AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
of natural numbers such that
be an acute scalene triangle. Let 
and 
be two distinct interior points of the segment 
such that 
. Suppose that:
and 
are the feet of the perpendiculars from from 
to the lines 
and 
respectively.
and 
are the feet of the perpendiculars from 
to the lines and respectively.
and 
intersect on the line .
be an odd prime and 
be integers so that the integers 
are divisible by . divides each of .
be a triangle with incenter 
, and let 
be a point on side . Points and are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side such that and are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
, the extension of 
past , and the extension of 
past . Prove that is tangent to the circumcircle of triangle .
decide whether there exist pairwise distinct positive integers 
such that for every 
, 
divides 
.
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
, consider the 
fractions 
The product of these fractions equals 
, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of for which this is possible and prove that you have found them all.
be a function satisfying
for all integers 
and 
. Show that there exist positive integers 
and such that 
for all integers .
be a rectangle with 
and the center the point 
. Perpendicular from on 
intersects lines 
and in points and respectively. Points 
and 
are midpoints of segments ![$[CD]$](http://latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
and ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
respectively. Prove that 
.
be a rectangle with and the center the point . Perpendicular from on intersects lines and in points and respectively. Points and are midpoints of segments and respectively. Prove that .
, thus 
and thence 
as homologous medians. But this is very easy, since 
and 
are cyclic, making 
.
midpoint of and to prove 
.
and 
.
and 
are similar (then since 
this will imply 
as desired).
it will suffice to prove that 
.
and 
we find that 
and similarly 
.![\[\frac{AE}{CF}=\frac{b-BE}{BF-a}=\frac{b-\frac{d^2}{2b}}{\frac{d^2}{2a}-a}=\frac{a}{b} \cdot \frac{2b^2-d^2}{d^2-2a^2}=\frac{a}{b}\]](http://latex.artofproblemsolving.com/a/3/f/a3fd4d9a8e4e263a09004a53180356900ea7182c.png)
using that 
.