A Short Existence Proof of the Polar
by djmathman, Jul 14, 2022, 10:05 PM
I derived this proof last year while teaching Geo 4 at AwesomeMath and realized it wasn't posted anywhere. I figured this should be public somewhere.
Before proceeding with the proof, we will need a fact that is interesting in its own right. I'm not sure how well-known this is, but it was certainly surprising to me when I learned it last year.
Lemma. Let
be a circle, and let 
be a line passing through its center . For each point 
, let 
be the image of 
under inversion about . Then 
is a projective map. In other words, for any points 
, 
, 
, and 
on ,
![\[
(ABCD) = (f(A)f(B)f(C)f(D)).
\]](//latex.artofproblemsolving.com/6/9/d/69dd6249418be48e72e5666a5538020071d6ec4f.png)
(Note that 
denotes the cross ratio. Ever since I first taught Geo 4, I write the cross ratio without commas. Oops.)
Proof. Relabel
as 
for ease of notation. It suffices to show that negative inversions are projective maps, since the difference between these two types of inversions is merely a reflection.
![[asy]
size(230);
defaultpen(linewidth(0.6)+fontsize(11));
pair P = (-0.6,0), Pp = (1/0.6,0), O = origin, Q = (0,1);
draw(unitcircle,red);
label("$\omega$",dir(110),dir(110),red);
draw((-1.2,0)--(2,0),blue);
label("$\ell$",(1.9,0),N,blue);
draw(P--Q--Pp^^rightanglemark(P,Q,Pp,3)^^O--Q^^rightanglemark(Q,O,Pp,3),gray);
dot("$P$",P,S,linewidth(3.3));
dot("$P'$",Pp,S,linewidth(3.3));
dot("$O$",O,S,linewidth(3.3));
dot("$Q$",Q,N);
[/asy]](//latex.artofproblemsolving.com/9/7/a/97a3cdcea74c06161aa3a90d9dc7ba6ae59fa25f.png)
Let
be the radius of , and let 
be a fixed point on such that 
. Note that 
, so 
. This means the map may be written as the composition of three maps
![\[
P \mapsto QP \mapsto QP' \mapsto P'.
\]](//latex.artofproblemsolving.com/5/c/4/5c4637e86e664d66df38ab2eb21c96bcdba970cf.png)
(In particular, the second map is a 
rotation in the pencil of lines through .) Each of these maps is projective, so is, too. 
.
We are now ready to move to our main result. Let be a circle and 
be any point. A variable line through intersects at two points and . Let be the unique point on for which 
.
Theorem (Existence of Polar). As varies, point lies on a fixed line 
. This line is called the polar of with respect to .
![[asy]
size(230);
defaultpen(linewidth(0.6)+fontsize(11));
real r = 0.35;
pair A = dir(140), B = dir(20), P = r*A + (1-r)*B, Q = 1/(1-r)*B-r/(1-r)*A, O = P/2, R = foot(Q,O,P), S = 1.3*R-0.3*Q, T = 1.1*Q-0.1*R;
draw(unitcircle,red);
draw(A--Q,blue);
draw(S--T,gray);
draw(circle(O,abs(O)),gray+linetype("3 3"));
dot("$A$",A,NW,linewidth(3.3));
dot("$B$",B,NE,linewidth(3.3));
dot("$P$",P,S,linewidth(3.3));
dot("$Q$",Q,NE,linewidth(3.3));
dot("$O$",origin,S,linewidth(3.3));
label("$\omega$",dir(100),dir(100),red);
label("$\ell$",(B+Q)/2,S,blue);
label("$p$",(Q+R)/2,NE,gray);
dot("$Q'$",(A+B)/2,NE,linewidth(3.3));
[/asy]](//latex.artofproblemsolving.com/e/5/8/e5862599a134b7c86d9ea8d0cd4a19844d4a4e87.png)
Proof. Consider the inversion
centered at with power 
. (Note that has negative power if lies inside the circle, as is the case in the diagram above.) Let 
be the image of with respect to . Observe that swaps and and sends to the infinity point 
in the direction of . Thus, since inversion is a projective map,
![\[
-1 = (ABPQ) \stackrel{\Phi}= (BA\infty_\ell Q') = \frac{\overrightarrow{BQ'}}{\overrightarrow{AQ'}}.
\]](//latex.artofproblemsolving.com/4/a/7/4a7342364f6cd8bcd3fd6b8634df6ea3c66835ed.png)
It follows that is the midpoint of 
. That is, 
.
it follows that, as the line varies, point moves along the circle with diameter 
. Inverting back gives that lies on a line, as desired. 
Before proceeding with the proof, we will need a fact that is interesting in its own right. I'm not sure how well-known this is, but it was certainly surprising to me when I learned it last year.
Lemma. Let
be a circle, and let 
be a line passing through its center . For each point 
, let 
be the image of 
under inversion about . Then 
is a projective map. In other words, for any points 
, 
, 
, and 
on ,![\[
(ABCD) = (f(A)f(B)f(C)f(D)).
\]](http://latex.artofproblemsolving.com/6/9/d/69dd6249418be48e72e5666a5538020071d6ec4f.png)
(Note that 
denotes the cross ratio. Ever since I first taught Geo 4, I write the cross ratio without commas. Oops.)Proof. Relabel
as 
for ease of notation. It suffices to show that negative inversions are projective maps, since the difference between these two types of inversions is merely a reflection.![[asy]
size(230);
defaultpen(linewidth(0.6)+fontsize(11));
pair P = (-0.6,0), Pp = (1/0.6,0), O = origin, Q = (0,1);
draw(unitcircle,red);
label("$\omega$",dir(110),dir(110),red);
draw((-1.2,0)--(2,0),blue);
label("$\ell$",(1.9,0),N,blue);
draw(P--Q--Pp^^rightanglemark(P,Q,Pp,3)^^O--Q^^rightanglemark(Q,O,Pp,3),gray);
dot("$P$",P,S,linewidth(3.3));
dot("$P'$",Pp,S,linewidth(3.3));
dot("$O$",O,S,linewidth(3.3));
dot("$Q$",Q,N);
[/asy]](http://latex.artofproblemsolving.com/9/7/a/97a3cdcea74c06161aa3a90d9dc7ba6ae59fa25f.png)
Let
be the radius of , and let 
be a fixed point on such that 
. Note that 
, so 
. This means the map may be written as the composition of three maps![\[
P \mapsto QP \mapsto QP' \mapsto P'.
\]](http://latex.artofproblemsolving.com/5/c/4/5c4637e86e664d66df38ab2eb21c96bcdba970cf.png)
(In particular, the second map is a 
rotation in the pencil of lines through .) Each of these maps is projective, so is, too. 
.We are now ready to move to our main result. Let
be a circle and 
be any point. A variable line through intersects at two points and . Let be the unique point on for which 
.Theorem (Existence of Polar). As
varies, point lies on a fixed line 
. This line is called the polar of with respect to .![[asy]
size(230);
defaultpen(linewidth(0.6)+fontsize(11));
real r = 0.35;
pair A = dir(140), B = dir(20), P = r*A + (1-r)*B, Q = 1/(1-r)*B-r/(1-r)*A, O = P/2, R = foot(Q,O,P), S = 1.3*R-0.3*Q, T = 1.1*Q-0.1*R;
draw(unitcircle,red);
draw(A--Q,blue);
draw(S--T,gray);
draw(circle(O,abs(O)),gray+linetype("3 3"));
dot("$A$",A,NW,linewidth(3.3));
dot("$B$",B,NE,linewidth(3.3));
dot("$P$",P,S,linewidth(3.3));
dot("$Q$",Q,NE,linewidth(3.3));
dot("$O$",origin,S,linewidth(3.3));
label("$\omega$",dir(100),dir(100),red);
label("$\ell$",(B+Q)/2,S,blue);
label("$p$",(Q+R)/2,NE,gray);
dot("$Q'$",(A+B)/2,NE,linewidth(3.3));
[/asy]](http://latex.artofproblemsolving.com/e/5/8/e5862599a134b7c86d9ea8d0cd4a19844d4a4e87.png)
Proof. Consider the inversion
centered at with power 
. (Note that has negative power if lies inside the circle, as is the case in the diagram above.) Let 
be the image of with respect to . Observe that swaps and and sends to the infinity point 
in the direction of . Thus, since inversion is a projective map,![\[
-1 = (ABPQ) \stackrel{\Phi}= (BA\infty_\ell Q') = \frac{\overrightarrow{BQ'}}{\overrightarrow{AQ'}}.
\]](http://latex.artofproblemsolving.com/4/a/7/4a7342364f6cd8bcd3fd6b8634df6ea3c66835ed.png)
It follows that is the midpoint of 
. That is, 
.it follows that, as the line
varies, point moves along the circle with diameter 
. Inverting back gives that lies on a line, as desired. 
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