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 set consisting of non-negative integers such that:
,
, both 
and 
..
such that![\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]](http://latex.artofproblemsolving.com/8/0/0/8009a4709ffd6708023d67db2b544e605d1fbe68.png)
for all real numbers 
satisfying 
.
and 
such that![\[ {a^2+b\over b^2-a}\quad\mbox{and}\quad{b^2+a\over a^2-b} \]](http://latex.artofproblemsolving.com/e/3/b/e3bc96af9f635e1c10e21dc91894c8afbf113a6d.png)
. Points 
and 
lie on segments 
and 
, respectively. Lines 
and 
intersect at point 
. The circumcircles of triangles 
and 
intersect at a second point 
. The circumcircles of triangles and 
intersect at a second point 
. Point 
lies on segment 
such that 
. Prove that triangles 
and 
are similar.
segments and 
are parallel. Angle bisectors of 
and 
meet at . Angle bisectors of 
and 
meet at . Prove that 
is cyclic
with incentre 
and excentres 
and 
, opposite the vertices 
and 
respectively. The incircle touches 
and at 
and 
respectively. Prove that the circles 
and 
have a common point other than .
and 
, then find the maximum value of ![$\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$](http://latex.artofproblemsolving.com/b/f/a/bfa17e8bb4cc05a84ed6a757f506d6d0ed2b0081.png)
be an acute triangle with orthocenter 
and circumcenter 
. Let and be points on the line segments and respectively such that 
is a parallelogram. Prove that 
.
, let 
be the midpoint of side 
. Let the line perpendicular to at point intersect 
at 
. If 
is tangent to at 
, find 
.
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line. be an acute triangle with incentre and 
. Let lines 
and 
intersect the circumcircle of at 
and 
, respectively. Consider points 
and 
such that 
and 
are parallelograms (with 
, and 
). Let 
be the point of intersection of lines 
and 
. Prove that points 
, and are concyclic.
be a scalene triangle with circumcenter , incenter , and incircle 
. Let touch the sides 
, 
, and 
at points , , and respectively. Let be the projection of to 
. The line 
intersects the circumcircle of again at point 
. The circumcircles of 
and 
intersect again at points 
and 
respectively. Prove that the lines 
, 
, and 
are concurrent.
, 
, is the point on such that 
. Let 
be the circle through tangent to 
at , and 
the circle through tangent to at . Let 
be the second intersection of and . Prove that lies on .
be a point on ray beyond is tangent to circle 
and is tangent to circle 
is cyclic.
, giving the required collinearity.
be the intersection of and . We will prove that 
, which implies![\[
(!) \quad AB \text{ is tangent to } (CF'B).
\]](http://latex.artofproblemsolving.com/0/9/1/0912a5d89c9b96991df593e9811fe2ccb3ad79db.png)
Due to the Power of a Point theorem,![\[
AD^2 = AF \cdot AC.
\]](http://latex.artofproblemsolving.com/c/7/c/c7ce7f6916003527f4d1812c1cfcaf61d3bf0bf6.png)
Since 
, we have![\[
AB^2 = AD^2 = AF \cdot AC.
\]](http://latex.artofproblemsolving.com/e/f/f/eff6902bf20790e7dfdb273eee89ca8dac4c58c8.png)
. Note that 
is the radical axis of 
, and so this implies that 
lies on extended, which then implies lies on .
be 
,
is tangent to 
and 
meet on