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Problems
Contests
National and Regional Contests
Poland Contests
Polish MO Finals
1998 Polish MO Finals
1998 Polish MO Finals
Part of
Polish MO Finals
Subcontests
(3)
3
2
Hide problems
Pyramid and convex pentagon
P
A
B
C
D
E
PABCDE
P
A
BC
D
E
is a pyramid with
A
B
C
D
E
ABCDE
A
BC
D
E
a convex pentagon. A plane meets the edges
P
A
,
P
B
,
P
C
,
P
D
,
P
E
PA, PB, PC, PD, PE
P
A
,
PB
,
PC
,
P
D
,
PE
in points
A
′
,
B
′
,
C
′
,
D
′
,
E
′
A', B', C', D', E'
A
′
,
B
′
,
C
′
,
D
′
,
E
′
distinct from
A
,
B
,
C
,
D
,
E
A, B, C, D, E
A
,
B
,
C
,
D
,
E
and
P
P
P
. For each of the quadrilaterals
A
B
B
′
A
′
,
B
C
C
′
B
,
C
D
D
′
C
′
,
D
E
E
′
D
′
,
E
A
A
′
E
′
ABB'A', BCC'B, CDD'C', DEE'D', EAA'E'
A
B
B
′
A
′
,
BC
C
′
B
,
C
D
D
′
C
′
,
D
E
E
′
D
′
,
E
A
A
′
E
′
take the intersection of the diagonals. Show that the five intersections are coplanar.
board containing all unit squares
S
S
S
is a board containing all unit squares in the
x
y
xy
x
y
plane whose vertices have integer coordinates and which lie entirely inside the circle
x
2
+
y
2
=
199
8
2
x^2 + y^2 = 1998^2
x
2
+
y
2
=
199
8
2
. In each square of
S
S
S
is written
+
1
+1
+
1
. An allowed move is to change the sign of every square in
S
S
S
in a given row, column or diagonal. Can we end up with exactly one
−
1
-1
−
1
and
+
1
+1
+
1
on the rest squares by a sequence of allowed moves?
2
2
Hide problems
Fibonacci and another sequence
F
n
F_n
F
n
is the Fibonacci sequence
F
0
=
F
1
=
1
F_0 = F_1 = 1
F
0
=
F
1
=
1
,
F
n
+
2
=
F
n
+
1
+
F
n
F_{n+2} = F_{n+1} + F_n
F
n
+
2
=
F
n
+
1
+
F
n
. Find all pairs
m
>
k
≥
0
m > k \geq 0
m
>
k
≥
0
such that the sequence
x
0
,
x
1
,
x
2
,
.
.
.
x_0, x_1, x_2, ...
x
0
,
x
1
,
x
2
,
...
defined by
x
0
=
F
k
F
m
x_0 = \frac{F_k}{F_m}
x
0
=
F
m
F
k
and
x
n
+
1
=
2
x
n
−
1
1
−
x
n
x_{n+1} = \frac{2x_n - 1}{1 - x_n}
x
n
+
1
=
1
−
x
n
2
x
n
−
1
for
x
n
≠
1
x_n \not = 1
x
n
=
1
, or
1
1
1
if
x
n
=
1
x_n = 1
x
n
=
1
, contains the number
1
1
1
Triangle and basic properties
The points
D
,
E
D, E
D
,
E
on the side
A
B
AB
A
B
of the triangle
A
B
C
ABC
A
BC
are such that
A
D
D
B
A
E
E
B
=
(
A
C
C
B
)
2
\frac{AD}{DB}\frac{AE}{EB} = \left(\frac{AC}{CB}\right)^2
D
B
A
D
EB
A
E
=
(
CB
A
C
)
2
. Show that
∠
A
C
D
=
∠
B
C
E
\angle ACD = \angle BCE
∠
A
C
D
=
∠
BCE
.
1
2
Hide problems
equations in a,b,c,x,y,z
Find all solutions in positive integers to: \begin{eqnarray*} a + b + c = xyz \\ x + y + z = abc \end{eqnarray*}
infinitely many muplitples of 7 in a sequence
Define the sequence
a
1
,
a
2
,
a
3
,
.
.
.
a_1, a_2, a_3, ...
a
1
,
a
2
,
a
3
,
...
by
a
1
=
1
a_1 = 1
a
1
=
1
,
a
n
=
a
n
−
1
+
a
[
n
/
2
]
a_n = a_{n-1} + a_{[n/2]}
a
n
=
a
n
−
1
+
a
[
n
/2
]
. Does the sequence contain infinitely many multiples of
7
7
7
?