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.
and 
be touch points of the incenter of 
at 
and 
, respectively. Let 
and 
be the circumcenter of triangles 
and 
, respectively. Show that 
and 
concurrent.
be a point inside a triangle 
. Let 
meet 
at 
, let 
meet 
at 
, and let 
meet at 
. Let 
be the point such that is the midpoint of 
, let 
be the point such that is the midpoint of 
, and let 
be the point such that is the midpoint of 
. Prove that points 
cannot all lie strictly inside the circumcircle of triangle .
with the property that ![\[f(x-f(y))=f(f(x))-f(y)-1\]](http://latex.artofproblemsolving.com/f/2/5/f25cc1e8ae0be1fd02b347fd94be4fab88af1d46.png)
holds for all 
.
are such that 
, find the minimum value of expression 
be an acute triangle such that 
with circumcircle 
and circumcentre 
. Let 
and 
be the tangents to at 
and 
respectively, which meet at 
. Let 
be the foot of the perpendicular from onto the line segment 
. The line through 
parallel to line meets at 
. Prove that the line 
passes through the midpoint of the line segment 
.
+1 w
, no matter how its elements are partitioned into two subsets, at least one of the subsets must contain three numbers 
(
is allowed) such that 
. Find the minimal 
.
differents subjects, each consist of 
students. Given that for any 
different subjects, there exists a student compete in both subjects. Prove that there exists a student who compete in at least 
different subjects.
, define 
and 
, where 
and 
denote the product and sum of the digits of , respectively. Find all solutions to 
for which the following statement holds: there exists at least one triple 
of integers such that
and all triples of real numbers, satisfying the equations, are such that 
are integers.
students and a teacher with 
marbles. The teacher then play a Marble distribution according to the following rules. At the start, the teacher distributed all her marbles to students, so that each student receives at least 
marbles from the teacher. Then, the teacher chooses a student , who has never been chosen before, such that the number of marbles that he owns in a multiple of 
. That chosen student then equally distribute half of his marbles to 
other students. The same goes on until the teacher is not able to choose anymore student., such that for some initial numbers of marbles that the students receive, the teacher can choose all the student(according to the rule above), so that each student receiving equal amount of marbles at the end.
satisfying the following condition: 
. We define some new points.
Let 
and 
be -Ex and Queue points respectively. Let , 
. Let 
be pole of 
. is polar of and by La Hire we have lies on Polar of . Also 
and so 
lies on Polar of . lies on Polar of or that 
. By PoP we have ![\[X_AZ \cdot X_AY=X_AB \cdot X_AC=X_AD \cdot X_AM=X_AQ_A \cdot X_AA=\lambda\]](http://latex.artofproblemsolving.com/4/2/2/42207c12ec87453057f273fc9e9b8bbc5359d12e.png)
Hence invert at with radius 
and since inversion preserves cross ratios we have ![\[(X_A,P;Y,Z)=(\infty,P^*;Y,Z)\]](http://latex.artofproblemsolving.com/4/3/5/4351dac5d2390b145f3fdfb759371f92f879ac70.png)
So we need to prove that 
is midpoint of 
but see that 
and it is well known that 
so , 
, collinear and obviously 
bisects and so we are done.
and 
Show that
Let 
Show that
. Prove that
and orthocenter 
and midpoint of 
then prove that 
passes through pole of 