The Nagel Line
by liberator, Aug 17, 2014, 8:40 PM
Problem: Prove that in 
, the incenter 
, the centroid 
, the Spieker center 
and the Nagel point 
are collinear.
Proof: We denote the reference triangle as, and its medial triangle as 
. Let 
denote the homothecy 
, and let 
denote the homothecy 
, i.e., the homothecy from the incircle to A-excircle. Let 
be the intouch triangle, and let the cevians defining the Nagel point be 
.
![[asy]
/* Free script by liberator, 17 August 2014 */
unitsize(3cm);
defaultpen(fontsize(10pt));
/* Initialize objects */
pair A = (-2.5, 2.5);
pair B = (-3.5, -0.5);
pair C = (0.5, -0.5);
pair P = (-2.04018, -0.5);
pair Pa = (-0.95982, -0.5);
pair A1 = midpoint(B--C);
pair B1 = midpoint(C--A);
pair C1 = midpoint(A--B);
pair I = incenter(A,B,C);
pair G = centroid(A,B,C);
pair X = rotate(180,I)*P;
pair S = incenter(A1,B1,C1);
pair N1 = rotate(180,S)*I;
pair N = intersectionpoint(A--Pa, I--N1);
/* Draw objects */
draw(incircle(A,B,C), rgb(0.9,0,0));
draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1));
draw(A--A1, rgb(0.4,0.6,0.8)+linewidth(1));
draw(A--Pa, rgb(0.4,0.6,0.8)+linewidth(1));
draw(X--P, rgb(0.4,0.6,0.8)+linewidth(1));
draw(I--A1, rgb(0.4,0.6,0.8)+linewidth(1));
draw(I--N, rgb(0.4,0.6,0.8)+linewidth(1));
/* Place dots on and label each point */
dot(A); label("$A$", A, dir(100));
dot(B); label("$B$", B, dir(200));
dot(C); label("$C$", C, dir(340));
dot(P); label("$P$", P, dir(-90));
dot(Pa); label("$P_A$", Pa, dir(-90));
dot(A1); label("$A_1$", A1, dir(-90));
dot(I); label("$I$", I, dir(150));
dot(G); label("$G$", G, dir(90));
dot(X); label("$X$", X, dir(70));
dot(S); label("$S$", S, dir(-90));
dot(N); label("$N$", N, dir(-90));
[/asy]](//latex.artofproblemsolving.com/8/0/b/80b6003d8b91eba29ad1c8d9453a8de7abe829a1.png)
Lemma: The incenter of is the Nagel point of its medial triangle.
Proof: Note that
is the external center of similitude of the incircle and the A-excircle, which is tangent to 
at 
. Let 
be the first point of intersection of 
and the incircle. From , it is evident that the tangent to 
at is parallel to ; it follows that 
is a diameter of the incircle.
is the midpoint of 
, so it follows 
is homothetic to 
; in particular, 
.
Considering the homothecy, the ray is mapped to the ray 
: this occurs for each cevian , so it follows that under , is mapped to , and the result follows. 
Corollary:
are collinear.
Proof: This follows from the fact that is the center of the homothecy , which maps to and to . Moreover, 
.
, the incenter 
, the centroid 
, the Spieker center 
and the Nagel point 
are collinear.Proof: We denote the reference triangle as
, and its medial triangle as 
. Let 
denote the homothecy 
, and let 
denote the homothecy 
, i.e., the homothecy from the incircle to A-excircle. Let 
be the intouch triangle, and let the cevians defining the Nagel point be 
.![[asy]
/* Free script by liberator, 17 August 2014 */
unitsize(3cm);
defaultpen(fontsize(10pt));
/* Initialize objects */
pair A = (-2.5, 2.5);
pair B = (-3.5, -0.5);
pair C = (0.5, -0.5);
pair P = (-2.04018, -0.5);
pair Pa = (-0.95982, -0.5);
pair A1 = midpoint(B--C);
pair B1 = midpoint(C--A);
pair C1 = midpoint(A--B);
pair I = incenter(A,B,C);
pair G = centroid(A,B,C);
pair X = rotate(180,I)*P;
pair S = incenter(A1,B1,C1);
pair N1 = rotate(180,S)*I;
pair N = intersectionpoint(A--Pa, I--N1);
/* Draw objects */
draw(incircle(A,B,C), rgb(0.9,0,0));
draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1));
draw(A--A1, rgb(0.4,0.6,0.8)+linewidth(1));
draw(A--Pa, rgb(0.4,0.6,0.8)+linewidth(1));
draw(X--P, rgb(0.4,0.6,0.8)+linewidth(1));
draw(I--A1, rgb(0.4,0.6,0.8)+linewidth(1));
draw(I--N, rgb(0.4,0.6,0.8)+linewidth(1));
/* Place dots on and label each point */
dot(A); label("$A$", A, dir(100));
dot(B); label("$B$", B, dir(200));
dot(C); label("$C$", C, dir(340));
dot(P); label("$P$", P, dir(-90));
dot(Pa); label("$P_A$", Pa, dir(-90));
dot(A1); label("$A_1$", A1, dir(-90));
dot(I); label("$I$", I, dir(150));
dot(G); label("$G$", G, dir(90));
dot(X); label("$X$", X, dir(70));
dot(S); label("$S$", S, dir(-90));
dot(N); label("$N$", N, dir(-90));
[/asy]](http://latex.artofproblemsolving.com/8/0/b/80b6003d8b91eba29ad1c8d9453a8de7abe829a1.png)
Lemma: The incenter of
is the Nagel point of its medial triangle.Proof: Note that
is the external center of similitude of the incircle and the A-excircle, which is tangent to 
at 
. Let 
be the first point of intersection of 
and the incircle. From , it is evident that the tangent to 
at is parallel to ; it follows that 
is a diameter of the incircle.
is the midpoint of 
, so it follows 
is homothetic to 
; in particular, 
.Considering the homothecy
, the ray is mapped to the ray 
: this occurs for each cevian , so it follows that under , is mapped to , and the result follows. 
Corollary:
are collinear.Proof: This follows from the fact that
is the center of the homothecy , which maps to and to . Moreover, 
. This post has been edited 3 times. Last edited by liberator, Aug 18, 2014, 10:11 PM
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