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Problems
Contests
National and Regional Contests
Russia Contests
Saint Petersburg Mathematical Olympiad
2022 Saint Petersburg Mathematical Olympiad
2022 Saint Petersburg Mathematical Olympiad
Part of
Saint Petersburg Mathematical Olympiad
Subcontests
(7)
6
1
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Representation as ratio if two fractional parts
Find all pairs of nonzero rational numbers
x
,
y
x, y
x
,
y
, such that every positive rational number can be written as
{
r
x
}
{
r
y
}
\frac{\{rx\}} {\{ry\}}
{
ry
}
{
r
x
}
for some positive rational
r
r
r
.
5
2
Hide problems
Orthocenter geo with tangencies
Altitudes
A
A
1
,
B
B
1
,
C
C
1
AA_1, BB_1, CC_1
A
A
1
,
B
B
1
,
C
C
1
of acute triangle
A
B
C
ABC
A
BC
intersect at point
H
H
H
. On the tangent drawn from point
C
C
C
to the circle
(
A
B
1
C
1
)
(AB_1C_1)
(
A
B
1
C
1
)
, the perpendicular
H
Q
HQ
H
Q
is drawn (the point
Q
Q
Q
lies inside the triangle
A
B
C
ABC
A
BC
). Prove that the circle passing through the point
B
1
B_1
B
1
and touching the line
A
B
AB
A
B
at point
A
A
A
is also tangent to line
A
1
Q
A_1Q
A
1
Q
.
Weird gcd function
Let
n
n
n
be a positive integer and let
a
1
,
a
2
,
⋯
a
k
a_1, a_2, \cdots a_k
a
1
,
a
2
,
⋯
a
k
be all numbers less than
n
n
n
and coprime to
n
n
n
in increasing order. Find the set of values the function
f
(
n
)
=
g
c
d
(
a
1
3
−
1
,
a
2
3
−
1
,
⋯
,
a
k
3
−
1
)
f(n)=gcd(a_1^3-1, a_2^3-1, \cdots, a_k^3-1)
f
(
n
)
=
g
c
d
(
a
1
3
−
1
,
a
2
3
−
1
,
⋯
,
a
k
3
−
1
)
.
7
2
Hide problems
Good sets of cards
Given is a set of
2
n
2n
2
n
cards numbered
1
,
2
,
⋯
,
n
1,2, \cdots, n
1
,
2
,
⋯
,
n
, each number appears twice. The cards are put on a table with the face down. A set of cards is called good if no card appears twice. Baron Munchausen claims that he can specify
80
80
80
sets of
n
n
n
cards, of which at least one is sure to be good. What is the maximal
n
n
n
for which the Baron's words could be true?
Permutations of digits and divisibilities
Given are
n
n
n
distinct natural numbers. For any two of them, the one is obtained from the other by permuting its digits (zero cannot be put in the first place). Find the largest
n
n
n
such that it is possible all these numbers to be divisible by the smallest of them?
4
3
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Strongly majorizing pairs of sequences
We will say that a set of real numbers
A
=
(
a
1
,
.
.
.
,
a
17
)
A = (a_1,... , a_{17})
A
=
(
a
1
,
...
,
a
17
)
is stronger than the set of real numbers
B
=
(
b
1
,
.
.
.
,
b
17
)
B = (b_1, . . . , b_{17})
B
=
(
b
1
,
...
,
b
17
)
, and write
A
>
B
A >B
A
>
B
if among all inequalities
a
i
>
b
j
a_i > b_j
a
i
>
b
j
the number of true inequalities is at least
3
3
3
times greater than the number of false. Prove that there is no chain of sets
A
1
,
A
2
,
.
.
.
,
A
N
A_1, A_2, . . . , A_N
A
1
,
A
2
,
...
,
A
N
such that
A
1
>
A
2
>
⋯
A
N
>
A
1
A_1>A_2> \cdots A_N>A_1
A
1
>
A
2
>
⋯
A
N
>
A
1
. Remark: For 11.4, the constant
3
3
3
is changed to
2
2
2
and
N
=
3
N=3
N
=
3
and
17
17
17
is changed to
m
m
m
and
n
n
n
in the definition (the number of elements don't have to be equal).
Game with piles of stones
There are two piles of stones:
1703
1703
1703
stones in one pile and
2022
2022
2022
in the other. Sasha and Olya play the game, making moves in turn, Sasha starts. Let before the player's move the heaps contain
a
a
a
and
b
b
b
stones, with
a
≥
b
a \geq b
a
≥
b
. Then, on his own move, the player is allowed take from the pile with
a
a
a
stones any number of stones from
1
1
1
to
b
b
b
. A player loses if he can't make a move. Who wins?Remark: For 10.4, the initial numbers are
(
444
,
999
)
(444,999)
(
444
,
999
)
Points between parabolas
We will say that a point of the plane
(
u
,
v
)
(u, v)
(
u
,
v
)
lies between the parabolas
y
=
f
(
x
)
y = f(x)
y
=
f
(
x
)
and
y
=
g
(
x
)
y = g(x)
y
=
g
(
x
)
if
f
(
u
)
≤
v
≤
g
(
u
)
f(u) \leq v \leq g(u)
f
(
u
)
≤
v
≤
g
(
u
)
. Find the smallest real
p
p
p
for which the following statement is true: for any segment, the ends and the midpoint of which lie between the parabolas
y
=
x
2
y = x^2
y
=
x
2
and
y
=
x
2
+
1
y=x^2+1
y
=
x
2
+
1
, then they lie entirely between the parabolas
y
=
x
2
y=x^2
y
=
x
2
and
y
=
x
2
+
p
y=x^2+p
y
=
x
2
+
p
.
3
3
Hide problems
Weird polynomial game
Ivan and Kolya play a game, Ivan starts. Initially, the polynomial
x
−
1
x-1
x
−
1
is written of the blackboard. On one move, the player deletes the current polynomial
f
(
x
)
f(x)
f
(
x
)
and replaces it with
a
x
n
+
1
−
f
(
−
x
)
−
2
ax^{n+1}-f(-x)-2
a
x
n
+
1
−
f
(
−
x
)
−
2
, where
deg
(
f
)
=
n
\deg(f)=n
de
g
(
f
)
=
n
and
a
a
a
is a real root of
f
f
f
. The player who writes a polynomial which does not have real roots loses. Can Ivan beat Kolya?
Isogonal lines wrt angle BIC
Given is a triangle
A
B
C
ABC
A
BC
with altitude
A
H
AH
A
H
, diameter of the circumcircle
A
D
AD
A
D
and incenter
I
I
I
. Prove that
∠
B
I
H
=
∠
D
I
C
\angle BIH = \angle DIC
∠
B
I
H
=
∠
D
I
C
.
Angle bisectors in a trapezoid and cyclic quads
Given is a trapezoid
A
B
C
D
ABCD
A
BC
D
,
A
D
∥
B
C
AD \parallel BC
A
D
∥
BC
. The angle bisectors of the two pairs of opposite angles meet at
X
,
Y
X, Y
X
,
Y
. Prove that
A
X
Y
D
AXYD
A
X
Y
D
and
B
X
Y
C
BXYC
BX
Y
C
are cyclic.
2
2
Hide problems
Conditional geo with angle 45°
Given is a triangle
A
B
C
ABC
A
BC
with
∠
B
A
C
=
45
\angle BAC=45
∠
B
A
C
=
45
;
A
D
,
B
E
,
C
F
AD, BE, CF
A
D
,
BE
,
CF
are altitudes and
E
F
∩
B
C
=
X
EF \cap BC=X
EF
∩
BC
=
X
. If
A
X
∥
D
E
AX \parallel DE
A
X
∥
D
E
, find the angles of the triangle.
Set Theory masked with children and songs
12
12
12
schoolchildren are engaged in a circle of patriotic songs, each of them knows a few songs (maybe none). We will say that a group of schoolchildren can sing a song if at least one member of the group knows it. Supervisor the circle noticed that any group of
10
10
10
circle members can sing exactly
20
20
20
songs, and any group of
8
8
8
circle members - exactly
16
16
16
songs. Prove that the group of all
12
12
12
circle members can sing exactly
24
24
24
songs.
1
1
Hide problems
a+k divides b+k for all k<b implies a-k|b-k for all k<b
The positive integers
a
a
a
and
b
b
b
are such that
a
+
k
a+k
a
+
k
is divisible by
b
+
k
b+k
b
+
k
for all positive integers numbers
k
<
b
k<b
k
<
b
. Prove that
a
−
k
a-k
a
−
k
is divisible by
b
−
k
b-k
b
−
k
for all positive integers
k
<
b
k<b
k
<
b
.