Proving collinearities with the Newton-Gauss line
by liberator, Apr 11, 2017, 10:29 PM
So apparently I'm being peer pressured into posting 
Here is the desired blog entry, a whole 364 days after the last! I hope that this satisfies your requests/demands or whatever you want to call it.
This is just to illustrate a nice little trick that I've used to solve several (read: three) problems.
To show that three points are collinear, we can use the converse of the Newton-Gauss line property, by proving that circles on appropriate diameters are coaxal (orthocentres often come in handy here), which can give us the fourth side of a complete quadrilateral. This is perhaps best demonstrated by three examples:
Problem 1: Let
be a triangle. Let 
and 
be the circumcevian triangles of points 
and 
respectively. Prove that 
and the triangle determined by lines 
are in perspective.
[Jerabek's theorem]
![[asy]
unitsize(2cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
path carc(pair A, pair B, pair C, real d=0, bool dir) {
pair O=circumcenter(A,B,C);
return arc(O,circumradius(A,B,C),degrees(A-O)+d,degrees(C-O)-d,dir);
}
pair A=dir(110), B=dir(200), C=dir(-20), Q=(-0.1,-0.05), P_a=-A, P_b=-B, P_c=-C, Q_a=IP(unitcircle,L(A,Q,-0.1,10)), Q_b=IP(unitcircle,L(B,Q,-0.1,10)), Q_c=IP(unitcircle,L(C,Q,-0.1,10)), X=extension(P_a,Q_a,B,C), Y=extension(P_b,Q_b,C,A), Z=extension(P_c,Q_c,A,B);
D(X--B--A--C);
D(unitcircle,heavygreen);
D(carc(A,foot(A,B,C),X,CCW),red);
D(X--Q_a--A--foot(A,B,C));
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("Q",Q,E);
D("Q_a",Q_a);
D("P_a",P_a,N);
D("X",X,E);
D("H",A+B+C,SSE);
[/asy]](//latex.artofproblemsolving.com/f/f/5/ff53bd829f6adc02545920abcdf07f47fb85d7c2.png)
Solution. Let
, and define 
analogously. Take a homography fixing 
and sending to the centre. It is easy to see that the circles on diameters 
are coaxal, with common radical axis 
. Then by the converse of the Newton-Gauss line, it follows that 
are collinear. 
[Sorry for the lack of complete diagrams for the next two problems. Some of the points are very far apart, meaning that the diagrams would probably not be very clear anyway]
Problem 2: A circle
with centre intersects the sides 
of triangle at 
and 
, 
and 
, 
and 
respectively. Let 
be the circumcentre of triangle 
, 
the circumcentre of triangle 
, 
the circumcentre of triangle 
. Prove that 
, 
and 
are concurrent.
[UK Training for RMM 2017]
![[asy]
unitsize(2cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
pair P=origin, A_1=dir(230), A_2=dir(-50), B_1=dir(-20), B_2=dir(95), C_1=dir(120), C_2=dir(185), A=extension(B_1,B_2,C_1,C_2), B=extension(C_1,C_2,A_1,A_2), C=extension(A_1,A_2,B_1,B_2), Ap=circumcenter(A_1,A_2,P), Bp=circumcenter(B_1,B_2,P), Cp=circumcenter(C_1,C_2,P), X=2/(conj(IP(A--Ap,unitcircle))+conj(OP(A--Ap,unitcircle)));
D(A--B--C--cycle);
DPA(A--Ap^^B--Bp^^C--Cp);
D(unitcircle,heavygreen);
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("A'",Ap,E);
D("B'",Bp);
D("C'",Cp,SW);
D("P",P,E);
[/asy]](//latex.artofproblemsolving.com/1/0/4/1040b46ebc7f74aadcc798bafb674c1c7977e396.png)
Solution. Let the polar of
in meet the polar of at 
, and define 
analogously. It suffices to show that 
are collinear. Let 
be the triangle formed by the polars of 
in , which is homothetic to with centre . Consider the circles on diameters 
. Easily, and the orthocentre of lie on a common radical axis of these circles. Thus the midpoints of are collinear, so the result follows by the converse of the Newton-Gauss line.
Problem 3: A circle with centre intersects the sides of triangle at and , and , and respectively. Let the lines tangent to at and meet at and define similarly and . Prove that the lines , , and are concurrent.
[Silouanas Brazitikos, UK IMO Training and Selection Camp April 2017]
![[asy]
unitsize(2cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
pair P=origin, A_1=dir(230), A_2=dir(-50), B_1=dir(-20), B_2=dir(95), C_1=dir(120), C_2=dir(185), A=extension(B_1,B_2,C_1,C_2), B=extension(C_1,C_2,A_1,A_2), C=extension(A_1,A_2,B_1,B_2), Ap=2/(conj(A_1)+conj(A_2)), Bp=2/(conj(B_1)+conj(B_2)), Cp=2/(conj(C_1)+conj(C_2)), X=extension(B,C,tangent(A,P,1),tangent(A,P,1,2));
D(A--B--C--cycle);
DPA(A--Ap^^B--Bp^^C--Cp);
D(unitcircle,heavygreen);
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("A'",Ap);
D("B'",Bp,NE);
D("C'",Cp,NW);
D("P",P,E);
[/asy]](//latex.artofproblemsolving.com/7/5/8/7589f309fd84c2af1fcad5b83f786616906562dc.png)
Solution. Let the polar of in meet 
at , and define analogously. It suffices to show that are collinear. Consider the circles on diameters . Easily, and the orthocentre of lie on a common radical axis of these circles. Thus the midpoints of are collinear, so the result follows by the converse of the Newton-Gauss line.
Here is the desired blog entry, a whole 364 days after the last! I hope that this satisfies your requests/demands or whatever you want to call it.This is just to illustrate a nice little trick that I've used to solve several (read: three) problems.
To show that three points are collinear, we can use the converse of the Newton-Gauss line property, by proving that circles on appropriate diameters are coaxal (orthocentres often come in handy here), which can give us the fourth side of a complete quadrilateral. This is perhaps best demonstrated by three examples:
Problem 1: Let
be a triangle. Let 
and 
be the circumcevian triangles of points 
and 
respectively. Prove that 
and the triangle determined by lines 
are in perspective.[Jerabek's theorem]
![[asy]
unitsize(2cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
path carc(pair A, pair B, pair C, real d=0, bool dir) {
pair O=circumcenter(A,B,C);
return arc(O,circumradius(A,B,C),degrees(A-O)+d,degrees(C-O)-d,dir);
}
pair A=dir(110), B=dir(200), C=dir(-20), Q=(-0.1,-0.05), P_a=-A, P_b=-B, P_c=-C, Q_a=IP(unitcircle,L(A,Q,-0.1,10)), Q_b=IP(unitcircle,L(B,Q,-0.1,10)), Q_c=IP(unitcircle,L(C,Q,-0.1,10)), X=extension(P_a,Q_a,B,C), Y=extension(P_b,Q_b,C,A), Z=extension(P_c,Q_c,A,B);
D(X--B--A--C);
D(unitcircle,heavygreen);
D(carc(A,foot(A,B,C),X,CCW),red);
D(X--Q_a--A--foot(A,B,C));
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("Q",Q,E);
D("Q_a",Q_a);
D("P_a",P_a,N);
D("X",X,E);
D("H",A+B+C,SSE);
[/asy]](http://latex.artofproblemsolving.com/f/f/5/ff53bd829f6adc02545920abcdf07f47fb85d7c2.png)
Solution. Let
, and define 
analogously. Take a homography fixing 
and sending to the centre. It is easy to see that the circles on diameters 
are coaxal, with common radical axis 
. Then by the converse of the Newton-Gauss line, it follows that 
are collinear. 
[Sorry for the lack of complete diagrams for the next two problems. Some of the points are very far apart, meaning that the diagrams would probably not be very clear anyway]
Problem 2: A circle
with centre intersects the sides 
of triangle at 
and 
, 
and 
, 
and 
respectively. Let 
be the circumcentre of triangle 
, 
the circumcentre of triangle 
, 
the circumcentre of triangle 
. Prove that 
, 
and 
are concurrent.[UK Training for RMM 2017]
![[asy]
unitsize(2cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
pair P=origin, A_1=dir(230), A_2=dir(-50), B_1=dir(-20), B_2=dir(95), C_1=dir(120), C_2=dir(185), A=extension(B_1,B_2,C_1,C_2), B=extension(C_1,C_2,A_1,A_2), C=extension(A_1,A_2,B_1,B_2), Ap=circumcenter(A_1,A_2,P), Bp=circumcenter(B_1,B_2,P), Cp=circumcenter(C_1,C_2,P), X=2/(conj(IP(A--Ap,unitcircle))+conj(OP(A--Ap,unitcircle)));
D(A--B--C--cycle);
DPA(A--Ap^^B--Bp^^C--Cp);
D(unitcircle,heavygreen);
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("A'",Ap,E);
D("B'",Bp);
D("C'",Cp,SW);
D("P",P,E);
[/asy]](http://latex.artofproblemsolving.com/1/0/4/1040b46ebc7f74aadcc798bafb674c1c7977e396.png)
Solution. Let the polar of
in meet the polar of at 
, and define 
analogously. It suffices to show that 
are collinear. Let 
be the triangle formed by the polars of 
in , which is homothetic to with centre . Consider the circles on diameters 
. Easily, and the orthocentre of lie on a common radical axis of these circles. Thus the midpoints of are collinear, so the result follows by the converse of the Newton-Gauss line.
Problem 3: A circle
with centre intersects the sides of triangle at and , and , and respectively. Let the lines tangent to at and meet at and define similarly and . Prove that the lines , , and are concurrent.[Silouanas Brazitikos, UK IMO Training and Selection Camp April 2017]
![[asy]
unitsize(2cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10);
pair P=origin, A_1=dir(230), A_2=dir(-50), B_1=dir(-20), B_2=dir(95), C_1=dir(120), C_2=dir(185), A=extension(B_1,B_2,C_1,C_2), B=extension(C_1,C_2,A_1,A_2), C=extension(A_1,A_2,B_1,B_2), Ap=2/(conj(A_1)+conj(A_2)), Bp=2/(conj(B_1)+conj(B_2)), Cp=2/(conj(C_1)+conj(C_2)), X=extension(B,C,tangent(A,P,1),tangent(A,P,1,2));
D(A--B--C--cycle);
DPA(A--Ap^^B--Bp^^C--Cp);
D(unitcircle,heavygreen);
D("A",A,NW);
D("B",B,SW);
D("C",C,SE);
D("A'",Ap);
D("B'",Bp,NE);
D("C'",Cp,NW);
D("P",P,E);
[/asy]](http://latex.artofproblemsolving.com/7/5/8/7589f309fd84c2af1fcad5b83f786616906562dc.png)
Solution. Let the polar of
in meet 
at , and define analogously. It suffices to show that are collinear. Consider the circles on diameters . Easily, and the orthocentre of lie on a common radical axis of these circles. Thus the midpoints of are collinear, so the result follows by the converse of the Newton-Gauss line. 
October 2018
April 2018