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Problems
Contests
National and Regional Contests
Germany Contests
German National Olympiad
2005 German National Olympiad
2005 German National Olympiad
Part of
German National Olympiad
Subcontests
(6)
6
1
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Sequence!
The sequence
x
0
,
x
1
,
x
2
,
.
.
.
.
.
x_0,x_1,x_2,.....
x
0
,
x
1
,
x
2
,
.....
of real numbers is called with period
p
p
p
,with
p
p
p
being a natural number, when for each
p
≥
2
p\ge2
p
≥
2
,
x
n
=
x
n
+
p
x_n=x_{n+p}
x
n
=
x
n
+
p
. Prove that,for each
p
≥
2
p\ge2
p
≥
2
there exists a sequence such that
p
p
p
is its least period and
x
n
+
1
=
x
n
−
1
x
n
x_{n+1}=x_n-\frac{1}{x_n}
x
n
+
1
=
x
n
−
x
n
1
(
n
=
0
,
1
,
.
.
.
.
)
(n=0,1,....)
(
n
=
0
,
1
,
....
)
4
1
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Q(Q(Q(2005^2005)))=7
I am not a spammer, at least, this is the way I use to think about myself, and thus I will not open a new thread for the following problem from today's DeMO exam: Let Q(n) denote the sum of the digits of a positive integer n. Prove that
Q
(
Q
(
Q
(
200
5
2005
)
)
)
=
7
Q\left(Q\left(Q\left(2005^{2005}\right)\right)\right)=7
Q
(
Q
(
Q
(
200
5
2005
)
)
)
=
7
. [EDIT: Since this post was split into a new thread, I comment: The problem is completely analogous to the problem posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=31409 , with the only difference that you have to consider the number
200
5
2005
2005^{2005}
200
5
2005
instead of
444
4
4444
4444^{4444}
444
4
4444
.] Darij
5
1
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1 / r = 1 / r_a + 1 / r_b + 1 / r_c for tetrahedron
(a) [Problem for class 11] Let r be the inradius and
r
a
r_a
r
a
,
r
b
r_b
r
b
,
r
c
r_c
r
c
the exradii of a triangle ABC. Prove that
1
r
=
1
r
a
+
1
r
b
+
1
r
c
\frac{1}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}
r
1
=
r
a
1
+
r
b
1
+
r
c
1
. (b) [Problem for classes 12/13] Let r be the radius of the insphere and let
r
a
r_a
r
a
,
r
b
r_b
r
b
,
r
c
r_c
r
c
,
r
d
r_d
r
d
the radii of the four exspheres of a tetrahedron ABCD. (An exsphere of a tetrahedron is a sphere touching one sideface and the extensions of the three other sidefaces.) Prove that
2
r
=
1
r
a
+
1
r
b
+
1
r
c
+
1
r
d
\frac{2}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}+\frac{1}{r_d}
r
2
=
r
a
1
+
r
b
1
+
r
c
1
+
r
d
1
. I am really sorry for posting these, but else, Orl will probably post them. This time, we really did not have any challenging problem on the DeMO. But at least, the problems were simple enough that I solved all of them. ;) Darij
3
1
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Turning on lamps
Let s be a positive real. Consider a two-dimensional Cartesian coordinate system. A lattice point is defined as a point whose coordinates in this system are both integers. At each lattice point of our coordinate system, there is a lamp. Initially, only the lamp in the origin of the Cartesian coordinate system is turned on; all other lamps are turned off. Each minute, we additionally turn on every lamp L for which there exists another lamp M such that - the lamp M is already turned on, and - the distance between the lamps L and M equals s. Prove that each lamp will be turned on after some time ... (a) ... if s = 13. [This was the problem for class 11.] (b) ... if s = 2005. [This was the problem for classes 12/13.] (c) ... if s is an integer of the form
s
=
p
1
p
2
.
.
.
p
k
s=p_1p_2...p_k
s
=
p
1
p
2
...
p
k
if
p
1
p_1
p
1
,
p
2
p_2
p
2
, ...,
p
k
p_k
p
k
are different primes which are all
≡
1
m
o
d
4
\equiv 1\mod 4
≡
1
mod
4
. [This is my extension of the problem, generalizing both parts (a) and (b).] (d) ... if s is an integer whose prime factors are all
≡
1
m
o
d
4
\equiv 1\mod 4
≡
1
mod
4
. [This is ZetaX's extension of the problem, and it is stronger than (c).] Darij
1
1
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No comment
Find all pairs (x; y) of real numbers satisfying the system of equations
x
3
+
1
−
x
y
2
−
y
2
=
0
x^3 + 1 - xy^2 - y^2 = 0
x
3
+
1
−
x
y
2
−
y
2
=
0
;
y
3
−
1
−
x
2
y
+
x
2
=
0
y^3 - 1 - x^2y + x^2 = 0
y
3
−
1
−
x
2
y
+
x
2
=
0
. Darij
2
1
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Triangle, circumcircle, perpendicular bisectors...
According to the estimated number of participants who gave a correct solution, this was the hardest (!) problem from today's paper. So here is this great German killer - be warned! Given a circle k and three pairwisely distinct points A, B, C on this circle. Let h and g be the perpendiculars to the line BC at the points B and C. The perpendicular bisector of the segment AB meets the line h at a point F; the perpendicular bisector of the segment AC meets the line g at a point G. Prove that the product
B
F
⋅
C
G
BF\cdot CG
BF
⋅
CG
is independent from the position of the point A, as long as the points B and C stay fixed. The actual problem behind the problem: Why on hell should the points B and C stay fixed? Darij