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.
with circumcenter 
and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line 
. Let 
be the point on the circumcircle of triangle such that 
is perpendicular to , and let 
be the point on line 
such that 
. Let 
and 
be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
. If this is still too hard assume in 
that 
is a prime.
sastifying 
such that 
be a monic polynomial with integer coefficient. And suppose there exist 4 distinct integer 
such that 
.
such that 
be a set of Natural numbers from 
to 
. (
) distinct elements from this set.
. If 
and 
, show that the area of 
is not less than the area of 
, and the pairwise comparison of the sums of elements of all its subsets (with the empty set defined as having a sum of 0), which amounts to 
inequalities, these given comparisons satisfy the following three constraints:
and 
, if the sums of elements of and are greater than those of subsets 
and 
respectively, then the sum of elements of the union 
is greater than that of 
.
that satisfies all these conditions?
be an acute triangle. Denote by 
the foot of the perpendicular line drawn from the point 
to the side 
, by the midpoint of , and by 
the orthocenter of . Let 
be the point of intersection of the circumcircle 
of the triangle and the half line 
, and 
be the point of intersection (other than ) of the line 
and the circle . Prove that 
must hold.
the length of the line segment .)
such that 
(where 
), and a functional equation 
that satisfies 
for all , has a finite number of solutions. Find such .
is the function obtained by composing 
times, that is, 
)
The angles of the triangle are in arithmetic progression.
The common tangent to the Nine-point circle and the Incircle is parallel to the Euler Line.
The angles of the triangle are in arithmetic progression. The common tangent to the Nine-point circle and the Incircle is parallel to the Euler Line.
The angles of the triangle are in arithmetic progression. The common tangent to the Nine-point circle and the Incircle is parallel to the Euler Line.![\[\angle A = 60^{\circ} \iff IO = IH\]](http://latex.artofproblemsolving.com/7/a/4/7a44310ac993baef4acd797cadec7d33563578a4.png)
. WLOG we chose that to be 
. The latter half comes from the fact that the common tangent to 
and 
is perpendicular to the line 
. Therefore, it is parallel to the euler line iff the euler line is perpendicular to , which is true iff is the perpendicular bisector of 
( is the circumcentre, and is the orthocentre of 
) which is in turn true iff 
.
are isogonal conjugates, so 
.
since 
is well known.
(coupled with ) is true iff 
.![\[ \triangle AIH \cong \triangle AIO \iff AH =AO\]](http://latex.artofproblemsolving.com/b/3/a/b3a2e7c02181120c714bca863b9004167361328d.png)
one of 
is 
at least one of is 
be the Feuerbach Point of triangle , be the center of the Euler circle. Note that tangent at to the incircle is perpendicular to . It follows that 
.
lies in the interior of at least one of the triangles 
; say 
. Then 
is the angle bisector of 
and so we get 
. All arguments here can be reversed. 
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
.
