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Problems
Contests
International Contests
Mediterranean Mathematics Olympiad
2019 Mediterranean Mathematics Olympiad
2019 Mediterranean Mathematics Olympiad
Part of
Mediterranean Mathematics Olympiad
Subcontests
(4)
4
1
Hide problems
x^2+y^2+z^2 >= x^3+y^3+z^3 +6xyz, inside an equilateral of height 1
Let
P
P
P
be a point in the interior of an equilateral triangle with height
1
1
1
, and let
x
,
y
,
z
x,y,z
x
,
y
,
z
denote the distances from
P
P
P
to the three sides of the triangle. Prove that
x
2
+
y
2
+
z
2
≥
x
3
+
y
3
+
z
3
+
6
x
y
z
x^2+y^2+z^2 ~\ge~ x^3+y^3+z^3 +6xyz
x
2
+
y
2
+
z
2
≥
x
3
+
y
3
+
z
3
+
6
x
yz
3
1
Hide problems
sum of digits in decimal represent. of 4x^4+y^4-z^2+4xyz at most 2
Prove that there exist infinitely many positive integers
x
,
y
,
z
x,y,z
x
,
y
,
z
for which the sum of the digits in the decimal representation of
4
x
4
+
y
4
−
z
2
+
4
x
y
z
~4x^4+y^4-z^2+4xyz
4
x
4
+
y
4
−
z
2
+
4
x
yz
~
is at most
2
2
2
.(Proposed by Gerhard Woeginger, Austria)
2
1
Hide problems
sequence inequality wanted, r\cdot m_r+m_s ~\ge~ (r+1)(s-1)
Let
m
1
<
m
2
<
⋯
<
m
s
m_1<m_2<\cdots<m_s
m
1
<
m
2
<
⋯
<
m
s
be a sequence of
s
≥
2
s\ge2
s
≥
2
positive integers, none of which can be written as the sum of (two or more) distinct other numbers in the sequence. For every integer
r
r
r
with
1
≤
r
<
s
1\le r<s
1
≤
r
<
s
, prove that
r
⋅
m
r
+
m
s
≥
(
r
+
1
)
(
s
−
1
)
.
r\cdot m_r+m_s ~\ge~ (r+1)(s-1).
r
⋅
m
r
+
m
s
≥
(
r
+
1
)
(
s
−
1
)
.
(Proposed by Gerhard Woeginger, Austria)
1
1
Hide problems
1/r_B +1/r_C= 2\cdot ( 1/r + 1/b +1/c), inradii in tirangle with <A=60
Let
Δ
A
B
C
\Delta ABC
Δ
A
BC
be a triangle with angle
∠
C
A
B
=
6
0
∘
\angle CAB=60^{\circ}
∠
C
A
B
=
6
0
∘
, let
D
D
D
be the intersection point of the angle bisector at
A
A
A
and the side
B
C
BC
BC
, and let
r
B
,
r
C
,
r
r_B,r_C,r
r
B
,
r
C
,
r
be the respective radii of the incircles of
A
B
D
ABD
A
B
D
,
A
D
C
ADC
A
D
C
,
A
B
C
ABC
A
BC
. Let
b
b
b
and
c
c
c
be the lengths of sides
A
C
AC
A
C
and
A
B
AB
A
B
of the triangle. Prove that
1
r
B
+
1
r
C
=
2
⋅
(
1
r
+
1
b
+
1
c
)
\frac{1}{r_B} +\frac{1}{r_C} ~=~ 2\cdot\left( \frac1r +\frac1b +\frac1c\right)
r
B
1
+
r
C
1
=
2
⋅
(
r
1
+
b
1
+
c
1
)