MathDB
Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
1989 Poland - Second Round
1989 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(6)
6
1
Hide problems
two tangential quads leadto a third tangential quad
In the triangle
A
B
C
ABC
A
BC
, the lines
C
P
CP
CP
,
A
P
AP
A
P
,
B
P
BP
BP
are drawn through the internal point
P
P
P
and intersect the sides
A
B
AB
A
B
,
B
C
BC
BC
,
C
A
CA
C
A
at points
K
K
K
,
L
L
L
,
M
M
M
, respectively. Prove that if circles can be inscribed in the quadrilaterals
A
K
P
M
AKPM
A
K
PM
and
K
B
L
P
KBLP
K
B
L
P
, then a circle can also be inscribed in the quadrilateral
L
C
M
P
LCMP
L
CMP
.
3
1
Hide problems
min vol of a tetrahedron
Given is a trihedral angle
O
A
B
C
OABC
O
A
BC
with a vertex
O
O
O
and a point
P
P
P
in its interior. Let
V
V
V
be the volume of a parallelepiped with two vertices at points
O
O
O
and
P
P
P
, whose three edges are contained in the rays
O
A
→
\overrightarrow{OA}
O
A
,
O
B
→
\overrightarrow{OB}
OB
,
O
C
→
\overrightarrow{OC}
OC
. Calculate the minimum volume of a tetrahedron whose three faces are contained in the faces of the trihedral angle
O
A
B
C
OABC
O
A
BC
and the fourth face contains the point
P
P
P
.
5
1
Hide problems
c_{n+1} = [ 3/2 c_n ]
Given a sequence
(
c
n
)
(c_n)
(
c
n
)
of natural numbers defined recursively:
c
1
=
2
c_1 = 2
c
1
=
2
,
c
n
+
1
=
[
3
2
c
n
]
c_{n+1} = \left[ \frac{3}{2}c_n\right]
c
n
+
1
=
[
2
3
c
n
]
. Prove that there are infinitely many even numbers and infinitely many odd numbers among the terms of this sequence.
4
1
Hide problems
x_1a_1 + ...x_{11}a_{ 11} is divisible by 1989.
The given integers are
a
1
,
a
2
,
…
,
a
11
a_1, a_2, \ldots , a_{11}
a
1
,
a
2
,
…
,
a
11
. Prove that there exists a non-zero sequence
x
1
,
x
2
,
…
,
x
11
x_1, x_2, \ldots, x_{11}
x
1
,
x
2
,
…
,
x
11
with terms from the set
{
−
1
,
0
,
1
}
\{-1,0,1\}
{
−
1
,
0
,
1
}
such that the number
x
1
a
1
+
…
x
11
a
11
x_1a_1 + \ldots x_{11}a_{ 11}
x
1
a
1
+
…
x
11
a
11
is divisible by 1989.
2
1
Hide problems
expected value of random variable X is \sum_{k=1}^n \frac{1}{k!}
For a randomly selected permutation
f
=
(
f
1
,
.
.
.
,
f
n
)
\mathbf{f} = (f_1,..., f_n)
f
=
(
f
1
,
...
,
f
n
)
of the set
{
1
,
…
,
n
}
\{1,\ldots, n\}
{
1
,
…
,
n
}
let us denote by
X
(
f
)
X(\mathbf{f})
X
(
f
)
the largest number
k
≤
n
k \leq n
k
≤
n
such that
f
i
<
f
i
+
1
f_i < f_{ i+1}
f
i
<
f
i
+
1
for all numbers
i
<
k
i < k
i
<
k
. Prove that the expected value of the random variable
X
X
X
is
∑
k
=
1
n
1
k
!
\sum_{k=1}^n \frac{1}{k!}
∑
k
=
1
n
k
!
1
.
1
1
Hide problems
tg 7x - sin 6x=cos 4x - ctg 7x
Solve the equation
t
g
7
x
−
sin
6
x
=
cos
4
x
−
c
t
g
7
x
.
tg 7x - \sin 6x=\cos 4x - ctg 7x.
t
g
7
x
−
sin
6
x
=
cos
4
x
−
c
t
g
7
x
.