such that:
with all 
is real number
,
,
be collinear points such that 
,
,
.
,
,
are the reflections of 
,
is a nondegenerate triangle, then its circumcenter lies on the circumcircle of 
be three positive real numbers satisfying 
Prove that
There is a very elegant proof :-D Could anyone think of it?
such that 
and has exactly 
Y by jhu08, PHSH
Let 
be the center of the equilateral triangle 
. Pick two points 
and 
other than 
, , 
on the circle 
so that on this circle , , , , are placed in this order. Extensions of 
and 
intersects respectively with side 
and 
at points 
and 
. Line 
and 
intersects at point 
. Analogously point 
is defined. Let 
and 
meet again at point 
other than .
Prove that
.
Remark.
denotes the circumcircle of triangle 
.
be the center of the equilateral triangle 
. Pick two points 
and 
other than 
, , 
on the circle 
so that on this circle , , , , are placed in this order. Extensions of 
and 
intersects respectively with side 
and 
at points 
and 
. Line 
and 
intersects at point 
. Analogously point 
is defined. Let 
and 
meet again at point 
other than .Prove that
.Remark.
denotes the circumcircle of triangle 
.This post has been edited 2 times. Last edited by Li4, Mar 29, 2020, 3:17 AM
Reason: typo
Reason: typo
lies on 
and 
so 
and therefore 
is the midpoint of arc 
in 
Let 
be the intersection of 
then 
so note that 
is the bisector of 
we get 
and 
hence 
lie on a circle 
with center 
.
.
and 
.
intersect 
again at 
.
.
and 
(Since 
)
.
.
and 
.![[asy]
size(300);
pair A,B,C,O,R,S,O_1,Q_1,P_0,Q;
A=dir(90);
B=dir(210);
C=dir(330);
O=circumcenter(A,B,C);
P_0=0.7*B+0.3*C;
path ooo = circumcircle(B,O,C);
pair P = IP(circumcircle(B,O,C), A--P_0);
dot("$P_1$",P,dir(270));
draw(unitcircle,blue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(350));
dot("$O$", O, dir(320));
pair T = circumcenter(A,O,P);
R = extension(B,P,A,C);
S = extension(C,P,A,B);
Q=foot(T,B,C);
Q_1=extension(A,P,T,Q);
dot("$R$",R,dir(30));
dot("$S$",S,dir(200));
draw(circumcircle(A,O,P),heavycyan);
dot("$O_1$",T,dir(230));
draw(A--B--C--cycle,darkcyan);
draw(C--S,heavymagenta);
draw(B--R,heavymagenta);
dot("$Q_1$",Q_1,dir(180));
draw(A--T--O,red);
draw(T--Q_1,orange);
draw(Q_1--O,orange);
draw(A--P,deepblue);
draw(A--O,fuchsia+dashed);
draw(Q--Q_1,fuchsia+dashed);
pair M_0=dir(270);
draw(arc(M_0,length(C-M_0),20,160),mediumblue);
[/asy]](http://latex.artofproblemsolving.com/9/1/2/912341617ff44de2140945de7a69b419e28d86bd.png)
lie on a circle. Note that 
.
.
.
.
.
.
,
.
,
.
.
at 
.
.
we have 
with transversal 
we get : 
.
,
bisects 
.
.
of 
,
and summing, we are done. 
.
(then that would analogously imply 
and we would be done).
lie on some circle 
so 
.
.![[asy]
size(200);
pair A=dir(90),B=dir(-150),C=dir(-30),O=(0,0),P1=waypoint(circumcircle(B,O,C),0.3),R=extension(B,P1,A,C),S=extension(C,P1,A,B),Q1=extension(R,S,A,P1),L=extension(A,O,R,S),T=2*foot(A,circumcenter(A,R,S),(0,0))-A;
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$O$",O,dir(-70));
dot("$P_1$",P1,dir(-90));
dot("$R$",R,dir(R));
dot("$S$",S,dir(S));
dot("$Q_1$",Q1,dir(160));
dot("$L$",L,dir(70));
dot("$T$",T,dir(T));
draw(CP(circumcenter(B,O,C),B,15,165)^^unitcircle,red);
draw(circumcircle(A,R,S),green);
draw(A--B--C--A^^B--R^^C--S^^O--A--P1^^R--S^^O--T,magenta);
[/asy]](http://latex.artofproblemsolving.com/3/e/1/3e1106b430ae5bcdf73b8cf7e542cd3188981dcd.png)
.
.
.
which implies our claim (as 
)
and 
coincide).
.
is the internal angle bisector of 
.
.
wrt points 
we obtain
This proves our claim. 
is the internal angle bisector of 
of 
.
then by Claim 3 we obtain 
is the reflection of 
.
This completes the proof of the problem. 
