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Problems
Contests
National and Regional Contests
Bulgaria Contests
Bulgaria National Olympiad
1966 Bulgaria National Olympiad
1966 Bulgaria National Olympiad
Part of
Bulgaria National Olympiad
Subcontests
(4)
Problem 4
1
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tetrahedron, three edges form triangle through vertex
It is given a tetrahedron with vertices
A
,
B
,
C
,
D
A,B,C,D
A
,
B
,
C
,
D
.(a) Prove that there exists a vertex of the tetrahedron with the following property: the three edges of that tetrahedron through that vertex can form a triangle. (b) On the edges
D
A
,
D
B
DA,DB
D
A
,
D
B
and
D
C
DC
D
C
there are given the points
M
,
N
M,N
M
,
N
and
P
P
P
for which:
D
M
=
D
A
n
,
D
N
=
D
B
n
+
1
D
P
=
D
C
n
+
2
DM=\frac{DA}n,\enspace DN=\frac{DB}{n+1}\enspace DP=\frac{DC}{n+2}
D
M
=
n
D
A
,
D
N
=
n
+
1
D
B
D
P
=
n
+
2
D
C
where
n
n
n
is a natural number. The plane defined by the points
M
,
N
M,N
M
,
N
and
P
P
P
is
α
n
\alpha_n
α
n
. Prove that all planes
α
n
\alpha_n
α
n
,
(
n
=
1
,
2
,
3
,
…
)
(n=1,2,3,\ldots)
(
n
=
1
,
2
,
3
,
…
)
pass through a single straight line.
Problem 3
1
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triangle given H, M(AB), midpoint of foot
(a) In the plane of the triangle
A
B
C
ABC
A
BC
, find a point with the following property: its symmetrical points with respect to the midpoints of the sides of the triangle lie on the circumscribed circle. (b) Construct the triangle
A
B
C
ABC
A
BC
if it is known the positions of the orthocenter
H
H
H
, midpoint of the side
A
B
AB
A
B
and the midpoint of the segment joining the feet of the heights through vertices
A
A
A
and
B
B
B
.
Problem 2
1
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inequality, four variables in R+ (Bulgaria 1966 P2)
Prove that for every four positive numbers
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
the following inequality is true:
a
2
+
b
2
+
c
2
+
d
2
4
≥
a
b
c
+
a
b
d
+
a
c
d
+
b
c
d
4
3
.
\sqrt{\frac{a^2+b^2+c^2+d^2}4}\ge\sqrt[3]{\frac{abc+abd+acd+bcd}4}.
4
a
2
+
b
2
+
c
2
+
d
2
≥
3
4
ab
c
+
ab
d
+
a
c
d
+
b
c
d
.
Problem 1
1
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3x(x-3y)=y^2+z^2 over Z (Bulgaria 1966 P1)
Prove that the equation
3
x
(
x
−
3
y
)
=
y
2
+
z
2
3x(x-3y)=y^2+z^2
3
x
(
x
−
3
y
)
=
y
2
+
z
2
doesn't have any integer solutions except
x
=
0
,
y
=
0
,
z
=
0
x=0,y=0,z=0
x
=
0
,
y
=
0
,
z
=
0
.