Six Point Circle Theorem
by yayups, Jun 28, 2017, 2:11 AM
Let us start by stating the theorem:
We need to start by defining the terminology in the theorem.
Let us now prove 2 lemmas:
Let us now prove the six point circle theorem.
Note that if we take
, the circumcenter, and use the fact that the isogonal conjugate of 
is the orthocenter 
, we retrieve the nine-point circle. Notice how the proofs are very similar as well. One thing to note is that the points on the nine-point circle that are the midpoints of the orthocenter to the vertices do not have an easy generalization in the more general six point circle setting.
Six Point Circle Theorem wrote:
Let be a point in the interior of triangle 
, and let 
be the isogonal conjugate of . Let the pedal triangle of be 
, and let the pedal triangle of be 
. Then, 
lie on a circle with center the midpoint of 
.
be a point in the interior of triangle 
, and let 
be the isogonal conjugate of . Let the pedal triangle of be 
, and let the pedal triangle of be 
. Then, 
lie on a circle with center the midpoint of 
.We need to start by defining the terminology in the theorem.
![[asy]
unitsize(0.4inch);
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[/asy]](http://latex.artofproblemsolving.com/d/a/d/dad25a06403db3b9d578b56df51168e5e50d3ef0.png)
![\[\angle PAB = \angle QAC, \angle PBA = \angle QBC, \angle PCB = \angle QCB.\]](http://latex.artofproblemsolving.com/0/c/6/0c69d3d7bea94dfb9aa5afa379cfbdedb00acc30.png)
Let us now justify the existence and uniqueness of 
be the reflection of ray 
across the 
-
and 
similarly. Then, by the definition of 
.![\[\frac{\sin\angle(\ell_A,AB)}{\sin\angle(\ell_A,AC)}\frac{\sin\angle(\ell_C,CA)}{\sin\angle(\ell_C,CB)}\frac{\sin\angle(\ell_B,BC)}{\sin\angle(\ell_B,BA)}=1.\]](http://latex.artofproblemsolving.com/4/4/4/444e2347b80612244c60c2634076b55fe370db44.png)
However, by the definition of ![\[\frac{\sin\angle(\ell_A,AB)}{\sin\angle(\ell_A,AC)}\frac{\sin\angle(\ell_C,CA)}{\sin\angle(\ell_C,CB)}\frac{\sin\angle(\ell_B,BC)}{\sin\angle(\ell_B,BA)}
=
\frac{\sin\angle PAC}{\sin\angle PAB}\frac{\sin\angle PBA}{\sin\angle PBC}\frac{\sin\angle PCB}{\sin\angle PCA}
=
1
\]](http://latex.artofproblemsolving.com/3/f/a/3fa6452025f9b0b7811cef06426fea0e10d70680.png)
where the last equality follows again from trig Ceva. This justifies the existence and uniqueness of 
and the orthocenter 
is the isogonal conjugate of itself.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/c/f/2/cf2ecd8706dac1598373d216c060d2f7b927ef2e.png)
Let us now prove 2 lemmas:
Lemma 1 wrote:
Let 
be the pedal triangle of with repsect to triangle , and let be the isogonal conjugate of . Then, 
, 
, and 
.
be the pedal triangle of with repsect to triangle , and let be the isogonal conjugate of . Then, 
, 
, and 
.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/1/c/7/1c7fd691e0515e1a6757b8114c4fe9e69359692e.png)
is cyclic, since 
,![\[\angle EAQ + \angle AEF = \angle PAF + \angle APF = 90\]](http://latex.artofproblemsolving.com/c/d/9/cd9d7a50b7eb83207d1aa6f5990323e3a9535698.png)
which implies that 
Lemma 2 wrote:
Consider the same setup as in Lemma 1. Define 
to be the reflection of in 
, 
to be the reflection of in 
, and 
to be the reflection of in 
. Then, the circumcenter of 
is .
to be the reflection of in 
, 
to be the reflection of in 
, and 
to be the reflection of in 
. Then, the circumcenter of 
is .![[asy]
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[/asy]](http://latex.artofproblemsolving.com/d/3/6/d36259ed3736bce1f1fab0e0939b84ec947786bf.png)
,
.
is the perpendicular bisector of 
,
.
since 
since 
.
.
,
is the perpendicular bisector of 
,
is the perpendicular bisector of 
.Let us now prove the six point circle theorem.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/e/a/d/ead8ec48e154e50ece7ed2f2b2f937770e610cb3.png)
in 
,
in 
,
in 
is 
centered at 
.
,
,
,
is 
,
is the midpoint of 
.
.
,
,
.
,
,
.Note that if we take
, the circumcenter, and use the fact that the isogonal conjugate of is the orthocenter 
, we retrieve the nine-point circle. Notice how the proofs are very similar as well. One thing to note is that the points on the nine-point circle that are the midpoints of the orthocenter to the vertices do not have an easy generalization in the more general six point circle setting.This post has been edited 15 times. Last edited by yayups, Oct 10, 2017, 9:08 PM
April 2021
September 2020