AC
A
C
and
B
L
BL
B
L
intersect at
R
R
R
,
B
L
BL
B
L
and
C
K
CK
C
K
intersect at
T
T
T
. Prove the radical centre of circumcircle of triangles
B
S
K
,
C
L
R
BSK, CLR
BS
K
,
C
L
R
and
B
T
C
BTC
BTC
is on the median of vertex
A
A
A
in triangle
A
B
C
ABC
A
BC
. Proposed by Ali Zamani
Iran combinatorial number thoery
Hesam chose
10
10
10
distinct positive integers and he gave all pairwise
gcd
\gcd
g
cd
's and pairwise
l
c
m
{\text lcm}
l
c
m
's (a total of
90
90
90
numbers) to Masoud. Can Masoud always find the first
10
10
10
numbers, just by knowing these
90
90
90
numbers?Proposed by Morteza Saghafian