More on The Kosnita Point
by XmL, Apr 25, 2016, 10:09 PM
In this post, we dive into the realm of Kosnita point to examine its properties and relations with other special points. We will look at many interpretations of the point and how it plays a role with respect to different triangles. In particular, we will disect the significance of the intersections of line
with the circumcircle of
.
We will make
our main reference triangle,
are respectively the circumcenter, orthocenter, nine-point center, and Kosnita point of
. I will also label the propositions with trivial or non-trivial. This is done for personal note so do not be alarmed if it does not appear that way.
Section 1 Key Configuration
Proposition 1(non-trivial): Let
denote the reflection triangle of the Kosnita point
of
. Reflect
about the midpoints of
to get
; define
similarly. Surprisingly, it is true that
.
Proof: Let
be the reflection of
over
. Define
. The thing that makes Kosnita special is that we have
. The perpendicular now follows because
. 
A immediate consequence of this is that triangles
are perspective at
. In addition, we have
and particularly
. These facts will become important later in discussions about the perpendiculars to
.
Of course, if you've read my previous blogs, then these results are nothing new and were used in my proof for the infamous
problem. In my collaborated solution with Ivan Zelich, one of our Lemmas was proving
, where
is the intersection of
with the side of the Ceva triangle of
that faces
. This fact, however, is immediate from our observations above by considering the harmonic pencil
(point at infinity is on
) and its perpendicular counterpart through
.
Section 2 Perspective at
For a moment we will take Kosnita out of the picture and look at somethings more intrinsic to
:
Proposition 2(non-trivial): Let
be on
such t hat
; define
similarly. Prove that
and the tangential triangle of
are orthologic to each other at
, the orthocenter of
.
Proof: It suffices to show that
, where
is the intersection of the tangents to
through
. Perhaps unexpectedly, I will call upon poles and polar to show this.
Let
be the orthic triangle of
, so
is the center of the incircle
of
. Let
denote the poles of
wrt
. The pole of
is a point
that satisfies
. It is not hard to see that
; thus
are homothetic through H
and we are done. 
Remark:
is also the Ceva triangle of the isotomic conjugate of
wrt
. Also, this might not be the easiest way to prove something about
since there are more natural constructions that are related to that line.
Let
be the reflection triangle of
wrt
. Suppose
. Since
is clearly a parallelogram(
are congruent and homothetic through
), thus
. We now show a result that can be generalized:
Proposition 3(Trivial):
.
Proof: Note that
, thus ![\[\angle O_BAJ=\angle ZO_CL, \angle AJO_B=\angle O_CLZ \implies AJO_B\sim O_CLZ \]](//latex.artofproblemsolving.com/c/d/3/cd38fe26cbf93100f8e55086817fc0407cd32efc.png)
. Similarly, we can show that
. Thus
, which means
. 
We now bring back the Kosnita point
. Recall in our first discussion that
, where
is defined as follows: Let
,
. Since
and
, by properties of harmonic division we have
. Because
, we have
are collinear. In other words, we have just proven the following:
Main Result 1(Non-trivial):
, the reflection triangle of Kosnita point
, is perspective with the tangential triangle of
at
.
Section 3
and 
We now go deeper into how Kosnita relates to various triangles. With just Proposition 1, we can actually prove:
Main Result 2(Non-trivial):The orthocenter of
, denoted by
, lies on
. In fact,
is the midpoint of
.
Proof: By properties of isogonal conjugates,
is the circumcenter of
. Let
denote the midpoint of
. Reflect
over
to get
. Thus
,
, and the corresponding parallels give us
are collinear and
. 
A corollary of this is, if
is the reflection of
over
, then
. The construction of
is analogous to the perspective approach in proving the Liang-Zelich Theorem, so it is safe to say that
is a very special point and we can show the following:
Proposition 4(Trivial):
are concyclic and the isogonal of
wrt
is parallel to
.
Proof: We showed that
, so
because
are isogonal wrt
. Thus the cyclic follows. Suppose
are isogonal wrt
. Since
are isogonals of the same angle, we have
because
. 
If we construct the analogs of
for
, then they all concur on
at
, specifically at the isogonal conjugate of the point at infinity on
(this then implies that the Simson's line of
wrt
is perpendicular to
.) Furthermore,
are collinear by Sondat's Theorem. To see this, relabel
as
and define
similarly. We showed that
and
are perspective at
(which is actually immediate by Liang-Zelich). Since
are the clearly the orthologic centers for the two triangles, Sondat's gives
. What a found! The following result summarizes this discussion:
Main result 3(non-trivial):
are collinear, where
is the isogonal conjugate of the point at infinity on
wrt
.
To prepare the construction for investigating the point
, we need the following:
Lemma(Trivial):
is the circumcenter of
. Let
be an arbitrary point. Define
on the circle
such that
. Prove that the orthologic center of
wrt
lies on
and is the isogonal conjugate of the point at infinity on
wrt
.
Proof: Let
denote the line through
and parallel to
, it suffices to show that the isogonal of
wrt
is perpendicular to
. Select a point
and let
be its feet of projection on
. By property of isogonals, it amounts to showing
, which is equivalent to
are homothetic. If we let
, then with
we can easily show that
and the homothety follows. 
We now introduce the crux construction:
Proposition 5(non-trivial): Reflect
over
to get
; similarly construct
. From our first discussion, it is clear that
. We will now show that
is a cyclic quadrilateral with center
.
Proof: It suffices to prove that
. Reflect
over
, the midpoint of
to get
. Since
is the circumcenter of
(isogonal conjugates), we know
is a diameter of
is the midpoint of
. Note that
is also the midpoint of
, so
and we are done. 
Now consider the Lemma with
. Observe that since
are reflexive about
, the corresponding triangle
in the Lemma is congruent and homothetic to
in the current diagram. Because
is the isogonal conjugate of the point ot infinity on
wrt
, therefore
is the orthologic center of
wrt
. In particular,
is the orthocenter of
.
It is not hard to see that
. Hence we have
Main result 4(Very Non-trivial):
From this we can obtain some pretty extraordinary facts:
1.
, where
is the circumradius of
.
2.
3. We can avoid Sondats and use this to deduce
are collinear
4.
is similar to well known configuration, in which
are tangent to
. It is surprising for a bunch of special points to exhibit this beautiful property!
To end this section, we reveal the identity of the second intersection between
and
:
Main Result 5(Non-trivial):Let
. We claim that
is the Euler reflection point of
, i.e.
are reflexive over
.
Proof: Since
is parallel to the reflection of
over
, thus it suffices to show that
.
We have shown before that
, thus
. By the tangent result (fact 4) above, we have
; hence our claim follows. 


We will make



Section 1 Key Configuration
Proposition 1(non-trivial): Let








Proof: Let







A immediate consequence of this is that triangles





Of course, if you've read my previous blogs, then these results are nothing new and were used in my proof for the infamous









Section 2 Perspective at

For a moment we will take Kosnita out of the picture and look at somethings more intrinsic to

Proposition 2(non-trivial): Let








Proof: It suffices to show that




Let















Remark:




Let








Proposition 3(Trivial):

Proof: Note that

![\[\angle O_BAJ=\angle ZO_CL, \angle AJO_B=\angle O_CLZ \implies AJO_B\sim O_CLZ \]](http://latex.artofproblemsolving.com/c/d/3/cd38fe26cbf93100f8e55086817fc0407cd32efc.png)





We now bring back the Kosnita point










Main Result 1(Non-trivial):




Section 3


We now go deeper into how Kosnita relates to various triangles. With just Proposition 1, we can actually prove:
Main Result 2(Non-trivial):The orthocenter of





Proof: By properties of isogonal conjugates,












A corollary of this is, if






Proposition 4(Trivial):




Proof: We showed that











If we construct the analogs of

















Main result 3(non-trivial):




To prepare the construction for investigating the point

Lemma(Trivial):











Proof: Let















We now introduce the crux construction:
Proposition 5(non-trivial): Reflect







Proof: It suffices to prove that













![\[DH_K=AT=K_AO_A=KO=KH_K\]](http://latex.artofproblemsolving.com/4/7/2/4727009935e42cde27bea3f172d9926552c0582f.png)

Now consider the Lemma with














It is not hard to see that

Main result 4(Very Non-trivial):

From this we can obtain some pretty extraordinary facts:
1.



2.

3. We can avoid Sondats and use this to deduce

4.



To end this section, we reveal the identity of the second intersection between


Main Result 5(Non-trivial):Let





Proof: Since




We have shown before that




This post has been edited 6 times. Last edited by XmL, May 19, 2016, 4:10 AM
Reason: Thank You Telv!
Reason: Thank You Telv!