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Contests
National and Regional Contests
Iran Contests
Iran Team Selection Test
2017 Iran Team Selection Test
2017 Iran Team Selection Test
Part of
Iran Team Selection Test
Subcontests
(6)
4
3
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Number Theory from Iranian TST 2017
We arranged all the prime numbers in the ascending order:
p
1
=
2
<
p
2
<
p
3
<
⋯
p_1=2<p_2<p_3<\cdots
p
1
=
2
<
p
2
<
p
3
<
⋯
. Also assume that
n
1
<
n
2
<
⋯
n_1<n_2<\cdots
n
1
<
n
2
<
⋯
is a sequence of positive integers that for all
i
=
1
,
2
,
3
,
⋯
i=1,2,3,\cdots
i
=
1
,
2
,
3
,
⋯
the equation
x
n
i
≡
2
(
m
o
d
p
i
)
x^{n_i} \equiv 2 \pmod {p_i}
x
n
i
≡
2
(
mod
p
i
)
has a solution for
x
x
x
. Is there always a number
x
x
x
that satisfies all the equations?Proposed by Mahyar Sefidgaran , Yahya Motevasel
2017 Iran TST2 day2 p4
A
n
+
1
n+1
n
+
1
-tuple
(
h
1
,
h
2
,
⋯
,
h
n
+
1
)
\left(h_1,h_2, \cdots, h_{n+1}\right)
(
h
1
,
h
2
,
⋯
,
h
n
+
1
)
where
h
i
(
x
1
,
x
2
,
⋯
,
x
n
)
h_i\left(x_1,x_2, \cdots , x_n\right)
h
i
(
x
1
,
x
2
,
⋯
,
x
n
)
are
n
n
n
variable polynomials with real coefficients is called good if the following condition holds: For any
n
n
n
functions
f
1
,
f
2
,
⋯
,
f
n
:
R
→
R
f_1,f_2, \cdots ,f_n : \mathbb R \to \mathbb R
f
1
,
f
2
,
⋯
,
f
n
:
R
→
R
if for all
1
≤
i
≤
n
+
1
1 \le i \le n+1
1
≤
i
≤
n
+
1
,
P
i
(
x
)
=
h
i
(
f
1
(
x
)
,
f
2
(
x
)
,
⋯
,
f
n
(
x
)
)
P_i(x)=h_i \left(f_1(x),f_2(x), \cdots, f_n(x) \right)
P
i
(
x
)
=
h
i
(
f
1
(
x
)
,
f
2
(
x
)
,
⋯
,
f
n
(
x
)
)
is a polynomial with variable
x
x
x
, then
f
1
(
x
)
,
f
2
(
x
)
,
⋯
,
f
n
(
x
)
f_1(x),f_2(x), \cdots, f_n(x)
f
1
(
x
)
,
f
2
(
x
)
,
⋯
,
f
n
(
x
)
are polynomials.
a
)
a)
a
)
Prove that for all positive integers
n
n
n
, there exists a good
n
+
1
n+1
n
+
1
-tuple
(
h
1
,
h
2
,
⋯
,
h
n
+
1
)
\left(h_1,h_2, \cdots, h_{n+1}\right)
(
h
1
,
h
2
,
⋯
,
h
n
+
1
)
such that the degree of all
h
i
h_i
h
i
is more than
1
1
1
.
b
)
b)
b
)
Prove that there doesn't exist any integer
n
>
1
n>1
n
>
1
that for which there is a good
n
+
1
n+1
n
+
1
-tuple
(
h
1
,
h
2
,
⋯
,
h
n
+
1
)
\left(h_1,h_2, \cdots, h_{n+1}\right)
(
h
1
,
h
2
,
⋯
,
h
n
+
1
)
such that all
h
i
h_i
h
i
are symmetric polynomials.Proposed by Alireza Shavali
Iran TST 2017 third exam
There are
6
6
6
points on the plane such that no three of them are collinear. It's known that between every
4
4
4
points of them, there exists a point that it's power with respect to the circle passing through the other three points is a constant value
k
k
k
.(Power of a point in the interior of a circle has a negative value.) Prove that
k
=
0
k=0
k
=
0
and all
6
6
6
points lie on a circle.Proposed by Morteza Saghafian[/I]
6
3
Hide problems
A nice combinatorics from Iranian TST 2017
In the unit squares of a transparent
1
×
100
1 \times 100
1
×
100
tape, numbers
1
,
2
,
⋯
,
100
1,2,\cdots,100
1
,
2
,
⋯
,
100
are written in the ascending order.We fold this tape on it's lines with arbitrary order and arbitrary directions until we reach a
1
×
1
1 \times1
1
×
1
tape with
100
100
100
layers.A permutation of the numbers
1
,
2
,
⋯
,
100
1,2,\cdots,100
1
,
2
,
⋯
,
100
can be seen on the tape, from the top to the bottom. Prove that the number of possible permutations is between
2
100
2^{100}
2
100
and
4
100
4^{100}
4
100
. (e.g. We can produce all permutations of numbers
1
,
2
,
3
1,2,3
1
,
2
,
3
with a
1
×
3
1\times3
1
×
3
tape)Proposed by Morteza Saghafian
2017 Iran TST2 day2 p6
Let
k
>
1
k>1
k
>
1
be an integer. The sequence
a
1
,
a
2
,
⋯
a_1,a_2, \cdots
a
1
,
a
2
,
⋯
is defined as:
a
1
=
1
,
a
2
=
k
a_1=1, a_2=k
a
1
=
1
,
a
2
=
k
and for all
n
>
1
n>1
n
>
1
we have:
a
n
+
1
−
(
k
+
1
)
a
n
+
a
n
−
1
=
0
a_{n+1}-(k+1)a_n+a_{n-1}=0
a
n
+
1
−
(
k
+
1
)
a
n
+
a
n
−
1
=
0
Find all positive integers
n
n
n
such that
a
n
a_n
a
n
is a power of
k
k
k
.Proposed by Amirhossein Pooya
Geometry from Iran TST 2017
In triangle
A
B
C
ABC
A
BC
let
O
O
O
and
H
H
H
be the circumcenter and the orthocenter. The point
P
P
P
is the reflection of
A
A
A
with respect to
O
H
OH
O
H
. Assume that
P
P
P
is not on the same side of
B
C
BC
BC
as
A
A
A
. Points
E
,
F
E,F
E
,
F
lie on
A
B
,
A
C
AB,AC
A
B
,
A
C
respectively such that
B
E
=
P
C
,
C
F
=
P
B
BE=PC \ , CF=PB
BE
=
PC
,
CF
=
PB
. Let
K
K
K
be the intersection point of
A
P
,
O
H
AP,OH
A
P
,
O
H
. Prove that
∠
E
K
F
=
9
0
∘
\angle EKF = 90 ^{\circ}
∠
E
K
F
=
9
0
∘
Proposed by Iman Maghsoudi
5
3
Hide problems
A difficult geometry from Iranian TST 2017
In triangle
A
B
C
ABC
A
BC
, arbitrary points
P
,
Q
P,Q
P
,
Q
lie on side
B
C
BC
BC
such that
B
P
=
C
Q
BP=CQ
BP
=
CQ
and
P
P
P
lies between
B
,
Q
B,Q
B
,
Q
.The circumcircle of triangle
A
P
Q
APQ
A
PQ
intersects sides
A
B
,
A
C
AB,AC
A
B
,
A
C
at
E
,
F
E,F
E
,
F
respectively.The point
T
T
T
is the intersection of
E
P
,
F
Q
EP,FQ
EP
,
FQ
.Two lines passing through the midpoint of
B
C
BC
BC
and parallel to
A
B
AB
A
B
and
A
C
AC
A
C
, intersect
E
P
EP
EP
and
F
Q
FQ
FQ
at points
X
,
Y
X,Y
X
,
Y
respectively. Prove that the circumcircle of triangle
T
X
Y
TXY
TX
Y
and triangle
A
P
Q
APQ
A
PQ
are tangent to each other.Proposed by Iman Maghsoudi
2017 Iran TST2 day2 p5
k
,
n
k,n
k
,
n
are two arbitrary positive integers. Prove that there exists at least
(
k
−
1
)
(
n
−
k
+
1
)
(k-1)(n-k+1)
(
k
−
1
)
(
n
−
k
+
1
)
positive integers that can be produced by
n
n
n
number of
k
k
k
's and using only
+
,
−
,
×
,
÷
+,-,\times, \div
+
,
−
,
×
,
÷
operations and adding parentheses between them, but cannot be produced using
n
−
1
n-1
n
−
1
number of
k
k
k
's.Proposed by Aryan Tajmir
Polynomial from Iran TST 2017
Let
{
c
i
}
i
=
0
∞
\left \{ c_i \right \}_{i=0}^{\infty}
{
c
i
}
i
=
0
∞
be a sequence of non-negative real numbers with
c
2017
>
0
c_{2017}>0
c
2017
>
0
. A sequence of polynomials is defined as
P
−
1
(
x
)
=
0
,
P
0
(
x
)
=
1
,
P
n
+
1
(
x
)
=
x
P
n
(
x
)
+
c
n
P
n
−
1
(
x
)
.
P_{-1}(x)=0 \ , \ P_0(x)=1 \ , \ P_{n+1}(x)=xP_n(x)+c_nP_{n-1}(x).
P
−
1
(
x
)
=
0
,
P
0
(
x
)
=
1
,
P
n
+
1
(
x
)
=
x
P
n
(
x
)
+
c
n
P
n
−
1
(
x
)
.
Prove that there doesn't exist any integer
n
>
2017
n>2017
n
>
2017
and some real number
c
c
c
such that
P
2
n
(
x
)
=
P
n
(
x
2
+
c
)
.
P_{2n}(x)=P_n(x^2+c).
P
2
n
(
x
)
=
P
n
(
x
2
+
c
)
.
Proposed by Navid Safaei
2
3
Hide problems
Combinatorics from Iranian TST 2017
In the country of Sugarland, there are
13
13
13
students in the IMO team selection camp.
6
6
6
team selection tests were taken and the results have came out. Assume that no students have the same score on the same test.To select the IMO team, the national committee of math Olympiad have decided to choose a permutation of these
6
6
6
tests and starting from the first test, the person with the highest score between the remaining students will become a member of the team.The committee is having a session to choose the permutation. Is it possible that all
13
13
13
students have a chance of being a team member?Proposed by Morteza Saghafian
2017 Iran TST2 P2
Find the largest number
n
n
n
that for which there exists
n
n
n
positive integers such that non of them divides another one, but between every three of them, one divides the sum of the other two.Proposed by Morteza Saghafian
Geometry from Iran TST 2017
Let
P
P
P
be a point in the interior of quadrilateral
A
B
C
D
ABCD
A
BC
D
such that:
∠
B
P
C
=
2
∠
B
A
C
,
∠
P
C
A
=
∠
P
A
D
,
∠
P
D
A
=
∠
P
A
C
\angle BPC=2\angle BAC \ \ ,\ \ \angle PCA = \angle PAD \ \ ,\ \ \angle PDA=\angle PAC
∠
BPC
=
2∠
B
A
C
,
∠
PC
A
=
∠
P
A
D
,
∠
P
D
A
=
∠
P
A
C
Prove that:
∠
P
B
D
=
∣
∠
B
C
A
−
∠
P
C
A
∣
\angle PBD= \left | \angle BCA - \angle PCA \right |
∠
PB
D
=
∣
∠
BC
A
−
∠
PC
A
∣
Proposed by Ali Zamani
3
3
Hide problems
Geometry from Iranian TST 2017
In triangle
A
B
C
ABC
A
BC
let
I
a
I_a
I
a
be the
A
A
A
-excenter. Let
ω
\omega
ω
be an arbitrary circle that passes through
A
,
I
a
A,I_a
A
,
I
a
and intersects the extensions of sides
A
B
,
A
C
AB,AC
A
B
,
A
C
(extended from
B
,
C
B,C
B
,
C
) at
X
,
Y
X,Y
X
,
Y
respectively. Let
S
,
T
S,T
S
,
T
be points on segments
I
a
B
,
I
a
C
I_aB,I_aC
I
a
B
,
I
a
C
respectively such that
∠
A
X
I
a
=
∠
B
T
I
a
\angle AXI_a=\angle BTI_a
∠
A
X
I
a
=
∠
BT
I
a
and
∠
A
Y
I
a
=
∠
C
S
I
a
\angle AYI_a=\angle CSI_a
∠
A
Y
I
a
=
∠
CS
I
a
.Lines
B
T
,
C
S
BT,CS
BT
,
CS
intersect at
K
K
K
. Lines
K
I
a
,
T
S
KI_a,TS
K
I
a
,
TS
intersect at
Z
Z
Z
. Prove that
X
,
Y
,
Z
X,Y,Z
X
,
Y
,
Z
are collinear.Proposed by Hooman Fattahi
2017 Iran TST2 p3
There are
27
27
27
cards, each has some amount of (
1
1
1
or
2
2
2
or
3
3
3
) shapes (a circle, a square or a triangle) with some color (white, grey or black) on them. We call a triple of cards a match such that all of them have the same amount of shapes or distinct amount of shapes, have the same shape or distinct shapes and have the same color or distinct colors. For instance, three cards shown in the figure are a match be cause they have distinct amount of shapes, distinct shapes but the same color of shapes. What is the maximum number of cards that we can choose such that non of the triples make a match?Proposed by Amin Bahjati
Functional Equation from Iran TST 2017
Find all functions
f
:
R
+
×
R
+
→
R
+
f: \mathbb {R}^+ \times \mathbb {R}^+ \to \mathbb {R}^+
f
:
R
+
×
R
+
→
R
+
that satisfy the following conditions for all positive real numbers
x
,
y
,
z
:
x,y,z:
x
,
y
,
z
:
f
(
f
(
x
,
y
)
,
z
)
=
x
2
y
2
f
(
x
,
z
)
f\left ( f(x,y),z \right )=x^2y^2f(x,z)
f
(
f
(
x
,
y
)
,
z
)
=
x
2
y
2
f
(
x
,
z
)
f
(
x
,
1
+
f
(
x
,
y
)
)
≥
x
2
+
x
y
f
(
x
,
x
)
f\left ( x,1+f(x,y) \right ) \ge x^2 + xyf(x,x)
f
(
x
,
1
+
f
(
x
,
y
)
)
≥
x
2
+
x
y
f
(
x
,
x
)
Proposed by Mojtaba Zare, Ali Daei Nabi
1
3
Hide problems
Inequality from Iranian TST 2017
Let
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
be positive real numbers with
a
+
b
+
c
+
d
=
2
a+b+c+d=2
a
+
b
+
c
+
d
=
2
. Prove the following inequality:
(
a
+
c
)
2
a
d
+
b
c
+
(
b
+
d
)
2
a
c
+
b
d
+
4
≥
4
(
a
+
b
+
1
c
+
d
+
1
+
c
+
d
+
1
a
+
b
+
1
)
.
\frac{(a+c)^{2}}{ad+bc}+\frac{(b+d)^{2}}{ac+bd}+4\geq 4\left ( \frac{a+b+1}{c+d+1}+\frac{c+d+1}{a+b+1} \right).
a
d
+
b
c
(
a
+
c
)
2
+
a
c
+
b
d
(
b
+
d
)
2
+
4
≥
4
(
c
+
d
+
1
a
+
b
+
1
+
a
+
b
+
1
c
+
d
+
1
)
.
Proposed by Mohammad Jafari
2017 Iran TST2 P1
A
B
C
D
ABCD
A
BC
D
is a trapezoid with
A
B
∥
C
D
AB \parallel CD
A
B
∥
C
D
. The diagonals intersect at
P
P
P
. Let
ω
1
\omega _1
ω
1
be a circle passing through
B
B
B
and tangent to
A
C
AC
A
C
at
A
A
A
. Let
ω
2
\omega _2
ω
2
be a circle passing through
C
C
C
and tangent to
B
D
BD
B
D
at
D
D
D
.
ω
3
\omega _3
ω
3
is the circumcircle of triangle
B
P
C
BPC
BPC
. Prove that the common chord of circles
ω
1
,
ω
3
\omega _1,\omega _3
ω
1
,
ω
3
and the common chord of circles
ω
2
,
ω
3
\omega _2, \omega _3
ω
2
,
ω
3
intersect each other on
A
D
AD
A
D
.Proposed by Kasra Ahmadi
Number Theory from Iran TST 2017
Let
n
>
1
n>1
n
>
1
be an integer. Prove that there exists an integer
n
−
1
≥
m
≥
⌊
n
2
⌋
n-1 \ge m \ge \left \lfloor \frac{n}{2} \right \rfloor
n
−
1
≥
m
≥
⌊
2
n
⌋
such that the following equation has integer solutions with
a
m
>
0
:
a_m>0:
a
m
>
0
:
a
m
m
+
1
+
a
m
+
1
m
+
2
+
⋯
+
a
n
−
1
n
=
1
lcm
(
1
,
2
,
⋯
,
n
)
\frac{a_{m}}{m+1}+\frac{a_{m+1}}{m+2}+ \cdots + \frac{a_{n-1}}{n}=\frac{1}{\textrm{lcm}\left ( 1,2, \cdots , n \right )}
m
+
1
a
m
+
m
+
2
a
m
+
1
+
⋯
+
n
a
n
−
1
=
lcm
(
1
,
2
,
⋯
,
n
)
1
Proposed by Navid Safaei