be positive numbers such that 
.![\[T = \dfrac{bc}{\sqrt{a}+3}+\dfrac{ca}{\sqrt{b}+3}+\dfrac{ab}{\sqrt{c}+3} \]](http://latex.artofproblemsolving.com/6/d/5/6d570d06679f17452790b90171428d645e97fe9f.png)
,
on 
,
,
,
,
concur at a point 
.
be a point on line 
,
be a point on line 
be a point on line 
,
,
concur at a single point.
be non-negative real numbers such that 
Prove that
Prove that 
Y by doxuanlong15052000, anhtaitran, myang, Mosquitall, mathsworm, Vrangr, xymking, kiyoras_2001, Aryan-23, Pluto1708, Adventure10
You are given a triangle 
. Let 
be the incenter of . 
meets 
at 
. Let the midpoint of 
be 
. Define 
and 
similarly. Prove that the orthocenter 
of 
lies on the Euler line 
of 
.
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.8711111111111105cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = 0.38222222222222174, xmax = 7.253333333333332, ymin = 1.3244444444444483, ymax = 5.76; /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0.);
draw((1.9555555555555553,5.1822222222222205)--(0.98,1.48)--(6.98,1.6)--cycle, zzttqq);
draw((2.6164393817611407,3.3540843431907774)--(2.4462605672897575,2.633493129380323)--(4.220492896507974,2.4526678031487092)--cycle, zzttqq);
/* draw figures */
draw((1.9555555555555553,5.1822222222222205)--(0.98,1.48), zzttqq);
draw((0.98,1.48)--(6.98,1.6), zzttqq);
draw((6.98,1.6)--(1.9555555555555553,5.1822222222222205), zzttqq);
draw((2.6164393817611407,3.3540843431907774)--(2.4462605672897575,2.633493129380323), zzttqq);
draw((2.4462605672897575,2.633493129380323)--(4.220492896507974,2.4526678031487092), zzttqq);
draw((4.220492896507974,2.4526678031487092)--(2.6164393817611407,3.3540843431907774), zzttqq);
draw((1.9555555555555553,5.1822222222222205)--(3.277323207966726,1.5259464641593345));
draw((0.98,1.48)--(3.912521134579515,3.7869862587606455));
draw((1.4609857930159464,3.305335606297419)--(6.98,1.6));
draw(circle((3.9572978173621425,2.6751091318929014), 3.20820637278888));
draw((xmin, -0.12109095620117256*xmin + 3.154302108570096)--(xmax, -0.12109095620117256*xmax + 3.154302108570096)); /* line */
/* dots and labels */
dot((1.9555555555555553,5.1822222222222205),dotstyle);
label("$A$", (1.9911111111111102,5.271111111111111), NE * labelscalefactor);
dot((0.98,1.48),dotstyle);
label("$B$", (1.0133333333333325,1.564444444444448), NE * labelscalefactor);
dot((6.98,1.6),dotstyle);
label("$C$", (7.013333333333332,1.6888888888888922), NE * labelscalefactor);
dot((2.781576170286085,2.897282699160009),linewidth(3.pt) + dotstyle);
label("$I$", (2.817777777777777,2.9511111111111132), NE * labelscalefactor);
dot((3.277323207966726,1.5259464641593345),linewidth(3.pt) + dotstyle);
label("$X$", (3.3155555555555543,1.5822222222222257), NE * labelscalefactor);
dot((3.912521134579515,3.7869862587606455),linewidth(3.pt) + dotstyle);
label("$Y$", (3.9466666666666654,3.84), NE * labelscalefactor);
dot((1.4609857930159464,3.305335606297419),linewidth(3.pt) + dotstyle);
label("$Z$", (1.4933333333333325,3.36), NE * labelscalefactor);
dot((2.6164393817611407,3.3540843431907774),linewidth(3.pt) + dotstyle);
label("$D$", (2.6488888888888877,3.4044444444444464), NE * labelscalefactor);
dot((2.4462605672897575,2.633493129380323),linewidth(3.pt) + dotstyle);
label("$E$", (2.426666666666666,2.4711111111111137), NE * labelscalefactor);
dot((4.220492896507974,2.4526678031487092),linewidth(3.pt) + dotstyle);
label("$F$", (4.186666666666666,2.275555555555558), NE * labelscalefactor);
dot((2.5644297106474463,2.8437728627971035),linewidth(3.pt) + dotstyle);
dot((3.9572978173621425,2.6751091318929014),linewidth(3.pt) + dotstyle);
dot((3.305185185185185,2.754074074074073),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](//latex.artofproblemsolving.com/9/0/0/900b6ab895c0d0f1121afc3772092ac88071947a.png)
. Let 
be the incenter of . 
meets 
at 
. Let the midpoint of 
be 
. Define 
and 
similarly. Prove that the orthocenter 
of 
lies on the Euler line 
of 
.![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.8711111111111105cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = 0.38222222222222174, xmax = 7.253333333333332, ymin = 1.3244444444444483, ymax = 5.76; /* image dimensions */
pen zzttqq = rgb(0.6,0.2,0.);
draw((1.9555555555555553,5.1822222222222205)--(0.98,1.48)--(6.98,1.6)--cycle, zzttqq);
draw((2.6164393817611407,3.3540843431907774)--(2.4462605672897575,2.633493129380323)--(4.220492896507974,2.4526678031487092)--cycle, zzttqq);
/* draw figures */
draw((1.9555555555555553,5.1822222222222205)--(0.98,1.48), zzttqq);
draw((0.98,1.48)--(6.98,1.6), zzttqq);
draw((6.98,1.6)--(1.9555555555555553,5.1822222222222205), zzttqq);
draw((2.6164393817611407,3.3540843431907774)--(2.4462605672897575,2.633493129380323), zzttqq);
draw((2.4462605672897575,2.633493129380323)--(4.220492896507974,2.4526678031487092), zzttqq);
draw((4.220492896507974,2.4526678031487092)--(2.6164393817611407,3.3540843431907774), zzttqq);
draw((1.9555555555555553,5.1822222222222205)--(3.277323207966726,1.5259464641593345));
draw((0.98,1.48)--(3.912521134579515,3.7869862587606455));
draw((1.4609857930159464,3.305335606297419)--(6.98,1.6));
draw(circle((3.9572978173621425,2.6751091318929014), 3.20820637278888));
draw((xmin, -0.12109095620117256*xmin + 3.154302108570096)--(xmax, -0.12109095620117256*xmax + 3.154302108570096)); /* line */
/* dots and labels */
dot((1.9555555555555553,5.1822222222222205),dotstyle);
label("$A$", (1.9911111111111102,5.271111111111111), NE * labelscalefactor);
dot((0.98,1.48),dotstyle);
label("$B$", (1.0133333333333325,1.564444444444448), NE * labelscalefactor);
dot((6.98,1.6),dotstyle);
label("$C$", (7.013333333333332,1.6888888888888922), NE * labelscalefactor);
dot((2.781576170286085,2.897282699160009),linewidth(3.pt) + dotstyle);
label("$I$", (2.817777777777777,2.9511111111111132), NE * labelscalefactor);
dot((3.277323207966726,1.5259464641593345),linewidth(3.pt) + dotstyle);
label("$X$", (3.3155555555555543,1.5822222222222257), NE * labelscalefactor);
dot((3.912521134579515,3.7869862587606455),linewidth(3.pt) + dotstyle);
label("$Y$", (3.9466666666666654,3.84), NE * labelscalefactor);
dot((1.4609857930159464,3.305335606297419),linewidth(3.pt) + dotstyle);
label("$Z$", (1.4933333333333325,3.36), NE * labelscalefactor);
dot((2.6164393817611407,3.3540843431907774),linewidth(3.pt) + dotstyle);
label("$D$", (2.6488888888888877,3.4044444444444464), NE * labelscalefactor);
dot((2.4462605672897575,2.633493129380323),linewidth(3.pt) + dotstyle);
label("$E$", (2.426666666666666,2.4711111111111137), NE * labelscalefactor);
dot((4.220492896507974,2.4526678031487092),linewidth(3.pt) + dotstyle);
label("$F$", (4.186666666666666,2.275555555555558), NE * labelscalefactor);
dot((2.5644297106474463,2.8437728627971035),linewidth(3.pt) + dotstyle);
dot((3.9572978173621425,2.6751091318929014),linewidth(3.pt) + dotstyle);
dot((3.305185185185185,2.754074074074073),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/9/0/0/900b6ab895c0d0f1121afc3772092ac88071947a.png)
This post has been edited 2 times. Last edited by WizardMath, May 30, 2017, 7:16 AM
Reason: Year added in source
Reason: Year added in source
and a point 
lying on the anticomplement of the Kiepert hyperbola of 
Let 
be the isotomic conjugate of 
be the cevian triangle of 
of 
be the circumcenter of 
be the intersection of 
with the 9-point circle of 
is the Poncelet point of 
Since 
is the Steiner line of 
WRT the medial triangle of 
so it suffices to prove 
of 
the cyclocevian conjugate 
of 
(WRT 
and 
similarly. Prove that the orthocenter 
of 
lies on the Euler line 
of 
be the midpoint of 
similarly. We have 
collinear and 
.
.
of 
.
be the locus of point H s.t. 
(
similarly.
,
similarly. It's obvious that 
are concurrent.
.
is the diameter of 
.
.
,
,
passes through 
.
be the circumcenter of 
be the center of 
.
.
,
.
be the locus of point H s.t. 
.
(
.
.
.
