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Contests
National and Regional Contests
Russia Contests
All-Russian Olympiad
2024 All-Russian Olympiad
2024 All-Russian Olympiad
Part of
All-Russian Olympiad
Subcontests
(8)
4
3
Hide problems
A line cutting segments and edges of a quadrilateral
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrilateral with
∠
A
+
∠
D
=
9
0
∘
\angle A+\angle D=90^\circ
∠
A
+
∠
D
=
9
0
∘
and
E
E
E
the point of intersection of its diagonals. The line
ℓ
\ell
ℓ
cuts the segments
A
B
AB
A
B
,
C
D
CD
C
D
,
A
E
AE
A
E
and
E
D
ED
E
D
in points
X
,
Y
,
Z
,
T
X,Y,Z,T
X
,
Y
,
Z
,
T
, respectively. Suppose that
A
Z
=
C
E
AZ=CE
A
Z
=
CE
and
B
E
=
D
T
BE=DT
BE
=
D
T
. Prove that the length of the segment
X
Y
XY
X
Y
is not larger than the diameter of the the circumcircle of
E
T
Z
ETZ
ETZ
. Proposed by A. Kuznetsov, I. Frolov
Equal lengths in cyclic quadrilateral
In cyclic quadrilateral
A
B
C
D
ABCD
A
BC
D
,
∠
A
+
∠
D
=
π
2
\angle A+ \angle D=\frac{\pi}{2}
∠
A
+
∠
D
=
2
π
.
A
C
AC
A
C
intersects
B
D
BD
B
D
at
E
{E}
E
. A line
l
{l}
l
cuts segment
A
B
,
C
D
,
A
E
,
D
E
AB, CD, AE, DE
A
B
,
C
D
,
A
E
,
D
E
at
X
,
Y
,
Z
,
T
X, Y, Z, T
X
,
Y
,
Z
,
T
respectively. If
A
Z
=
C
E
AZ=CE
A
Z
=
CE
and
B
E
=
D
T
BE=DT
BE
=
D
T
, prove that the diameter of the circumcircle of
△
E
Z
T
\triangle EZT
△
EZT
equals
X
Y
XY
X
Y
.
New cyclic quadrilaterals from old ones, and a beautiful concurrency
A quadrilateral
A
B
C
D
ABCD
A
BC
D
without parallel sides is inscribed in a circle
ω
\omega
ω
. We draw a line
ℓ
a
∥
B
C
\ell_a \parallel BC
ℓ
a
∥
BC
through the point
A
A
A
, a line
ℓ
b
∥
C
D
\ell_b \parallel CD
ℓ
b
∥
C
D
through the point
B
B
B
, a line
ℓ
c
∥
D
A
\ell_c \parallel DA
ℓ
c
∥
D
A
through the point
C
C
C
, and a line
ℓ
d
∥
A
B
\ell_d \parallel AB
ℓ
d
∥
A
B
through the point
D
D
D
. Suppose that the quadrilateral whose successive sides lie on these four straight lines is inscribed in a circle
γ
\gamma
γ
and that
ω
\omega
ω
and
γ
\gamma
γ
intersect in points
E
E
E
and
F
F
F
. Show that the lines
A
C
,
B
D
AC, BD
A
C
,
B
D
and
E
F
EF
EF
intersect in one point. Proposed by A. Kuznetsov
8
3
Hide problems
A huge group of children compare their heights
1000
1000
1000
children, no two of the same height, lined up. Let us call a pair of different children
(
a
,
b
)
(a,b)
(
a
,
b
)
good if between them there is no child whose height is greater than the height of one of
a
a
a
and
b
b
b
, but less than the height of the other. What is the greatest number of good pairs that could be formed? (Here,
(
a
,
b
)
(a,b)
(
a
,
b
)
and
(
b
,
a
)
(b,a)
(
b
,
a
)
are considered the same pair.) Proposed by I. Bogdanov
Replacing n numbers around the circle by divisors
Let
n
>
2
n>2
n
>
2
be a positive integer. Masha writes down
n
n
n
natural numbers along a circle. Next, Taya performs the following operation: Between any two adjacent numbers
a
a
a
and
b
b
b
, she writes a divisor of the number
a
+
b
a+b
a
+
b
greater than
1
1
1
, then Taya erases the original numbers and obtains a new set of
n
n
n
numbers along the circle. Can Taya always perform these operations in such a way that after some number of operations, all the numbers are equal? Proposed by T. Korotchenko
The numbers 1^0,2^1,...,k^(k-1) give many residues modulo p!
Prove that there exists
c
>
0
c>0
c
>
0
such that for any odd prime
p
=
2
k
+
1
p=2k+1
p
=
2
k
+
1
, the numbers
1
0
,
2
1
,
3
2
,
…
,
k
k
−
1
1^0, 2^1,3^2,\dots,k^{k-1}
1
0
,
2
1
,
3
2
,
…
,
k
k
−
1
give at least
c
p
c\sqrt{p}
c
p
distinct residues modulo
p
p
p
. Proposed by M. Turevsky, I. Bogdanov
7
3
Hide problems
Eight trinomials on a board, do they have common roots?
There are
8
8
8
different quadratic trinomials written on the board, among them there are no two that add up to a zero polynomial. It turns out that if we choose any two trinomials
g
1
(
x
)
,
g
2
(
X
)
g_1(x), g_2(X)
g
1
(
x
)
,
g
2
(
X
)
from the board, then the remaining
6
6
6
trinomials can be denoted as
g
3
(
x
)
,
g
4
(
x
)
,
…
,
g
8
(
x
)
g_3(x),g_4(x),\dots,g_8(x)
g
3
(
x
)
,
g
4
(
x
)
,
…
,
g
8
(
x
)
so that all four polynomials
g
1
(
x
)
+
g
2
(
x
)
,
g
3
(
x
)
+
g
4
(
x
)
,
g
5
(
x
)
+
g
6
(
x
)
g_1(x)+g_2(x),g_3(x)+g_4(x),g_5(x)+g_6(x)
g
1
(
x
)
+
g
2
(
x
)
,
g
3
(
x
)
+
g
4
(
x
)
,
g
5
(
x
)
+
g
6
(
x
)
and
g
7
(
x
)
+
g
8
(
x
)
g_7(x)+g_8(x)
g
7
(
x
)
+
g
8
(
x
)
have a common root. Do all trinomials on the board necessarily have a common root? Proposed by S. Berlov
What should we expect from globalization?
In a country there are
n
>
100
n>100
n
>
100
cities and initially no roads. The government randomly determined the cost of building a two-way road between any two cities, using all amounts from
1
1
1
to
n
(
n
−
1
)
2
\frac{n(n-1)}{2}
2
n
(
n
−
1
)
thalers once (all options are equally likely). The mayor of each city chooses the cheapest of the
n
−
1
n-1
n
−
1
roads emanating from that city and it is built (this may be the mutual desired of the mayors of both cities being connected, or only one of the two). After the construction of these roads, the cities are divided into
M
M
M
connected components (between cities of the same connected component, you can get along the constructed roads, possibly via other cities, but this is not possible for cities of different components). Find the expected value of the random variable
M
M
M
. Proposed by F. Petrov
Placing segments on a line with certain intersection relations
Let
x
1
x_1
x
1
and
x
2
x_2
x
2
be positive integers. On a straight line,
y
1
y_1
y
1
white segments and
y
2
y_2
y
2
black segments are given, with
y
1
≥
x
1
y_1 \ge x_1
y
1
≥
x
1
and
y
2
≥
x
2
y_2 \ge x_2
y
2
≥
x
2
. Suppose that no two segments of the same colour intersect (and do not have common ends). Moreover, suppose that for any choice of
x
1
x_1
x
1
white segments and
x
2
x_2
x
2
black segments, some pair of selected segments will intersect. Prove that
(
y
1
−
x
1
)
(
y
2
−
x
2
)
<
x
1
x
2
(y_1-x_1)(y_2-x_2)<x_1x_2
(
y
1
−
x
1
)
(
y
2
−
x
2
)
<
x
1
x
2
. Proposed by G. Chelnokov
6
2
Hide problems
Intersecting OH with the circumcircle of BHC
The altitudes of an acute triangle
A
B
C
ABC
A
BC
with
A
B
<
A
C
AB<AC
A
B
<
A
C
intersect at a point
H
H
H
, and
O
O
O
is the center of the circumcircle
Ω
\Omega
Ω
. The segment
O
H
OH
O
H
intersects the circumcircle of
B
H
C
BHC
B
H
C
at a point
X
X
X
, different from
O
O
O
and
H
H
H
. The circumcircle of
A
O
X
AOX
A
OX
intersects the smaller arc
A
B
AB
A
B
of
Ω
\Omega
Ω
at point
Y
Y
Y
. Prove that the line
X
Y
XY
X
Y
bisects the segment
B
C
BC
BC
. Proposed by A. Tereshin
Yet another OH-symmetry
Let
A
B
C
ABC
A
BC
be an acute non-isosceles triangle with circumcircle
ω
\omega
ω
, circumcenter
O
O
O
and orthocenter
H
H
H
. We draw a line perpendicular to
A
H
AH
A
H
through
O
O
O
and a line perpendicular to
A
O
AO
A
O
through
H
H
H
. Prove that the points of intersection of these lines with sides
A
B
AB
A
B
and
A
C
AC
A
C
lie on a circle, which is tangent to
ω
\omega
ω
. Proposed by A. Kuznetsov
5
1
Hide problems
Is the winter still not over? Let&#039;s get rid of the snow!
A neighborhood consists of
10
×
10
10 \times 10
10
×
10
squares. On New Year's Eve it snowed for the first time and since then exactly
10
10
10
cm of snow fell on each square every night (and snow fell only at night). Every morning, the janitor selects one row or column and shovels all the snow from there onto one of the adjacent rows or columns (from each cell to the adjacent side). For example, he can select the seventh column and from each of its cells shovel all the snow into the cell of the left of it. You cannot shovel snow outside the neighborhood. On the evening of the 100th day of the year, an inspector will come to the city and find the cell with the snowdrift of maximal height. The goal of the janitor is to ensure that this height is minimal. What height of snowdrift will the inspector find? Proposed by A. Solynin
3
1
Hide problems
Two boys move potatoes between bags
Two boys are given a bag of potatoes, each bag containing
150
150
150
tubers. They take turns transferring the potatoes, where in each turn they transfer a non-zero tubers from their bag to the other boy's bag. Their moves must satisfy the following condition: In each move, a boy must move more tubers than he had in his bag before any of his previous moves (if there were such moves). So, with his first move, a boy can move any non-zero quantity, and with his fifth move, a boy can move
200
200
200
tubers, if before his first, second, third and fourth move, the numbers of tubers in his bag was less than
200
200
200
. What is the maximal total number of moves the two boys can do? Proposed by E. Molchanov
2
3
Hide problems
Can the 50 divisors of a number be distinct mod 100?
A positive integer has exactly
50
50
50
divisors. Is it possible that no difference of two different divisors is divisible by
100
100
100
? Proposed by A. Chironov
Cutting three-square corners from a coloured board
Let
n
≥
3
n \ge 3
n
≥
3
be an odd integer. In a
2
n
×
2
n
2n \times 2n
2
n
×
2
n
board, we colour
2
(
n
−
1
)
2
2(n-1)^2
2
(
n
−
1
)
2
cells. What is the largest number of three-square corners that can surely be cut out of the uncoloured figure? Proposed by G. Sharafetdinova
A mysteriously complicated equation with a surprising symmetry
Call a triple
(
a
,
b
,
c
)
(a,b,c)
(
a
,
b
,
c
)
of positive numbers mysterious if
a
2
+
1
a
2
c
2
+
2
a
b
+
b
2
+
1
b
2
a
2
+
2
b
c
+
c
2
+
1
c
2
b
2
+
2
c
a
=
2
(
a
+
b
+
c
)
.
\sqrt{a^2+\frac{1}{a^2c^2}+2ab}+\sqrt{b^2+\frac{1}{b^2a^2}+2bc}+\sqrt{c^2+\frac{1}{c^2b^2}+2ca}=2(a+b+c).
a
2
+
a
2
c
2
1
+
2
ab
+
b
2
+
b
2
a
2
1
+
2
b
c
+
c
2
+
c
2
b
2
1
+
2
c
a
=
2
(
a
+
b
+
c
)
.
Prove that if the triple
(
a
,
b
,
c
)
(a,b,c)
(
a
,
b
,
c
)
is mysterious, then so is the triple
(
c
,
b
,
a
)
(c,b,a)
(
c
,
b
,
a
)
. Proposed by A. Kuznetsov, K. Sukhov
1
3
Hide problems
If p^23, p^24, q^23, q^24 are in AP, then it also includes p and q
Let
p
p
p
and
q
q
q
be different prime numbers. We are given an infinite decreasing arithmetic progression in which each of the numbers
p
23
,
p
24
,
q
23
p^{23}, p^{24}, q^{23}
p
23
,
p
24
,
q
23
and
q
24
q^{24}
q
24
occurs. Show that the numbers
p
p
p
and
q
q
q
also occur in this progression. Proposed by A. Kuznetsov
Who has more good numbers?
Petya and Vasya only know positive integers not exceeding
1
0
9
−
4000
10^9-4000
1
0
9
−
4000
. Petya considers numbers as good which are representable in the form
a
b
c
+
a
b
+
a
c
+
b
c
abc+ab+ac+bc
ab
c
+
ab
+
a
c
+
b
c
, where
a
,
b
a,b
a
,
b
and
c
c
c
are natural numbers not less than
100
100
100
. Vasya considers numbers as good which are representable in the form
x
y
z
−
x
−
y
−
z
xyz-x-y-z
x
yz
−
x
−
y
−
z
, where
x
,
y
x,y
x
,
y
and
z
z
z
are natural numbers strictly bigger than
100
100
100
. For which of them are there more good numbers? Proposed by I. Bogdanov
Can you wrap your head (or a tetrahedron) around an infinite cylinder?
We are given an infinite cylinder in space (i.e. the locus of points of a given distance
R
>
0
R>0
R
>
0
from a given straight line). Can six straight lines containing the edges of a tetrahedron all have exactly one common point with this cylinder? Proposed by A. Kuznetsov