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Problems
Contests
National and Regional Contests
Russia Contests
All-Russian Olympiad
2017 All-Russian Olympiad
2017 All-Russian Olympiad
Part of
All-Russian Olympiad
Subcontests
(8)
2
3
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Collinearity in Isosceles Trapezoid
A
B
C
D
ABCD
A
BC
D
is an isosceles trapezoid with
B
C
∣
∣
A
D
BC || AD
BC
∣∣
A
D
. A circle
ω
\omega
ω
passing through
B
B
B
and
C
C
C
intersects the side
A
B
AB
A
B
and the diagonal
B
D
BD
B
D
at points
X
X
X
and
Y
Y
Y
respectively. Tangent to
ω
\omega
ω
at
C
C
C
intersects the line
A
D
AD
A
D
at
Z
Z
Z
. Prove that the points
X
X
X
,
Y
Y
Y
, and
Z
Z
Z
are collinear.
Integer polynomial
a
,
b
,
c
a,b,c
a
,
b
,
c
- different natural numbers. Can we build quadratic polynomial
P
(
x
)
=
k
x
2
+
l
x
+
m
P(x)=kx^2+lx+m
P
(
x
)
=
k
x
2
+
l
x
+
m
, with
k
,
l
,
m
k,l,m
k
,
l
,
m
are integer,
k
>
0
k>0
k
>
0
that for some integer points it get values
a
3
,
b
3
,
c
3
a^3,b^3,c^3
a
3
,
b
3
,
c
3
?
Tangent in Isosceles Triangle
Let
A
B
C
ABC
A
BC
be an acute angled isosceles triangle with
A
B
=
A
C
AB=AC
A
B
=
A
C
and circumcentre
O
O
O
. Lines
B
O
BO
BO
and
C
O
CO
CO
intersect
A
C
,
A
B
AC, AB
A
C
,
A
B
respectively at
B
′
,
C
′
B', C'
B
′
,
C
′
. A straight line
l
l
l
is drawn through
C
′
C'
C
′
parallel to
A
C
AC
A
C
. Prove that the line
l
l
l
is tangent to the circumcircle of
△
B
′
O
C
\triangle B'OC
△
B
′
OC
.
7
1
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Divisors of number
There is number
N
N
N
on the board. Every minute Ivan makes next operation: takes any number
a
a
a
written on the board, erases it, then writes all divisors of
a
a
a
except
a
a
a
( Can be same numbers on the board). After some time on the board there are
N
2
N^2
N
2
numbers. For which
N
N
N
is it possible?
6
1
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Chips in the table
In the
200
×
200
200\times 200
200
×
200
table in some cells lays red or blue chip. Every chip "see" other chip, if they lay in same row or column. Every chip "see" exactly
5
5
5
chips of other color. Find maximum number of chips in the table.
5
2
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Divisors on number
n
n
n
is composite.
1
<
a
1
<
a
2
<
.
.
.
<
a
k
<
n
1<a_1<a_2<...<a_k<n
1
<
a
1
<
a
2
<
...
<
a
k
<
n
- all divisors of
n
n
n
. It is known, that
a
1
+
1
,
.
.
.
,
a
k
+
1
a_1+1,...,a_k+1
a
1
+
1
,
...
,
a
k
+
1
are all divisors for some
m
m
m
(except
1
,
m
1,m
1
,
m
). Find all such
n
n
n
.
Triangles from polynoms
P
(
x
)
P(x)
P
(
x
)
is polynomial with degree
n
≥
2
n\geq 2
n
≥
2
and nonnegative coefficients.
a
,
b
,
c
a,b,c
a
,
b
,
c
- sides for some triangle. Prove, that
P
(
a
)
n
,
P
(
b
)
n
,
P
(
c
)
n
\sqrt[n]{P(a)},\sqrt[n]{P(b)},\sqrt[n]{P(c)}
n
P
(
a
)
,
n
P
(
b
)
,
n
P
(
c
)
are sides for some triangle too.
4
3
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Table with cells
Every cell of
100
×
100
100\times 100
100
×
100
table is colored black or white. Every cell on table border is black. It is known, that in every
2
×
2
2\times 2
2
×
2
square there are cells of two colors. Prove, that exist
2
×
2
2\times 2
2
×
2
square that is colored in chess order.
Relatively prime sums
Are there infinite increasing sequence of natural numbers, such that sum of every 2 different numbers are relatively prime with sum of every 3 different numbers?
Magic trick
Magicman and his helper want to do some magic trick. They have special card desk. Back of all cards is common color and face is one of
2017
2017
2017
colors. Magic trick: magicman go away from scene. Then viewers should put on the table
n
>
1
n>1
n
>
1
cards in the row face up. Helper looks at these cards, then he turn all cards face down, except one, without changing order in row. Then magicman returns on the scene, looks at cards, then show on the one card, that lays face down and names it face color. What is minimal
n
n
n
such that magicman and his helper can has strategy to make magic trick successfully?
1
4
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8
2
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Tangents intersect on AC
In a non-isosceles triangle
A
B
C
ABC
A
BC
,
O
O
O
and
I
I
I
are circumcenter and incenter,respectively.
B
′
B^\prime
B
′
is reflection of
B
B
B
with respect to
O
I
OI
O
I
and lies inside the angle
A
B
I
ABI
A
B
I
.Prove that the tangents to circumcirle of
△
B
B
′
I
\triangle BB^\prime I
△
B
B
′
I
at
B
′
B^\prime
B
′
,
I
I
I
intersect on
A
C
AC
A
C
. (A. Kuznetsov)
A nice(Hard) problem from Russia!
Given a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
. We denote
I
A
,
I
B
,
I
C
I_A,I_B, I_C
I
A
,
I
B
,
I
C
and
I
D
I_D
I
D
centers of
ω
A
,
ω
B
,
ω
C
\omega_A, \omega_B,\omega_C
ω
A
,
ω
B
,
ω
C
and
ω
D
\omega_D
ω
D
,inscribed In the triangles
D
A
B
,
A
B
C
,
B
C
D
DAB, ABC, BCD
D
A
B
,
A
BC
,
BC
D
and
C
D
A
CDA
C
D
A
, respectively.It turned out that
∠
B
I
A
A
+
∠
I
C
I
A
I
D
=
18
0
∘
\angle BI_AA + \angle I_CI_AI_D = 180^\circ
∠
B
I
A
A
+
∠
I
C
I
A
I
D
=
18
0
∘
. Prove that
∠
B
I
B
A
+
∠
I
C
I
B
I
D
=
18
0
∘
\angle BI_BA + \angle I_CI_BI_D = 180^{\circ}
∠
B
I
B
A
+
∠
I
C
I
B
I
D
=
18
0
∘
. (A. Kuznetsov)
3
4
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