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Problems
Contests
International Contests
Tournament Of Towns
1990 Tournament Of Towns
1990 Tournament Of Towns
Part of
Tournament Of Towns
Subcontests
(38)
(276) 4
1
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TOT 276 1990 Autumn O S4 bricks from 11x12x13 box
We have “bricks” made in the following way: we take a unit cube and glue to three of its faces which have a common vertex three more cubes in such a way that the faces glued together coincide. Is it possible to construct from these bricks an
11
×
12
×
13
11 \times 12 \times 13
11
×
12
×
13
box?(A Andjans, Riga )
(275) 3
1
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TOT 275 1990 Autumn O S3 two identical clocks on wall
There are two identical clocks on the wall, one showing the current Moscow time and the other showing current local time. The minimum distance between the ends of their hour hands equals
m
m
m
and the maximum distance equals
M
M
M
. Find the distance between the centres of the clocks. (S Fomin, Leningrad)
(274) 2
1
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TOT 274 1990 Autumn O S2 3 infinite sets of // lines
The plane is divided by three infinite sets of parallel lines into equilateral triangles of equal area. Let
M
M
M
be the set of their vertices, and
A
A
A
and
B
B
B
be two vertices of such an equilateral triangle. One may rotate the plane through
12
0
o
120^o
12
0
o
around any vertex of the set
M
M
M
. Is it possible to move the point
A
A
A
to the point
B
B
B
by a number of such rotations(N Vasiliev, Moscow)
(273) 1
1
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TOT 273 1990 Autumn O S1 pos/ integers 1-n^2 in nxn chessboard
The positive integers from
1
1
1
to
n
2
n^2
n
2
are placed arbitrarily on the squares of a chess board with dimensions
n
×
n
n\times n
n
×
n
. Prove that there are two adjacent squares (having a common vertex or a common side) such that the difference between the numbers placed on them is not less than
n
+
1
n + 1
n
+
1
.(N Sedrakyan, Yerevan)
(280) 5
1
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TOT 280 1990 Autumn A S5 exradius wanted starting with isosceles
In triangle
A
B
C
ABC
A
BC
we have
A
C
=
C
B
AC = CB
A
C
=
CB
. On side
A
B
AB
A
B
is a point
D
D
D
such that the radius of the incircle of triangle
A
C
D
ACD
A
C
D
is equal to the radius of the circle tangent to the segment
D
B
DB
D
B
and to the extensions of the lines
C
D
CD
C
D
and
C
B
CB
CB
. Prove that this radius equals a quarter of either of the two equal altitudes of triangle
A
B
C
ABC
A
BC
.
(279) 4
1
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TOT 279 1990 Autumn A S4 10 red and 10 white points
There are
20
20
20
points in the plane and no three of them are collinear. Of these points
10
10
10
are red while the other
10
10
10
are blue. Prove that there exists a straight line such that there are
5
5
5
red points and
5
5
5
blue points on either side of this line.(A Kushnirenko, Moscow)
(278) 3
1
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TOT 278 1990 Autumn A S3 finite set of unit squares
A finite set
M
M
M
of unit squares on the plane is considered. The sides of the squares are parallel to the coordinate axes and the squares are allowed to intersect. It is known that the distance between the centres of any pair of squares is no greater than
2
2
2
. Prove that there exists a unit square (not necessarily belonging to
M
M
M
) with sides parallel to the coordinate axes and which has at least one common point with each of the squares in
M
M
M
.(A Andjans, Riga)
(277) 2
1
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TOT 273 1990 Autumn A S2 equailateral givena and wanted
A point
M
M
M
is chosen on the arc
A
C
AC
A
C
of the circumcircle of the equilateral triangle
A
B
C
ABC
A
BC
.
P
P
P
is the midpoint of this arc,
N
N
N
is the midpoint of the chord
B
M
BM
BM
and
K
K
K
is the foot of the perpendicular drawn from
P
P
P
to
M
C
MC
MC
. Prove that the triangle
A
N
K
ANK
A
N
K
is equilateral.(I Nagel, Yevpatoria)
(267) 1
1
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TOT 267 1990 Autumn A J1 compare nested fractions
Given
a
=
1
2
+
1
3
+
1
.
.
.
+
.
.
.
99
,
a
n
d
b
=
1
2
+
1
3
+
1
.
.
.
+
.
.
.
99
+
1
100
a=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{...+\dfrac{...}{99}}}}, \,\,and\,\,\, b=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{...+\dfrac{...}{99+\dfrac{1}{100}}}}}
a
=
2
+
3
+
...
+
99
...
1
1
1
,
an
d
b
=
2
+
3
+
...
+
99
+
100
1
...
1
1
1
Prove that
∣
a
−
b
∣
<
1
99
!
100
!
|a-b| <\frac{1}{99! 100!}
∣
a
−
b
∣
<
99
!
100
!
1
(G Galperin, Moscow)
(272) 6
1
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TOT 272 1990 Autumn A J6 n different cards
A deck consists of
n
n
n
different cards. A move consists of taking out a group of cards in sequence from some place in the deck, and putting it back someplace else without changing the order within the group or turning any cards over. We are required to reverse the order of cards in the deck by such moves.(a) Prove that for
n
=
9
n = 9
n
=
9
, this can be done in
5
5
5
moves. (b) Prove that for
n
=
52
n = 52
n
=
52
, this i. can be done in
27
27
27
moves, ii. can’t be done in
17
17
17
moves, iii. can’t be done in
26
26
26
moves. (SM Voronin, Tchelyabinsk)
(271) 5
1
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TOT 271 1990 Autumn A J5 x_{n+1}=|x_n|-x_{n-1} has period 9
The numerical sequence
{
x
n
}
\{x_n\}
{
x
n
}
satisfies the condition
x
n
+
1
=
∣
x
n
∣
−
x
n
−
1
x_{n+1}=|x_n|-x_{n-1}
x
n
+
1
=
∣
x
n
∣
−
x
n
−
1
for all
n
>
1
n > 1
n
>
1
. Prove that the sequence is periodic with period
9
9
9
, i.e. for any
n
>
1
n > 1
n
>
1
we have
x
n
=
x
n
+
9
x_n = x_{n+9}
x
n
=
x
n
+
9
.(M Kontsevich, Moscow)
(270) 4
1
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TOT 270 1990 Autumn A J4 quadrs with equal sides and 2 // are #
The sides
A
B
AB
A
B
,
B
C
BC
BC
,
C
D
CD
C
D
and
D
A
DA
D
A
of the quadrilateral
A
B
C
D
ABCD
A
BC
D
are respectively equal to the sides
A
′
B
′
A'B'
A
′
B
′
,
B
′
C
′
B'C'
B
′
C
′
,
C
′
D
′
C'D'
C
′
D
′
and
D
′
A
′
D'A'
D
′
A
′
of the quadrilateral
A
′
B
′
C
D
A'B'CD
A
′
B
′
C
D
' and it is known that
A
B
∥
C
D
AB \parallel CD
A
B
∥
C
D
and
B
′
C
′
∥
D
′
A
′
B'C' \parallel D'A'
B
′
C
′
∥
D
′
A
′
. Prove that both quadrilaterals are parallelograms.(V Proizvolov, Moscow)
(269) 3
1
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TOT 269 1990 Autumn A J3 bw in 8x8 board
An
8
8
8
by
8
8
8
board (with
64
64
64
1
1
1
by
1
1
1
squares) is painted white. We are allowed to choose any rectangle consisting of
3
3
3
of the
64
64
64
squares and paint each of the
3
3
3
squares in the opposite colour (the white ones black, the black ones white). Is it possible to paint the entire board black by means of such operations? (IS Rubanov, Kirov)
(268) 2
1
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TOT 268 1990 Autumn A J2 locus of midpoints, semicircle
A semicircle
S
S
S
is drawn on
A
B
AB
A
B
as diameter. For an arbitrary point
C
C
C
in
S
S
S
(
C
≠
A
C\ne A
C
=
A
,
C
≠
B
C \ne B
C
=
B
), squares are attached to sides
A
C
AC
A
C
and
B
C
BC
BC
of triangle
A
B
C
ABC
A
BC
outside the triangle. Find the locus of the midpoint of the segment joining the centres of the squares as
C
C
C
moves along
S
S
S
.(J Tabov, Sofia)
(266) 4
1
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TOT 266 1990 Autumn O J4 right isosceles tiles100 x 100 square board
A square board with dimensions
100
×
100
100 \times 100
100
×
100
is divided into
10000
10 000
10000
unit squares. One of the squares is cut out. Is it possible to cover the rest of the board by isosceles right angled triangles which have hypotenuses of length
2
2
2
, and in such a way that their hypotenuses lie on sides of the squares and their other two sides lie on diagonals? The triangles must not overlap each other or extend beyond the edges of the board.(S Fomin, Leningrad)
(265) 3
1
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TOT 265 1990 Autumn O J3 10 pos. integers , each divdes their sum
Find
10
10
10
different positive integers such that each of them is a divisor of their sum(S Fomin, Leningrad)
(264) 2
1
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TOT 264 1990 Autumn O J2 equilateral and regular hexagon have cmoomon center
The vertices of an equilateral triangle lie on sides
A
B
AB
A
B
,
C
D
CD
C
D
and
E
F
EF
EF
of a regular hexagon
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
. Prove that the triangle and the hexagon have a common centre. (N Sedrakyan, Yerevan )
(263) 1
1
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TOT 263 1990 Autumn O J1 two positive numbers with sum < product
Suppose two positive real numbers are given. Prove that if their sum is less than their product then their sum is greater than four.(N Vasiliev, Moscow)
(262) 6
1
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TOT 262 1990 Spring A S6 ink-blots on a white paper square
There are some ink-blots on a white paper square with side length
a
a
a
. The area of each blot is not greater than
1
1
1
and every line parallel to any one of the sides of the square intersects no more than one blot. Prove that the total area of the blots is not greater than
a
a
a
.(A. Razborov, Moscow)
(261) 5
1
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TOT 261 1990 Spring A S5 triangular section if convex polyhedron
Does there exist a convex polyhedron which has a triangular section (by a plane not passing through the vertices) and each vertex of the polyhedron belonging to (a) no less than
5
5
5
faces? (b) exactly
5
5
5
faces?(G. Galperin)
(260) 4
1
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TOT 260 1990 Spring A S4 <ACP =<QCB or <ACP+ <QCB =180^o
Let
A
B
C
D
ABCD
A
BC
D
be a trapezium with
A
C
=
B
C
AC = BC
A
C
=
BC
. Let
H
H
H
be the midpoint of the base
A
B
AB
A
B
and let
ℓ
\ell
ℓ
be a line passing through
H
H
H
. Let
ℓ
\ell
ℓ
meet
A
D
AD
A
D
at
P
P
P
and
B
D
BD
B
D
at
Q
Q
Q
. Prove that the angles
A
C
P
ACP
A
CP
and
Q
C
B
QCB
QCB
are either equal or have a sum of
18
0
o
180^o
18
0
o
.(I. Sharygin, Moscow)
(259) 3
1
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TOT 259 1990 Spring A S3 cake for p or q persons
A cake is prepared for a dinner party to which only
p
p
p
or
q
q
q
persons will come (
p
p
p
and
q
q
q
are given co-prime integers). Find the minimum number of pieces (not necessarily equal) into which the cake must be cut in advance so that the cake may be equally shared between the persons in either case. (D. Fomin, Leningrad)
(258) 2
1
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TOT 258 1990 Spring A S2 different basic collections of weights, senior version
We call a collection of weights (each weighing an integer value) basic if their total weight equals
500
500
500
and each object of integer weight not greater than
500
500
500
can be balanced exactly with a uniquely determined set of weights from the collection. (Uniquely means that we are not concerned with order or which weights of equal value are chosen to balance against a particular object, if in fact there is a choice.) (a) Find an example of a basic collection other than the collection of
500
500
500
weights each of value
1
1
1
. (b) How many different basic collections are there?(D. Fomin, Leningrad)
(257) 1
1
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TOT 257 1990 Spring A S1 P(x) divisible by (x-1)^n
Prove that for all natural
n
n
n
there exists a polynomial
P
(
x
)
P(x)
P
(
x
)
divisible by
(
x
−
1
)
n
(x-1)^n
(
x
−
1
)
n
such that its degree is not greater than
2
n
2^n
2
n
and each of its coefficients is equal to
1
1
1
,
0
0
0
or
−
1
-1
−
1
.(D. Fomin, Leningrad)
(256) 4
1
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TOT 256 1990 Spring O S4 103 coins
A set of
103
103
103
coins that look alike is given. Two coins (whose weights are equal) are counterfeit. The other
101
101
101
(genuine) coins also have the same weight, but a different weight from that of the counterfeit coins. However it is not known whether it is the genuine coins or the counterfeit coins which are heavier. How can this question be resolved by three weighings on the one balance? (It is not required to separate the counterfeit coins from the genuine ones.)(D. Fomin, Leningrad)
(255) 3
1
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TOT 255 1990 Spring O S3 combo geo with dodecahedron , icosahedron
(a) Some vertices of a dodecahedron are to be marked so that each face contains a marked vertex. What is the smallest number of marked vertices for which this is possible?(b) Answer the same question, but for an icosahedron.(G. Galperin, Moscow)(Recall that a dodecahedron has
12
12
12
pentagonal faces which meet in threes at each vertex, while an icosahedron has
20
20
20
triangular faces which meet in fives at each vertex.)
(254) 2
1
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TOT 254 1990 Spring O S2 n =4k+1 , exist n odds whose sum = product
Prove that (a) if the natural number
n
n
n
can be represented as
n
=
4
k
+
1
n =4k+1
n
=
4
k
+
1
(where
k
k
k
is an integer), then there exist
n
n
n
odd positive integers whose sum is equal to their product, (b) if
n
n
n
cannot be represented in this form then such a set does not exist.(M. Kontsevich)
(253) 1
1
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TOT 253 1990 Spring O S1 triangl construction
Construct a triangle given two of its side lengths if it is known that the median drawn from their common vertex divides the angle between them in the ratio
1
:
2
1:2
1
:
2
. (V. Chikin)
(252) 6
1
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TOT 252 1990 Spring A J6 different basic collections of weights
We call a collection of weights (each weighing an integer value) basic if their total weight equals
200
200
200
and each object of integer weight not greater than
200
200
200
can be balanced exactly with a uniquely determined set of weights from the collection. (Uniquely means that we are not concerned with order or which weights of equalc value are chosen to balance against a particular object, if in fact there is a choice.) (a) Find an example of a basic collection other than the collection of
200
200
200
weights each of value
1
1
1
. (b) How many different basic collections are there?(D. Fomin, Leningrad)
(251) 5
1
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TOT 251 1990 Spring A J5 no (m, n) <= 1000 m/(n+1)< \sqrt2 < (m+1)/n
Find the number of pairs
(
m
,
n
)
(m, n)
(
m
,
n
)
of positive integers, both of which are
≤
1000
\le 1000
≤
1000
, such that
m
n
+
1
<
2
<
m
+
1
n
\frac{m}{n+1}< \sqrt2 < \frac{m+1}{n}
n
+
1
m
<
2
<
n
m
+
1
(recalling that
2
=
1.414213..
\sqrt2 = 1.414213..
2
=
1.414213..
.).(D. Fomin, Leningrad)
(250) 4
1
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TOT 250 1990 Spring A J4 collinear wanted, rhombus
Let
A
B
C
D
ABCD
A
BC
D
be a rhombus and
P
P
P
be a point on its side
B
C
BC
BC
. The circle passing through
A
,
B
A, B
A
,
B
, and
P
P
P
intersects
B
D
BD
B
D
once more at the point
Q
Q
Q
and the circle passing through
C
,
P
C,P
C
,
P
and
Q
Q
Q
intersects
B
D
BD
B
D
once more at the point
R
R
R
. Prove that
A
,
R
A, R
A
,
R
and
P
P
P
lie on the one straight line. (D. Fomin, Leningrad)
(249) 3
1
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TOT 249 1990 Spring A J3 integer weights of 15 elephants stand in a row
Fifteen elephants stand in a row. Their weights are expressed by integer numbers of kilograms. The sum of the weight of each elephant (except the one on the extreme right) and the doubled weight of its right neighbour is exactly
15
15
15
tonnes. Determine the weight of each elephant. (F.L. Nazarov)
(248) 2
1
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TOT 248 1990 Spring A J2 intersection of 2 equal squares, ratio, equal areas
If a square is intersected by another square equal to it but rotated by
4
5
o
45^o
4
5
o
around its centre, each side is divided into three parts in a certain ratio
a
:
b
:
a
a : b : a
a
:
b
:
a
(which one can compute). Make the following construction for an arbitrary convex quadrilateral: divide each of its sides into three parts in this same ratio
a
:
b
:
a
a : b : a
a
:
b
:
a
, and draw a line through the two division points neighbouring each vertex. Prove that the new quadrilateral bounded by the four drawn lines has the same area as the original one. (A. Savin, Moscow)
(247) 1
1
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TOT 247 1990 Spring A J1 100$ graphs of different quadratic in plane
Find the maximum number of parts into which the
O
x
y
Oxy
O
x
y
-plane can be divided by
100
100
100
graphs of different quadratic functions of the form
y
=
a
x
2
+
b
x
+
c
y = ax^2 + bx + c
y
=
a
x
2
+
b
x
+
c
. (N.B. Vasiliev, Moscow)
(246) 4
1
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TOT 246 1990 Spring O J4 counterfeit 2 of 61coins
A set of
61
61
61
coins that look alike is given. Two coins (whose weights are equal) are counterfeit. The other
59
59
59
(genuine) coins also have the same weight, but a different weight from that of the counterfeit coins. However it is not known whether it is the genuine coins or the counterfeit coins which are heavier. How can this question be resolved by three weighings on the one balance? (It is not required to separate the counterfeit coins from the genuine ones.)(D. Fomin, Leningrad)
(245) 3
1
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TOT 245 1990 Spring O J3 27 equal cubes, 9 red, 9 blue , 9 white
Is it possible to put together
27
27
27
equal cubes,
9
9
9
red,
9
9
9
blue and
9
9
9
white, so as to obtain a big cube in which each row (parallel to an arbitrary edge of the cube) contains three cubes with exactly two different colours? (S. Fomin, Leningrad)
(244) 2
1
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TOT 244 1990 Spring O J2 equal circles wanted, 2 other disjoint circles
Two circles
c
c
c
and
d
d
d
are situated in the plane each outside the other. The points
C
C
C
and
D
D
D
are located on circles
c
c
c
and
d
d
d
respectively, so as to be as far apart as possible. Two smaller circles are constructed inside
c
c
c
and
d
d
d
. Of these the first circle touches
c
c
c
and the two tangents drawn from
C
C
C
to
d
d
d
, while the second circle touches
d
d
d
and the two tangents from
D
D
D
to
c
c
c
. Prove that the small circles are equal. (J. Tabov, Sofia)
(243) 1
1
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TOT 243 1990 Spring O J1 identity with sum ( 1+ 1/2 + ...+ 1/n)^2
For every natural number
n
n
n
prove that
(
1
+
1
2
+
.
.
.
+
1
n
)
2
+
(
1
2
+
.
.
.
+
1
n
)
2
+
.
.
.
+
(
1
n
−
1
+
1
2
)
2
+
(
1
n
)
2
=
2
n
−
(
1
+
1
2
+
.
.
.
+
1
n
)
\left( 1+ \frac12 + ...+ \frac1n \right)^2+ \left( \frac12 + ...+ \frac1n \right)^2+...+ \left( \frac{1}{n-1} + \frac12 \right)^2+ \left( \frac1n \right)^2=2n- \left( 1+ \frac12 + ...+ \frac1n \right)
(
1
+
2
1
+
...
+
n
1
)
2
+
(
2
1
+
...
+
n
1
)
2
+
...
+
(
n
−
1
1
+
2
1
)
2
+
(
n
1
)
2
=
2
n
−
(
1
+
2
1
+
...
+
n
1
)
(S. Manukian, Yerevan)