[/quote]
,
be a strictly decreasing functio
denote the set of positive integers. Consider a function 
.
we write 
.
has the following two properties:
;
is finite.
is periodic.
be a triangle with circumcircle 
and incentre 
.
intersects the lines 
,
,
at points 
,
,
,
,
,
,
,
,
of the segments 
,
,
,
.
and 
Prove that
,
,
be the midpoints of sides BC, CA, and AB, respectively. Point 
is located on the ray 
so that 
is similar to 
.
and 
on rays 
and 
,
,
,
are concurrent, i.e. these three lines intersect at a point.
is drawn. An arbitrary circle passing through points 
meets the segments 
,
and 
for the second time at points 
,
and 
respectively. Segments 
and 
meet at point 
.
or the sum of areas of triangles 
and 
?
be reals such that 
and 
Prove that
is divisibly by 
.
and 
be two infinite sequences of integers. Prove that there exists an infinite sequence of integers 
such that for any positive integer 
,![\[
\sum_{k|n} k \cdot z_k^{\frac{n}{k}} = \left( \sum_{k|n} k \cdot x_k^{\frac{n}{k}} \right) \cdot \left( \sum_{k|n} k \cdot y_k^{\frac{n}{k}} \right).
\]](http://latex.artofproblemsolving.com/2/5/9/259ee32c89befe7a3ee2797e0d5b1df1c3066cf6.png)
such that for all reals 
Proposed by Aritra12, India
and 
imply that either 
or 
.
implies that 
.
also implies that 
implies that for any 
or 
.
,
is injective. 
and 
imply that 
for all 
.
.
such that 
,
gives 
for all such 
.
then gives 
and thus 
.
which means that 
.
for all 
for some 
,
which is impossible. Thus 
in this case.
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Periodically we'll post difficult challenge problems, which will appear both in this forum and in the forum top bar above.
Here's the first one:
Here's the first one:
This post has been edited 2 times. Last edited by rrusczyk, Jun 2, 2003, 4:17 PM
(we toss out the 
case because that wouldn’t work out in the initial equation. Plugging in 
and 
gives 
is strictly less than or greater than 
.
.
where k is a non-square integer and 
which are well known, studied and solved. But I havent seen applications of them in such problems. My usual way to solve one of these is by Weierstrass theorem of elliptic curves if its a Mordell equation, or Fermat Christmas Theorem after some manipulation of given equation.
is well known (BMO 1998 I believe)
first, if 
then both 
so supposing 
and 
by taking 
we get
,
congruent to 
such that 
or 
then 
which means 
meaning 
,
then 
and since 
meaning 
so 
,
