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Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
1998 Poland - Second Round
1998 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(6)
1
1
Hide problems
f(f(k)) = g(g(k)) = k , g(f(k)) = k +1, prove or disprove
Let
A
n
=
{
1
,
2
,
.
.
.
,
n
}
A_n = \{1,2,...,n\}
A
n
=
{
1
,
2
,
...
,
n
}
. Prove or disprove: For all integers
n
≥
2
n \ge 2
n
≥
2
there exist functions
f
,
g
:
A
n
→
A
n
f,g : A_n \to A_n
f
,
g
:
A
n
→
A
n
which satisfy
f
(
f
(
k
)
)
=
g
(
g
(
k
)
)
=
k
f(f(k)) = g(g(k)) = k
f
(
f
(
k
))
=
g
(
g
(
k
))
=
k
for
1
≤
k
≤
n
1 \le k \le n
1
≤
k
≤
n
, and
g
(
f
(
k
)
)
=
k
+
1
g(f(k)) = k +1
g
(
f
(
k
))
=
k
+
1
for
1
≤
k
≤
n
−
1
1 \le k \le n -1
1
≤
k
≤
n
−
1
.
6
1
Hide problems
criterion for perpendicularity of edges AB and CD of tetrahedron ABCD
Prove that the edges
A
B
AB
A
B
and
C
D
CD
C
D
of a tetrahedron
A
B
C
D
ABCD
A
BC
D
are perpendicular if and only if there exists a parallelogram
C
D
P
Q
CDPQ
C
D
PQ
such that
P
A
=
P
B
=
P
D
PA = PB = PD
P
A
=
PB
=
P
D
and
Q
A
=
Q
B
=
Q
C
QA = QB = QC
Q
A
=
QB
=
QC
.
4
1
Hide problems
x^2 +3y^2 = 1998x , diophantine
Find all pairs of integers
(
x
,
y
)
(x,y)
(
x
,
y
)
satisfying
x
2
+
3
y
2
=
1998
x
x^2 +3y^2 = 1998x
x
2
+
3
y
2
=
1998
x
.
2
1
Hide problems
triangle geometry
In triangle
A
B
C
ABC
A
BC
, the angle
∠
B
C
A
\angle BCA
∠
BC
A
is obtuse and
∠
B
A
C
=
2
∠
A
B
C
.
\angle BAC = 2\angle ABC\,.
∠
B
A
C
=
2∠
A
BC
.
The line through
B
B
B
and perpendicular to
B
C
BC
BC
intersects line
A
C
AC
A
C
in
D
D
D
. Let
M
M
M
be the midpoint of
A
B
AB
A
B
. Prove that
∠
A
M
C
=
∠
B
M
D
\angle AMC=\angle BMD
∠
A
MC
=
∠
BM
D
.source : http://cage.ugent.be/~hvernaev/Olympiade/PMO982.pdf
5
1
Hide problems
Poland 1998
Let
a
1
,
a
2
,
…
,
a
7
,
b
1
,
b
2
,
…
,
b
7
≥
0
a_1,a_2,\ldots,a_7, b_1,b_2,\ldots,b_7\geq 0
a
1
,
a
2
,
…
,
a
7
,
b
1
,
b
2
,
…
,
b
7
≥
0
be real numbers satisfying
a
i
+
b
i
≤
2
a_i+b_i\le 2
a
i
+
b
i
≤
2
for all
i
=
1
,
7
‾
i=\overline{1,7}
i
=
1
,
7
. Prove that there exist
k
≠
m
k\ne m
k
=
m
such that
∣
a
k
−
a
m
∣
+
∣
b
k
−
b
m
∣
≤
1
|a_k-a_m|+|b_k-b_m|\le 1
∣
a
k
−
a
m
∣
+
∣
b
k
−
b
m
∣
≤
1
.Thanks for show me the mistake typing
3
1
Hide problems
Poland again
If
a
a
a
,
b
b
b
,
c
c
c
,
d
d
d
,
e
e
e
,
f
f
f
are nonnegative real numbers satisfying a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f \equal{} 1 and ace \plus{} bdf \geq \frac {1}{108}, then prove that abc \plus{} bcd \plus{} cde \plus{} de f \plus{} efa \plus{} fab \leq \frac {1}{36}