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Problems
Contests
National and Regional Contests
Czech Republic Contests
Czech and Slovak Olympiad III A
1980 Czech And Slovak Olympiad IIIA
1980 Czech And Slovak Olympiad IIIA
Part of
Czech and Slovak Olympiad III A
Subcontests
(6)
6
1
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set of 7 planes
Let
M
M
M
be the set of five points in space, none of which four do not lie in a plane. Let
R
R
R
be a set of seven planes with properties: a) Each plane from the set
R
R
R
contains at least one point of the set
M
M
M
. b) None of the points of the set M lie in the five planes of the set
R
R
R
. Prove that there are also two distinct points
P
P
P
,
Q
Q
Q
,
P
∈
M
P \in M
P
∈
M
,
Q
∈
M
Q \in M
Q
∈
M
, that the line
P
Q
PQ
PQ
is not the intersection of any two planes from the set
R
R
R
.
5
1
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3x3 ineq system in Z, 3x^2 +2yz <= 1+y^2
Solve a set of inequalities in the domain of integer numbers:
3
x
2
+
2
y
z
≤
1
+
y
2
3x^2 +2yz \le 1+y^2
3
x
2
+
2
yz
≤
1
+
y
2
3
y
2
+
2
z
x
≤
1
+
z
2
3y^2 +2zx \le 1+z^2
3
y
2
+
2
z
x
≤
1
+
z
2
3
z
2
+
2
x
y
≤
1
+
x
2
3z^2 +2xy \le 1+x^2
3
z
2
+
2
x
y
≤
1
+
x
2
4
1
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min f(x) = sum |x-a_i| , n even
Let
a
1
<
a
2
<
.
.
.
<
a
n
a_1 < a_2< ...< a_n
a
1
<
a
2
<
...
<
a
n
are real numbers,
f
(
x
)
=
∑
i
=
1
n
∣
x
−
a
i
∣
,
f(x) = \sum_{i=1}^n|x-a_i|,
f
(
x
)
=
i
=
1
∑
n
∣
x
−
a
i
∣
,
for
n
n
n
even. Find the minimum of this function.
3
1
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min no of convex sets whose union is M (plane - 3 vertices of triangle)
The set
M
M
M
was formed from the plane by removing three points
A
,
B
,
C
A, B, C
A
,
B
,
C
, which are vertices of the triangle. What is the smallest number of convex sets whose union is
M
M
M
?[hide=original wording] Množina M Vznikla z roviny vyjmutím tří bodů A, B, C, které jsou vrcholy trojúhelníka. Jaký je nejmenší počet konvexních množin, jejichž sjednocením je M?
2
1
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isosceles trapezoid
Find the side sizes of an isosceles trapezoid that has longest side
13
13
13
cm, perimeter
28
28
28
cm and area
27
27
27
cm
2
^2
2
. Is there such a trapezoid, if we we ask for area
27.001
27.001
27.001
cm
2
^2
2
?
1
1
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1980^{197} | (k + 1)(k + 2)...(k + 1980)
Prove that for every nonnegative integer
k
k
k
there is a product
(
k
+
1
)
(
k
+
2
)
.
.
.
(
k
+
1980
)
(k + 1)(k + 2)...(k + 1980)
(
k
+
1
)
(
k
+
2
)
...
(
k
+
1980
)
divisible by
198
0
197
1980^{197}
198
0
197
.