[/quote]
,
and 
.
Where 
satisfies 
.
,
.
and a constant 
,
and 
are also friendly polynomials. Prove that 
.
lying outside the circle 
,
and 
.
is chosen on the segment 
.
be the midpoints of segments 
and 
respectively. The circumcircle of triangle 
intersects 
)
passes through the centroid of triangle 
.
and let 
be the foot of the
-
.
in line 
and 
respectively. A line through 
and 
respectively such that 
while 
and 
and 
are tangent to each other.
with circumcircle 
.
is the angle bisector of 
.
intersects 
.
angle bisector of 
if and only if 
.
be real numbers satisfying
.
,
past 
.
such that ![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of 
and 
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
and 
intersect 
and 
at 
intersect 
.
with orthocenter 
intersects 
at 
intersect 
.
intersects 
and 
intersect on the circumcircle of triangle 
.
.![$D \in [AB]$](http://latex.artofproblemsolving.com/0/7/9/07914568284f3822cf15730a6a0e868f0bd79996.png)
is a point. Perpendiculars from ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
intersects with ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
intersects with 
and 
.![$$[E_1F_1]=\frac{3}{4}$$](http://latex.artofproblemsolving.com/3/e/f/3efdf53cb3ff4b2a7fa60691f5aac9b3eeecf00c.png)
and 
.
.
Y by Adventure10, Mango247, Rounak_iitr
Squares 
and 
are constructed on the sides of acute-angled triangle 
, outside of the triangle. Line 
intersects line segment 
at 
, while line 
intersects line segment 
at 
. Point 
, lying inside triangle , is an intersection of the circumcircles of triangles 
and 
. Point 
is the midpoint of 
. Prove that angle 
.
and 
are constructed on the sides of acute-angled triangle 
, outside of the triangle. Line 
intersects line segment 
at 
, while line 
intersects line segment 
at 
. Point 
, lying inside triangle , is an intersection of the circumcircles of triangles 
and 
. Point 
is the midpoint of 
. Prove that angle 
.
which is clearly the 2nd intersection of 
and 
Since 
then 
is cyclic 
of 
is the reflection of 
then 
by SAS, because 
and 
Therefore, 
but since 
then 
and 
are then the altitude and circumdiameter of 
issuing from 
which in turn are isogonals WRT 
are isogonals WRT 
is cyclic, and so is 
.
,
,
and 
is the radical axis of the two circles. As 
symmedian of 
and we are done (any point of symmedian has distances to the adjacent sides proportional to the lengths of the sides).
,
,
.
,
,
is a symmedian. Thus, we are done.
.
and 
are cyclic. From 
we deduce that 
,
,
,
,
.
about ![\[ \{ \begin{gathered} L \to A \\ N \to N' \\ T \to T'. \end{gathered}\]](http://latex.artofproblemsolving.com/7/c/0/7c01f9f0d9c6ff882f5bfb5a40714e4d98c6baa1.png)
From midpoints, 
,
,
,
.
,
.
are isogonal: i.e. the circumcenter 
is a diameter of 
,
and 
,
,
to intersect 
at 
and 
are cyclic. Indeed, for ![\[ \angle{MQK} = 180 - \angle{NCL} = \angle{ACB} = 90 - \angle{YCB} = \angle{CYB} = 180 - \angle{MYL} \]](http://latex.artofproblemsolving.com/5/6/0/560c4d56f005435c320b401de1e3ea9bc8df8729.png)
, and similar for 
and 
intersect on the 
.
be its isogonal conjugate in 
so 
is tangent to 
.
is also tangent to 
is a median; hence 
.
we obtain 
so 
just as we desired. 
![[asy]
size(9cm);
pair A=(0,0),B=(10,0),C=(2,7),K,L,M,Nn,X,Y,P,Ss,Q,O;
K=C*(0,1);L=C+(A-C)*(0,-1);M=B+(C-B)*(0,-1);Nn=C+(B-C)*(0,1);
X=extension(C,Nn,A,K);Y=extension(C,L,B,M);
pair[] x=intersectionpoints(circumcircle(K,X,Nn),circumcircle(L,Y,M));
P=x[1];Q=x[0];
O=(C+Q)/2;Ss=(A+B)/2;
D(MP("A",A,S)--MP("S",Ss,S)--MP("B",B,S)--MP("C",C,N)--A);
draw(circumcircle(K,X,Nn)^^circumcircle(L,Y,M)^^circumcircle(L,C,Nn),green);
D(A--MP("X",X,S)--MP("K",K,S)--MP("L",L,W)--C--MP("N",Nn,E)--MP("M",M,SE)--MP("Y",Y,S)--B,magenta);D(MP("Q",Q,N)--MP("O",O,NE)--MP("P",P,S),heavycyan+dashed);
draw(X--C--Y^^L--Q--Nn,magenta);
draw((C+L)/2--O--foot(O,C,A)--C,red+dotted);
dot(A^^B^^C^^K^^L^^M^^Nn^^X^^Y^^P^^Q^^O^^Ss);[/asy]](http://latex.artofproblemsolving.com/3/5/8/3583fd95654c494933de1afbf0b5a019e537b775.png)
.
is cyclic (since 
)
and 
,
and 
are cyclic too.
and 
,
Therefore 
.
is a diameter of 
is 
from the red dotted rectangle above. Similarly, 
Therefore from a well-known property of symmedians, 
So 
is actually the 
