MathDB
Problems
Contests
National and Regional Contests
Kosovo Contests
Kosovo Team Selection Test
2015 Kosovo Team Selection Test
2015 Kosovo Team Selection Test
Part of
Kosovo Team Selection Test
Subcontests
(5)
5
1
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Convex quadrilateral
In convex quadrilateral ABCD,diagonals AC and BD intersect at S and are perpendicular. a)Prove that midpoints M,N,P,Q of AD,AB,BC,CD form a rectangular b)If diagonals of MNPQ intersect O and AD=5,BC=10,AC=10,BD=11 find value of SO
4
1
Hide problems
Prove that
Let
P
1
,
P
2
,
.
.
.
,
P
2556
P_1,P_2,...,P_{2556}
P
1
,
P
2
,
...
,
P
2556
be distinct points inside a regular hexagon
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
of side
1
1
1
. If any three points from the set
S
=
{
A
,
B
,
C
,
D
,
E
,
F
,
P
1
,
P
2
.
.
.
,
P
2556
}
S=\{A,B,C,D,E,F,P_1,P_2...,P_{2556}\}
S
=
{
A
,
B
,
C
,
D
,
E
,
F
,
P
1
,
P
2
...
,
P
2556
}
aren't collinear, prove that there exists a triangle with area smaller than
1
1700
\frac{1}{1700}
1700
1
, with vertices from the set
S
S
S
.
3
1
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Find solution of system
It's given system of equations
a
11
x
1
+
a
12
x
2
+
a
1
n
x
n
=
b
1
a_{11}x_1+a_{12}x_2+a_{1n}x_n=b_1
a
11
x
1
+
a
12
x
2
+
a
1
n
x
n
=
b
1
a
21
x
1
+
a
22
x
2
+
a
2
n
x
n
=
b
2
a_{21}x_1+a_{22}x_2+a_{2n}x_n=b_2
a
21
x
1
+
a
22
x
2
+
a
2
n
x
n
=
b
2
..........
a
n
1
x
1
+
a
n
2
x
2
+
a
n
n
x
n
=
b
n
a_{n1}x_1+a_{n2}x_2+a_{nn}x_n=b_n
a
n
1
x
1
+
a
n
2
x
2
+
a
nn
x
n
=
b
n
such that
a
11
,
a
12
,
.
.
.
,
a
1
n
,
b
1
,
a
21
,
a
22
,
.
.
.
,
a
2
n
,
b
2
,
.
.
.
,
a
n
1
,
a
n
2
,
.
.
.
,
a
n
n
,
b
n
,
a_{11},a_{12},...,a_{1n},b_1,a_{21},a_{22},...,a_{2n},b_2,...,a_{n1},a_{n2},...,a_{nn},b_n,
a
11
,
a
12
,
...
,
a
1
n
,
b
1
,
a
21
,
a
22
,
...
,
a
2
n
,
b
2
,
...
,
a
n
1
,
a
n
2
,
...
,
a
nn
,
b
n
,
form an arithmetic sequence.If system has one solution find it
2
1
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Infinite points
Prove that circle l(0,2) with equation
x
2
+
y
2
=
4
x^2+y^2=4
x
2
+
y
2
=
4
contains infinite points with rational coordinates
1
1
Hide problems
Prove that for every n
a)Prove that for every n,natural number exist natural numbers a and b such that
(
1
−
2
)
n
=
a
−
b
2
(1-\sqrt{2})^n=a-b\sqrt{2}
(
1
−
2
)
n
=
a
−
b
2
and
a
2
−
2
b
2
=
(
−
1
)
n
a^2-2b^2=(-1)^n
a
2
−
2
b
2
=
(
−
1
)
n
b)Using first equation prove that for every n exist m such that
(
2
−
1
)
n
=
m
−
m
−
1
(\sqrt{2}-1)^n=\sqrt{m}-\sqrt{m-1}
(
2
−
1
)
n
=
m
−
m
−
1