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 non-negative real numbers such that 
Prove that
be circumscribed about circle 
, and let 
be the orthocenter of 
. The circle touches line 
at 
. The tangent to the circle 
at meets at 
. Let 
be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
such that![\[ f(x)f\left(x + 4f(y)\right) = xf\left(x + 3y\right) + f(x)f(y) \]](http://latex.artofproblemsolving.com/d/b/1/db14b64aa71cb3c8dbf53707f719f5d9a43bd7e4.png)
for any positive real numbers 
.
and 
be coprime integers. Call a set 
balanced if the numbers 
form an arithmetic progression with difference 
, and the numbers 
form an arithmetic progression with difference 
.
and 
, determine the maximum size of a collection of balanced sets such that every two of them have a non-empty intersection.
be an integer. The sequence 
is defined via 
, 
and![\[ a_{n+2} = ka_{n+1} + a_n \]](http://latex.artofproblemsolving.com/b/5/e/b5e3e6f9f0b0709113498266e41b372299a2986f.png)
for any positive integer 
. Prove that there are finitely many pairs 
of positive integers such that 
.
be a circumscribed quadrilateral with incircle 
and no two opposite angles equal. Let 
be an arbitrary point on the diagonal 
, which is inside . The segments 
and 
intersect at 
and 
. The tangents to at and intersect at . Prove that lies on the line .
on one variable with real coefficients such that the equation 
has exactly 
ordered pairs 
as solutions?
be positive reals such that 
. Prove that 
(Shining Sun, USA)
be an acute triangle with 
, midpoint 
of side , altitude 
(
), and orthocenter 
. A circle passes through points 
and 
, is tangent to line 
, and intersects the circumcircle of triangle at a second point 
. The circumcircle of triangle 
intersects line 
at a second point 
. Prove that the lines 
and 
are perpendicular.
be a positive integer. Prove that there exists a positive integer 
such that exactly 
of the integers 
are squarefree.
be a non-constant monic polynomial with integer coefficients and let 
be an infinite sequence. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
.
cells of an 
square grid are colored black, and the remaining cells are white. The cost of such a coloring is the minimum number of white cells that need to be recolored black so that from any black cell 
, one can reach any other black cell 
through a sequence 
of black cells where each consecutive pair 
are adjacent (sharing a common side) for every 
. Let 
denote the maximum possible cost over all initial colorings with exactly black cells. Determine a constant 
such that![\[
\frac{1}{3}n^{\alpha} \leq f(n) \leq 3n^{\alpha}
\]](http://latex.artofproblemsolving.com/0/a/0/0a06d4ff52479c8cb785c6450cea8748bf7769b6.png)
for any 
.
are two fixed points on a circle centered at 
, and is an interior point of the circle that differs from . 
are concyclic. Prove that the bisector of 
is perpendicular to line 
.
where 
is the circumference circumscripted to OPQ which is; using chord-angles, we note that and 
are fixed: so also 
is fixed: now prove the case of 
m=o, and see that it's 
denote the circumcircle of 
, and let 
be the intersection of the angle bisector of 
and . Let 
. Note that 
. Let 
be the point diametrically opposite 
. Then 
. Since 
is isosceles, 
, so 
. Thus 
, so 
and we are done.
are two fixed points on a circle centered at , and is an interior point of the circle that differs from . are concyclic. Prove that the bisector of is perpendicular to line .
.
the circle with centre and by 
the other circle .Drop the perpendicular from point to segment 
and extend it to the point diametrically opposite , naming them 
and respectively. Clearly, since 
, it suffices to prove that 
are concyclic. Now for the best part, consider an inversion with respect to ( is the centre of inversion), This it suffices to prove that 
are collinear but this clearly follows from the fact that both 
and 
are diameters and thus 
. Thus done .
of (PMOQ).Rest is trivial.
.
The problem becomes a known theorem 
,
,
.
.
.
.
.
.
.
be the circumcircle of 
,
,
is the bisector of 