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Bulgaria Contests
Bulgaria National Olympiad
1996 Bulgaria National Olympiad
1996 Bulgaria National Olympiad
Part of
Bulgaria National Olympiad
Subcontests
(3)
3
2
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Bulgaria National Olympiad 1996
The quadratic polynomials
f
f
f
and
g
g
g
with real coefficients are such that if
g
(
x
)
g(x)
g
(
x
)
is an integer for some
x
>
0
x>0
x
>
0
, then so is
f
(
x
)
f(x)
f
(
x
)
. Prove that there exist integers
m
,
n
m,n
m
,
n
such that
f
(
x
)
=
m
g
(
x
)
+
n
f(x)=mg(x)+n
f
(
x
)
=
m
g
(
x
)
+
n
for all
x
x
x
.
Bulgaria National Olympiad 1996
A square table of size
7
×
7
7\times 7
7
×
7
with the four corner squares deleted is given.[*] What is the smallest number of squares which need to be colored black so that a
5
−
5-
5
−
square entirely uncolored Greek cross (Figure 1) cannot be found on the table? [*] Prove that it is possible to write integers in each square in a way that the sum of the integers in each Greek cross is negative while the sum of all integers in the square table is positive.[asy] size(3.5cm); usepackage("amsmath"); MP("\text{Figure }1.", (1.5, 3.5), N); DPA(box((0,1),(3,2))^^box((1,0),(2,3)), black); [/asy]
2
2
Hide problems
Bulgaria National Olympiad 1996
Find the side length of the smallest equilateral triangle in which three discs of radii
2
,
3
,
4
2,3,4
2
,
3
,
4
can be placed without overlap.
Bulgaria National Olympiad
The quadrilateral
A
B
C
D
ABCD
A
BC
D
is inscribed in a circle. The lines
A
B
AB
A
B
and
C
D
CD
C
D
meet each other in the point
E
E
E
, while the diagonals
A
C
AC
A
C
and
B
D
BD
B
D
in the point
F
F
F
. The circumcircles of the triangles
A
F
D
AFD
A
F
D
and
B
F
C
BFC
BFC
have a second common point, which is denoted by
H
H
H
. Prove that
∠
E
H
F
=
9
0
∘
\angle EHF=90^\circ
∠
E
H
F
=
9
0
∘
.
1
2
Hide problems
Bulgaria National Olympiad 1996
Find all prime numbers
p
,
q
p,q
p
,
q
for which
p
q
pq
pq
divides
(
5
p
−
2
p
)
(
5
q
−
2
q
)
(5^p-2^p)(5^q-2^q)
(
5
p
−
2
p
)
(
5
q
−
2
q
)
.
Sequence $\{a_n\}$
Sequence
{
a
n
}
\{a_n\}
{
a
n
}
it define
a
1
=
1
a_1=1
a
1
=
1
and
a
n
+
1
=
a
n
n
+
n
a
n
a_{n+1}=\frac{a_n}{n}+\frac{n}{a_n}
a
n
+
1
=
n
a
n
+
a
n
n
for all
n
≥
1
n\ge 1
n
≥
1
\\ Prove that
⌊
a
n
2
⌋
=
n
\lfloor a_n^2\rfloor=n
⌊
a
n
2
⌋
=
n
for all
n
≥
4.
n\ge 4.
n
≥
4.