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National and Regional Contests
Russia Contests
All-Russian Olympiad
1976 All Soviet Union Mathematical Olympiad
1976 All Soviet Union Mathematical Olympiad
Part of
All-Russian Olympiad
Subcontests
(15)
234
1
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ASU 234 All Soviet Union MO 1976 g(x_1)^2 + g(x_2)^2 + g(x_3)^2 = 1
Given a sphere of unit radius with the big circle (i.e of unit radius) that will be called "equator". We shall use the words "pole", "parallel","meridian" as self-explanatory. a) Let
g
(
x
)
g(x)
g
(
x
)
, where
x
x
x
is a point on the sphere, be the distance from this point to the equator plane. Prove that
g
(
x
)
g(x)
g
(
x
)
has the property if
x
1
,
x
2
,
x
3
x_1, x_2, x_3
x
1
,
x
2
,
x
3
are the ends of the pairwise orthogonal radiuses, then
g
(
x
1
)
2
+
g
(
x
2
)
2
+
g
(
x
3
)
2
=
1
(
∗
)
g(x_1)^2 + g(x_2)^2 + g(x_3)^2 = 1 \,\,\,\, (*)
g
(
x
1
)
2
+
g
(
x
2
)
2
+
g
(
x
3
)
2
=
1
(
∗
)
Let function
f
(
x
)
f(x)
f
(
x
)
be an arbitrary nonnegative function on a sphere that satisfies (*) property. b) Let
x
1
x_1
x
1
and
x
2
x_2
x
2
points be on the same meridian between the north pole and equator, and
x
1
x_1
x
1
is closer to the pole than
x
2
x_2
x
2
. Prove that
f
(
x
1
)
>
f
(
x
2
)
f(x_1) > f(x_2)
f
(
x
1
)
>
f
(
x
2
)
. c) Let
y
1
y_1
y
1
be closer to the pole than
y
2
y_2
y
2
. Prove that
f
(
y
1
)
>
f
(
y
2
)
f(y_1) > f(y_2)
f
(
y
1
)
>
f
(
y
2
)
. d) Let
z
1
z_1
z
1
and
z
2
z_2
z
2
be on the same parallel. Prove that
f
(
z
1
)
=
f
(
z
2
)
f(z_1) = f(z_2)
f
(
z
1
)
=
f
(
z
2
)
. e) Prove that for all
x
,
f
(
x
)
=
g
(
x
)
x , f(x) = g(x)
x
,
f
(
x
)
=
g
(
x
)
.
233
1
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ASU 233 All Soviet Union MO 1976 +1,-1 on vertices of n-gon
Given right
n
n
n
-gon wit the point
O
O
O
-- its centre. All the vertices are marked either with
+
1
+1
+
1
or
−
1
-1
−
1
. We may change all the signs in the vertices of regular
k
k
k
-gon (
2
≤
k
≤
n
2 \le k \le n
2
≤
k
≤
n
) with the same centre
O
O
O
. (By
2
2
2
-gon we understand a segment, being halved by
O
O
O
.) Prove that in a), b) and c) cases there exists such a set of
(
+
1
)
(+1)
(
+
1
)
s and
(
−
1
)
(-1)
(
−
1
)
s, that we can never obtain a set of
(
+
1
)
(+1)
(
+
1
)
s only. a)
n
=
15
n = 15
n
=
15
,b)
n
=
30
n = 30
n
=
30
,c)
n
>
2
n > 2
n
>
2
,d) Let us denote
K
(
n
)
K(n)
K
(
n
)
the maximal number of
(
+
1
)
(+1)
(
+
1
)
and
(
−
1
)
(-1)
(
−
1
)
sets such, that it is impossible to obtain one set from another. Prove, for example, that
K
(
200
)
=
2
80
K(200) = 2^{80}
K
(
200
)
=
2
80
232
1
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ASU 232 All Soviet Union MO 1976 n numbers around a circle with zero sum
n
n
n
numbers are written down along the circumference. Their sum equals to zero, and one of them equals
1
1
1
. a) Prove that there are two neighbours with their difference not less than
n
/
4
n/4
n
/4
. b) Prove that there is a number that differs from the arithmetic mean of its two neighbours not less than on
8
/
(
n
2
)
8/(n^2)
8/
(
n
2
)
. c) Try to improve the previous estimation, i.e what number can be used instead of
8
8
8
? d) Prove that for
n
=
30
n=30
n
=
30
there is a number that differs from the arithmetic mean of its two neighbours not less than on
2
/
113
2/113
2/113
, give an example of such
30
30
30
numbers along the circumference, that not a single number differs from the arithmetic mean of its two neighbours more than on
2
/
113
2/113
2/113
.
231
1
Hide problems
ASU 231 All Soviet Union MO 1976 sequence of universal numbers
Given natural
n
n
n
. We shall call "universal" such a sequence of natural number
a
1
,
a
2
,
.
.
.
,
a
k
,
k
≥
n
a_1, a_2, ... , a_k, k\ge n
a
1
,
a
2
,
...
,
a
k
,
k
≥
n
, if we can obtain every transposition of the first
n
n
n
natural numbers (i.e such a sequence of
n
n
n
numbers, that every one is encountered only once) by deleting some its members. (Examples: (
1
,
2
,
3
,
1
,
2
,
1
,
3
1,2,3,1,2,1,3
1
,
2
,
3
,
1
,
2
,
1
,
3
) is universal for
n
=
3
n=3
n
=
3
, and (
1
,
2
,
3
,
2
,
1
,
3
,
1
1,2,3,2,1,3,1
1
,
2
,
3
,
2
,
1
,
3
,
1
) -- not, because you can't obtain (
3
,
1
,
2
3,1,2
3
,
1
,
2
) from it.) The goal is to estimate the length of the shortest universal sequence for given
n
n
n
. a) Give an example of the universal sequence of
n
2
n2
n
2
members. b) Give an example of the universal sequence of
(
n
2
−
n
+
1
)
(n^2 - n + 1)
(
n
2
−
n
+
1
)
members. c) Prove that every universal sequence contains not less than
n
(
n
+
1
)
/
2
n(n + 1)/2
n
(
n
+
1
)
/2
members d) Prove that the shortest universal sequence for
n
=
4
n=4
n
=
4
contains 12 members e) Find as short universal sequence, as you can. The Organising Committee knows the method for
(
n
2
−
2
n
+
4
)
(n^2 - 2n +4)
(
n
2
−
2
n
+
4
)
members.
230
1
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ASU 230 All Soviet Union MO 1976 dividing an equilateral into big triangles
Let us call "big" a triangle with all sides longer than
1
1
1
. Given a equilateral triangle with all the sides equal to
5
5
5
. Prove that: a) You can cut
100
100
100
big triangles out of given one. b) You can divide the given triangle onto
100
100
100
big nonintersecting ones fully covering the initial one. c) The same as b), but the triangles either do not have common points, or have one common side, or one common vertex. d) The same as c), but the initial triangle has the side
3
3
3
.
229
1
Hide problems
ASU 229 All Soviet Union MO 1976 beetle on a 99x99 chessboard
Given a chess-board
99
×
99
99\times 99
99
×
99
with a set
F
F
F
of fields marked on it (the set is different in three tasks). There is a beetle sitting on every field of the set
F
F
F
. Suddenly all the beetles have raised into the air and flied to another fields of the same set. The beetles from the neighbouring fields have landed either on the same field or on the neighbouring ones (may be far from their starting point). (We consider the fields to be neighbouring if they have at least one common vertex.) Consider a statement:"There is a beetle, that either stayed on the same field or moved to the neighbouring one". Is it always valid if the figure
F
F
F
is: a) A central cross, i.e. the union of the
50
50
50
-th row and the
50
50
50
-th column? b) A window frame, i.e. the union of the
1
1
1
-st,
50
50
50
-th and
99
99
99
-th rows and the
1
1
1
-st,
50
50
50
-th and
99
99
99
-th columns? c) All the chess-board?
222
1
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ASU 222 All Soviet Union MO 1976 sum of arcs =180^o , 3 equal circles
Given three circumferences of the same radius in a plane. a) All three are crossing in one point
K
K
K
. Consider three arcs
A
K
,
C
K
,
E
K
AK,CK,EK
A
K
,
C
K
,
E
K
: the
A
,
C
,
E
A,C,E
A
,
C
,
E
are the points of the circumferences intersection and the arcs are taken in the clockwise direction. Every arc is inside one circle, outside the second and on the border of the third one. Prove that the sum of the arcs is
180
180
180
degrees. b) Consider the case, when the three circles give a curvilinear triangle
B
D
F
BDF
B
D
F
as their intersection (instead of one point
K
K
K
). The arcs are taken in the clockwise direction. Every arc is inside one circle, outside the second and on the border of the third one. Prove that the sum of the
A
B
,
C
D
AB, CD
A
B
,
C
D
and
E
F
EF
EF
arcs is
180
180
180
degrees.
221
1
Hide problems
ASU 221 All Soviet Union MO 1976 rows of 1000 numbers on blacboard
A row of
1000
1000
1000
numbers is written on the blackboard. We write a new row, below the first according to the rule: We write under every number
a
a
a
the natural number, indicating how many times the number
a
a
a
is encountered in the first line. Then we write down the third line: under every number
b
b
b
-- the natural number, indicating how many times the number
b
b
b
is encountered in the second line, and so on. a) Prove that there is a line that coincides with the preceding one. b) Prove that the eleventh line coincides with the twelfth. c) Give an example of the initial line such, that the tenth row differs from the eleventh.
223
1
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ASU 223 All Soviet Union MO 1976 x_k = min{ |x_i-x_j|,0<i<j< k}, x_{21} = 0
The natural numbers
x
1
x_1
x
1
and
x
2
x_2
x
2
are less than
1000
1000
1000
. We construct a sequence:
x
3
=
∣
x
1
−
x
2
∣
x_3 = |x_1 - x_2|
x
3
=
∣
x
1
−
x
2
∣
x
4
=
m
i
n
{
∣
x
1
−
x
2
∣
,
∣
x
1
−
x
3
∣
,
∣
x
2
−
x
3
∣
}
x_4 = min \{ |x_1 - x_2|, |x_1 - x_3|, |x_2 - x_3|\}
x
4
=
min
{
∣
x
1
−
x
2
∣
,
∣
x
1
−
x
3
∣
,
∣
x
2
−
x
3
∣
}
.
.
.
...
...
x
k
=
m
i
n
{
∣
x
i
−
x
j
∣
,
0
<
i
<
j
<
k
}
x_k = min \{ |x_i - x_j|, 0 <i < j < k\}
x
k
=
min
{
∣
x
i
−
x
j
∣
,
0
<
i
<
j
<
k
}
.
.
.
...
...
Prove that
x
21
=
0
x_{21} = 0
x
21
=
0
.
224
1
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ASU 224 All Soviet Union MO 1976 3-digit binary no on cube vertices
Can you mark the cube's vertices with the three-digit binary numbers in such a way, that the numbers at all the possible couples of neighbouring vertices differ in at least two digits?
227
1
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ASU 227 All Soviet Union MO 1976 rectangles cur from paper
There are
n
n
n
rectangles drawn on the rectangular sheet of paper with the sides of the rectangles parallel to the sheet sides. The rectangles do not have pairwise common interior points. Prove that after cutting out the rectangles the sheet will split into not more than
n
+
1
n+1
n
+
1
part.
228
1
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ASU 228 All Soviet Union MO 1976 3 pedestrians and 3 roads
There are three straight roads. Three pedestrians are moving along those roads, and they are NOT on one line in the initial moment. Prove that they will be one line not more than twice
226
1
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ASU 226 All Soviet Union MO 1976 max no of concyclic points, mids of 1976-gon
Given regular
1976
1976
1976
-gon. The midpoints of all the sides and diagonals are marked. What is the greatest number of the marked points lying on one circumference?
225
1
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ASU 225 All Soviet Union MO 1976 |a|+|b|+|c|+|d| \ge |a+d|+|b+d|+|c+d|
Given
4
4
4
vectors
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
in the plane, such that
a
+
b
+
c
+
d
=
0
a+b+c+d=0
a
+
b
+
c
+
d
=
0
. Prove the following inequality:
∣
a
∣
+
∣
b
∣
+
∣
c
∣
+
∣
d
∣
≥
∣
a
+
d
∣
+
∣
b
+
d
∣
+
∣
c
+
d
∣
|a|+|b|+|c|+|d| \ge |a+d|+|b+d|+|c+d|
∣
a
∣
+
∣
b
∣
+
∣
c
∣
+
∣
d
∣
≥
∣
a
+
d
∣
+
∣
b
+
d
∣
+
∣
c
+
d
∣
220
1
Hide problems
ASU 220 All Soviet Union MO 1976 50 exact watches in a table
There are
50
50
50
exact watches lying on a table. Prove that there exist a certain moment, when the sum of the distances from the centre of the table to the ends of the minute hands is more than the sum of the distances from the centre of the table to the centres of the watches.