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)
be an acute-angled triangle. Let be a point on the segment 
the incentre of 
The circumcircle of 
meets 
at and the circumcircle of 
meets 
at 
If the area of 
and the area of 
are equal, prove that 
be a point on such that 
The condition that ![$[PID]=[QID]$](http://latex.artofproblemsolving.com/5/5/9/559a53df3c7705bef486261f81b010aac3a71975.png)
implies that 
is median. Notice that 
Now we cliam that is the foot of on 
![[asy]
size(7cm); pointpen=black; pathpen=black; defaultpen(fontsize(9pt));
pair A=dir(30);
pair B=dir(-147);
pair C=dir(-33);
pair I=incenter(A,B,C);
pair D=foot(I,B,C);
pair E=extension(I,foot(I,A,D),B,C);
pair P=OP(circumcircle(A,B,D),L(B,I,5,5));
pair Q=IP(circumcircle(A,C,D),L(C,I,5,5));
D(circumcircle(A,B,D),dotted);
D(circumcircle(A,C,D),dotted);
D(A--B--C--cycle,purple+linewidth(1.2));
D(A--D);
D(B--P);
D(C--Q);
D(I--E,dashed+red);
D(P--Q,dashed+red);
D(C--E);
D(I--D);
dot("A",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(-75));
dot("$I$",I,dir(90));
dot("$D$",D,dir(-90)*1.5);
dot("$E$",E,dir(E));
dot("$P$",P,dir(110)*1.5);
dot("$Q$",Q,dir(180));
[/asy]](http://latex.artofproblemsolving.com/8/5/1/8517eefbb5c2191e4e1c31b165d2e8af677e138a.png)
We use barycentric coordinate. Set 
and 
where 
and 
Thus![\[\overrightarrow{DA}=(a,-m,m-a),\quad \overrightarrow{EI}=(a,b-t,t-a-b).\]](http://latex.artofproblemsolving.com/3/9/a/39a560143fd9b6f1fa48f6b3d23b5313c1e83afb.png)
By the we get![\[0=a^2(m(a+b)-mt+mb-mt-ab+at)+b^2(at-a^2-ab+ma-a^2)+c^2(ab-at-ma),\]](http://latex.artofproblemsolving.com/9/c/5/9c51e95785127b01132ba5b6b5e14b025d7320af.png)
collect the 
terms we get![\[(2ma-a^2+c^2-b^2)t=am(a+2b)-b(a^2+b^2+2ab-c^2)+m(b^2-c^2).\]](http://latex.artofproblemsolving.com/0/2/c/02c8db82da97400f7eedd4642f5f0bece2af81a5.png)
Substitute 
we get![\[\begin{aligned}
(c^2-b^2)t&=am(b-c)+m(b^2-c^2)-b((a+b)^2-c^2)\\
&=m(b-c)(a+b+c)-b(a+b+c)(a+b-c)\\
&=(a+b+c)(mb-mc-ab-b^2+bc).
\end{aligned}\]](http://latex.artofproblemsolving.com/3/7/6/37691d89ce296d4903c73578b652dccf019222ca.png)
Using 
again we get![\[(c^2-b^2)m=(2m-a)(mb-mc-ab-b^2+bc),\]](http://latex.artofproblemsolving.com/b/d/b/bdb317d09f0cda16b4517661f6cdf66ff16b50a9.png)
which is the following quadratic in 
:![\[2(b-c)m^2-m(c^2+b^2-2bc+3ab-ac)+ab(a+b-c)=0.\]](http://latex.artofproblemsolving.com/0/0/4/004a063af0ac0470ff58252a7ac408df7b52d5ff.png)
Since so we conclude that 
as desired.
and 
so 
are concyclic. Notice that 
is median and 
is isogonal to with respect to 
it follows that 
is harmonic quadrilateral. Therefore 
) and without bary, complex 
be the intersections of 
with 
and 
with 
.
thus 
is cyclic .
then 
since and have the same area and 
and 
are supplementary then 
.
be the intersections of with and with , then is cyclic.
and 
have the same area, will pass through the midpoint of , then by Ceva's theorem we have 
so 
so 
.
and 
.
since is the -median of 
. And so we are done.
means that is on the median of .
means that should (not yet proved) lie on the Apollonian circle of . should be the 
-point of , which is what we are going to prove. and are reflections across the line , from 
and 
., , , are concyclic by angle chasing, 
. lies on the reflection of circumcircle 
across , which implies that is the -point of as desired.-point is discussed.
and 
.
and 
.
Where 
.
are equal means 
besides 
so 
so if 
ie 
but 
implie
therefore 