Paralogic Triangles
by yayups, Jun 13, 2019, 4:00 AM
In this post, we consider Paralogic Triangles. The content of this post comes from a discussion with tastymath75025 and pinetree1. It's also mostly from Lemmas in Olympiad Geometry.
![[asy]
unitsize(1.5inches);
pair A,B,C,D,E,F,X,Y,Z,P,Q,H,HH,R;
A=dir(105);
B=dir(210);
C=dir(330);
F=(0.3)*A+(0.7)*B;
E=(3/5)*A+(2/5)*C;
D=extension(E,F,B,C);
X=2*circumcenter(A,E,F)-A;
Y=2*circumcenter(B,D,F)-B;
Z=2*circumcenter(C,D,E)-C;
P=extension(A,X,B,Y);
Q=2*foot(P,circumcenter(A,B,C),circumcenter(X,Y,Z))-P;
H=orthocenter(A,B,C);
HH=orthocenter(X,Y,Z);
R=2*foot(Q,D,E)-Q;
draw(A--B--C--cycle,linewidth(1.5)+red);
draw(D--B);
draw(X--Y--Z--cycle,linewidth(1.5)+orange);
draw(X--E,dotted);
draw(circumcircle(A,B,C));
draw(D--F--E,blue);
draw(A--X--P,green);
draw(P--B--Y,green);
draw(Z--P--C,green);
draw(circumcircle(X,Y,Z));
dot("$A$",A,dir(A));
dot("$B$",B,dir(230));
dot("$C$",C,dir(C));
dot("$F$",F,1.5*dir(105));
dot("$E$",E,dir(0));
dot("$D$",D,dir(180));
dot("$X$",X,dir(-10));
dot("$Y$",Y,dir(160));
dot("$Z$",Z,dir(200));
dot("$P$",P,1.3*dir(270));
dot("$Q$",Q,1.3*dir(110));
dot("$H$",H,1.3*dir(0));
dot("$H'$",HH,1.3*dir(180));
dot("$R$",R,1.3*dir(-30));
[/asy]](//latex.artofproblemsolving.com/8/4/2/84257186b6f8aa72f72176e4e9c53b6fadd2c304.png)
Consider
, and let 
be an arbitrary line that intersects the sides at 
, 
and 
. Let 
be the the perpendicular to 
through , and define 
and 
similarly. Finally, let 
be the triangle determined by , and .
It's clear that
because the repsective sides are perpendicular. In fact, the triangles are perspective through 
which lies on 
. Firstly, by Desargues's theorem, the triangles are perspective. To show that 
, note that
![\[\angle BCP=\angle DCZ=\angle DEZ=\angle FEX=\angle FAX=\angle BAP,\]](//latex.artofproblemsolving.com/b/9/8/b984b441625c2c151d95668b5059e6a289069cc1.png)
since 
cyclic, so 
. We also see that 
since the setup is symmetric under 
interchange.
It turns out that
and 
are in fact orthogonal. The key is to constuct the point 
which is the other intersection of the two circles. Note that 
and 
, so is the spiral center sending 
. Thus, is the center of the unique spiral similarity sending 
. Now we see that any such spiral similarity must have angle equal to 
since the corresponding sides of and 
are perpendicular. Thus, the unique spiral similarity sending to has angle , which implies the circles are orthogonal.
The point turns out to be quite nice, and in fact it is the Miquel point of 
and 
. To see this, note that by the spiral similarity, 
, so 
is cyclic, which implies it's the Miquel point of both quadrilaterals, since is already on and . In particular, this means that has a Simson line with respect to both complete quadrilaterals, which will be useful soon.
We now prove Sondat's Theorem for Paralogic Triangles. Let
be the orthocenter of and 
the orthocenter of . The theorem states that the line 
bisects 
. Note that when 
is a Simson line, then 
, and the theorem is the classic statement that the Simson line bisects 
.
Let
be the Steiner line of with respect to , which is just the Simson line scaled up by a factor of 
at . By the lemma above, we see that it passes through . But since is the Miquel point, the line also passes through the reflection of in , which we'll call 
. Thus, line is 
.
Let
be the corresponding Steiner line of with respect to . We see that is sent to under the spiral similarity at , so 
By the above, we also have 
. This implies that 
. Thus, is the perpendicular bisector of 
, which implies that it passes through the center 
, which is the midpoint of , as desired. This completes the proof. 
![[asy]
unitsize(1.5inches);
pair A,B,C,D,E,F,X,Y,Z,P,Q,H,HH,R;
A=dir(105);
B=dir(210);
C=dir(330);
F=(0.3)*A+(0.7)*B;
E=(3/5)*A+(2/5)*C;
D=extension(E,F,B,C);
X=2*circumcenter(A,E,F)-A;
Y=2*circumcenter(B,D,F)-B;
Z=2*circumcenter(C,D,E)-C;
P=extension(A,X,B,Y);
Q=2*foot(P,circumcenter(A,B,C),circumcenter(X,Y,Z))-P;
H=orthocenter(A,B,C);
HH=orthocenter(X,Y,Z);
R=2*foot(Q,D,E)-Q;
draw(A--B--C--cycle,linewidth(1.5)+red);
draw(D--B);
draw(X--Y--Z--cycle,linewidth(1.5)+orange);
draw(X--E,dotted);
draw(circumcircle(A,B,C));
draw(D--F--E,blue);
draw(A--X--P,green);
draw(P--B--Y,green);
draw(Z--P--C,green);
draw(circumcircle(X,Y,Z));
dot("$A$",A,dir(A));
dot("$B$",B,dir(230));
dot("$C$",C,dir(C));
dot("$F$",F,1.5*dir(105));
dot("$E$",E,dir(0));
dot("$D$",D,dir(180));
dot("$X$",X,dir(-10));
dot("$Y$",Y,dir(160));
dot("$Z$",Z,dir(200));
dot("$P$",P,1.3*dir(270));
dot("$Q$",Q,1.3*dir(110));
dot("$H$",H,1.3*dir(0));
dot("$H'$",HH,1.3*dir(180));
dot("$R$",R,1.3*dir(-30));
[/asy]](http://latex.artofproblemsolving.com/8/4/2/84257186b6f8aa72f72176e4e9c53b6fadd2c304.png)
Consider
, and let 
be an arbitrary line that intersects the sides at 
, 
and 
. Let 
be the the perpendicular to 
through , and define 
and 
similarly. Finally, let 
be the triangle determined by , and .It's clear that
because the repsective sides are perpendicular. In fact, the triangles are perspective through 
which lies on 
. Firstly, by Desargues's theorem, the triangles are perspective. To show that 
, note that![\[\angle BCP=\angle DCZ=\angle DEZ=\angle FEX=\angle FAX=\angle BAP,\]](http://latex.artofproblemsolving.com/b/9/8/b984b441625c2c151d95668b5059e6a289069cc1.png)
since 
cyclic, so 
. We also see that 
since the setup is symmetric under 
interchange.It turns out that
and 
are in fact orthogonal. The key is to constuct the point 
which is the other intersection of the two circles. Note that 
and 
, so is the spiral center sending 
. Thus, is the center of the unique spiral similarity sending 
. Now we see that any such spiral similarity must have angle equal to 
since the corresponding sides of and 
are perpendicular. Thus, the unique spiral similarity sending to has angle , which implies the circles are orthogonal.The point
turns out to be quite nice, and in fact it is the Miquel point of 
and 
. To see this, note that by the spiral similarity, 
, so 
is cyclic, which implies it's the Miquel point of both quadrilaterals, since is already on and . In particular, this means that has a Simson line with respect to both complete quadrilaterals, which will be useful soon.We now prove Sondat's Theorem for Paralogic Triangles. Let
be the orthocenter of and 
the orthocenter of . The theorem states that the line 
bisects 
. Note that when 
is a Simson line, then 
, and the theorem is the classic statement that the Simson line bisects 
.Let
be the Steiner line of with respect to , which is just the Simson line scaled up by a factor of 
at . By the lemma above, we see that it passes through . But since is the Miquel point, the line also passes through the reflection of in , which we'll call 
. Thus, line is 
.Let
be the corresponding Steiner line of with respect to . We see that is sent to under the spiral similarity at , so 
By the above, we also have 
. This implies that 
. Thus, is the perpendicular bisector of 
, which implies that it passes through the center 
, which is the midpoint of , as desired. This completes the proof. 
be the isogonal conjugate of 
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