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Problems
Contests
International Contests
Tuymaada Olympiad
1998 Tuymaada Olympiad
1998 Tuymaada Olympiad
Part of
Tuymaada Olympiad
Subcontests
(8)
7
1
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average of sum^2 of all possible sequences of numbers -1,+1 of length 100
All possible sequences of numbers
−
1
-1
−
1
and
+
1
+1
+
1
of length
100
100
100
are considered. For each of them, the square of the sum of the terms is calculated. Find the arithmetic average of the resulting values.
1
1
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1997/1998 as a sum of reciprocal to naturals
Write the number
1997
1998
\frac{1997}{1998}
1998
1997
as a sum of different numbers, inverse to naturals.
4
1
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6 planes divide surface and volume of a tetrahedron with equal opposite sides
Given the tetrahedron
A
B
C
D
ABCD
A
BC
D
, whose opposite edges are equal, that is,
A
B
=
C
D
,
A
C
=
B
D
AB=CD, AC=BD
A
B
=
C
D
,
A
C
=
B
D
and
B
C
=
A
D
BC=AD
BC
=
A
D
. Prove that exist exactly
6
6
6
planes intersecting the triangular angles of the tetrahedron and dividing the total surface and volume of this tetrahedron in half.
8
1
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cut a pyramid into 8 equal and similar to it pyramids.
Given the pyramid
A
B
C
D
ABCD
A
BC
D
. Let
O
O
O
be the midpoint of edge
A
C
AC
A
C
. Given that
D
O
DO
D
O
is the height of the pyramid,
A
B
=
B
C
=
2
D
O
AB=BC=2DO
A
B
=
BC
=
2
D
O
and the angle
A
B
C
ABC
A
BC
is right. Cut this pyramid into
8
8
8
equal and similar to it pyramids.
6
1
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sequence of first digits of numbers in the form 2^n+3^n nonperiodic
Prove that the sequence of the first digits of the numbers in the form
2
n
+
3
n
2^n+3^n
2
n
+
3
n
is nonperiodic.
5
1
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hypotenuse in a right triangle inscibed in y=x^2 not less that 2
A right triangle is inscribed in parabola
y
=
x
2
y=x^2
y
=
x
2
. Prove that it's hypotenuse is not less than
2
2
2
.
3
1
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inequality with a segment that splits a triangle area in half and inradius
The segment of length
ℓ
\ell
ℓ
with the ends on the border of a triangle divides the area of that triangle in half. Prove that
ℓ
>
r
2
\ell >r\sqrt2
ℓ
>
r
2
, where
r
r
r
is the radius of the inscribed circle of the triangle.
2
1
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(x^3-1000)^{1/2}=(x^2+100)^{1/3}
Solve the equation
(
x
3
−
1000
)
1
/
2
=
(
x
2
+
100
)
1
/
3
(x^3-1000)^{1/2}=(x^2+100)^{1/3}
(
x
3
−
1000
)
1/2
=
(
x
2
+
100
)
1/3