322. Incenter, Gergonne, Nagel a.s.o. of ABC.
by Virgil Nicula, Oct 14, 2011, 11:48 AM
PP1. Let 
be a triangle with the centriod 
, the circumcircle 
, the incircle 
and the medial 
. The circle touches the sides
of
in 
, 
and 
. Prove that the orthocenter of 
belongs to 
and the incircle of belongs to 
.
Proof. Note that
etc. We also know that 
is the orthocenter of the excenter triangle. Thus, 
etc. Therefore the intouch 
is homothetic to
the excenter
. So there is a point 
that takes the intouch triangle to the excenter triangle, i.e. 
and the Euler's lines of these triangles
are parallelly. But note that is the circumcenter of and is also the orthocenter of . Therefore, these triangles share the same Euler's line.
But since
is the nine-point center of the excenter triangle and also lies on the Euler line of the excenter triangle. Thus we have that , and the orthocenter
of are collinear and lie on the common Euler's line of and 
.
PP2. Let be a triangle with incircle 
. Denote 
and 
(Gergonne) , 
(Nagel)
points of . Prov that 
, where 
.
Proof 1 (own). Suppose w.l.o.g. that
and denote : the orthocenter 
and 
for which
. Show easily that 
and 
. Therefore, 
![$\left[(b+c)-a\right]^2=\left[a+(b+c)\right]\left[3a-(b+c)\right]\iff$](//latex.artofproblemsolving.com/4/2/5/425d6bca65820ba3f77ceb12e93c98f16b7c3d84.png)
.
Proof 2 (own). Suppose w.l.o.g. that . Is well-known that 
. Show easily that 
. Thus 
the division
is harmonically 
the division 
is harmonically 
.
Proof 3 (Yetti).
is foot of perpendicular from 
to 
. 
cuts at 
. Excircle 
in 
touches at 
and 
. 
cuts again at 
.
is common midpoint of , 
. 
are centrally similar with center and coefficient 
and 
. Since 
bisects 
cross ratio 
is harmonic 
. Cross ratio 
is always harmonic.
See here for one way to show it. Therefore, 
.
Proof 4 (Luisgeometra). Let
be the reflection of about 
Thus, 
bisects 
and are collinear. Barycentric coordinates of 
and
w.r.t. are given by 
and 
These points are collinear ![$\left [\begin {array}{ccc} -a^2 & a^2+b^2-c^2 & a^2+c^2-b^2\\ b+c-a& c+a-b & a+b-c\\ 0& a+b-c& c+a-b \end{array}\right]=0$](//latex.artofproblemsolving.com/c/7/a/c7a550ed518de84f95fa9343a5f175153ef8ee70.png)
![$\Longleftrightarrow 2a(c-b)[(b+c-a)^2-2a^2]=0 \Longleftrightarrow$](//latex.artofproblemsolving.com/3/1/d/31d591739070268cc4814b04070ff6a9bd666143.png)
Either
or 
be a triangle with the centriod 
, the circumcircle 
, the incircle 
and the medial 
. The circle touches the sidesof
in 
, 
and 
. Prove that the orthocenter of 
belongs to 
and the incircle of belongs to 
.Proof. Note that
etc. We also know that 
is the orthocenter of the excenter triangle. Thus, 
etc. Therefore the intouch 
is homothetic tothe excenter
. So there is a point 
that takes the intouch triangle to the excenter triangle, i.e. 
and the Euler's lines of these trianglesare parallelly. But note that
is the circumcenter of and is also the orthocenter of . Therefore, these triangles share the same Euler's line.But since
is the nine-point center of the excenter triangle and also lies on the Euler line of the excenter triangle. Thus we have that , and the orthocenterof
are collinear and lie on the common Euler's line of and 
.PP2. Let
be a triangle with incircle 
. Denote 
and 
(Gergonne) , 
(Nagel)points of
. Prov that 
, where 
.Proof 1 (own). Suppose w.l.o.g. that
and denote : the orthocenter 
and 
for which
. Show easily that 
and 
. Therefore, 
![$\left[(b+c)-a\right]^2=\left[a+(b+c)\right]\left[3a-(b+c)\right]\iff$](http://latex.artofproblemsolving.com/4/2/5/425d6bca65820ba3f77ceb12e93c98f16b7c3d84.png)
.Proof 2 (own). Suppose w.l.o.g. that
. Is well-known that 
. Show easily that 
. Thus 
the division
is harmonically 
the division 
is harmonically 
.Proof 3 (Yetti).
is foot of perpendicular from 
to 
. 
cuts at 
. Excircle 
in 
touches at 
and 
. 
cuts again at 
.
is common midpoint of , 
. 
are centrally similar with center and coefficient 
and 
. Since 
bisects 
cross ratio 
is harmonic 
. Cross ratio 
is always harmonic.See here for one way to show it. Therefore,
.Proof 4 (Luisgeometra). Let
be the reflection of about 
Thus, 
bisects 
and are collinear. Barycentric coordinates of 
and w.r.t.
are given by 
and 
These points are collinear
![$\left [\begin {array}{ccc} -a^2 & a^2+b^2-c^2 & a^2+c^2-b^2\\ b+c-a& c+a-b & a+b-c\\ 0& a+b-c& c+a-b \end{array}\right]=0$](http://latex.artofproblemsolving.com/c/7/a/c7a550ed518de84f95fa9343a5f175153ef8ee70.png)
![$\Longleftrightarrow 2a(c-b)[(b+c-a)^2-2a^2]=0 \Longleftrightarrow$](http://latex.artofproblemsolving.com/3/1/d/31d591739070268cc4814b04070ff6a9bd666143.png)
Either
or 
This post has been edited 26 times. Last edited by Virgil Nicula, Apr 20, 2016, 1:28 PM
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