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Contests
National and Regional Contests
Lithuania Contests
Grand Duchy of Lithuania
2019 Grand Duchy of Lithuania
2019 Grand Duchy of Lithuania
Part of
Grand Duchy of Lithuania
Subcontests
(4)
1
1
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sum \sqrt{z + xy} >=\sqrt{xyz}+\sqrt{x }+\sqrt{y} +\sqrt{z} if 1/x+1/y+1/z=1
Let
x
,
y
,
z
x, y, z
x
,
y
,
z
be positive numbers such that
1
x
+
1
y
+
1
z
=
1
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1
x
1
+
y
1
+
z
1
=
1
. Prove that
x
+
y
z
+
y
+
z
x
+
z
+
x
y
≥
x
y
z
+
x
+
y
+
z
\sqrt{x + yz} +\sqrt{y + zx} +\sqrt{z + xy} \ge\sqrt{xyz}+\sqrt{x }+\sqrt{y} +\sqrt{z}
x
+
yz
+
y
+
z
x
+
z
+
x
y
≥
x
yz
+
x
+
y
+
z
2
1
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max of f(P) as P ranges over all colourings, bw in 20x20 table
Every cell of a
20
×
20
20 \times 20
20
×
20
table has to be coloured black or white (there are
2
400
2^{400}
2
400
such colourings in total). Given any colouring
P
P
P
, we consider division of the table into rectangles with sides in the grid lines where no rectangle contains more than two black cells and where the number of rectangles containing at most one black cell is the least possible. We denote this smallest possible number of rectangles containing at most one black cell by
f
(
P
)
f(P)
f
(
P
)
. Determine the maximum value of
f
(
P
)
f(P)
f
(
P
)
as
P
P
P
ranges over all colourings.
4
1
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p^2 + 5pq + 4q^2 is perfect square
Determine all pairs of prime numbers
(
p
,
q
)
(p, q)
(
p
,
q
)
such that
p
2
+
5
p
q
+
4
q
2
p^2 + 5pq + 4q^2
p
2
+
5
pq
+
4
q
2
is a square of an integer.
3
1
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<OHC = 90^o wanted, XP + Y Q = AB + XY ,orthocenter, circumcenter
Let
A
B
C
ABC
A
BC
be an acute triangle with orthocenter
H
H
H
and circumcenter
O
O
O
. The perpendicular bisector of segment
C
H
CH
C
H
intersects the sides
A
C
AC
A
C
and
B
C
BC
BC
in points
X
X
X
and
Y
Y
Y
, respectively. The lines
X
O
XO
XO
and
Y
O
YO
Y
O
intersect the side
A
B
AB
A
B
in points
P
P
P
and
Q
Q
Q
, respectively. Prove that if
X
P
+
Y
Q
=
A
B
+
X
Y
XP + Y Q = AB + XY
XP
+
Y
Q
=
A
B
+
X
Y
then
∠
O
H
C
=
9
0
o
\angle OHC = 90^o
∠
O
H
C
=
9
0
o
.