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National and Regional Contests
Russia Contests
239 Open Math Olympiad
2018 239 Open Mathematical Olympiad
2018 239 Open Mathematical Olympiad
Part of
239 Open Math Olympiad
Subcontests
(15)
10-11.8
1
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239 Open MO, 2018, Senior League, Problem 8
Graph
G
G
G
becomes planar when any vertex is removed. Prove that its vertices can be properly colored with 5 colors. (Using the four-color theorem without proof is not allowed!)Proposed by D. Karpov
10-11.7
1
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239 Open MO, 2018, Senior League, Problem 7
In a triangle, each median forms an angle with the side it is drawn to, which is less than
α
\alpha
α
. Prove that one of the angles of the triangle is greater than
18
0
∘
−
4
3
α
180^\circ - \frac{4}{3}\alpha
18
0
∘
−
3
4
α
. Proposed by S. Berlov
10-11.6
1
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239 Open MO, 2018, Senior League, Problem 6
For which positive integers
n
n
n
,
m
m
m
does there exist a polynomial of degree
n
n
n
, all coefficients of which are powers of
m
m
m
with integer exponents, having
n
n
n
rational roots, counting multiplicities?Proposed by Fedor Petrov
10-11.5
1
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239 Open M0, 2018, Senior League, Problem 5
Given a trapezoid
A
B
C
D
ABCD
A
BC
D
, with
A
B
∥
C
D
AB\parallel CD
A
B
∥
C
D
. Lines
A
C
AC
A
C
and
B
D
BD
B
D
intersect at point
E
E
E
, and lines
A
D
AD
A
D
and
B
C
BC
BC
intersect at point
F
F
F
. It turns out that the circle with diameter
E
F
EF
EF
is tangent to the midline of the trapezoid. Prove that there exists a square such that there is a mutual correspondence between all six lines containing pairs of its vertices, and points
A
A
A
,
B
B
B
,
C
C
C
,
D
D
D
,
E
E
E
, and
F
F
F
: each line corresponds to a point lying on it. Proposed by V. Mokin
10-11.4
1
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Open 239 MO, 2018, Senior League, Problem 4
In a
9
×
9
9\times 9
9
×
9
table, all cells contain zeros. The following operations can be performed on the table:1. Choose an arbitrary row, add one to all the numbers in that row, and shift all these numbers one cell to the right (and place the last number in the first position).2. Choose an arbitrary column, subtract one from all its numbers, and shift all these numbers one cell down (and place the bottommost number in the top cell).Is it possible to obtain a table in which all cells, except two, contain zeros, with 1 in the bottom-left cell and -1 in the top-right cell after several such operations?Proposed by N. Vlasova
10-11.3
1
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239 Open MO, 2018, Senior League, Problem 3
Given a prime number
p
>
5
p>5
p
>
5
. It is known that the length of the smallest period of the fraction
1
/
p
1/p
1/
p
is a multiple of three. This period (including possible leading zeros) was written on a strip of paper and cut into three equal-length parts
a
a
a
,
b
b
b
,
c
c
c
(they may also have leading zeros). What could be the sum of the three periodic fractions:
0.
(
a
)
0.(a)
0.
(
a
)
,
0.
(
b
)
0.(b)
0.
(
b
)
, and
0.
(
c
)
0.(c)
0.
(
c
)
?Proposed by A. Khrabrov
10-11.1
1
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239 Open MO, 2018, Senior League, Problem 1
Prove that in any convex polygon where all pairwise distances between vertices are distinct, there exists a vertex such that the closest vertex of the polygon is adjacent to it.Proposed by D. Shiryayev, S. Berlov
8-9.8
1
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Cars on the road
On a straight road, points
1
,
2
,
…
,
n
1, 2, \ldots, n
1
,
2
,
…
,
n
are marked. The distance between any two adjacent points is 1. A "placement" refers to the arrangement of
n
n
n
cars, numbered with the same numbers, at the marked points (there can be multiple cars at one point). The "distance" between two placements is defined as the minimum total length of sections that need to be paved so that cars from the first placement can drive on the asphalt, forming the second one (cars can change places on the road). Prove that for any
α
<
1
\alpha<1
α
<
1
, there exists an integer number
n
n
n
for which there are
10
0
n
100^n
10
0
n
placements, the pairwise distances between which are greater than
α
n
\alpha n
α
n
.Proposed by Ilya Bogdanov
8-9.7
1
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Sequence with strange recurrency
The sequence
a
n
a_n
a
n
is defined by the following conditions:
a
1
=
1
a_1=1
a
1
=
1
, and for any
n
∈
N
n\in \mathbb N
n
∈
N
, the number
a
n
+
1
a_{n+1}
a
n
+
1
is obtained from
a
n
a_n
a
n
by adding three if
n
n
n
is a member of this sequence, and two if it is not. Prove that
a
n
<
(
1
+
2
)
n
a_n<(1+\sqrt 2)n
a
n
<
(
1
+
2
)
n
for all
n
n
n
.Proposed by Mikhail Ivanov
8-9.6
1
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beautiful number theory inequality
Petya wrote down 100 positive integers
n
,
n
+
1
,
…
,
n
+
99
n, n+1, \ldots, n+99
n
,
n
+
1
,
…
,
n
+
99
, and Vasya wrote down 99 positive integers
m
,
m
−
1
,
…
,
m
−
98
m, m-1, \ldots, m-98
m
,
m
−
1
,
…
,
m
−
98
. It turned out that for each of Petya's numbers, there is a number from Vasya that divides it. Prove that
m
>
n
3
/
10
,
000
,
000
m>n^3/10, 000, 000
m
>
n
3
/10
,
000
,
000
. Proposed by Ilya Bogdanov
8-9.5
1
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Coverings of triangle net
An equilateral triangle with side 101 is placed on a plane so that one of its sides is horizontal and the triangle is above it. It is divided into smaller equilateral triangles with side 1 by segments parallel to its sides. All sides of these smaller triangles are colored red (including the entire border of the large triangle). An equilateral triangle on a plane is called a "mirror" triangle if its sides are parallel to the sides of the original triangle, but it lies below its horizontal side. What is the smallest number of contours of mirror triangles needed to cover all the red segments? (Mirror triangles may overlap and extend beyond the original triangle.)Proposed by Dmitry Shiryayev
8-9.4
1
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Inequality of angles
In a triangle, each median forms an angle with the side it is drawn to, which is less than
α
\alpha
α
. Prove that one of the angles of the triangle is greater than
18
0
∘
−
3
2
α
180^\circ-\frac{3}{2}\alpha
18
0
∘
−
2
3
α
. Proposed by Sergey Berlov
8-9.3
1
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All subsets into triples
Is it possible to divide all non-empty subsets of a set of 10 elements into triples so that in each triple, two of the subsets do not intersect and in their union give the third? Proposed by Vladislav Frank
8-9.2
1
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Right-angled triangle
On the hypotenuse
A
B
AB
A
B
of a right-angled triangle
A
B
C
ABC
A
BC
, point
R
R
R
is chosen, on the cathetus
B
C
BC
BC
a point
T
T
T
, and on the segment
A
T
AT
A
T
a point
S
S
S
are chosen in such a way that the angles
∠
A
R
T
\angle ART
∠
A
RT
and
∠
A
S
C
\angle ASC
∠
A
SC
are right angles. Points
M
M
M
and
N
N
N
are the midpoints of the segments
C
B
CB
CB
and
C
R
CR
CR
, respectively. Prove that points
M
M
M
,
T
T
T
,
S
S
S
, and
N
N
N
lie on the same circle.Proposed by S. Berlov
8-9.1
1
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remainders are equal
Given a prime number
p
p
p
. A positive integer
x
x
x
is divided by
p
p
p
with a remainder, and the number
p
2
p^2
p
2
is divided by
x
x
x
with a remainder. The remainders turned out to be equal. Find themProposed by Sergey Berlov