MathDB
Problems
Contests
National and Regional Contests
Bulgaria Contests
Bulgaria National Olympiad
1992 Bulgaria National Olympiad
1992 Bulgaria National Olympiad
Part of
Bulgaria National Olympiad
Subcontests
(6)
Problem 6
1
Hide problems
balls in boxes, number of balls based on box number
There are given one black box and
n
n
n
white boxes (
n
n
n
is a random natural number). White boxes are numbered with the numbers
1
,
2
,
…
,
n
1,2,\ldots,n
1
,
2
,
…
,
n
. In them are put
n
n
n
balls. It is allowed the following rearrangement of the balls: if in the box with number
k
k
k
there are exactly
k
k
k
balls, that box is made empty - one of the balls is put in the black box and the other
k
−
1
k-1
k
−
1
balls are put in the boxes with numbers:
1
,
2
,
…
,
k
−
1
1,2,\ldots,k-1
1
,
2
,
…
,
k
−
1
. (Ivan Tonov)
Problem 5
1
Hide problems
angle bisectors in triangle, segments equal
Points
D
,
E
,
F
D,E,F
D
,
E
,
F
are midpoints of the sides
A
B
,
B
C
,
C
A
AB,BC,CA
A
B
,
BC
,
C
A
of triangle
A
B
C
ABC
A
BC
. Angle bisectors of the angles
B
D
C
BDC
B
D
C
and
A
D
C
ADC
A
D
C
intersect the lines
B
C
BC
BC
and
A
C
AC
A
C
respectively at the points
M
M
M
and
N
N
N
, and the line
M
N
MN
MN
intersects the line
C
D
CD
C
D
at the point
O
O
O
. Let the lines
E
O
EO
EO
and
F
O
FO
FO
intersect respectively the lines
A
C
AC
A
C
and
B
C
BC
BC
at the points
P
P
P
and
Q
Q
Q
. Prove that
C
D
=
P
Q
CD=PQ
C
D
=
PQ
. (Plamen Koshlukov)
Problem 4
1
Hide problems
x^(2p)+y^(2p)+z^(2p)=t^(2p) implies p|xyzt
Let
p
p
p
be a prime number in the form
p
=
4
k
+
3
p=4k+3
p
=
4
k
+
3
. Prove that if the numbers
x
0
,
y
0
,
z
0
,
t
0
x_0,y_0,z_0,t_0
x
0
,
y
0
,
z
0
,
t
0
are solutions of the equation
x
2
p
+
y
2
p
+
z
2
p
=
t
2
p
x^{2p}+y^{2p}+z^{2p}=t^{2p}
x
2
p
+
y
2
p
+
z
2
p
=
t
2
p
, then at least one of them is divisible by
p
p
p
. (Plamen Koshlukov)
Problem 3
1
Hide problems
counting lattice point sequences
Let
m
m
m
and
n
n
n
are fixed natural numbers and
O
x
y
Oxy
O
x
y
is a coordinate system in the plane. Find the total count of all possible situations of
n
+
m
−
1
n+m-1
n
+
m
−
1
points
P
1
(
x
1
,
y
1
)
,
P
2
(
x
2
,
y
2
)
,
…
,
P
n
+
m
−
1
(
x
n
+
m
−
1
,
y
n
+
m
−
1
)
P_1(x_1,y_1),P_2(x_2,y_2),\ldots,P_{n+m-1}(x_{n+m-1},y_{n+m-1})
P
1
(
x
1
,
y
1
)
,
P
2
(
x
2
,
y
2
)
,
…
,
P
n
+
m
−
1
(
x
n
+
m
−
1
,
y
n
+
m
−
1
)
in the plane for which the following conditions are satisfied:(i) The numbers
x
i
x_i
x
i
and
y
i
(
i
=
1
,
2
,
…
,
n
+
m
−
1
)
y_i~(i=1,2,\ldots,n+m-1)
y
i
(
i
=
1
,
2
,
…
,
n
+
m
−
1
)
are integers and
1
≤
x
i
≤
n
,
1
≤
y
i
≤
m
1\le x_i\le n,1\le y_i\le m
1
≤
x
i
≤
n
,
1
≤
y
i
≤
m
. (ii) Every one of the numbers
1
,
2
,
…
,
n
1,2,\ldots,n
1
,
2
,
…
,
n
can be found in the sequence
x
1
,
x
2
,
…
,
x
n
+
m
−
1
x_1,x_2,\ldots,x_{n+m-1}
x
1
,
x
2
,
…
,
x
n
+
m
−
1
and every one of the numbers
1
,
2
,
…
,
m
1,2,\ldots,m
1
,
2
,
…
,
m
can be found in the sequence
y
1
,
y
2
,
…
,
y
n
+
m
−
1
y_1,y_2,\ldots,y_{n+m-1}
y
1
,
y
2
,
…
,
y
n
+
m
−
1
. (iii) For every
i
=
1
,
2
,
…
,
n
+
m
−
2
i=1,2,\ldots,n+m-2
i
=
1
,
2
,
…
,
n
+
m
−
2
the line
P
i
P
i
+
1
P_iP_{i+1}
P
i
P
i
+
1
is parallel to one of the coordinate axes. (Ivan Gochev, Hristo Minchev)
Problem 2
1
Hide problems
no arith. seq. of 41 terms in 1904-subset of [1992]
Prove that there exists
1904
1904
1904
-element subset of the set
{
1
,
2
,
…
,
1992
}
\{1,2,\ldots,1992\}
{
1
,
2
,
…
,
1992
}
, which doesn’t contain an arithmetic progression consisting of
41
41
41
terms. (Ivan Tonov)
Problem 1
1
Hide problems
tetrahedron, barycenter-related, angle doesn't depend on point
Through a random point
C
1
C_1
C
1
from the edge
D
C
DC
D
C
of the regular tetrahedron
A
B
C
D
ABCD
A
BC
D
is drawn a plane, parallel to the plane
A
B
C
ABC
A
BC
. The plane constructed intersects the edges
D
A
DA
D
A
and
D
B
DB
D
B
at the points
A
1
,
B
1
A_1,B_1
A
1
,
B
1
respectively. Let the point
H
H
H
is the midpoint of the altitude through the vertex
D
D
D
of the tetrahedron
D
A
1
B
1
C
1
DA_1B_1C_1
D
A
1
B
1
C
1
and
M
M
M
is the center of gravity (barycenter) of the triangle
A
B
C
1
ABC_1
A
B
C
1
. Prove that the measure of the angle
H
M
C
HMC
H
MC
doesn’t depend on the position of the point
C
1
C_1
C
1
. (Ivan Tonov)