Nine-point Center is awesome
by XmL, Aug 15, 2014, 9:43 PM
As we all know, a lot of properties concerning the nine-point center are hard to prove synthetically. An example of this can be found here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=550000&p=3549742#p3549742. Anyway I've stumble upon a few more while working on a problem that Arab gave me(Correlary 3) and here they are:
We start with a pretty awesome Generalization:
Generalization 1:
are the medial, altitude triangle of
, it's well known that the nine-point circle is their circumcircle. Let
with
the Euler line of
,
is the circumcevian triangle of
wrt
. Prove that
is the circumcevian triangle of some point on
wrt
.
Proof: Let
be the circumcevian triangle. It's well known that
the Euler line of
, hence by Pascal's theorem
concur on the Euler line.
.
With this Generalization in mind, we examine the special case where
becomes the nine-point center of
. If we let
denote the nine-point center of
, then
. It's easy to prove by Ceva's theorem that
concur implies
concur.
Correlary 1: The reflection of the
-altitude over
concur, where
.
Proof: Let
be the foot of
-altitude on
, we will show that
and our assertion follows. Let
denote the reflection of the circumcenter of
over
, hence
is the midpoint of
and that
are reflesive over the perpendicular bisector of
, therefore
and we are done.
.
What a beautiful property regarding the nine-point center!
Correlary 2: Let
denote the circumcenter of
, let
denote the reflection of
over
, similarly define
. Prove that
concur.
Proof: Since
, we reflect
over
to get
, hence
. Since
perpendicularly bisects
, hence
. Moreover it's easy to prove that
, hence
is parallelogram
If we similarly define
, then
are homothetic and
concur
concur and we are done.
.
Note that
is homothetic to the pedal triangle of the Konita's point of
and
corresponds to the Konita's points, therefore we now have a cool property regarding Konita's point!
Also if we let
denote the point at the concurrence, then the circumcenter of
lies on
. To prove this, just note that
lies on the Neuberg Cubic and the assertion is simply a property of the cubic.
Correlary 3: Reflect
over
to obatin
, similarly define
, then
concur.
Proof: Construct
such that its medial triangle is
, hence now we just have to prove
concur. By Desargue's theorem this is equivalent to
and the cyclic contructions are collinear. Apply an inversion at
with an arbitrary radius, Let
. Hence the pole of
wrt the inversion circle is the reflection of
over
and the pole of
is the interestion of the polar of
with
where
are the poles of
. Together the pole of
is the reflection of
over
. Since it's easy to prove that
is homothetic to
and
are obviously two corresponding points, therefore we have the poles of
and the cyclic contructions concur
the intersection points are collinear and we are done.
.
We start with a pretty awesome Generalization:
Generalization 1:











Proof: Let





With this Generalization in mind, we examine the special case where







Correlary 1: The reflection of the



Proof: Let













What a beautiful property regarding the nine-point center!
Correlary 2: Let







Proof: Since
















Note that



Also if we let




Correlary 3: Reflect





Proof: Construct























This post has been edited 2 times. Last edited by XmL, Aug 16, 2014, 2:22 AM