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Problems
Contests
National and Regional Contests
Russia Contests
Moscow Mathematical Olympiad
1950 Moscow Mathematical Olympiad
1950 Moscow Mathematical Olympiad
Part of
Moscow Mathematical Olympiad
Subcontests
(15)
187
1
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MMO 187 Moscow MO 1950 draw 10 bus routes with stops
Is it possible to draw
10
10
10
bus routes with stops such that for any
8
8
8
routes there is a stop that does not belong to any of the routes, but any
9
9
9
routes pass through all the stops?
186
1
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MMO 186 Moscow MO 1950 all the tangent points lie in 1 plane
A spatial quadrilateral is circumscribed around a sphere. Prove that all the tangent points lie in one plane.
185
1
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MMO 185 Moscow MO 1950 erase 90 from 1,2,3,...,101 increasing / decreasing
The numbers
1
,
2
,
3
,
.
.
.
,
101
1, 2, 3, . . . , 101
1
,
2
,
3
,
...
,
101
are written in a row in some order. Prove that it is always possible to erase
90
90
90
of the numbers so that the remaining
11
11
11
numbers remain arranged in either increasing or decreasing order.
184
1
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MMO 184 Moscow MO 1950 20 points on a circles
* On a circle,
20
20
20
points are chosen. Ten non-intersecting chords without mutual endpoints connect some of the points chosen. How many distinct such arrangements are there?
183
1
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MMO 183 Moscow MO 1950 < 1/2 square’s perimeter lies outside the triangle.
A circle is inscribed in a triangle and a square is circumscribed around this circle so that no side of the square is parallel to any side of the triangle. Prove that less than half of the square’s perimeter lies outside the triangle.
182
1
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MMO 182 Moscow MO 1950 product 1/2 3/4 5/6 7/8 ... 99/100 <1/10
Prove that
1
2
3
4
5
6
7
8
.
.
.
99
100
<
1
10
\frac{1}{2} \frac{3}{4} \frac{5}{6} \frac{7}{8} ... \frac{99}{100 } <\frac{1}{10}
2
1
4
3
6
5
8
7
...
100
99
<
10
1
.
181
1
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MMO 181 Moscow MO 1950 polygons by all diagonals in 1950-gon
a) In a convex
13
13
13
-gon all diagonals are drawn, dividing it into smaller polygons. What is the greatest number of sides can these polygons have?b) In a convex
1950
1950
1950
-gon all diagonals are drawn, dividing it into smaller polygons. What is the greatest number of sides can these polygons have?
180
1
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MMO 180 Moscow MO 1950\sqrt{x+3−4\sqrt{x−1}}+\sqrt{x+8−6\sqrt{x−1}}=1
Solve the equation
x
+
3
−
4
x
−
1
+
x
+
8
−
6
x
−
1
=
1
\sqrt {x + 3 - 4 \sqrt{x -1}} +\sqrt{x + 8 - 6 \sqrt{x - 1}}= 1
x
+
3
−
4
x
−
1
+
x
+
8
−
6
x
−
1
=
1
.
179
1
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MMO 179 Moscow MO 1950 1 triangular pyramid inside another 1, common base
Two triangular pyramids have common base. One pyramid contains the other. Can the sum of the lengths of the edges of the inner pyramid be longer than that of the outer one?
178
1
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MMO 178 Moscow MO 1950 sinx =(sinBsinC)/(1 - cosB cosCcosA)
Let
A
A
A
be an arbitrary angle,let
B
B
B
and
C
C
C
be acute angles. Is there an angle
x
x
x
such that
sin
x
=
sin
B
⋅
sin
C
1
−
cos
B
⋅
cos
C
⋅
cos
A
?
\sin x =\frac{\sin B \cdot \sin C}{1 - \cos B \cdot \cos C \cdot \cos A} ?
sin
x
=
1
−
cos
B
⋅
cos
C
⋅
cos
A
sin
B
⋅
sin
C
?
177
1
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MMO 177 Moscow MO 1950 6 kms from point A, walking or using a cab
In a country, one can get from some point
A
A
A
to any other point either by walking, or by calling a cab, waiting for it, and then being driven. Every citizen always chooses the method of transportation that requires the least time. It turns out that the distances and the traveling times are as follows:
1
1
1
km takes
10
10
10
min,
2
2
2
km takes
15
15
15
min,
3
3
3
km takes
17.5
17.5
17.5
min. We assume that the speeds of the pedestrian and the cab, and the time spent waiting for cabs, are all constants. How long does it take to reach a point which is
6
6
6
km from
A
A
A
?
176
1
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MMO 176 Moscow MO 1950 Aa+Bb+Cc>=(Ab+Ac+Ba+Bc+Ca+Cb)/2
Let
a
,
b
,
c
a, b, c
a
,
b
,
c
be the lengths of the sides of a triangle and
A
,
B
,
C
A, B, C
A
,
B
,
C
, the opposite angles. Prove that
A
a
+
B
b
+
C
c
≥
A
b
+
A
c
+
B
a
+
B
c
+
C
a
+
C
b
2
Aa + Bb + Cc \ge \frac{Ab + Ac + Ba + Bc + Ca + Cb}{2}
A
a
+
B
b
+
C
c
≥
2
A
b
+
A
c
+
B
a
+
B
c
+
C
a
+
C
b
175
1
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MMO 175 Moscow MO 1950 n circles concurrent given, points coincide wanted
a) We are given
n
n
n
circles
O
1
,
O
2
,
.
.
.
,
O
n
O_1, O_2, . . . , O_n
O
1
,
O
2
,
...
,
O
n
, passing through one point
O
O
O
. Let
A
1
,
.
.
.
,
A
n
A_1, . . . , A_n
A
1
,
...
,
A
n
denote the second intersection points of
O
1
O_1
O
1
with
O
2
,
O
2
O_2, O_2
O
2
,
O
2
with
O
3
O_3
O
3
, etc.,
O
n
O_n
O
n
with
O
1
O_1
O
1
, respectively. We choose an arbitrary point
B
1
B_1
B
1
on
O
1
O_1
O
1
and draw a line segment through
A
1
A_1
A
1
and
B
1
B_1
B
1
to the second intersection with
O
2
O_2
O
2
at
B
2
B_2
B
2
, then draw a line segment through
A
2
A_2
A
2
and
B
2
B_2
B
2
to the second intersection with
O
3
O_3
O
3
at
B
3
B_3
B
3
, etc., until we get a point
B
n
B_n
B
n
on
O
n
O_n
O
n
. We draw the line segment through
B
n
B_n
B
n
and
A
n
A_n
A
n
to the second intersection with
O
1
O_1
O
1
at
B
n
+
1
B_{n+1}
B
n
+
1
. If
B
k
B_k
B
k
and
A
k
A_k
A
k
coincide for some
k
k
k
, we draw the tangent to
O
k
O_k
O
k
through
A
k
A_k
A
k
until this tangent intersects
O
k
+
1
O_{k+1}
O
k
+
1
at
B
k
+
1
B_{k+1}
B
k
+
1
. Prove that
B
n
+
1
B_{n+1}
B
n
+
1
coincides with
B
1
B_1
B
1
.b) for
n
=
3
n=3
n
=
3
the same problem.
174
1
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MMO 174 Moscow MO 1950 3 piles from 81 weights of 1^2, 2^2, . . . , 81^2
a) Given
555
555
555
weights: of
1
1
1
g,
2
2
2
g,
3
3
3
g, . . . ,
555
555
555
g, divide them into three piles of equal mass.b) Arrange
81
81
81
weights of
1
2
,
2
2
,
.
.
.
,
8
1
2
1^2, 2^2, . . . , 81^2
1
2
,
2
2
,
...
,
8
1
2
(all in grams) into three piles of equal mass.
173
1
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MMO 173 Moscow MO 1950 largest circle in chessboard
On a chess board, the boundaries of the squares are assumed to be black. Draw a circle of the greatest possible radius lying entirely on the black squares.