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.
be a hyperboloid of one sheet and let 
be another quadric that intersects the hyperboloid at the curve 
. Let 
be a point on , and let 
be a line through in one specific ruling of the hyperboloid. Let this line intersect again at 
, now define 
to be the line through in the opposite ruling. Similarily define 
. Prove that if 
then this is true for all initial choices of .
doesn't have any solution on integers.
be positive integers such that 
is a square. Prove that 
is not prime.
be a real polynomial of degree 
that has only simple roots. It is possible to determine a positive quantity 
so that for every pair of positive real numbers 
, 
with 
, every transformed polynomial of the form 
has exactly 
or 
sign variations.
such that 
![$P\in \mathbb{R}[x] $](http://latex.artofproblemsolving.com/1/1/9/1194a1382b5ae7a9945cf8d9a6281954cf8cdb3d.png)
with even-degree
be a real numbers such that 
Prove that
Where 
be a cyclic quadrilateral with 
and let 
and 
be the feet of altitudes from 
to the lines 
and 
,.Prove 
bisects 
.
is strictly increasing : 
such that:
, let 
be the set of positive integers that do not exceed and are coprime to . Define 
as the smallest positive integer that allows to be partitioned into disjoint subsets, each forming an arithmetic progression.
satisfying 
, 
, and 
.
be a real numbers such that 
Prove that
Where
be natural number and 
be the set of distinct natural numbers. Define function 
with 
. Prove that have at least distinct values.
with 
. Let point be orthocenter 
and 
midpoint 
. It is also known that 
perpendicular to 
. Let 
be a point on the segment 
such that 
and point 
be the intersection of 
and 
. Prove that 
concyclic
be 
.
passes the midpoint of 
.
is a harmonic quad, so 
meets at one point. However, 
, so 
is the midpoint of .
is the midpoint of 
, and 
is the midpoint og so 
.
is cyclic, so by Reims 
is cyclic.
.
for which 
is a perfect square.
