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Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
1964 Poland - Second Round
1964 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(6)
6
1
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4 from any 5 points , are not vertices of an acute triangle.
Prove that from any five points in the plane it is possible to choose three points that are not vertices of an acute triangle.
5
1
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section of triangle with any plane is right triangle.
Given is a trihedral angle with edges
S
A
SA
S
A
,
S
B
SB
SB
,
S
C
SC
SC
, all plane angles of which are acute, and the dihedral angle at edge
S
A
SA
S
A
is right. Prove that the section of this triangle with any plane perpendicular to any edge, at a point different from the vertex
S
S
S
, is a right triangle.
4
1
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(z - x)(x - y) = a, (x - y)(y - z) = b, (y - z)(z - x) = c
Find the real numbers
x
,
y
,
z
x, y, z
x
,
y
,
z
satisfying the system of equations
(
z
−
x
)
(
x
−
y
)
=
a
(z - x)(x - y) = a
(
z
−
x
)
(
x
−
y
)
=
a
(
x
−
y
)
(
y
−
z
)
=
b
(x - y)(y - z) = b
(
x
−
y
)
(
y
−
z
)
=
b
(
y
−
z
)
(
z
−
x
)
=
c
(y - z)(z - x) = c
(
y
−
z
)
(
z
−
x
)
=
c
where
a
,
b
,
c
a, b, c
a
,
b
,
c
are given real numbers.
3
1
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3 primes form an arithmetic progression with ifference inot divisible by 6
Prove that if three prime numbers form an arithmetic progression whose difference is not divisible by 6, then the smallest of these numbers is
3
3
3
.
2
1
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perpenidicularity, joining arc midpoints
The circle is divided into four non-overlapping gaps
A
B
AB
A
B
,
B
C
BC
BC
,
C
D
CD
C
D
and
D
A
DA
D
A
. Prove that the segment joining the midpoints of the arcs
A
B
AB
A
B
and
C
D
CD
C
D
is perpendicular to the segment joining the midpoints of the arcs
B
C
BC
BC
and
D
A
DA
D
A
.
1
1
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sum 1/sin 2na = ...
Prove that if
n
n
n
is a natural number and the angle
α
\alpha
α
is not a multiple of
18
0
∘
2
n
\frac{180^{\circ}}{2^n}
2
n
18
0
∘
, then
1
sin
2
α
+
1
sin
4
α
+
1
sin
8
α
+
.
.
.
+
=
c
t
g
α
−
c
t
g
2
n
α
.
\frac{1}{\sin 2\alpha} + \frac{1}{\sin 4\alpha} + \frac{1}{\sin 8\alpha} + ... + = ctg \alpha - ctg 2^n \alpha.
sin
2
α
1
+
sin
4
α
1
+
sin
8
α
1
+
...
+
=
c
t
gα
−
c
t
g
2
n
α
.