Line Combining Fermat Point, Orthocenter, and Centroid
by cooljoseph, Apr 6, 2025, 9:22 PM
On triangle 
, draw exterior equilateral triangles on sides 
and 
to obtain 
and 
, respectively. Let 
be the intersection of the altitude through 
and the median through 
. Let 
be the intersection of the altitude through 
and line 
. Let 
be the intersection of the median through and the line 
. Prove that , , and lie on a common line.
![[asy]
size(481.245,258.036);
label("$A$", (-1.5,7.8));
label("$B$", (5.2, 0.7));
label("$C$", (-5.3,0.7));
label("$B'$", (-9.1,7.4));
label("$C'$", (7.8,9.8));
path line = (0,1)--(-1.40861,7.40089);
draw(line, gray+linewidth(1.5)+solid );
path line = (-5,1)--(1.7957,4.20044);
draw(line, gray+linewidth(1.5)+solid );
path line = (-1.40861,1)--(-1.40861,7.40089);
draw(line, orange+linewidth(1.5)+solid );
path line = (-2.60567,5.26737)--(5,1);
draw(line, orange+linewidth(1.5)+solid );
path line = (-8.74763,7.31068)--(5,1);
draw(line, blue+linewidth(1.5)+solid );
path line = (-5,1)--(7.33903,9.75046);
draw(line, blue+linewidth(1.5)+solid );
path line = (5,1)--(-1.40861,7.40089);
draw(line, black+linewidth(2.5)+solid );
path line = (-8.74763,7.31068)--(-1.40861,7.40089);
draw(line, black+linewidth(2.5)+solid );
path line = (-1.40861,7.40089)--(7.33903,9.75046);
draw(line, black+linewidth(2.5)+solid );
path line = (-8.74763,7.31068)--(-5,1);
draw(line, black+linewidth(2.5)+solid );
path line = (-5,1)--(-1.40861,7.40089);
draw(line, black+linewidth(2.5)+solid );
path line = (-5,1)--(5,1);
draw(line, black+linewidth(2.5)+solid );
path line = (5,1)--(7.33903,9.75046);
draw(line, black+linewidth(2.5)+solid );
path line = (-10.3821,3.4813)--(8.86775,3.62203);
draw(line, rgb(0, 0.666, 0)+linewidth(1)+dashed);
dot((0.436645,3.5604), linewidth(3)+solid);
dot((-1.40861,3.54691), linewidth(3)+solid);
dot((-0.561846,3.5531), linewidth(3)+solid);
path frame = (-10.3821,-0.0943394)--(-10.3821,10.2271)--(8.86775,10.2271)--(8.86775,-0.0943394)--cycle;
clip(frame);
[/asy]](//latex.artofproblemsolving.com/4/6/9/469fd874cf0fb80a65093eb5bb02d383b06b4bcb.png)
, draw exterior equilateral triangles on sides 
and 
to obtain 
and 
, respectively. Let 
be the intersection of the altitude through 
and the median through 
. Let 
be the intersection of the altitude through 
and line 
. Let 
be the intersection of the median through and the line 
. Prove that , , and lie on a common line.![[asy]
size(481.245,258.036);
label("$A$", (-1.5,7.8));
label("$B$", (5.2, 0.7));
label("$C$", (-5.3,0.7));
label("$B'$", (-9.1,7.4));
label("$C'$", (7.8,9.8));
path line = (0,1)--(-1.40861,7.40089);
draw(line, gray+linewidth(1.5)+solid );
path line = (-5,1)--(1.7957,4.20044);
draw(line, gray+linewidth(1.5)+solid );
path line = (-1.40861,1)--(-1.40861,7.40089);
draw(line, orange+linewidth(1.5)+solid );
path line = (-2.60567,5.26737)--(5,1);
draw(line, orange+linewidth(1.5)+solid );
path line = (-8.74763,7.31068)--(5,1);
draw(line, blue+linewidth(1.5)+solid );
path line = (-5,1)--(7.33903,9.75046);
draw(line, blue+linewidth(1.5)+solid );
path line = (5,1)--(-1.40861,7.40089);
draw(line, black+linewidth(2.5)+solid );
path line = (-8.74763,7.31068)--(-1.40861,7.40089);
draw(line, black+linewidth(2.5)+solid );
path line = (-1.40861,7.40089)--(7.33903,9.75046);
draw(line, black+linewidth(2.5)+solid );
path line = (-8.74763,7.31068)--(-5,1);
draw(line, black+linewidth(2.5)+solid );
path line = (-5,1)--(-1.40861,7.40089);
draw(line, black+linewidth(2.5)+solid );
path line = (-5,1)--(5,1);
draw(line, black+linewidth(2.5)+solid );
path line = (5,1)--(7.33903,9.75046);
draw(line, black+linewidth(2.5)+solid );
path line = (-10.3821,3.4813)--(8.86775,3.62203);
draw(line, rgb(0, 0.666, 0)+linewidth(1)+dashed);
dot((0.436645,3.5604), linewidth(3)+solid);
dot((-1.40861,3.54691), linewidth(3)+solid);
dot((-0.561846,3.5531), linewidth(3)+solid);
path frame = (-10.3821,-0.0943394)--(-10.3821,10.2271)--(8.86775,10.2271)--(8.86775,-0.0943394)--cycle;
clip(frame);
[/asy]](http://latex.artofproblemsolving.com/4/6/9/469fd874cf0fb80a65093eb5bb02d383b06b4bcb.png)
This post has been edited 4 times. Last edited by cooljoseph, 2 hours ago
Reason: Fixed first sentence
Reason: Fixed first sentence
so that the sum of its edges' lengths is minumum.
with 
be the area of the triangle of sides 
,
and 
.
where 
is the circumradius and 
is the volume of the tetrahedron.
which moves on line 
towards 
.
be the intersections of 
with the sides of 
are the midpoints of 
,
,
,
and 
.
in the plane of a square 
for which ![\[\max (PA, PC)=\frac1{\sqrt2}(PB+PD).\]](http://latex.artofproblemsolving.com/3/9/8/3984b1970837a85af0bac93fd711843630a1fd99.png)
be a triangle, and let 
,
is the circumcenter of 
and 
as the orthocenters of triangles 
and 
.
passes through a fixed point which is independent of the choice of 
,
be the orthocenter of 
.
.
at 
be the midpoint of 
,
meet 
at 
.
is tangent to 
,
at 
.
meets 
.
.
.
be a path within 
with 
.
.
and the length of the part of 
.
September 2017
February 2017