451. ABC with C(O,R), C(I,r) and N-Nagel point ==> ON=R-2r.
by Virgil Nicula, Nov 26, 2016, 11:34 AM
P1. Prove that 
with the circumcircle 
the incircle 
the centroid 
the orthocenter 
and the
Nagel point
there is the relations: 
so that 
and 
Proof. Prove easily that
and 
is the Euler's line. Denote the midpoint 
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
Prove easily that 
and apply the Van Aubel's relations 
Apply an well known identity 
Is well known that
Apply the Stewart's relation to 
in 
![$\left\|IG^2\cdot \cancel{m_a}+\cancel{m_a}\cdot \frac {2m_a}3\cdot \frac {m_a}3=(bc-4Rr)\cdot \frac {\cancel{m_a}}3+\left[r^2 +\left(\frac {b-c}2\right)^2\right]\cdot \frac{2\cancel{m_a}}3\right\|\ \odot 18\iff$](//latex.artofproblemsolving.com/6/4/7/64777be1de70054cfbdfa2aeb6ba371708714282.png)
![$18\cdot IG^2+\cancel{2\left(b^2+c^2\right)}-a^2=\cancel{6bc}-24Rr+12r^2+\cancel 3\left[\left(b^2+c^2\right)-\cancel{2bc}\right].$](//latex.artofproblemsolving.com/6/b/3/6b310b5fe0f6afb8e916be7b342e0b652d096017.png)
Thus, 
From the relation 
obtain the remarkable inequality 
P2. Prove that with the incircle 
the circumcircle 
and the Nagel point there is the relation 
.
Proof. Let
. Is well-known that 
, 
and identity 
.
I denoted
. Apply the Stewart's relation to the cevian 
in 
.
Since
and 
obtain that 
. Therefore, 
![$\frac {s-a}{s^2}\cdot \left[c^2(s-b)+b^2(s-c)-a(s-b)(s-c)\right]+$](//latex.artofproblemsolving.com/9/9/a/99af5d03d6365738d1cf14e59d0cf1fbf0ab402b.png)
. In conclusion, 
and from the relation 
obtain that 
.
Remark. ![$\sum a^2(s-b)(s-c)=\sum a^2\left[bc-s(s-a)\right]=$](//latex.artofproblemsolving.com/7/2/1/72164697fbe130a813bb87d80392c0b85de58f00.png)
![$s\left[8s^3-6s\left(s^2+r^2+4Rr\right)+12Rsr\right]=$](//latex.artofproblemsolving.com/5/d/0/5d0516385585bb0785253b6fe03a95498fa94e0f.png)
. Otherwise. 
P3. Prove that with the orthocenter 
the incircle 
and the circumcircle there is the relation 
.
Proof. Denote
and apply the Stewart's relation to the cevian in 
Remark. Apply the Euler's relation to the trapezoid
Since 
obtain that 
P4 (Leonard Giugiuc & Abdikadir Altintas). Let an
right-angle 
with the centroid 
and the Lemoine's point 
Prove that 
(standard notations).
Proof. Denote
and the midpoint of the side ![$[BC]\ .$](//latex.artofproblemsolving.com/5/f/a/5fad78e281930919485d791e012363fda8c76507.png)
Is well-known or prove easily that in an 
-right-angle the -symmedian is the -altitude. From the Aubel's relation obtain
that
i.e. 
is the midpoint of the altitude 
Denote 
so that 
Thus, 
i.e.
Therefore, 
where 
Hence 
Similarly obtain
Therefore, 
In conclusion,
what is true.
with the circumcircle 
the incircle 
the centroid 
the orthocenter 
and the Nagel point
there is the relations: 
so that 
and 
Proof. Prove easily that
and 
is the Euler's line. Denote the midpoint 
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and 
Prove easily that 
and apply the Van Aubel's relations 
Apply an well known identity 
Is well known that
Apply the Stewart's relation to 
in 
![$\left\|IG^2\cdot \cancel{m_a}+\cancel{m_a}\cdot \frac {2m_a}3\cdot \frac {m_a}3=(bc-4Rr)\cdot \frac {\cancel{m_a}}3+\left[r^2 +\left(\frac {b-c}2\right)^2\right]\cdot \frac{2\cancel{m_a}}3\right\|\ \odot 18\iff$](http://latex.artofproblemsolving.com/6/4/7/64777be1de70054cfbdfa2aeb6ba371708714282.png)
![$18\cdot IG^2+\cancel{2\left(b^2+c^2\right)}-a^2=\cancel{6bc}-24Rr+12r^2+\cancel 3\left[\left(b^2+c^2\right)-\cancel{2bc}\right].$](http://latex.artofproblemsolving.com/6/b/3/6b310b5fe0f6afb8e916be7b342e0b652d096017.png)
Thus, 
From the relation 
obtain the remarkable inequality 
P2. Prove that
with the incircle 
the circumcircle 
and the Nagel point there is the relation 
.Proof. Let
. Is well-known that 
, 
and identity 
.I denoted
. Apply the Stewart's relation to the cevian 
in 
.Since
and 
obtain that 
. Therefore, 
![$\frac {s-a}{s^2}\cdot \left[c^2(s-b)+b^2(s-c)-a(s-b)(s-c)\right]+$](http://latex.artofproblemsolving.com/9/9/a/99af5d03d6365738d1cf14e59d0cf1fbf0ab402b.png)
. In conclusion, 
and from the relation 
obtain that 
.
Remark. ![$\sum a^2(s-b)(s-c)=\sum a^2\left[bc-s(s-a)\right]=$](http://latex.artofproblemsolving.com/7/2/1/72164697fbe130a813bb87d80392c0b85de58f00.png)
![$s\left[8s^3-6s\left(s^2+r^2+4Rr\right)+12Rsr\right]=$](http://latex.artofproblemsolving.com/5/d/0/5d0516385585bb0785253b6fe03a95498fa94e0f.png)
. Otherwise. 
P3. Prove that
with the orthocenter 
the incircle 
and the circumcircle there is the relation 
.Proof. Denote
and apply the Stewart's relation to the cevian in 
Remark. Apply the Euler's relation to the trapezoid
Since 
obtain that 
P4 (Leonard Giugiuc & Abdikadir Altintas). Let an
right-angle 
with the centroid 
and the Lemoine's point 
Prove that 
(standard notations).Proof. Denote
and the midpoint of the side ![$[BC]\ .$](http://latex.artofproblemsolving.com/5/f/a/5fad78e281930919485d791e012363fda8c76507.png)
Is well-known or prove easily that in an 
-right-angle the -symmedian is the -altitude. From the Aubel's relation obtainthat
i.e. 
is the midpoint of the altitude 
Denote 
so that 
Thus, 
i.e.
Therefore, 
where 
Hence 
Similarly obtain
Therefore, 
In conclusion,
what is true. This post has been edited 85 times. Last edited by Virgil Nicula, Mar 14, 2019, 2:57 PM
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