[/quote]
,
such that 
for all 
.
for which, for all 
,![\[
f(x) f(yf(x)) + y f(xy) = \frac{f\left(\frac{x}{y}\right)}{y} + \frac{f\left(\frac{y}{x}\right)}{x}
\]](http://latex.artofproblemsolving.com/9/0/c/90c180110402e1a32b70edb2b0a03a28727457d1.png)
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
,![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
with 
right and 
,
.
we take two points 
and 
so that 
and 
.
and 
containing inside 
and 
,
and 
,
and 
,
and that 
,
and 
are collinear.
inside a square 
we raise a segment 
perpendicular to the plane of the square. Show that the projections of 
,
,
and 
are coplanar if and only if ![$O\in [AC]\cup [BD]$](http://latex.artofproblemsolving.com/6/6/6/666df8f2afa87f15c79f9c96d9d1cea17a2a7af0.png)
.
.![\[
(2^x - 1)(5^x - 1) = y^n
\]](http://latex.artofproblemsolving.com/f/8/f/f8f013a96863bc5f39aa23edb98eed9aec54a17f.png)
have no positive integer solution 
.
![\[
f : \mathbb{R}^+ \to \mathbb{R}
\]](http://latex.artofproblemsolving.com/e/7/b/e7bfc5b236fb6f00ef328f850b1b14632cbf8416.png)
satisfying the condition that for all 
,![\[
f(x^2 + y) \geq f(x) + y.
\]](http://latex.artofproblemsolving.com/c/7/c/c7cd7c9a4d4372afc23e4503bb1c05fc7d07e788.png)
such that:
be a positive integer, and let 
denote the integers modulo 
satisfying 
for all 
.
Y by nguyendangkhoa17112003, Adventure10, Mango247
Let 
be a triangle, and 
its orthocenter. Let 
be a point on the circumcircle of triangle (distinct from the vertices 
, 
, 
), and let 
be the foot of the altitude of triangle from the vertex . Let the parallel to the line 
through the point meet the parallel to the line 
through the point at a point 
. Let the parallel to the line 
through the point meet the parallel to the line through the point at a point 
. The lines 
and 
intersect at some point 
. Prove that the lines 
and are parallel.
be a triangle, and 
its orthocenter. Let 
be a point on the circumcircle of triangle (distinct from the vertices 
, 
, 
), and let 
be the foot of the altitude of triangle from the vertex . Let the parallel to the line 
through the point meet the parallel to the line 
through the point at a point 
. Let the parallel to the line 
through the point meet the parallel to the line through the point at a point 
. The lines 
and 
intersect at some point 
. Prove that the lines 
and are parallel.
,
and 
,
.
.
and 
are the reflections of the point 
w.r.t. the midpoints of the sides 
,
repectively ! I think I make no mistakes. I wish you a happy New Year - 2000 !
.
and 
.
,
,
.
be the foot of altitudes from 
respectively and wlog let 
(the proof is exactly same in all the cases).Now 
is a parallelogram so 
.
are concyclic 
.
are concyclic.So 
.
so from 
we get that 
.
are concyclic 
.
across 
be 
and the reflection of 
.
is cyclic 
(
,
(simple transformation geometry tricks)
and 
are cyclic(simple angle chasing)
f
and some simple angle chasing)
and 
solves the problem.
,
cyclic.
,
and 
.
is cyclic.
.
.
.
,
& similarly 
are cyclic. 
and then note that,
Hence, 
is cyclic and By Reim's Theorem 
.
,
is a projective map. Similarly 
is also projective. Also, taking perspectivity from 
is a projective map. Let 
As 
are fixed, so we get that the points 
are all projective related to 
for three positions of 
and when 
(It requires some work, but I am really not in the mood of writing all that 
). Hence, done. 
![[asy]
size(10cm);
defaultpen(fontsize(10pt));
pair A = dir(125), B = dir(210), C = dir(330), P = dir(280), Q = A+B-P, R = A+C-P, H = orthocenter(A,B,C), E = foot(B,C,A), X = extension(A,Q,R,H);
dot("$A$", A, dir(90));
dot("$B$", B, dir(260));
dot("$C$", C, dir(330));
dot("$P$", P, dir(280));
dot("$Q$", Q, dir(90));
dot("$R$", R, dir(90));
dot("$X$", X, dir(0));
dot("$H$", H, dir(270));
dot("$E$", E, dir(45));
draw(unitcircle);
draw(A--B--C--A, linewidth(1.2));
draw(R--H--B--P--C--H--A--X--E--H^^C--R--A--Q--B^^A--P);
draw(circumcircle(A,B,H)^^circumcircle(A,C,H)^^circumcircle(A,E,H), dotted);
[/asy]](http://latex.artofproblemsolving.com/9/3/9/939a4fb451694e35f03503c067717c1e1a97608f.png)
denote a directed angle modulo 
.
and 
so 
.
,
as desired. 
is cyclic. Therefore 
and we are done. 
Now, since 
and 
,
But remark that 
so it follows that 
.
is cyclic, so 
which means 
,
and 
are congruent and corresponding sides are parallel. Let the orthocenter of 
;
so that a translation bringing 
and 
are concyclic. So 
as desired.
be the foot of the perpendicular of 
are concyclic. This implies that 
so 
which implies the result.
