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 .
, let be the circumcenter. Let be the midpoint of . to be . Let the circumcircle of 
hit 
at 
.
are colinear.
and 
, we have 
.
.
, so we have 
, as desired 
as 
.
is concyclic. & 
be midpoints of and .
=
- $ \angle(90°- A)$
is concyclic , hence the conclusion follows.
and 
which we will call 
lies on 
The central claim is that 
is cyclic.
and also note that 
Note that 
since 
and 
and 
Thus 
Note 
Thus is cyclic.
Since 
as well, we are done.![[asy]
size(6cm);
pair A = dir(110);
pair B = dir(-40);
pair C = dir(220);
pair O = circumcenter(A,B,C);
pair M;
M=(A+C)/2;
pair D;
D=foot(A,B,C);
pair P=extension(D,M,B,O);
draw(A--B--C--cycle);
draw(D--P);
draw(P--B,dotted);
draw(circumcircle(P,A,B),blue);
draw(circumcircle(A,B,C));
draw(circumcircle(O,A,M));
dot("$A$",A,N);
dot("$B$",B,SE);
dot("$C$",C,SW);
dot("$D$",D,S);
dot("$M$",M,W);
dot("$O$",O,S);
dot("$P$",P,dir(190));
[/asy]](http://latex.artofproblemsolving.com/7/f/7/7f7a6ad21ac8fbaecb89fcbeafa8c84810dcc928.png)
and 
.
Where 
.
From here it follows that 
.
,
be 
and 
,
be the point opposite 
.
Therefore, 
is cyclic, and 
.
Hence, 
collinear 
so 
. Thus 
is cyclic with a center 
then 
It follows that 
lie on a circle whence we get 
finishing the problem. 
.![[asy]
size(7cm);
import geometry;
pen c1=brown+red+orange;
pair A,B,C,D,M,O,P;
A=(-6,5); B=(-8,-1); C=(3,-1); D=foot(A,B,C); M=midpoint(A--C); O=circumcenter(A,B,C); P=intersectionpoints(circle(A,M,O),D--M)[0];
draw(A--B--C--cycle); draw(A--D); draw(A--O); draw(D--M); draw(circle(A,M,O)); draw(A--P); draw(O--M); draw(circle(A,B,D),dotted+c1); draw(B--O,dashed);
dot(A^^B^^C^^D^^M^^O^^P);
label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,NE); label("$O$",O,SE); label("$P$",P,SE);
[/asy]](http://latex.artofproblemsolving.com/e/a/6/ea67acb4366439164c23c1bbfbbd65161d15f148.png)
so 
is cyclic. Hence, 
