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Problems
Contests
National and Regional Contests
Russia Contests
Moscow Mathematical Olympiad
1953 Moscow Mathematical Olympiad
1953 Moscow Mathematical Olympiad
Part of
Moscow Mathematical Olympiad
Subcontests
(26)
253
1
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MMO 253 Moscow MO 1953 ax^2+bx+c=0, -ax^2+bx+c =0, a/2x^2 + bx + c = 0
Given the equations (1)
a
x
2
+
b
x
+
c
=
0
ax^2 + bx + c = 0
a
x
2
+
b
x
+
c
=
0
(2)
−
a
x
2
+
b
x
+
c
=
0
-ax^2 + bx + c = 0
−
a
x
2
+
b
x
+
c
=
0
prove that if
x
1
x_1
x
1
and
x
2
x_2
x
2
are some roots of equations (1) and (2), respectively, then there is a root
x
3
x_3
x
3
of the equation
a
2
x
2
+
b
x
+
c
=
0
\frac{a}{2}x^2 + bx + c = 0
2
a
x
2
+
b
x
+
c
=
0
such that either
x
1
≤
x
3
≤
x
2
x_1 \le x_3 \le x_2
x
1
≤
x
3
≤
x
2
or
x
1
≥
x
3
≥
x
2
x_1 \ge x_3 \ge x_2
x
1
≥
x
3
≥
x
2
.
257
1
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MMO 257 Moscow MO 1953 x_n=(x^2_{n-1}+2)/(2x_(n-1))
Let
x
0
=
1
0
9
x_0 = 10^9
x
0
=
1
0
9
,
x
n
=
x
n
−
1
2
+
2
2
x
n
−
1
x_n = \frac{x^2_{n-1}+2}{2x_{n-1}}
x
n
=
2
x
n
−
1
x
n
−
1
2
+
2
for
n
>
0
n > 0
n
>
0
. Prove that
0
<
x
36
−
2
<
1
0
−
9
0 < x_{36} - \sqrt2 < 10^{-9}
0
<
x
36
−
2
<
1
0
−
9
.
258
1
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MMO 258 Moscow MO 1953 2n moves of a knight on an infinite chess board
A knight stands on an infinite chess board. Find all places it can reach in exactly
2
n
2n
2
n
moves.
256
1
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MMO 256 Moscow MO 1953 solve sum (-1)^nx(x-1)...(x - n + 1)}{n!}=0
Find roots of the equation
1
−
x
1
+
x
(
x
−
1
)
2
!
−
.
.
.
+
(
−
1
)
n
x
(
x
−
1
)
.
.
.
(
x
−
n
+
1
)
n
!
=
0
1 -\frac{x}{1}+ \frac{x(x - 1)}{2!} -... +\frac{ (-1)^nx(x-1)...(x - n + 1)}{n!}= 0
1
−
1
x
+
2
!
x
(
x
−
1
)
−
...
+
n
!
(
−
1
)
n
x
(
x
−
1
)
...
(
x
−
n
+
1
)
=
0
255
1
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MMO 255 Moscow MO 1953 divide a cube into 3 equal pyramids
Divide a cube into three equal pyramids.
254
1
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MMO 254 Moscow MO 1953 moves in a 101&times;200 sheet of graph paper
Given a
101
×
200
101\times 200
101
×
200
sheet of graph paper, we start moving from a corner square in the direction of the square’s diagonal (not the sheet’s diagonal) to the border of the sheet, then change direction obeying the laws of light’s reflection. Will we ever reach a corner square? https://cdn.artofproblemsolving.com/attachments/b/8/4ec2f4583f406feda004c7fb4f11a424c9b9ae.png
252
1
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MMO 252 Moscow MO 1953 concurrency, angles \pi - (given to a line)
Given triangle
△
A
1
A
2
A
3
\vartriangle A_1A_2A_3
△
A
1
A
2
A
3
and a straight line
ℓ
\ell
ℓ
outside it. The angles between the lines
A
1
A
2
A_1A_2
A
1
A
2
and
A
2
A
3
,
A
1
A
2
A_2A_3, A_1A_2
A
2
A
3
,
A
1
A
2
and
A
2
A
3
,
A
2
A
3
A_2A_3, A_2A_3
A
2
A
3
,
A
2
A
3
and
A
3
A
1
A_3A_1
A
3
A
1
are equal to
a
3
,
a
1
a_3, a_1
a
3
,
a
1
and
a
2
a_2
a
2
, respectively. The straight lines are drawn through points
A
1
,
A
2
,
A
3
A_1, A_2, A_3
A
1
,
A
2
,
A
3
forming with
ℓ
\ell
ℓ
angles of
π
−
a
1
,
π
−
a
2
,
π
−
a
3
\pi -a_1, \pi -a_2, \pi -a_3
π
−
a
1
,
π
−
a
2
,
π
−
a
3
, respectively. All angles are counted in the same direction from
ℓ
\ell
ℓ
. Prove that these new lines meet at one point.
251
1
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MMO 251 Moscow MO 1953 2 groups of convex polygons from 16 concyclic points
On a circle, distinct points
A
1
,
.
.
.
,
A
16
A_1, ... , A_{16}
A
1
,
...
,
A
16
are chosen. Consider all possible convex polygons all of whose vertices are among
A
1
,
.
.
.
,
A
16
A_1, ... , A_{16}
A
1
,
...
,
A
16
. These polygons are divided into
2
2
2
groups, the first group comprising all polygons with
A
1
A_1
A
1
as a vertex, the second group comprising the remaining polygons. Which group is more numerous?
250
1
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MMO 250 Moscow MO 1953 1953 digits on a circle, no divisible by 27
Somebody wrote
1953
1953
1953
digits on a circle. The
1953
1953
1953
-digit number obtained by reading these figures clockwise, beginning at a certain point, is divisible by
27
27
27
. Prove that if one begins reading the figures at any other place, one gets another
1953
1953
1953
-digit number also divisible by
27
27
27
.
249
1
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MMO 249 Moscow MO 1953 quadrilateral area S <= (a + c)(b + d)/4}
Let
a
,
b
,
c
,
d
a, b, c, d
a
,
b
,
c
,
d
be the lengths of consecutive sides of a quadrilateral, and
S
S
S
its area. Prove that
S
≤
(
a
+
b
)
(
c
+
d
)
4
S \le \frac{ (a + b)(c + d)}{4}
S
≤
4
(
a
+
b
)
(
c
+
d
)
248
1
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MMO 248 Moscow MO 1953 5x5 linear system + generalization
a) Solve the system
{
x
1
+
2
x
2
+
2
x
3
+
2
x
4
+
2
x
5
=
1
x
1
+
3
x
2
+
4
x
3
+
4
x
4
+
4
x
5
=
2
x
1
+
3
x
2
+
5
x
3
+
6
x
4
+
6
x
5
=
3
x
1
+
3
x
2
+
5
x
3
+
7
x
4
+
8
x
5
=
4
x
1
+
3
x
2
+
5
x
3
+
7
x
4
+
9
x
5
=
5
\begin{cases} x_1 + 2x_2 + 2x_3 + 2x_4 + 2x_5 = 1 \\ x_1 + 3x_2 + 4x_3 + 4x_4 + 4x_5 = 2 \\ x_1 + 3x_2 + 5x_3 + 6x_4 + 6x_5 = 3 \\ x_1 + 3x_2 + 5x_3 + 7x_4 + 8x_5 = 4 \\ x_1 + 3x_2 + 5x_3 + 7x_4 + 9x_5 = 5 \end{cases}
⎩
⎨
⎧
x
1
+
2
x
2
+
2
x
3
+
2
x
4
+
2
x
5
=
1
x
1
+
3
x
2
+
4
x
3
+
4
x
4
+
4
x
5
=
2
x
1
+
3
x
2
+
5
x
3
+
6
x
4
+
6
x
5
=
3
x
1
+
3
x
2
+
5
x
3
+
7
x
4
+
8
x
5
=
4
x
1
+
3
x
2
+
5
x
3
+
7
x
4
+
9
x
5
=
5
b) Solve the system
{
x
1
+
2
x
2
+
2
x
3
+
2
x
4
+
2
x
5
+
.
.
.
+
2
x
100
=
1
x
1
+
3
x
2
+
4
x
3
+
4
x
4
+
4
x
5
+
.
.
.
+
4
x
100
=
2
x
1
+
3
x
2
+
5
x
3
+
6
x
4
+
6
x
5
+
.
.
.
+
6
x
100
=
3
x
1
+
3
x
2
+
5
x
3
+
7
x
4
+
8
x
5
+
.
.
.
+
8
x
100
=
4
.
.
.
x
1
+
3
x
2
+
5
x
3
+
7
x
4
+
9
x
5
+
.
.
.
+
199
x
100
=
100
\begin{cases} x_1 + 2x_2 + 2x_3 + 2x_4 + 2x_5 +...+ 2x_{100}= 1 \\ x_1 + 3x_2 + 4x_3 + 4x_4 + 4x_5 +...+ 4x_{100}= 2 \\ x_1 + 3x_2 + 5x_3 + 6x_4 + 6x_5 +...+ 6x_{100}= 3 \\ x_1 + 3x_2 + 5x_3 + 7x_4 + 8x_5 +...+ 8x_{100}= 4 \\ ... \\ x_1 + 3x_2 + 5x_3 + 7x_4 + 9x_5 +...+ 199x_{100}= 100 \end{cases}
⎩
⎨
⎧
x
1
+
2
x
2
+
2
x
3
+
2
x
4
+
2
x
5
+
...
+
2
x
100
=
1
x
1
+
3
x
2
+
4
x
3
+
4
x
4
+
4
x
5
+
...
+
4
x
100
=
2
x
1
+
3
x
2
+
5
x
3
+
6
x
4
+
6
x
5
+
...
+
6
x
100
=
3
x
1
+
3
x
2
+
5
x
3
+
7
x
4
+
8
x
5
+
...
+
8
x
100
=
4
...
x
1
+
3
x
2
+
5
x
3
+
7
x
4
+
9
x
5
+
...
+
199
x
100
=
100
247
1
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MMO 247 Moscow MO 1953 500 points in a convex 1000-gon, triangles
Inside a convex
1000
1000
1000
-gon,
500
500
500
points are selected so that no three of the
1500
1500
1500
points — the ones selected and the vertices of the polygon — lie on the same straight line. This
1000
1000
1000
-gon is then divided into triangles so that all
1500
1500
1500
points are vertices of the triangles, and so that these triangles have no other vertices. How many triangles will there be?
246
1
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MMO 246 Moscow MO 1953 11 gears on a plane
a) On a plane,
11
11
11
gears are arranged so that the teeth of the first gear mesh with the teeth of the second gear, the teeth of the second gear with those of the third gear, etc., and the teeth of the last gear mesh with those of the first gear. Can the gears rotate?b) On a plane,
n
n
n
gears are arranged so that the teeth of the first gear mesh with the teeth of the second gear, the teeth of the second gear with those of the third gear, etc., and the teeth of the last gear mesh with those of the first gear. Can the gears rotate?
245
1
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MMO 245 Moscow MO 1953 tangential quadrilateral is a rhombus
A quadrilateral is circumscribed around a circle. Its diagonals intersect at the center of the circle. Prove that the quadrilateral is a rhombus.
244
1
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MMO 244 Moscow MO 1953 gcd (a + b, lcm (a, b)) = gcd (a, b)
Prove that
g
c
d
(
a
+
b
,
l
c
m
(
a
,
b
)
)
=
g
c
d
(
a
,
b
)
gcd (a + b, lcm(a, b)) = gcd (a, b)
g
c
d
(
a
+
b
,
l
c
m
(
a
,
b
))
=
g
c
d
(
a
,
b
)
for any
a
,
b
a, b
a
,
b
.
243
1
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MMO 243 Moscow MO 1953 equal cones to a given right circular one
Given a right circular cone and a point
A
A
A
. Find the set of vertices of cones equal to the given one, with axes parallel to that of the given one, and with
A
A
A
inside them. We shall assume that the cone is infinite in one side.
242
1
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MMO 242 Moscow MO 1953 regular star-shaped pentagon, areas
Let
A
A
A
be a vertex of a regular star-shaped pentagon, the angle at
A
A
A
being less than
18
0
o
180^o
18
0
o
and the broken line
A
A
1
B
B
1
C
C
1
D
D
1
E
E
1
AA_1BB_1CC_1DD_1EE_1
A
A
1
B
B
1
C
C
1
D
D
1
E
E
1
being its contour. Lines
A
B
AB
A
B
and
D
E
DE
D
E
meet at
F
F
F
. Prove that polygon
A
B
B
1
C
C
1
D
E
D
1
ABB_1CC_1DED_1
A
B
B
1
C
C
1
D
E
D
1
has the same area as the quadrilateral
A
D
1
E
F
AD_1EF
A
D
1
EF
.Note: A regular star pentagon is a figure formed along the diagonals of a regular pentagon.
241
1
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MMO 241 Moscow MO 1953 polynomial x^{200}y^{200}+1 cannot be f(x) g(y)
Prove that the polynomial
x
200
y
200
+
1
x^{200} y^{200} +1
x
200
y
200
+
1
cannot be represented in the form
f
(
x
)
g
(
y
)
f(x)g(y)
f
(
x
)
g
(
y
)
, where
f
f
f
and
g
g
g
are polynomials of only
x
x
x
and
y
y
y
, respectively.
240
1
Hide problems
MMO 240 Moscow MO 1953 skew segments comparison
Let
A
B
AB
A
B
and
A
1
B
1
A_1B_1
A
1
B
1
be two skew segments,
O
O
O
and
O
1
O_1
O
1
their respective midpoints. Prove that
O
O
1
OO_1
O
O
1
is shorter than a half sum of
A
A
1
AA_1
A
A
1
and
B
B
1
BB_1
B
B
1
.
239
1
Hide problems
MMO 239 Moscow MO 1953 locus in plane sin(x + y) = 0
On the plane find the locus of points whose coordinates satisfy
s
i
n
(
x
+
y
)
=
0
sin(x + y) = 0
s
in
(
x
+
y
)
=
0
.
238
1
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MMO 238 Moscow MO 1953 inequality with 2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}
Prove that if in the following fraction we have
n
n
n
radicals in the numerator and
n
−
1
n - 1
n
−
1
in the denominator, then
2
−
2
+
2
+
.
.
.
+
2
2
−
2
+
2
+
.
.
.
+
2
>
1
4
\frac{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}{2-\sqrt{2+\sqrt{2+...+\sqrt{2}}}}>\frac14
2
−
2
+
2
+
...
+
2
2
−
2
+
2
+
...
+
2
>
4
1
237
1
Hide problems
MMO 237 Moscow MO 1953 3circles are pair-wise tangent to each other
Three circles are pair-wise tangent to each other. Prove that the circle passing through the three tangent points is perpendicular to each of the initial three circles.
236
1
Hide problems
MMO 236 Moscow MO 1953 (n^2 + 8n + 15) not divisible by (n + 4)
Prove that
n
2
+
8
n
+
15
n^2 + 8n + 15
n
2
+
8
n
+
15
is not divisible by
n
+
4
n + 4
n
+
4
for any positive integer
n
n
n
.
235
1
Hide problems
MMO 235 Moscow MO 1953 Divide segment in halves using right triangle
Divide a segment in halves using a right triangle. (With a right triangle one can draw straight lines and erect perpendiculars but cannot draw perpendiculars.)
234
1
Hide problems
MMO 234 Moscow MO 1953 min no in form 1...1 divisible by 3...3
Find the smallest number of the form
1...1
1...1
1...1
in its decimal expression which is divisible by
3...3
⏟
100
\underbrace{\hbox{3...3}}_{\hbox{100}}
100
3...3
,.
233
1
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MMO 233 Moscow MO 1953 compare sum of angles in trapezoid
Prove that the sum of angles at the longer base of a trapezoid is less than the sum of angles at the shorter base.