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Contests
National and Regional Contests
Iran Contests
Iran Team Selection Test
2024 Iran Team Selection Test
2024 Iran Team Selection Test
Part of
Iran Team Selection Test
Subcontests
(12)
12
1
Hide problems
Tangency of circles with "135 degree" angles
For a triangle
△
A
B
C
\triangle ABC
△
A
BC
with an obtuse angle
∠
A
\angle A
∠
A
, let
E
,
F
E , F
E
,
F
be feet of altitudes from
B
,
C
B , C
B
,
C
on sides
A
C
,
A
B
AC , AB
A
C
,
A
B
respectively. The tangents from
B
,
C
B , C
B
,
C
to circumcircle of triangle
△
A
B
C
\triangle ABC
△
A
BC
intersect line
E
F
EF
EF
at points
K
,
L
K , L
K
,
L
respectively and we know that
∠
C
L
B
=
135
\angle CLB=135
∠
C
L
B
=
135
. Point
R
R
R
lies on segment
B
K
BK
B
K
in such a way that
K
R
=
K
L
KR=KL
K
R
=
K
L
and let
S
S
S
be a point on line
B
K
BK
B
K
such that
K
K
K
is between
B
,
S
B , S
B
,
S
and
∠
B
L
S
=
135
\angle BLS=135
∠
B
L
S
=
135
. Prove that the circle with diameter
R
S
RS
RS
is tangent to circumcircle of triangle
△
A
B
C
\triangle ABC
△
A
BC
.Proposed by Mehran Talaei
11
1
Hide problems
Sepehr the chemist
Let
n
<
k
n<k
n
<
k
be two natural numbers and suppose that Sepehr has
n
n
n
chemical elements ,
2
k
2k
2
k
grams from each , divided arbitrarily in
2
k
2k
2
k
cups.Find the smallest number
b
b
b
such that there is always possible for Sepehr to choose
b
b
b
cups , containing at least
2
2
2
grams from each element in total.Proposed by Josef Tkadlec & Morteza Saghafian
10
1
Hide problems
Another infinite prime divisors problem
Let
{
a
n
}
\{a_n\}
{
a
n
}
be a sequence of natural numbers such that each prime number greater than
1402
1402
1402
divides a member of that. Prove that the set of prime divisors of members of sequence
{
b
n
}
\{b_n\}
{
b
n
}
which
b
n
=
a
1
a
2
.
.
.
a
n
−
1
b_n=a_1a_2...a_n-1
b
n
=
a
1
a
2
...
a
n
−
1
, is infinite.Proposed by Navid Safaei
9
1
Hide problems
Complicated binomial divisibility
Prove that for any natural numbers
a
,
b
,
c
a , b , c
a
,
b
,
c
that
b
>
a
>
1
b>a>1
b
>
a
>
1
and
g
c
d
(
c
,
a
b
)
=
1
gcd(c,ab)=1
g
c
d
(
c
,
ab
)
=
1
, there exist a natural number
n
n
n
such that :
c
∣
(
b
n
a
n
)
c | \binom{b^n}{a^n}
c
∣
(
a
n
b
n
)
Proposed by Navid Safaei
8
1
Hide problems
A function on Q[x]
Find all functions
f
:
Q
[
x
]
→
Q
[
x
]
f : \mathbb{Q}[x] \to \mathbb{Q}[x]
f
:
Q
[
x
]
→
Q
[
x
]
such that two following conditions holds :
∀
P
,
Q
∈
Q
[
x
]
:
f
(
P
+
Q
)
=
f
(
P
)
+
f
(
Q
)
\forall P , Q \in \mathbb{Q}[x] : f(P+Q)=f(P)+f(Q)
∀
P
,
Q
∈
Q
[
x
]
:
f
(
P
+
Q
)
=
f
(
P
)
+
f
(
Q
)
∀
P
∈
Q
[
x
]
:
g
c
d
(
P
,
f
(
P
)
)
=
1
⟺
\forall P \in \mathbb{Q}[x] : gcd(P , f(P))=1 \iff
∀
P
∈
Q
[
x
]
:
g
c
d
(
P
,
f
(
P
))
=
1
⟺
P
P
P
is square-free. Which a square-free polynomial with rational coefficients is a polynomial such that there doesn't exist square of a non-constant polynomial with rational coefficients that divides it.Proposed by Sina Azizedin
7
1
Hide problems
Not a nice geo , so well ...
Let
△
A
B
C
\triangle ABC
△
A
BC
and
△
C
′
B
′
A
\triangle C'B'A
△
C
′
B
′
A
be two congruent triangles ( with this order and orient. ). Define point
M
M
M
as the midpoint of segment
A
B
AB
A
B
and suppose that the extension of
C
B
′
CB'
C
B
′
from
B
′
B'
B
′
passes trough
M
M
M
, if
F
F
F
be a point on the smaller arc
M
C
MC
MC
of circumcircle of triangle
△
B
M
C
\triangle BMC
△
BMC
such that
∠
F
B
′
A
=
90
\angle FB'A=90
∠
F
B
′
A
=
90
and
∠
C
′
C
B
′
≠
90
\angle C'CB' \neq 90
∠
C
′
C
B
′
=
90
, then prove that
∠
B
′
C
′
C
=
∠
C
A
F
\angle B'C'C=\angle CAF
∠
B
′
C
′
C
=
∠
C
A
F
.Proposed by Alireza Dadgarnia
6
1
Hide problems
Parsa and diagonals
Let
A
1
A
2
.
.
.
A
99
A_1A_2...A_{99}
A
1
A
2
...
A
99
be a regular
99
−
99-
99
−
gon and point
A
100
A_{100}
A
100
be its center. find the smallest possible natural number
n
n
n
, such that Parsa can color all segments
A
i
A
j
A_iA_j
A
i
A
j
(
1
≤
i
<
j
≤
100
1 \le i < j \le 100
1
≤
i
<
j
≤
100
) with one of
n
n
n
colors in such a way that no two homochromatic segments intersect each other or share a vertex.Proposed by Josef Tkadlec - Czech Republic
5
1
Hide problems
Phi function and number of divisors
Suppose that we have two natural numbers
x
,
y
≤
100
!
x , y \le 100!
x
,
y
≤
100
!
with undetermined values. Prove that there exist natural numbers
m
,
n
m , n
m
,
n
such that values of
x
,
y
x , y
x
,
y
get uniquely determined according to value of
φ
(
d
(
m
y
)
)
+
d
(
φ
(
n
x
)
)
\varphi(d(my))+d(\varphi(nx))
φ
(
d
(
m
y
))
+
d
(
φ
(
n
x
))
. ( for each natural number
n
n
n
,
d
(
n
)
d(n)
d
(
n
)
is number of its positive divisors and
φ
(
n
)
\varphi(n)
φ
(
n
)
is the number of the numbers less that
n
n
n
which are relatively prime to
n
n
n
. )Proposed by Mehran Talaei
4
1
Hide problems
Classical FE
Find all functions
f
:
R
→
R
f : \mathbb{R} \to \mathbb{R}
f
:
R
→
R
such that for any real numbers
x
,
y
x , y
x
,
y
this equality holds :
f
(
y
f
(
x
)
+
f
(
x
)
f
(
y
)
)
=
x
f
(
y
)
+
f
(
x
y
)
f(yf(x)+f(x)f(y))=xf(y)+f(xy)
f
(
y
f
(
x
)
+
f
(
x
)
f
(
y
))
=
x
f
(
y
)
+
f
(
x
y
)
Proposed by Navid Safaei
3
1
Hide problems
Absoulute value in inequality
For any real numbers
x
,
y
,
z
x , y ,z
x
,
y
,
z
prove that :
(
x
+
y
+
z
)
2
+
∑
c
y
c
(
x
+
y
)
(
y
+
z
)
1
+
∣
x
−
z
∣
≥
x
y
+
y
z
+
z
x
(x+y+z)^2 + \sum_{cyc}{\frac{(x+y)(y+z)}{1+|x-z|}} \ge xy+yz+zx
(
x
+
y
+
z
)
2
+
cyc
∑
1
+
∣
x
−
z
∣
(
x
+
y
)
(
y
+
z
)
≥
x
y
+
yz
+
z
x
Proposed by Navid Safaei
2
1
Hide problems
A right angled triangle with AC=2AB
For a right angled triangle
△
A
B
C
\triangle ABC
△
A
BC
with
∠
A
=
90
\angle A=90
∠
A
=
90
we have
A
C
=
2
A
B
AC=2AB
A
C
=
2
A
B
. Point
M
M
M
is the midpoint of side
B
C
BC
BC
and
I
I
I
is incenter of triangle
△
A
B
C
\triangle ABC
△
A
BC
. The line passing trough
M
M
M
and perpendicular to
B
I
BI
B
I
intersect with lines
B
I
BI
B
I
and
A
C
AC
A
C
at points
H
H
H
and
K
K
K
respectively. If the semi-line
I
K
IK
I
K
cuts circumcircle of triangle
△
A
B
C
\triangle ABC
△
A
BC
at
F
F
F
and
S
S
S
be the second intersection point of line
F
H
FH
F
H
with circumcircle of triangle
△
A
B
C
\triangle ABC
△
A
BC
, then prove that
S
M
SM
SM
is tangent to the incircle of triangle
△
A
B
C
\triangle ABC
△
A
BC
.Proposed by Mahdi Etesami Fard
1
1
Hide problems
۰Not so charming graph with degrees 2 and 3
Let
G
G
G
be a simple graph with
11
11
11
vertices labeled as
v
1
,
v
2
,
.
.
.
,
v
11
v_{1} , v_{2} , ... , v_{11}
v
1
,
v
2
,
...
,
v
11
such that the degree of
v
1
v_1
v
1
equals to
2
2
2
and the degree of other vertices are equal to
3
3
3
.If for any set
A
A
A
of these vertices which
∣
A
∣
≤
4
|A| \le 4
∣
A
∣
≤
4
, the number of vertices which are adjacent to at least one verex in
A
A
A
and are not in
A
A
A
themselves is at least equal to
∣
A
∣
|A|
∣
A
∣
, then find the maximum possible number for the diameter of
G
G
G
. (The distance between two vertices of graph is the number of edges of the shortest path between them and the diameter of a graph , is the largest distance between arbitrary pairs in
V
(
G
)
V(G)
V
(
G
)
. )Proposed by Alireza Haqi