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Problems
Contests
National and Regional Contests
Azerbaijan Contests
Azerbaijan Senior National Olympiad
2016 Azerbaijan National Mathematical Olympiad
2016 Azerbaijan National Mathematical Olympiad
Part of
Azerbaijan Senior National Olympiad
Subcontests
(4)
1
1
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Find the perimeter of the convex polygon
Find the perimeter of the convex polygon whose coordinates of the vertices are the set of pairs of the integer solutions of the equation
x
2
+
x
y
=
x
+
2
y
+
9
x^2+xy = x + 2y + 9
x
2
+
x
y
=
x
+
2
y
+
9
.
2
1
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Geometry with 18 degree
On the extension of the hypotenuse
A
B
AB
A
B
of the right-angled triangle
A
B
C
ABC
A
BC
, the point
D
D
D
after the point B is marked so that
D
C
=
2
B
C
DC = 2BC
D
C
=
2
BC
. Let the point
H
H
H
be the foot of the altitude dropped from the vertex
C
C
C
. If the distance from the point
H
H
H
to the side
B
C
BC
BC
is equal to the length of the segment
H
A
HA
H
A
, prove that
∠
B
D
C
=
18
\angle BDC = 18
∠
B
D
C
=
18
.
3
1
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Very prime number
Let's call any natural number "very prime" if any number of consecutive digits (in particular, a digit or number itself) is a prime number. For example,
23
23
23
and
37
37
37
are "very prime" numbers, but
237
237
237
and
357
357
357
are not. Find the largest "prime" number (with justification!).
4
2
Hide problems
Eventually integers will be observed
Let
A
=
1
⋅
3
⋅
5
⋅
.
.
.
⋅
(
2
n
−
1
)
2
⋅
4
⋅
6
⋅
.
.
.
⋅
(
2
n
)
A = \frac{1 \cdot 3 \cdot 5\cdot ... \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot (2n)}
A
=
2
⋅
4
⋅
6
⋅
...
⋅
(
2
n
)
1
⋅
3
⋅
5
⋅
...
⋅
(
2
n
−
1
)
Prove that in the infinite sequence
A
,
2
A
,
4
A
,
8
A
,
.
.
.
,
2
k
A
,
…
.
A, 2A, 4A, 8A, ..., 2^k A, ….
A
,
2
A
,
4
A
,
8
A
,
...
,
2
k
A
,
…
.
only integers will be observed, eventually.
Function with 2016 variables
Let
R
\mathbb R
R
be the set of real numbers. Determine all functions
f
:
R
→
R
f:\mathbb R\to\mathbb R
f
:
R
→
R
that satisfy the equation
∑
i
=
1
2015
f
(
x
i
+
x
i
+
1
)
+
f
(
∑
i
=
1
2016
x
i
)
≤
∑
i
=
1
2016
f
(
2
x
i
)
\sum_{i=1}^{2015} f(x_i + x_{i+1}) + f\left( \sum_{i=1}^{2016} x_i \right) \le \sum_{i=1}^{2016} f(2x_i)
i
=
1
∑
2015
f
(
x
i
+
x
i
+
1
)
+
f
(
i
=
1
∑
2016
x
i
)
≤
i
=
1
∑
2016
f
(
2
x
i
)
for all real numbers
x
1
,
x
2
,
.
.
.
,
x
2016
.
x_1, x_2, ... , x_{2016}.
x
1
,
x
2
,
...
,
x
2016
.