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Problems
Contests
National and Regional Contests
Czech Republic Contests
Czech and Slovak Olympiad III A
1995 Czech And Slovak Olympiad IIIA
1995 Czech And Slovak Olympiad IIIA
Part of
Czech and Slovak Olympiad III A
Subcontests
(6)
2
1
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(x+y)/2,\sqrt{xy}, 2xy/(x+y),\sqrt{((x^2 +y^2)/2} integers with sum 66
Find the positive real numbers
x
,
y
x,y
x
,
y
for which
x
+
y
2
,
x
y
,
2
x
y
x
+
y
,
x
2
+
y
2
2
\frac{x+y}{2},\sqrt{xy},\frac{2xy}{x+y},\sqrt{\frac{x^2 +y^2}{2}}
2
x
+
y
,
x
y
,
x
+
y
2
x
y
,
2
x
2
+
y
2
are integers whose sum is
66
66
66
.
5
1
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3 tangent circles, angles chasing
Let
A
,
B
A,B
A
,
B
be points on a circle
k
k
k
with center
S
S
S
such that
∠
A
S
B
=
9
0
o
\angle ASB = 90^o
∠
A
SB
=
9
0
o
. Circles
k
1
k_1
k
1
and
k
2
k_2
k
2
are tangent to each other at
Z
Z
Z
and touch
k
k
k
at
A
A
A
and
B
B
B
respectively. Circle
k
3
k_3
k
3
inside
∠
A
S
B
\angle ASB
∠
A
SB
is internally tangent to
k
k
k
at
C
C
C
and externally tangent to
k
1
k_1
k
1
and
k
2
k_2
k
2
at
X
X
X
and
Y
Y
Y
, respectively. Prove that
∠
X
C
Y
=
4
5
o
\angle XCY = 45^o
∠
XC
Y
=
4
5
o
6
1
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x^3 -2p(p+1)x^2+(p^4 +4p^3 -1)x-3p^3, roots are sidelengths
Find all real parameters
p
p
p
for which the equation
x
3
−
2
p
(
p
+
1
)
x
2
+
(
p
4
+
4
p
3
−
1
)
x
−
3
p
3
=
0
x^3 -2p(p+1)x^2+(p^4 +4p^3 -1)x-3p^3 = 0
x
3
−
2
p
(
p
+
1
)
x
2
+
(
p
4
+
4
p
3
−
1
)
x
−
3
p
3
=
0
has three distinct real roots which are sides of a right triangle.
4
1
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10000 ten-digit numbers divisible by 7, by reordering their digits
Do there exist
10000
10000
10000
ten-digit numbers divisible by
7
7
7
, all of which can be obtained from one another by a reordering of their digits?
3
1
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5 distinct points and 5 distinct lines are given in the plane
Five distinct points and five distinct lines are given in the plane. Prove that one can select two of the points and two of the lines so that none of the selected lines contains any of the selected points.
1
1
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<BAC+<CAD+<DAB = <ABC+<CBD+,DBA = 180^o in tetrahedron
Suppose that tetrahedron
A
B
C
D
ABCD
A
BC
D
satisfies
∠
B
A
C
+
∠
C
A
D
+
∠
D
A
B
=
∠
A
B
C
+
∠
C
B
D
+
∠
D
B
A
=
18
0
o
\angle BAC+\angle CAD+\angle DAB = \angle ABC+\angle CBD+\angle DBA = 180^o
∠
B
A
C
+
∠
C
A
D
+
∠
D
A
B
=
∠
A
BC
+
∠
CB
D
+
∠
D
B
A
=
18
0
o
. Prove that
C
D
≥
A
B
CD \ge AB
C
D
≥
A
B
.