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Problems
Contests
National and Regional Contests
Russia Contests
All-Russian Olympiad
2016 All-Russian Olympiad
2016 All-Russian Olympiad
Part of
All-Russian Olympiad
Subcontests
(8)
6
2
Hide problems
Partition of rectangle
A square is partitioned in
n
2
≥
4
n^2\geq 4
n
2
≥
4
rectanles using
2
(
n
−
1
)
2(n-1)
2
(
n
−
1
)
lines,
n
−
1
n-1
n
−
1
of which,are parallel to the one side of the square,
n
−
1
n-1
n
−
1
are parallel to the other side.Prove that we can choose
2
n
2n
2
n
rectangles of the partition,such that,for each two of them,we can place the one inside the other (possibly with rotation).
Aviaroutes in country
There are
n
>
1
n>1
n
>
1
cities in the country, some pairs of cities linked two-way through straight flight. For every pair of cities there is exactly one aviaroute (can have interchanges). Major of every city X counted amount of such numberings of all cities from
1
1
1
to
n
n
n
, such that on every aviaroute with the beginning in X, numbers of cities are in ascending order. Every major, except one, noticed that results of counting are multiple of
2016
2016
2016
. Prove, that result of last major is multiple of
2016
2016
2016
too.
4
1
Hide problems
Points in octahedrons
There is three-dimensional space. For every integer
n
n
n
we build planes
x
±
y
±
z
=
n
x \pm y\pm z = n
x
±
y
±
z
=
n
. All space is divided on octahedrons and tetrahedrons. Point
(
x
0
,
y
0
,
z
0
)
(x_0,y_0,z_0)
(
x
0
,
y
0
,
z
0
)
has rational coordinates but not lies on any plane. Prove, that there is such natural
k
k
k
, that point
(
k
x
0
,
k
y
0
,
k
z
0
)
(kx_0,ky_0,kz_0)
(
k
x
0
,
k
y
0
,
k
z
0
)
lies strictly inside the octahedron of partition.
3
2
Hide problems
Divisors of...divisors
Alexander has chosen a natural number
N
>
1
N>1
N
>
1
and has written down in a line,and in increasing order,all his positive divisors
d
1
<
d
2
<
…
<
d
s
d_1<d_2<\ldots <d_s
d
1
<
d
2
<
…
<
d
s
(where
d
1
=
1
d_1=1
d
1
=
1
and
d
s
=
N
d_s=N
d
s
=
N
).For each pair of neighbouring numbers,he has found their greater common divisor.The sum of all these
s
−
1
s-1
s
−
1
numbers (the greatest common divisors) is equal to
N
−
2
N-2
N
−
2
.Find all possible values of
N
N
N
.
Triangles on the paper
We have sheet of paper, divided on
100
×
100
100\times 100
100
×
100
unit squares. In some squares we put rightangled isosceles triangles with leg =
1
1
1
( Every triangle lies in one unit square and is half of this square). Every unit grid segment( boundary too) is under one leg of triangle. Find maximal number of unit squares, that don`t contains triangles.
1
2
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Exchanging Carpets
A carpet dealer,who has a lot of carpets in the market,is available to exchange a carpet of dimensions
a
⋅
b
a\cdot b
a
⋅
b
either with a carpet with dimensions
1
a
⋅
1
b
\frac{1}{a}\cdot \frac{1}{b}
a
1
⋅
b
1
or with two carpets with dimensions
c
⋅
b
c\cdot b
c
⋅
b
and
a
c
⋅
b
\frac{a}{c}\cdot b
c
a
⋅
b
(the customer can select the number
c
c
c
).The dealer supports that,at the beginning he had a carpet with dimensions greater than
1
1
1
and,after some exchanges like the ones we described above,he ended up with a set of carpets,each one having one dimension greater than
1
1
1
and one smaller than
1
1
1
.Is this possible?Note:The customer can demand from the dealer to consider a carpet of dimensions
a
⋅
b
a\cdot b
a
⋅
b
as one with dimensions
b
⋅
a
b\cdot a
b
⋅
a
.
Math in NBA
There are
30
30
30
teams in NBA and every team play
82
82
82
games in the year. Bosses of NBA want to divide all teams on Western and Eastern Conferences (not necessary equally), such that number of games between teams from different conferences is half of number of all games. Can they do it?
5
2
Hide problems
The last digits are zero...
Using each of the digits
1
,
2
,
3
,
…
,
8
,
9
1,2,3,\ldots ,8,9
1
,
2
,
3
,
…
,
8
,
9
exactly once,we form nine,not necassarily distinct,nine-digit numbers.Their sum ends in
n
n
n
zeroes,where
n
n
n
is a non-negative integer.Determine the maximum possible value of
n
n
n
.
Polynomial with integer roots.
Let
n
n
n
be a positive integer and let
k
0
,
k
1
,
…
,
k
2
n
k_0,k_1, \dots,k_{2n}
k
0
,
k
1
,
…
,
k
2
n
be nonzero integers such that
k
0
+
k
1
+
⋯
+
k
2
n
≠
0
k_0+k_1 +\dots+k_{2n}\neq 0
k
0
+
k
1
+
⋯
+
k
2
n
=
0
. Is it always possible to a permutation
(
a
0
,
a
1
,
…
,
a
2
n
)
(a_0,a_1,\dots,a_{2n})
(
a
0
,
a
1
,
…
,
a
2
n
)
of
(
k
0
,
k
1
,
…
,
k
2
n
)
(k_0,k_1,\dots,k_{2n})
(
k
0
,
k
1
,
…
,
k
2
n
)
so that the equation \begin{align*} a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+\dots+a_0=0 \end{align*} has not integer roots?
2
3
Hide problems
Concyclic points
ω
\omega
ω
is a circle inside angle
∡
B
A
C
\measuredangle BAC
∡
B
A
C
and it is tangent to sides of this angle at
B
,
C
B,C
B
,
C
.An arbitrary line
ℓ
\ell
ℓ
intersects with
A
B
,
A
C
AB,AC
A
B
,
A
C
at
K
,
L
K,L
K
,
L
,respectively and intersect with
ω
\omega
ω
at
P
,
Q
P,Q
P
,
Q
.Points
S
,
T
S,T
S
,
T
are on
B
C
BC
BC
such that
K
S
∥
A
C
KS \parallel AC
K
S
∥
A
C
and
T
L
∥
A
B
TL \parallel AB
T
L
∥
A
B
.Prove that
P
,
Q
,
S
,
T
P,Q,S,T
P
,
Q
,
S
,
T
are concyclic.(I.Bogdanov,P.Kozhevnikov)
Line passes through circumcenters is parallel to AD
Diagonals
A
C
,
B
D
AC,BD
A
C
,
B
D
of cyclic quadrilateral
A
B
C
D
ABCD
A
BC
D
intersect at
P
P
P
.Point
Q
Q
Q
is on
B
C
BC
BC
(between
B
B
B
and
C
C
C
) such that
P
Q
⊥
A
C
PQ \perp AC
PQ
⊥
A
C
.Prove that the line passes through the circumcenters of triangles
A
P
D
APD
A
P
D
and
B
Q
D
BQD
BQ
D
is parallel to
A
D
AD
A
D
.(A.Kuznetsov)
circumspheres intersect at one point
In the space given three segments
A
1
A
2
,
B
1
B
2
A_1A_2, B_1B_2
A
1
A
2
,
B
1
B
2
and
C
1
C
2
C_1C_2
C
1
C
2
, do not lie in one plane and intersect at a point
P
P
P
. Let
O
i
j
k
O_{ijk}
O
ijk
be center of sphere that passes through the points
A
i
,
B
j
,
C
k
A_i, B_j, C_k
A
i
,
B
j
,
C
k
and
P
P
P
. Prove that
O
111
O
222
,
O
112
O
221
,
O
121
O
212
O_{111}O_{222}, O_{112}O_{221}, O_{121}O_{212}
O
111
O
222
,
O
112
O
221
,
O
121
O
212
and
O
211
O
122
O_{211}O_{122}
O
211
O
122
intersect at one point. (P.Kozhevnikov)
8
2
Hide problems
Circumcircles are tangent
In acute triangle
A
B
C
ABC
A
BC
,
A
C
<
B
C
AC<BC
A
C
<
BC
,
M
M
M
is midpoint of
A
B
AB
A
B
and
Ω
\Omega
Ω
is it's circumcircle.Let
C
′
C^\prime
C
′
be antipode of
C
C
C
in
Ω
\Omega
Ω
.
A
C
′
AC^\prime
A
C
′
and
B
C
′
BC^\prime
B
C
′
intersect with
C
M
CM
CM
at
K
,
L
K,L
K
,
L
,respectively.The perpendicular drawn from
K
K
K
to
A
C
′
AC^\prime
A
C
′
and perpendicular drawn from
L
L
L
to
B
C
′
BC^\prime
B
C
′
intersect with
A
B
AB
A
B
and each other and form a triangle
Δ
\Delta
Δ
.Prove that circumcircles of
Δ
\Delta
Δ
and
Ω
\Omega
Ω
are tangent.(M.Kungozhin)
Circumcircles intersect at one point
Medians
A
M
A
,
B
M
B
,
C
M
C
AM_A,BM_B,CM_C
A
M
A
,
B
M
B
,
C
M
C
of triangle
A
B
C
ABC
A
BC
intersect at
M
M
M
.Let
Ω
A
\Omega_A
Ω
A
be circumcircle of triangle passes through midpoint of
A
M
AM
A
M
and tangent to
B
C
BC
BC
at
M
A
M_A
M
A
.Define
Ω
B
\Omega_B
Ω
B
and
Ω
C
\Omega_C
Ω
C
analogusly.Prove that
Ω
A
,
Ω
B
\Omega_A,\Omega_B
Ω
A
,
Ω
B
and
Ω
C
\Omega_C
Ω
C
intersect at one point.(A.Yakubov) [hide=P.S]sorry for my mistake in translation :blush: :whistling: .thank you jred for your help :coolspeak:
7
2
Hide problems
The sum not depends to x
In triangle
A
B
C
ABC
A
BC
,
A
B
<
A
C
AB<AC
A
B
<
A
C
and
ω
\omega
ω
is incirle.The
A
A
A
-excircle is tangent to
B
C
BC
BC
at
A
′
A^\prime
A
′
.Point
X
X
X
lies on
A
A
′
AA^\prime
A
A
′
such that segment
A
′
X
A^\prime X
A
′
X
doesn't intersect with
ω
\omega
ω
.The tangents from
X
X
X
to
ω
\omega
ω
intersect with
B
C
BC
BC
at
Y
,
Z
Y,Z
Y
,
Z
.Prove that the sum
X
Y
+
X
Z
XY+XZ
X
Y
+
XZ
not depends to point
X
X
X
.(Mitrofanov)
Two inequalities from Russia 2016
All russian olympiad 2016,Day 2 ,grade 9,P8 : Let
a
,
b
,
c
,
d
a, b, c, d
a
,
b
,
c
,
d
be are positive numbers such that
a
+
b
+
c
+
d
=
3
a+b+c+d=3
a
+
b
+
c
+
d
=
3
.Prove that
1
a
2
+
1
b
2
+
1
c
2
+
1
d
2
≤
1
a
2
b
2
c
2
d
2
\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\le\frac{1}{a^2b^2c^2d^2}
a
2
1
+
b
2
1
+
c
2
1
+
d
2
1
≤
a
2
b
2
c
2
d
2
1
All russian olympiad 2016,Day 2,grade 11,P7 : Let
a
,
b
,
c
,
d
a, b, c, d
a
,
b
,
c
,
d
be are positive numbers such that
a
+
b
+
c
+
d
=
3
a+b+c+d=3
a
+
b
+
c
+
d
=
3
.Prove that
1
a
3
+
1
b
3
+
1
c
3
+
1
d
3
≤
1
a
3
b
3
c
3
d
3
\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\le\frac{1}{a^3b^3c^3d^3}
a
3
1
+
b
3
1
+
c
3
1
+
d
3
1
≤
a
3
b
3
c
3
d
3
1
Russia national 2016