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National and Regional Contests
Iran Contests
Iran Team Selection Test
2018 Iran Team Selection Test
2018 Iran Team Selection Test
Part of
Iran Team Selection Test
Subcontests
(6)
6
3
Hide problems
iran tst 2018 combinatorics
a
1
,
a
2
,
…
,
a
n
a_1,a_2,\ldots,a_n
a
1
,
a
2
,
…
,
a
n
is a sequence of positive integers that has at least
2
n
3
+
1
\frac {2n}{3}+1
3
2
n
+
1
distinct numbers and each positive integer has occurred at most three times in it. Prove that there exists a permutation
b
1
,
b
2
,
…
,
b
n
b_1,b_2,\ldots,b_n
b
1
,
b
2
,
…
,
b
n
of
a
i
a_i
a
i
's such that all the
n
n
n
sums
b
i
+
b
i
+
1
b_i+b_{i+1}
b
i
+
b
i
+
1
are distinct (
1
≤
i
≤
n
1\le i\le n
1
≤
i
≤
n
,
b
n
+
1
≡
b
1
b_{n+1}\equiv b_1
b
n
+
1
≡
b
1
)Proposed by Mohsen Jamali
Iran combinatorics
A simple graph is called "divisibility", if it's possible to put distinct numbers on its vertices such that there is an edge between two vertices if and only if number of one of its vertices is divisible by another one.A simple graph is called "permutationary", if it's possible to put numbers
1
,
2
,
.
.
.
,
n
1,2,...,n
1
,
2
,
...
,
n
on its vertices and there is a permutation
π
\pi
π
such that there is an edge between vertices
i
,
j
i,j
i
,
j
if and only if
i
>
j
i>j
i
>
j
and
π
(
i
)
<
π
(
j
)
\pi(i)< \pi(j)
π
(
i
)
<
π
(
j
)
(it's not directed!)Prove that a simple graph is permutationary if and only if its complement and itself are divisibility.Proposed by Morteza Saghafian .
Geometry from Iran TST
Consider quadrilateral
A
B
C
D
ABCD
A
BC
D
inscribed in circle
ω
\omega
ω
.
P
≡
A
C
∩
B
D
P\equiv AC\cap BD
P
≡
A
C
∩
B
D
.
E
E
E
,
F
F
F
lie on sides
A
B
AB
A
B
,
C
D
CD
C
D
respectively such that
A
P
E
^
=
D
P
F
^
\hat {APE}=\hat {DPF}
A
PE
^
=
D
PF
^
. Circles
ω
1
\omega_1
ω
1
,
ω
2
\omega_2
ω
2
are tangent to
ω
\omega
ω
at
X
X
X
,
Y
Y
Y
respectively and also both tangent to the circumcircle of
△
P
E
F
\triangle PEF
△
PEF
at
P
P
P
. Prove that:
E
X
E
Y
=
F
X
F
Y
\frac {EX}{EY}=\frac {FX}{FY}
E
Y
EX
=
F
Y
FX
Proposed by Ali Zamani
1
3
Hide problems
Iran combinatorial number theory
Let
A
1
,
A
2
,
.
.
.
,
A
k
A_1, A_2, ... , A_k
A
1
,
A
2
,
...
,
A
k
be the subsets of
{
1
,
2
,
3
,
.
.
.
,
n
}
\left\{1,2,3,...,n\right\}
{
1
,
2
,
3
,
...
,
n
}
such that for all
1
≤
i
,
j
≤
k
1\leq i,j\leq k
1
≤
i
,
j
≤
k
:
A
i
∩
A
j
≠
∅
A_i\cap A_j \neq \varnothing
A
i
∩
A
j
=
∅
. Prove that there are
n
n
n
distinct positive integers
x
1
,
x
2
,
.
.
.
,
x
n
x_1,x_2,...,x_n
x
1
,
x
2
,
...
,
x
n
such that for each
1
≤
j
≤
k
1\leq j\leq k
1
≤
j
≤
k
:
l
c
m
i
∈
A
j
{
x
i
}
>
l
c
m
i
∉
A
j
{
x
i
}
lcm_{i \in A_j}\left\{x_i\right\}>lcm_{i \notin A_j}\left\{x_i\right\}
l
c
m
i
∈
A
j
{
x
i
}
>
l
c
m
i
∈
/
A
j
{
x
i
}
Proposed by Morteza Saghafian, Mahyar Sefidgaran
Functional equation
Find all functions
f
:
R
→
R
f:\mathbb{R}\rightarrow \mathbb{R}
f
:
R
→
R
that satisfy the following conditions: a. x+f(y+f(x))=y+f(x+f(y)) \forall x,y \in \mathbb{R} b. The set
I
=
{
f
(
x
)
−
f
(
y
)
x
−
y
∣
x
,
y
∈
R
,
x
≠
y
}
I=\left\{\frac{f(x)-f(y)}{x-y}\mid x,y\in \mathbb{R},x\neq y \right\}
I
=
{
x
−
y
f
(
x
)
−
f
(
y
)
∣
x
,
y
∈
R
,
x
=
y
}
is an interval.Proposed by Navid Safaei
an easy geometry from iran tst
Two circles
ω
1
(
O
)
\omega_1(O)
ω
1
(
O
)
and
ω
2
\omega_2
ω
2
intersect each other at
A
,
B
A,B
A
,
B
,and
O
O
O
lies on
ω
2
\omega_2
ω
2
. Let
S
S
S
be a point on
A
B
AB
A
B
such that
O
S
⊥
A
B
OS\perp AB
OS
⊥
A
B
. Line
O
S
OS
OS
intersects
ω
2
\omega_2
ω
2
at
P
P
P
(other than
O
O
O
). The bisector of
A
S
P
^
\hat{ASP}
A
SP
^
intersects
ω
1
\omega_1
ω
1
at
L
L
L
(
A
A
A
and
L
L
L
are on the same side of the line
O
P
OP
OP
). Let
K
K
K
be a point on
ω
2
\omega_2
ω
2
such that
P
S
=
P
K
PS=PK
PS
=
P
K
(
A
A
A
and
K
K
K
are on the same side of the line
O
P
OP
OP
). Prove that
S
L
=
K
L
SL=KL
S
L
=
K
L
.Proposed by Ali Zamani
5
3
Hide problems
Iran number theory
Prove that for each positive integer
m
m
m
, one can find
m
m
m
consecutive positive integers like
n
n
n
such that the following phrase doesn't be a perfect power:
(
1
3
+
201
8
3
)
(
2
3
+
201
8
3
)
⋯
(
n
3
+
201
8
3
)
\left(1^3+2018^3\right)\left(2^3+2018^3\right)\cdots \left(n^3+2018^3\right)
(
1
3
+
201
8
3
)
(
2
3
+
201
8
3
)
⋯
(
n
3
+
201
8
3
)
Proposed by Navid Safaei
iran tst 2018 geometry
Let
ω
\omega
ω
be the circumcircle of isosceles triangle
A
B
C
ABC
A
BC
(
A
B
=
A
C
AB=AC
A
B
=
A
C
). Points
P
P
P
and
Q
Q
Q
lie on
ω
\omega
ω
and
B
C
BC
BC
respectively such that
A
P
=
A
Q
AP=AQ
A
P
=
A
Q
.
A
P
AP
A
P
and
B
C
BC
BC
intersect at
R
R
R
. Prove that the tangents from
B
B
B
and
C
C
C
to the incircle of
△
A
Q
R
\triangle AQR
△
A
QR
(different from
B
C
BC
BC
) are concurrent on
ω
\omega
ω
.Proposed by Ali Zamani, Hooman Fattahi
Combinatorics from Iran TST
2
n
−
1
2n-1
2
n
−
1
distinct positive real numbers with sum
S
S
S
are given. Prove that there are at least
(
2
n
−
2
n
−
1
)
\binom {2n-2}{n-1}
(
n
−
1
2
n
−
2
)
different ways to choose
n
n
n
numbers among them such that their sum is at least
S
2
\frac {S}{2}
2
S
.Proposed by Amirhossein Gorzi
4
3
Hide problems
Iran geometry
Let
A
B
C
ABC
A
BC
be a triangle (
∠
A
≠
9
0
∘
\angle A\neq 90^\circ
∠
A
=
9
0
∘
).
B
E
,
C
F
BE,CF
BE
,
CF
are the altitudes of the triangle. The bisector of
∠
A
\angle A
∠
A
intersects
E
F
,
B
C
EF,BC
EF
,
BC
at
M
,
N
M,N
M
,
N
. Let
P
P
P
be a point such that
M
P
⊥
E
F
MP\perp EF
MP
⊥
EF
and
N
P
⊥
B
C
NP\perp BC
NP
⊥
BC
. Prove that
A
P
AP
A
P
passes through the midpoint of
B
C
BC
BC
.Proposed by Iman Maghsoudi, Hooman Fattahi
iran tst 2018 number theory
Call a positive integer "useful but not optimized " (!), if it can be written as a sum of distinct powers of
3
3
3
and powers of
5
5
5
. Prove that there exist infinitely many positive integers which they are not "useful but not optimized". (e.g.
37
=
(
3
0
+
3
1
+
3
3
)
+
(
5
0
+
5
1
)
37=(3^0+3^1+3^3)+(5^0+5^1)
37
=
(
3
0
+
3
1
+
3
3
)
+
(
5
0
+
5
1
)
is a " useful but not optimized" number)Proposed by Mohsen Jamali
A number theory problem from Iran TST
We say distinct positive integers
a
1
,
a
2
,
…
,
a
n
a_1,a_2,\ldots ,a_n
a
1
,
a
2
,
…
,
a
n
are "good" if their sum is equal to the sum of all pairwise
gcd
\gcd
g
cd
's among them. Prove that there are infinitely many
n
n
n
s such that
n
n
n
good numbers exist.Proposed by Morteza Saghafian
2
3
Hide problems
Inequality
Determine the least real number
k
k
k
such that the inequality
(
2
a
a
−
b
)
2
+
(
2
b
b
−
c
)
2
+
(
2
c
c
−
a
)
2
+
k
≥
4
(
2
a
a
−
b
+
2
b
b
−
c
+
2
c
c
−
a
)
\left(\frac{2a}{a-b}\right)^2+\left(\frac{2b}{b-c}\right)^2+\left(\frac{2c}{c-a}\right)^2+k \geq 4\left(\frac{2a}{a-b}+\frac{2b}{b-c}+\frac{2c}{c-a}\right)
(
a
−
b
2
a
)
2
+
(
b
−
c
2
b
)
2
+
(
c
−
a
2
c
)
2
+
k
≥
4
(
a
−
b
2
a
+
b
−
c
2
b
+
c
−
a
2
c
)
holds for all real numbers
a
,
b
,
c
a,b,c
a
,
b
,
c
.Proposed by Mohammad Jafari
iran tst 2018 combinatorics
Mojtaba and Hooman are playing a game. Initially Mojtaba draws
2018
2018
2018
vectors with zero sum. Then in each turn, starting with Mojtaba, the player takes a vector and puts it on the plane. After the first move, the players must put their vector next to the previous vector (the beginning of the vector must lie on the end of the previous vector). At last, there will be a closed polygon. If this polygon is not self-intersecting, Mojtaba wins. Otherwise Hooman. Who has the winning strategy?Proposed by Mahyar Sefidgaran, Jafar Namdar
a nice number theory problem from iran tst
Find the maximum possible value of
k
k
k
for which there exist distinct reals
x
1
,
x
2
,
…
,
x
k
x_1,x_2,\ldots ,x_k
x
1
,
x
2
,
…
,
x
k
greater than
1
1
1
such that for all
1
≤
i
,
j
≤
k
1 \leq i, j \leq k
1
≤
i
,
j
≤
k
,
x
i
⌊
x
j
⌋
=
x
j
⌊
x
i
⌋
.
x_i^{\lfloor x_j \rfloor }= x_j^{\lfloor x_i\rfloor}.
x
i
⌊
x
j
⌋
=
x
j
⌊
x
i
⌋
.
Proposed by Morteza Saghafian
3
3
Hide problems
Iran geometry
In triangle
A
B
C
ABC
A
BC
let
M
M
M
be the midpoint of
B
C
BC
BC
. Let
ω
\omega
ω
be a circle inside of
A
B
C
ABC
A
BC
and is tangent to
A
B
,
A
C
AB,AC
A
B
,
A
C
at
E
,
F
E,F
E
,
F
, respectively. The tangents from
M
M
M
to
ω
\omega
ω
meet
ω
\omega
ω
at
P
,
Q
P,Q
P
,
Q
such that
P
P
P
and
B
B
B
lie on the same side of
A
M
AM
A
M
. Let
X
≡
P
M
∩
B
F
X \equiv PM \cap BF
X
≡
PM
∩
BF
and
Y
≡
Q
M
∩
C
E
Y \equiv QM \cap CE
Y
≡
QM
∩
CE
. If
2
P
M
=
B
C
2PM=BC
2
PM
=
BC
prove that
X
Y
XY
X
Y
is tangent to
ω
\omega
ω
.Proposed by Iman Maghsoudi
Number theory
Let
a
1
,
a
2
,
a
3
,
⋯
a_1,a_2,a_3,\cdots
a
1
,
a
2
,
a
3
,
⋯
be an infinite sequence of distinct integers. Prove that there are infinitely many primes
p
p
p
that distinct positive integers
i
,
j
,
k
i,j,k
i
,
j
,
k
can be found such that
p
∣
a
i
a
j
a
k
−
1
p\mid a_ia_ja_k-1
p
∣
a
i
a
j
a
k
−
1
.Proposed by Mohsen Jamali
a really nice polynomial problem
n
>
1
n>1
n
>
1
and distinct positive integers
a
1
,
a
2
,
…
,
a
n
+
1
a_1,a_2,\ldots,a_{n+1}
a
1
,
a
2
,
…
,
a
n
+
1
are given. Does there exist a polynomial
p
(
x
)
∈
Z
[
x
]
p(x)\in\Bbb{Z}[x]
p
(
x
)
∈
Z
[
x
]
of degree
≤
n
\le n
≤
n
that satisfies the following conditions? a.
∀
1
≤
i
<
j
≤
n
+
1
:
gcd
(
p
(
a
i
)
,
p
(
a
j
)
)
>
1
\forall_{1\le i < j\le n+1}: \gcd(p(a_i),p(a_j))>1
∀
1
≤
i
<
j
≤
n
+
1
:
g
cd
(
p
(
a
i
)
,
p
(
a
j
))
>
1
b.
∀
1
≤
i
<
j
<
k
≤
n
+
1
:
gcd
(
p
(
a
i
)
,
p
(
a
j
)
,
p
(
a
k
)
)
=
1
\forall_{1\le i < j < k\le n+1}: \gcd(p(a_i),p(a_j),p(a_k))=1
∀
1
≤
i
<
j
<
k
≤
n
+
1
:
g
cd
(
p
(
a
i
)
,
p
(
a
j
)
,
p
(
a
k
))
=
1
Proposed by Mojtaba Zare