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National and Regional Contests
Czech Republic Contests
Czech and Slovak Olympiad III A
2003 Czech And Slovak Olympiad III A
2003 Czech And Slovak Olympiad III A
Part of
Czech and Slovak Olympiad III A
Subcontests
(6)
3
1
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x_n = \pm (n−1)x_{n−1} \pm (n−2)x_{n−2} \pm ... \pm 2x_2 \pm x_1
A sequence
(
x
n
)
n
=
1
∞
(x_n)_{n= 1}^{\infty}
(
x
n
)
n
=
1
∞
satisfies
x
1
=
1
x_1 = 1
x
1
=
1
and for each
n
>
1
,
x
n
=
±
(
n
−
1
)
x
n
−
1
±
(
n
−
2
)
x
n
−
2
±
.
.
.
±
2
x
2
±
x
1
n > 1, x_n = \pm (n-1)x_{n-1} \pm (n-2)x_{n-2} \pm ... \pm 2x_2 \pm x_1
n
>
1
,
x
n
=
±
(
n
−
1
)
x
n
−
1
±
(
n
−
2
)
x
n
−
2
±
...
±
2
x
2
±
x
1
. Prove that the signs ”
±
\pm
±
” can be chosen so that
x
n
≠
12
x_n \ne 12
x
n
=
12
holds only for finitely many
n
n
n
.
5
1
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2 two-digit nos in base z, one equal to the other one read in reverse order
Show that, for each integer
z
≥
3
z \ge 3
z
≥
3
, there exist two two-digit numbers
A
A
A
and
B
B
B
in base
z
z
z
, one equal to the other one read in reverse order, such that the equation
x
2
−
A
x
+
B
x^2 -Ax+B
x
2
−
A
x
+
B
has one double root. Prove that this pair is unique for a given
z
z
z
. For instance, in base
10
10
10
these numbers are
A
=
18
,
B
=
81
A = 18, B = 81
A
=
18
,
B
=
81
.
1
1
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2x2 system, x^2 -xy+y^2 = 7, x^2y+xy^2 = -2
Solve the following system in the set of real numbers:
x
2
−
x
y
+
y
2
=
7
x^2 -xy+y^2 = 7
x
2
−
x
y
+
y
2
=
7
,
x
2
y
+
x
y
2
=
−
2
x^2y+xy^2 = -2
x
2
y
+
x
y
2
=
−
2
.
2
1
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3 tangential quadrilaterals given, equilateral triangle wanted
On sides
B
C
,
C
A
,
A
B
BC,CA,AB
BC
,
C
A
,
A
B
of a triangle
A
B
C
ABC
A
BC
points
D
,
E
,
F
D,E,F
D
,
E
,
F
respectively are chosen so that
A
D
,
B
E
,
C
F
AD,BE,CF
A
D
,
BE
,
CF
have a common point, say
G
G
G
. Suppose that one can inscribe circles in the quadrilaterals
A
F
G
E
,
B
D
G
F
,
C
E
G
D
AFGE,BDGF,CEGD
A
FGE
,
B
D
GF
,
CEG
D
so that each two of them have a common point. Prove that triangle
A
B
C
ABC
A
BC
is equilateral.
4
1
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czech slovak triangle construction
Let be given an obtuse angle
A
K
S
AKS
A
K
S
in the plane. Construct a triangle
A
B
C
ABC
A
BC
such that
S
S
S
is the midpoint of
B
C
BC
BC
and
K
K
K
is the intersection point of
B
C
BC
BC
with the bisector of
∠
B
A
C
\angle BAC
∠
B
A
C
.
6
1
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52-nd Czech and Slovak Mathematical Olympiad 2003
a，b，c>0，abc=1，prove that(a/b)+(b/c)+(c/a)≥a+b+c.