be the orthocenter of an acute-angled triangle 
.
centered at the midpoint of 
and passing through 
and 
.
,
,
and 
.
with real coefficient. It is known that 
and 
.
can't have 
distinct real roots.
be an acute triangle. Let 
be a point on side 
and 
be a point on side 
are parallel. Let 
be an interior point of 
.
and 
meet side 
and 
,
and 
.
and 
intersect at a point 
.
,
are collinear.
.
Y by Davi-8191, tenplusten, Adventure10, Mango247, Rounak_iitr
A acute triangle 
has circumcenter 
. The circumcircle of 
, called 
, and the circumcircle of 
, called 
, meets 
again at 
and 
respectively. The perpendicular bisector of hits 
again at 
. Prove that the circumcenter of 
lies on if and only if the centers of 
and are colinear.
has circumcenter 
. The circumcircle of 
, called 
, and the circumcircle of 
, called 
, meets 
again at 
and 
respectively. The perpendicular bisector of hits 
again at 
. Prove that the circumcenter of 
lies on if and only if the centers of 
and are colinear.
be the circumcenter of 
at 
.
collinear.
be 
.
.
and 
are collinear. Since the radical axis of 
we conclude that the perpendicular bisectors of 
intersect on 
at 
with 
.
with 
.
,
lie on 
.
by simple angke chase while 
.
,
.
.
,
bisects 
.
,
as the unitcircle. Let 
be the complex numbers of 
.
,
and 
.
;
.
so 
,
,
of 
.
bisects 
.
and 
of 
,![\[\measuredangle AOO_1 = 90^{\circ} + \measuredangle OAB = \measuredangle ACB = \measuredangle AOE\]](http://latex.artofproblemsolving.com/8/2/d/82dcc43b512c4054e00ac749d28bad9d6bdf582f.png)
So, 
are collinear. Along with this, 
,![[asy]
unitsize(2.5 cm);
import geometry; import olympiad;
pair A = dir(100), B = dir(200), C = -1/B, O = (0, 0);
circle c_1 = circle((point) A,(point) B,(point) O), c_2 = circle((point) A, (point) C,(point) O);
pair D = intersectionpoints(line(B, C), c_1)[1], E = intersectionpoints(line(B, C), c_2)[0];
pair O_1 = circumcenter(A, B, O), O_2 = circumcenter(A, C, O);
pair K = extension(O_1, O_2, (B+C)/2, O);
pair T = circumcenter(A, D, E);
draw(A--B--C--cycle); dot(A^^B^^C); dot(O^^O_1^^O_2); draw(c_1^^c_2, linewidth(0.4)+dashed); dot(D^^E);
draw(D--A--E--cycle, linewidth(1));
dot(K); draw(circumcircle(A, D, E), dashed+linewidth(0.7));
dot(T); draw(O_1--O_2);
draw(A--K--T, linewidth(1.1));
draw(K--(B+C)/2);
label("$A$", A, A);
label("$B$", B, B);
label("$C$", C, C);
label("$D$", D, dir(240));
label("$E$", E, dir(-60));
label("$O$", O, dir(0));
label("$O_1$", O_1, dir(150));
label("$O_2$", O_2, dir(60));
label("$K$", K, dir(150));
label("$T$", T, dir(-60));
[/asy]](http://latex.artofproblemsolving.com/6/6/3/663fcd839aae18ca72884621b216c10594d3da25.png)
and circumcenter 
are the intersection points of 
with 
and 
,
.
.
and 
.
.![\[\measuredangle KIA = \measuredangle IAK = \measuredangle XAO = \measuredangle AXO\]](http://latex.artofproblemsolving.com/d/0/4/d044a4bae5516a14792d1b4d9931cca4f5c33086.png)
Therefore, 
.
,
.
.![[asy]
unitsize(2.5 cm);
import geometry;
pair A = dir(110), B = dir(210), C = -1/B, O = (0, 0), I = incenter(A, B, C);
pair X = circumcenter(I, B, C), Y = circumcenter(I, C, A), Z = circumcenter(I, A, B);
pair K = extension(A, O, Y, Z), D = projection(B, C)*I;
draw(unitcircle); draw(A--B--C--cycle); draw(X--Y--Z--cycle);
dot(A^^B^^C); dot(X^^Y^^Z); dot(O^^I); dot(K);
draw(O--A--X--O); draw(K--D, linewidth(1.2));
markangle(I, A, K, n=2, radius=0.7 cm); markangle(K, I, A, n=2, radius=0.7 cm); markangle(O, X, A, n=2, radius=0.7 cm);
markrightangle(K, D, B); markrightangle(O, (B+C)/2, C);
label("$A$", A, A);
label("$B$", B, B);
label("$C$", C, C);
label("$X$", X, X);
label("$Y$", Y, Y);
label("$Z$", Z, Z);
label("$O$", O, dir(0));
label("$I$", I, -dir(0));
label("$K$", K, dir(-30));
[/asy]](http://latex.artofproblemsolving.com/9/0/1/90153b90b9824eca9501e87111e81feb2aefce0b.png)
![[asy]
size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.13839182578881, xmax = 24.678453195216232, ymin = -20.864724143538304, ymax = 16.934536603715053; /* image dimensions */pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw(arc((3.9048088393944984,3.417712349159264),1.196179137571309,-151.68988080175083,-114.63464561747348)--(3.9048088393944984,3.417712349159264)--cycle, linewidth(2) + qqwuqq); draw(arc((3.9048088393944984,3.417712349159264),1.196179137571309,-88.16226320961216,-51.10702802533482)--(3.9048088393944984,3.417712349159264)--cycle, linewidth(2) + qqwuqq); /* draw figures */draw(circle((4.1440446669087585,-4.038470941701886), 7.460020311506091), linewidth(0.8) + zzttff); draw(circle((-3.4620030175639,-0.5505855990750161), 8.367634405361127), linewidth(0.8) + qqwuqq); draw((3.9048088393944984,3.417712349159264)--(-1.662239610071297,-8.722375930381483), linewidth(0.8) + fuqqzz); draw((11.24983850229547,-1.766996306786579)--(-1.662239610071297,-8.722375930381483), linewidth(0.8)); draw((11.24983850229547,-1.766996306786579)--(3.9048088393944984,3.417712349159264), linewidth(0.8)); draw(circle((6.839384262002882,-0.22005980400870417), 4.673876244547078), linewidth(0.8) + qqwuqq); draw(circle((1.7195675443454639,-1.347653441134675), 5.24251758536787), linewidth(0.8) + linetype("4 4") + blue); draw((-3.4620030175639,-0.5505855990750161)--(6.839384262002882,-0.22005980400870417), linewidth(0.8)); draw((4.1440446669087585,-4.038470941701886)--(3.9048088393944984,3.417712349159264), linewidth(0.8) + fuqqzz); draw((-3.4620030175639,-0.5505855990750161)--(3.9048088393944984,3.417712349159264), linewidth(0.8)); draw((6.839384262002882,-0.22005980400870417)--(3.9048088393944984,3.417712349159264), linewidth(0.8)); draw((-3.4620030175639,-0.5505855990750161)--(5.704572246887101,-4.754077982147205), linewidth(0.8)); draw((6.839384262002882,-0.22005980400870417)--(2.3715829157948893,-6.549467219835816), linewidth(0.8)); /* dots and labels */dot((4.1440446669087585,-4.038470941701886),dotstyle); label("$O$", (2.987738167256496,-4.3175794071352005), NE * labelscalefactor); dot((3.9048088393944984,3.417712349159264),dotstyle); label("$A$", (3.9845541152325867,4.733509400487703), NE * labelscalefactor); dot((-3.4620030175639,-0.5505855990750161),dotstyle); label("$O_1$", (-5.0665346923903165,-0.21069770147370676), NE * labelscalefactor); dot((-1.662239610071297,-8.722375930381483),linewidth(4pt) + dotstyle); label("$C$", (-2.35519531389535,-9.979493991639396), NE * labelscalefactor); dot((2.16791200013861,-0.369946614549803),linewidth(4pt) + dotstyle); label("$F$", (1.0339789092233584,0.1481560397976859), NE * labelscalefactor); dot((11.24983850229547,-1.766996306786579),linewidth(4pt) + dotstyle); label("$B$", (12.676789181584098,-1.7657305803164083), NE * labelscalefactor); dot((6.839384262002882,-0.22005980400870417),linewidth(4pt) + dotstyle); label("$O_{2}$", (7.493346252108426,-1.0878957356926666), NE * labelscalefactor); dot((2.3715829157948893,-6.549467219835816),linewidth(4pt) + dotstyle); label("$E$", (2.1105401330375364,-7.906116819849127), NE * labelscalefactor); dot((5.704572246887101,-4.754077982147205),linewidth(4pt) + dotstyle); label("$D$", (5.579459631994332,-6.191593389330251), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/b/d/f/bdfec09548e0a9544bfe9dc4df9fb5a815805af9.png)
are concyclic.
and 
.
and 
are collinear.
.
as desired. 
as desired.
.
.
is a perpendicular bisector of 
,
.
and 
,
of 
.
Because 
,
Thus both statements are equivalent to the same statement, and they must be equivalent.
