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.
such that
holds for all 
.
such that:
with all 
is real number
be a triangle, and let 
, 
, and 
be collinear points such that 
, 
, and 
. Points 
, 
, and 
are the reflections of , , and over 
, 
, and 
, respectively. Prove that if 
is a nondegenerate triangle, then its circumcenter lies on the circumcircle of .
be three positive real numbers satisfying 
Prove that
There is a very elegant proof :-D Could anyone think of it?
such that is a multiple of 
and has exactly positive factors.
opposite the vertex 
be tangent to the side at the point 
. Define the points 
on and 
on analogously, using the excircles opposite 
and 
, respectively. Suppose that the circumcentre of triangle 
lies on the circumcircle of triangle . Prove that triangle is right-angled.
be the circumcircle of triangle , 
is the midpoint of and 
be the second intersection of 
and . Tangent line of at intersects at 
, let 
be a transversal of and 
be intersections of 
with . Let 
be a tangent line of . Prove that 
is cyclic
be positive integers such that 
and 
. Find the maximum possible value of 
.
, let the incircle touch sides 
at 
, respectively. Let lie on the line through 
perpendicular to . Let 
be the intersections of 
with 
, respectively. Let 
be the projections of 
onto line . Let 
be the second intersections of 
with the incircle . Let 
be the intersection of 
and 
. Prove that the line 
bisects segment 
.
be an acute triangle. Its incircle touches the sides , and at the points , and 
, respectively. Let , 
and 
be the circumcenters of triangles 
, 
and 
, respectively. Prove that triangles and 
are similar.
be the incenter of 
. Line 
is the radical axis of 
and 
, so 
; consequently, 
. We can obtain analogous parallelisms for 
and 
, thus 
and are homothetic and hence similar. 
be the incenter of 
we have 
and 
, so 
it means that 
is cyclic with diameter 
and since is the cincuncenter of 
we'll have 
. Similarly 
is cyclic so we must have 
therefore 
by the 
case so with that we have 
and since 
it implies 
.
and 
, hence they're similar. 
![$[(x-2)^2+(y-4)^2][ (x-1)^2 + (y-2)^2] = 36$](http://latex.artofproblemsolving.com/b/1/0/b108dc2cac5b7491fb83418eb36c7184f6d5d30c.png)
then
and 
Prove that
- diameters, hence the homothety.