Projective Gauss Line
by yayups, May 13, 2019, 6:43 AM
The inspiration (and much of the content) of this post came from a discussion I was having recently with some people.
In this post, we'll outline a projective viewpoint on the Gauss Line, but first, let's give the standard presentation.
![[asy]
unitsize(1inches);
pair A=1.1*dir(110);
pair B=1.5*dir(65);
pair C=dir(-50);
pair D=0.7*dir(180);
pair P=extension(A,D,B,C);
pair Q=extension(A,B,C,D);
pair H=orthocenter(B,C,Q);
pair U=foot(B,Q,C);
pair V=foot(C,B,Q);
pair W=foot(Q,B,C);
pair X=IP(CP((A+C)/2,A),CP((D+B)/2,B));
pair Y=OP(CP((A+C)/2,A),CP((D+B)/2,B));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(230));
dot("$P$",P,dir(90));
dot("$Q$",Q,dir(180));
dot("$H$",H,1.9*dir(150));
dot("$U$",U,dir(260));
dot("$V$",V,dir(90));
dot("$W$",W,dir(-20));
draw(C--P);
draw(C--Q);
draw(D--P);
draw(B--Q);
draw(B--U,dashed+red);
draw(C--V,dashed+red);
draw(Q--W,dashed+red);
draw(CP((A+C)/2,A),dotted+blue);
draw(CP((D+B)/2,B),dotted+blue);
draw(CP((P+Q)/2,P),dotted+blue);
draw(X--Y,green);
[/asy]](//latex.artofproblemsolving.com/9/3/e/93ea3115a9832a30c9f31c63318bd43790ce0f9a.png)
The proof goes as follows. Let
be the orthocenter of 
, and let 
be the orthic triangle of . Note that the power of with respect to 
is 
, the power with respect to 
is 
, and the power with respect to 
is 
. These are all equal, so has equal power with respect to , , and . We may repeat a similar argument for the orthocenters of 
, 
, and 
, and we see then that , , and are coaxial. Thus, their centers are collinear, and this proves the existence of the Gauss line.
There are two things to check. Firstly, by taking a limit of appropriate thin ellipses inscribed in
, one that approaches segment 
, one for segment 
, and one for 
, we see that these midpoints are on this locus. However, the main thing we need to check is that this locus is a line. Note that projectively, the center of a conic is the pole of the line at infinity with respect to the conic.
So the problem reduces to the following:
It will be much easier to prove the dual statement, which we can get by letting 
, 
, 
, and 
.
We'll prove this version. Let
be the line tangent to 
at 
.
Let
be some random conic passing through 
. By Desargue's involution theorem with line , we see that there is some fixed involution (dependent only on ) that swaps the two points 
. By construction, is a fixed point of this involution, so there must be a second fixed point, call it 
. We see that and are fixed by this involution, so this involution must be harmonic conjugation in 
. Thus,
![\[(LL';UV)=-1\]](//latex.artofproblemsolving.com/5/b/f/5bf10e704961b4ff8dd339906bc1ee05f1031fb9.png)
so lies on the polar of with respect to , as desired (note that is only dependent on and ). 
This concludes the projective classification of the Gauss Line. Note that this then trivializes the theorem that the incenter of (if it exists) lies on the Gauss Line.
In this post, we'll outline a projective viewpoint on the Gauss Line, but first, let's give the standard presentation.
Gauss Line wrote:
For a complete quadrilateral 
(
and 
), the midpoints of , and are all collinear.
(
and 
), the midpoints of 
, 
and 
are all collinear.![[asy]
unitsize(1inches);
pair A=1.1*dir(110);
pair B=1.5*dir(65);
pair C=dir(-50);
pair D=0.7*dir(180);
pair P=extension(A,D,B,C);
pair Q=extension(A,B,C,D);
pair H=orthocenter(B,C,Q);
pair U=foot(B,Q,C);
pair V=foot(C,B,Q);
pair W=foot(Q,B,C);
pair X=IP(CP((A+C)/2,A),CP((D+B)/2,B));
pair Y=OP(CP((A+C)/2,A),CP((D+B)/2,B));
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(230));
dot("$P$",P,dir(90));
dot("$Q$",Q,dir(180));
dot("$H$",H,1.9*dir(150));
dot("$U$",U,dir(260));
dot("$V$",V,dir(90));
dot("$W$",W,dir(-20));
draw(C--P);
draw(C--Q);
draw(D--P);
draw(B--Q);
draw(B--U,dashed+red);
draw(C--V,dashed+red);
draw(Q--W,dashed+red);
draw(CP((A+C)/2,A),dotted+blue);
draw(CP((D+B)/2,B),dotted+blue);
draw(CP((P+Q)/2,P),dotted+blue);
draw(X--Y,green);
[/asy]](http://latex.artofproblemsolving.com/9/3/e/93ea3115a9832a30c9f31c63318bd43790ce0f9a.png)
The proof goes as follows. Let
be the orthocenter of 
, and let 
be the orthic triangle of . Note that the power of with respect to 
is 
, the power with respect to 
is 
, and the power with respect to 
is 
. These are all equal, so has equal power with respect to , , and . We may repeat a similar argument for the orthocenters of 
, 
, and 
, and we see then that , , and are coaxial. Thus, their centers are collinear, and this proves the existence of the Gauss line.Projective Gauss Line wrote:
The Gauss line is the locus of the centers of all inscribed conics to 
.
.There are two things to check. Firstly, by taking a limit of appropriate thin ellipses inscribed in
, one that approaches segment , one for segment , and one for , we see that these midpoints are on this locus. However, the main thing we need to check is that this locus is a line. Note that projectively, the center of a conic is the pole of the line at infinity with respect to the conic.So the problem reduces to the following:
Rephrased Problem v1 wrote:
Fix some line 
, and consider the family of conics 
inscribed in . The locus of the pole of in is a line.
, and consider the family of conics 
inscribed in . The locus of the pole of in is a line.
, 
, 
, and 
.Rephrased Problem v2 wrote:
Fix some point , and consider the family of conics that pass through some fixed points . The polar of in passes through a fixed point.
, and consider the family of conics 
that pass through some fixed points 
. The polar of in passes through a fixed point.We'll prove this version. Let
be the line tangent to 
at .Let
be some random conic passing through . By Desargue's involution theorem with line , we see that there is some fixed involution (dependent only on ) that swaps the two points 
. By construction, is a fixed point of this involution, so there must be a second fixed point, call it 
. We see that and are fixed by this involution, so this involution must be harmonic conjugation in 
. Thus,![\[(LL';UV)=-1\]](http://latex.artofproblemsolving.com/5/b/f/5bf10e704961b4ff8dd339906bc1ee05f1031fb9.png)
so lies on the polar of with respect to , as desired (note that is only dependent on and ). 
This concludes the projective classification of the Gauss Line. Note that this then trivializes the theorem that the incenter of
(if it exists) lies on the Gauss Line.This post has been edited 2 times. Last edited by yayups, Aug 4, 2019, 6:59 AM
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