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.
satisfies 
. If 
, find the maximum possible value of 
.
and a constant 
, it is given that 
and 
are also friendly polynomials. Prove that 
.
lying outside the circle 
, two tangents are drawn touching at points 
and 
. A point 
is chosen on the segment 
. Let points 
and 
be the midpoints of segments 
and 
respectively. The circumcircle of triangle 
intersects again at point 
(
). Prove that the line 
passes through the centroid of triangle 
.
be an acute-angled triangle with 
and let 
be the foot of the
-angle bisector on 
. The reflections of lines 
and 
in line meet and at points
and 
respectively. A line through meets and at 
and 
respectively such that and 
while lies strictly between 
and . Prove that the circumcircles of
and 
are tangent to each other.
with circumcircle 
. Point in such that 
is the angle bisector of 
. Circle with center and radius 
intersects and at and respectively, 
. Let be the midpoint of arc in that didn't have . Prove that 
angle bisector of 
if and only if 
.
be real numbers satisfying
. Prove that:
be an acute triangle with 
, and let be the projection from onto . Let be a point on the extension of 
past such that 
. Let be on the perpendicular bisector of 
such that and are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across and be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and , respectively. Let 
and 
intersect 
and 
at and . Let 
intersect at . Prove that 
.
be an acute triangle inscribed in a circle 
with orthocenter and altitude . The line passing through perpendicular to 
intersects at . The perpendicular bisector of intersects 
at . Let 
intersect at . Let be the reflection of across 
. The circumcircle of triangle 
intersects at different from . Prove that 
and 
intersect on the circumcircle of triangle 
. be an equilateral triangle with side lenght is 
.Let ![$D \in [AB]$](http://latex.artofproblemsolving.com/0/7/9/07914568284f3822cf15730a6a0e868f0bd79996.png)
is a point. Perpendiculars from to ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
intersects with and at points and respectively. Perpendiculars from and to ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
intersects with at points 
and 
. Prove that![$$[E_1F_1]=\frac{3}{4}$$](http://latex.artofproblemsolving.com/3/e/f/3efdf53cb3ff4b2a7fa60691f5aac9b3eeecf00c.png)
with incircle . Let 
be the point on such that the circumcircle of 
is tangent to and let the -excircle be tangent to at 
. Prove that the tangent from to and the tangent from to (distinct from ) meet on the line parallel to and passing through .
it. 
( is tangent to )
= Radius of 
is equilateral
is perpendicular to 
be intersection of 
.
be circumradius of triangle 
is tangent to 
is perpendicular to .
to meet at 
.
is equilateral, and 
is isosceles.
, 
thus is perpendicular to .
as the midpoint of 
and 
to be the circle 
then 
because 
is the centre of and is tangent to at . Therefore 
. Also 
, where 
is the intersection of with . Hence MN is perpendicular to BC.
and 
.
Where 
and 
.
and 
.
be the midpoint of 
be the center of 
.
.
.
.
is equilateral. So, 
.
.
.
and 
,
is equilateral.
showing 
.
.
.
.
.
.
be the center of 
be that of 
( as 
and 
)
.
as 
is the diameter of 
then we are done. This is easy to prove. For this I used coordinate...
let 
.
and that of 
,
we find the slope of this radical axis and see that it is equal to 
slope of 
or 
,
- equilateral and 
and 
as midline in 
.