MathDB
Problems
Contests
National and Regional Contests
Azerbaijan Contests
JBMO TST - Azerbaijan
2018 Azerbaijan JBMO TST
2018 Azerbaijan JBMO TST
Part of
JBMO TST - Azerbaijan
Subcontests
(4)
4
1
Hide problems
100 card with 43 having odd integers on them
In the beginning, there are
100
100
100
cards on the table, and each card has a positive integer written on it. An odd number is written on exactly
43
43
43
cards. Every minute, the following operation is performed: for all possible sets of
3
3
3
cards on the table, the product of the numbers on these three cards is calculated, all the obtained results are summed, and this sum is written on a new card and placed on the table. A day later, it turns out that there is a card on the table, the number written on this card is divisible by
2
2018
.
2^{2018}.
2
2018
.
Prove that one hour after the start of the process, there was a card on the table that the number written on that card is divisible by
2
2018
.
2^{2018}.
2
2018
.
1
1
Hide problems
Geometry with orthocenter and perpendiculars
Let
△
A
B
C
\triangle ABC
△
A
BC
be an acute triangle. Let us denote the foot of the altitudes from the vertices
A
,
B
A, B
A
,
B
and
C
C
C
to the opposite sides by
D
,
E
D, E
D
,
E
and
F
,
F,
F
,
respectively, and the intersection point of the altitudes of the triangle
A
B
C
ABC
A
BC
by
H
.
H.
H
.
Let
P
P
P
be the intersection of the line
B
E
BE
BE
and the segment
D
F
.
DF.
D
F
.
A straight line passing through
P
P
P
and perpendicular to
B
C
BC
BC
intersects
A
B
AB
A
B
at
Q
.
Q.
Q
.
Let
N
N
N
be the intersection of the segment
E
Q
EQ
EQ
with the perpendicular drawn from
A
.
A.
A
.
Prove that
N
N
N
is the midpoint of segment
A
H
.
AH.
A
H
.
3
2
Hide problems
2^x + x^2 + 25 is cube of a prime
Determine the integers
x
x
x
such that
2
x
+
x
2
+
25
2^x + x^2 + 25
2
x
+
x
2
+
25
is the cube of a prime number
powers of 2,3,5, and 7
Find all nonnegative integers
(
x
,
y
,
z
,
u
)
(x,y,z,u)
(
x
,
y
,
z
,
u
)
with satisfy the following equation:
2
x
+
3
y
+
5
z
=
7
u
.
2^x + 3^y + 5^z = 7^u.
2
x
+
3
y
+
5
z
=
7
u
.
2
1
Hide problems
Inequality
a) Find :
A
=
{
(
a
,
b
,
c
)
∈
R
3
∣
a
+
b
+
c
=
3
,
(
6
a
+
b
2
+
c
2
)
(
6
b
+
c
2
+
a
2
)
(
6
c
+
a
2
+
b
2
)
≠
0
}
A=\{(a,b,c) \in \mathbb{R}^{3} | a+b+c=3 , (6a+b^2+c^2)(6b+c^2+a^2)(6c+a^2+b^2) \neq 0\}
A
=
{(
a
,
b
,
c
)
∈
R
3
∣
a
+
b
+
c
=
3
,
(
6
a
+
b
2
+
c
2
)
(
6
b
+
c
2
+
a
2
)
(
6
c
+
a
2
+
b
2
)
=
0
}
b) Prove that for any
(
a
,
b
,
c
)
∈
A
(a,b,c) \in A
(
a
,
b
,
c
)
∈
A
next inequality hold :\begin{align*} \frac{a}{6a+b^2+c^2}+\frac{b}{6b+c^2+a^2}+\frac{c}{6c+a^2+b^2} \le \frac{3}{8} \end{align*}