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 reals such that 
and 
Prove that
+1 w
be a cyclic quadrilateral. The point 
is the reflection of 
the intersection of 
and 
, the point 
is the reflection of midpoint of 
. Also let 
be the reflection of 
midpoint of 
. Show that 
concyclic.
. Prove that: 
be a triangle, and let 
be a point on side . A line through intersects side 
at 
and ray 
at 
. The circumcircle of triangle 
intersects the circumcircle 
of triangle again at point 
distinct from point . The lines 
and 
intersect again at 
and 
respectively.
such that 
for all real number 
and 
be a scalene triangle with point 
and 
on the plane such that 
is an equilateral . Let intersect 
and 
at and respectively and intersect 
and 
at and respectively .
, point 
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds. is an isosceles triangle.
be a positive integer such that 
.
.
be lines in the plane such that no two of them are parallel and no three of them are concurrent. Let be the intersection point of lines 
. We call an "Interior Point" if there are points 
on 
and 
on 
such that is between and . Prove that there are at least 
Interior points.(
)
.
, 
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 the set of positive real numbers.
such that there exists a function 
with the property:![\[
f(y^2f(x) + y + c) = xf(x+y^2)
\]](http://latex.artofproblemsolving.com/e/3/0/e30ee6d9f69cc43422a5c2b9813192a632d359cb.png)
for all 
.
be non-negative real numbers such that 
Prove that
such that 
and let 
be the incenter of 
Prove that the circumcenter of 
are collinear.
are concurrent by jacobi theorem Let 
then the common radical axis of 
is the isogonal conjugate of the conic through 
are concurrent by jacobi theorem Let then the common radical axis of is the isogonal conjugate of the conic through
are concurrent by jacobi theorem Let then the common radical axis of is the isogonal conjugate of the conic through 
and let 
be the isogonal conjugate of 
wrt 
and 
be the excentral triangle and let 
,now it is not hard to show that 
,now by 
inversion at this problem reduces to showing that 
are concurrent which is not hard ,also it is easy to show that 
are perspective(actually prespector is 
) and pedal triangle of wrt is similar to those three mentioned triangle(this is kinda generalization of the fact that pedal triangle of isodynamic point is equilateral)
be the excenters of 
against 
and let 
be the 2nd intersection of 
and 
and 
and moreover 
are collinear. Thus if is the isogonal conjugate of WRT 
we get 
and 
But we know that 
together with 
it follows that 
and 
Hence if 
it follows that 
and 
are concylic 
and 
Similarly 
and 
cut 
again at the same point and the conclusion follows.
and 
are coxial with radical axis is 
and 
are coaxial with with radical axis is 
Define similarly the lines 
and 
and 
are concurrent at point 
and 
are also coaxial.
with orthic triangle 
Let 
be the point such that 
Then 
are concurrent.
be the point such that 
and define 
similarly, then by Jacobi's theorem 
are concurrent at 
so the isogonal conjugate of 
WRT lies on 
be the second intersection of 
then 
so 
Furthermore, from 
Similarly, lies on 
Let 
be the excentral triangle of and let 
be the isogonal conjugate of WRT From Lemma for 
and 
we get 
are concurrent. Analogously, 
are concurrent and 
are concurrent, so 
are concurrent at 
we get 
so 
and 
hence 
Similarly, 
lies on 
and a point 
Let 
be the 2nd intersection of 
with 
resp. and be a point on the circular isogonal pivotal cubic of through Then 
are coaxial and the radical axis of these three circle passes through the isogonal conjugate of WRT 
concur at , and by spiral similarity, 
are cyclic.
. We get a new problem: in with orthocenter 
let be a point. Let 
, and define 
similarly. Then 
concur.
be the orthic triangle. Let 
be the circular points at infinity and construct the cubic 
through 
which is possible since the first 
points are cayley bacharach. If 
then 
, but 
so 
, so 
and 
. Similarly 
.
, so concur on as desired.
and 
then prove 
