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.
amazing, if one can partition the set 
into nice subsets.
be an integer and 
is a sequence of real numbers for which 
and:
Prove that all elements of the sequence are positive.
satisfies 
. Let the incenter of triangle be 
, which touches 
at 
, respectively. Let 
be the midpoint of 
. Let the circle centered at passing through 
intersect 
at 
, respecively. Let line 
meet at 
, line 
meet 
at 
. Prove that the three lines 
are concurrent.
be the real polonomial such that all roots are real and distinct. Prove that there is a rational number 
that all roots of 
are real numbers
have altitudes 
circumcenter 
and orthocenter 
. Let 
be a point on line 
. The points 
are on lines 
respectively such that 
and 
. Prove that 
is perpendicular to 
.
other knights where friendship is mutual. Prove that King Arthur can place some of his knights around a round table in such a way that every knight is friends with the 
people adjacent to him and between them there are at least 
friendships of knights that are not adjacent to each other.
and 
intersect each other at points 
and 
. Point 
lies on the tangent line from to such that
. Arbitrary line 
passes through and cuts at points 
and 
. Lines and 
cut for the second time at points 
and 
respectively. Let 
be the foot of altitude from to . Prove that points 
and are collinear.
be real numbers satisfying 
Prove that 
and determine all cases of equality.
be an acute angled triangle with orthocenter and 
. Let 
be any other point on the arc 
of the circumcircle of 
and let line 
intersect line 
at 
and let line 
intersect line 
at 
. Let the circumcircles of 
and intersect again at (
). Let the lines 
intersect the circumcircle of 
again at 
(
). Prove that the lines 
concur.
such that:
for any positive integer 
, the numbers 
are all distinct, and for any positive integers and 
, 
?, consider points 
on line such that 
are in that order and 
and 
. Similarly consider points 
on line , and 
on line . Prove that orthocenters of triangles 
and 
are equidistant from the circumcenter of .
be an acute triangle inscribed in its circumcircle 
, and let 
be its incircle. Let 
be the point where the 
incircle of triangle touches 
. Suppose line 
intersects segment 
at 
, and intersects line 
at 
. Let the line through 
perpendicular to intersect line 
at 
. Prove that: ![\[AI \parallel PA'.\]](http://latex.artofproblemsolving.com/0/2/6/0264e64b5edff4aca1e35037c5b0f86763c3b4e7.png)
.
, so solutions come from the second one.
.
.
or from 
has nonnegative real roots (
. However, that equation in 
has no real roots.
does have two real roots. This becomes clear when you note that at 
the polynomial is negative, but for very positive or very negative 
, the polynomial is positive.
and 
easily.
and easily.
. We manipulate the polynomial as follows:
, and 
, and thus the answer is 
, as claimed.
The solutions to 
are 
so our answer is 
.
. By Difference of Cubes, 
, so our expression is equivalent to 
. It is clear that 
cannot have real roots since for all reals 
, so we are looking for the roots of 
. The root in the desired form is 
.
If 
then 
or 
where w is a cube root of unity. to be real w=1 but contradiction since that would make 10x^2-1 in the denominator 0. Hence instead it's just solution to first factor which is 
so solution is 
.
there exists positive integers 
such that
has exactly two real roots, one of which is 
where 
and 
are integers, 
and 
Find 
.
,
.
.
.
.
.
.
and rearranging, we get
,
and 
and subtracting 
terms must cancel in order to get this polynomial equal to 
,
into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying 
,
and we get
.
since we have to solve a quadratic with a negative discriminant, then multiply by 
.
.
(using 
)
,
)
for a final answer of 
Thus, the two real roots are 
Hence, the desired answer is 
has no real roots since it's 
