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Contests
National and Regional Contests
Russia Contests
All-Russian Olympiad
1972 All Soviet Union Mathematical Olympiad
1972 All Soviet Union Mathematical Olympiad
Part of
All-Russian Olympiad
Subcontests
(15)
168
1
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ASU 168 All Soviet Union MO 1972 game with digits **** - **** =
A game for two. One gives a digit and the second substitutes it instead of a star in the following difference:
∗
∗
∗
∗
−
∗
∗
∗
∗
=
**** - **** =
∗
∗
∗
∗
−
∗
∗
∗
∗
=
Then the first gives the next digit, and so on
8
8
8
times. The first wants to obtain the greatest possible difference, the second -- the least. Prove that:1. The first can operate in such a way that the difference would be not less than
4000
4000
4000
, not depending on the second's behaviour. 2. The second can operate in such a way that the difference would be not greater than
4000
4000
4000
, not depending on the first's behaviour.
172
1
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ASU 172 All Soviet Union MO 1972 min of max of ... , x_n/(1+x_1+...+x_n)
Let the sum of positive numbers
x
1
,
x
2
,
.
.
.
,
x
n
x_1, x_2, ... , x_n
x
1
,
x
2
,
...
,
x
n
be
1
1
1
. Let
s
s
s
be the greatest of the numbers
{
x
1
1
+
x
1
,
x
2
1
+
x
1
+
x
2
,
.
.
.
,
x
n
1
+
x
1
+
.
.
.
+
x
n
}
\left\{\frac{x_1}{1+x_1}, \frac{x_2}{1+x_1+x_2}, ..., \frac{x_n}{1+x_1+...+x_n}\right\}
{
1
+
x
1
x
1
,
1
+
x
1
+
x
2
x
2
,
...
,
1
+
x
1
+
...
+
x
n
x
n
}
What is the minimal possible
s
s
s
? What
x
i
x_i
x
i
correspond it?
173
1
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ASU 173 All Soviet Union MO 1972 1-round hockey tournament, odd points
One-round hockey tournament is finished (each plays with each one time, the winner gets
2
2
2
points, looser --
0
0
0
, and
1
1
1
point for draw). For arbitrary subgroup of teams there exists a team (may be from that subgroup) that has got an odd number of points in the games with the teams of the subgroup. Prove that there was even number of the participants.
171
1
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ASU 171 All Soviet Union MO 1972 put numbers 0,1,2 in 100x100 table
Is it possible to put the numbers
0
,
1
0,1
0
,
1
or
2
2
2
in the unit squares of the cross-lined paper
100
×
100
100\times 100
100
×
100
in such a way, that every rectangle
3
×
4
3\times 4
3
×
4
(and
4
×
3
4\times 3
4
×
3
) would contain three zeros, four ones and five twos?
170
1
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ASU 170 All Soviet Union MO interior point isosceles triangles with all vertices
The point
O
O
O
inside the convex polygon makes isosceles triangle with all the pairs of its vertices. Prove that
O
O
O
is the centre of the circumscribed circle.other formulation:
P
P
P
is a convex polygon and
X
X
X
is an interior point such that for every pair of vertices
A
,
B
A, B
A
,
B
, the triangle
X
A
B
XAB
X
A
B
is isosceles. Prove that all the vertices of
P
P
P
lie on a circle with center
X
X
X
.
169
1
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ASU 169 All Soviet Union MO 1972 min of least of {x, (y+ 1/x), 1/y}, for x,y>0
Let
x
,
y
x,y
x
,
y
be positive numbers,
s
s
s
-- the least of
{
x
,
(
y
+
1
/
x
)
,
1
/
y
}
\{ x, (y+ 1/x), 1/y\}
{
x
,
(
y
+
1/
x
)
,
1/
y
}
What is the greatest possible value of
s
s
s
? To what
x
x
x
and
y
y
y
does it correspond?
167
1
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ASU 167 All Soviet Union MO 1972 angles in inscribed heptagon
The
7
7
7
-gon
A
1
A
2
A
3
A
4
A
5
A
6
A
7
A_1A_2A_3A_4A_5A_6A_7
A
1
A
2
A
3
A
4
A
5
A
6
A
7
is inscribed in a circle. Prove that if the centre of the circle is inside the
7
7
7
-gon , than
∠
A
1
+
∠
A
2
+
∠
A
3
<
45
0
o
\angle A_1+ \angle A_2 + \angle A_3 < 450^o
∠
A
1
+
∠
A
2
+
∠
A
3
<
45
0
o
166
1
Hide problems
ASU 166 All Soviet Union MO 1972 3 out of 9 lines are concurrent
Each of the
9
9
9
straight lines divides the given square onto two quadrangles with the areas ratio as
2
:
3
2:3
2
:
3
. Prove that there exist three of them intersecting in one point
165
1
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ASU 165 All Soviet Union MO 1972 perpendiculars inside a convex ABCD
Let
O
O
O
be the intersection point of the diagonals of the convex quadrangle
A
B
C
D
ABCD
A
BC
D
. Prove that the line drawn through the points of intersection of the medians of triangles
A
O
B
AOB
A
OB
and
C
O
D
COD
CO
D
is orthogonal to the line drawn through the points of intersection of the heights of triangles
B
O
C
BOC
BOC
and
A
O
D
AOD
A
O
D
.
164
1
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ASU 164 All Soviet Union MO 1972 squares inside square of area 2
Given several squares with the total area
1
1
1
. Prove that you can pose them in the square of the area
2
2
2
without any intersections.
163
1
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ASU 163 All Soviet Union MO 1972 numbers in a triangle table
The triangle table is constructed according to the rule: You put the natural number
a
>
1
a>1
a
>
1
in the upper row, and then you write under the number
k
k
k
from the left side
k
2
k^2
k
2
, and from the right side --
(
k
+
1
)
(k+1)
(
k
+
1
)
. For example, if
a
=
2
a = 2
a
=
2
, you get the table on the picture. Prove that all the numbers on each particular line are different. 2 / \ / \ 4 3 / \ / \ 16 5 9 4 / \ / \ /\ / \
162
1
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ASU 162 All Soviet Union MO 1972 (a^n +1) /(a^m +1) => n/m
a) Let
a
,
n
,
m
a,n,m
a
,
n
,
m
be natural numbers,
a
>
1
a > 1
a
>
1
. Prove that if
(
a
m
+
1
)
(a^m + 1)
(
a
m
+
1
)
is divisible by
(
a
n
+
1
)
(a^n + 1)
(
a
n
+
1
)
than
m
m
m
is divisible by
n
n
n
. b) Let
a
,
b
,
n
,
m
a,b,n,m
a
,
b
,
n
,
m
be natural numbers,
a
>
1
,
a
a>1, a
a
>
1
,
a
and
b
b
b
are relatively prime. Prove that if
(
a
m
+
b
m
)
(a^m+b^m)
(
a
m
+
b
m
)
is divisible by
(
a
n
+
b
n
)
(a^n+b^n)
(
a
n
+
b
n
)
than
m
m
m
is divisible by
n
n
n
.
161
1
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ASU 161 All Soviet Union MO 1972 4^{27}+4^{1000}+4^x, perfect square, max x
Find the maximal
x
x
x
such that the expression
4
27
+
4
1000
+
4
x
4^{27} + 4^{1000} + 4^x
4
27
+
4
1000
+
4
x
is the exact square.
160
1
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ASU 160 All Soviet Union MO 1972 50 segments on the line
Given
50
50
50
segments on the line. Prove that one of the following statements is valid:1. Some
8
8
8
segments have the common point.2. Some
8
8
8
segments do not intersect each other.
159
1
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ASU 159 All Soviet Union MO 1972 rectangle <QNM=<MNP wanted
Given a rectangle
A
B
C
D
ABCD
A
BC
D
, points
M
M
M
-- the midpoint of
[
A
D
]
[AD]
[
A
D
]
side,
N
N
N
-- the midpoint of
[
B
C
]
[BC]
[
BC
]
side. Let us take a point
P
P
P
on the extension of the
[
D
C
]
[DC]
[
D
C
]
segment over the point
D
D
D
. Let us denote the intersection point of lines
(
P
M
)
(PM)
(
PM
)
and
(
A
C
)
(AC)
(
A
C
)
as
Q
Q
Q
. Prove that the
∠
Q
N
M
=
∠
M
N
P
\angle QNM= \angle MNP
∠
QNM
=
∠
MNP