MathDB
Problems
Contests
National and Regional Contests
Austria Contests
Austrian MO National Competition
2001 Federal Competition For Advanced Students, Part 2
2001 Federal Competition For Advanced Students, Part 2
Part of
Austrian MO National Competition
Subcontests
(3)
3
2
Hide problems
Prove that the three lines have a common point.
A triangle
A
B
C
ABC
A
BC
is inscribed in a circle with center
U
U
U
and radius
r
r
r
. A tangent
c
′
c'
c
′
to a larger circle
K
(
U
,
2
r
)
K(U, 2r)
K
(
U
,
2
r
)
is drawn so that C lies between the lines
c
=
A
B
c = AB
c
=
A
B
and
C
′
C'
C
′
. Lines
a
′
a'
a
′
and
b
′
b'
b
′
are analogously defined. The triangle formed by
a
′
,
b
′
,
c
′
a', b', c'
a
′
,
b
′
,
c
′
is denoted
A
′
B
′
C
′
A'B'C'
A
′
B
′
C
′
. Prove that the three lines, joining the midpoints of pairs of parallel sides of the two triangles, have a common point.
Prove that CF < FD
Let be given a semicircle with the diameter
A
B
AB
A
B
, and points
C
,
D
C,D
C
,
D
on it such that
A
C
=
C
D
AC = CD
A
C
=
C
D
. The tangent at
C
C
C
intersects the line
B
D
BD
B
D
at
E
E
E
. The line
A
E
AE
A
E
intersects the arc of the semicircle at
F
F
F
. Prove that
C
F
<
F
D
CF < FD
CF
<
F
D
.
2
2
Hide problems
Solve the system: x+y+z=6, 1/x+1/y+1/z=2-4/(xyz)
Determine all triples of positive real numbers
(
x
,
y
,
z
)
(x, y, z)
(
x
,
y
,
z
)
such that
x
+
y
+
z
=
6
,
x+y+z=6,
x
+
y
+
z
=
6
,
1
x
+
1
y
+
1
z
=
2
−
4
x
y
z
.
\frac 1x + \frac 1y + \frac 1z = 2 - \frac{4}{xyz}.
x
1
+
y
1
+
z
1
=
2
−
x
yz
4
.
All integers m that roots of 3x^3-3x^2+m = 0 are rational
Determine all integers
m
m
m
for which all solutions of the equation
3
x
3
−
3
x
2
+
m
=
0
3x^3-3x^2+m = 0
3
x
3
−
3
x
2
+
m
=
0
are rational.
1
2
Hide problems
Show that the sum is an integer divisible by 25
Prove that
1
25
∑
k
=
0
2001
[
2
k
25
]
\frac{1}{25} \sum_{k=0}^{2001} \left[ \frac{2^k}{25}\right]
25
1
∑
k
=
0
2001
[
25
2
k
]
is a positive integer.
Find all R to R functions with f(f(x)^2 + f(y)) = xf(x) + y
Find all functions
f
:
R
→
R
f :\mathbb R \to \mathbb R
f
:
R
→
R
such that for all real
x
,
y
x, y
x
,
y
f
(
f
(
x
)
2
+
f
(
y
)
)
=
x
f
(
x
)
+
y
.
f(f(x)^2 + f(y)) = xf(x) + y.
f
(
f
(
x
)
2
+
f
(
y
))
=
x
f
(
x
)
+
y
.