5. Theoretical preliminary for inequalities.
by Virgil Nicula, Apr 19, 2010, 2:27 PM
I. Theoretical preliminary for inequalities.
Are well-known (or prove easily) some relations in
w.r.t. the centroid 
, the orthocenter 
, the Nagel's point 
, the circumcircle 
, the incircle 
,
the
-excircle 
, 
, 
, the Gergonne's point 
and the symmedian center 
(Lemoine's point). I"ll use the standard notations. For example, 
,
, 
, semiperimeter 
, where 
, area ![$ S= [ABC]$](//latex.artofproblemsolving.com/d/c/f/dcfc48a99617713cb1e96f86c4d89f9bcd6444cf.png)
, 
- the length of 
-median, 
- length of -interior (exterior) bisector , 
- length
of-altitude, 
- length -symmedian, distance 
of the point 
to the line 
, the power 
of the point 
w.r.t. the circumcircle 
of a.s.o.
Linear identities.
Let and its interior point 
, where 
, 
, 
. Then 
, where 
is the cevian trangle of w.r.t.
, i.e. (Van Aubel's relation) 
. Is evidently that 
.
![$\bullet\ \ \ (\forall )\ x_k\in (0,1]\ ,\ k\in \overline{1,n}$](//latex.artofproblemsolving.com/2/4/9/24906ff7f51fc9bd6dc55bd9af9392bb2c1a0c4b.png)
there is the relation 
![$ \bullet\ \left\|\ \begin{array}{c}
\sum a^3=3abc+(a+b+c)\cdot \left[\ \left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\ \right]\\\\
\boxed{\ (a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)\ }\end{array}\ \right\|$](//latex.artofproblemsolving.com/6/4/b/64bb53c093e397069c877bf740b29e9054ba33e7.png)
and 
.
; ![$ \sum a^2(p-b)(p-c)=p\left[abc-4\prod (p-a)\right]=4p^2r(R-r)\ .$](//latex.artofproblemsolving.com/0/e/2/0e2a5d49f2e49ab7b537059e75d6e6359dc2be1a.png)
; 
; 
; 
![$ \bullet\ \boxed{\sum bc(s-b)(s-c)=r^2\left[s^2+(4R+r)^2\right]}\ ;\ \sum $](//latex.artofproblemsolving.com/e/7/6/e7635aae57ba41705d0d1eb6a04b6fe369326be4.png)
![$ \bullet\ (p-b)(p-c)(b-c)^2\ +\ p(p-a)(b+c-2a)^2\ =\ 4\cdot\left[\ bcm_a^2\ -\ 2ap(p-a)^2\ \right]\ \Longrightarrow$](//latex.artofproblemsolving.com/a/a/6/aa6f23abd9c5c8fa41e520ad07477fb6a7127a68.png)
Lagrange's identity 
Linear-angled identities/inequalities.
. Thus, 
, i.e. 
.
![$\implies\boxed{\begin{array}{c}
\sum a\cdot\cos A=2a\sin B\sin C=\frac {2S}{R}\le s\\\\
\sum a^2\cos A=\frac {r\left[3p^2-(2R+r)(4R+r)\right]}{R}\end{array}}$](//latex.artofproblemsolving.com/a/f/f/aff6d10463779d3e1daa71a73742d907a0832321.png)
.
; 
; 
Prove easily that
.
and 
.
If the triangle 
is acute, then there are the inequalities 
and 
.
Area 
of the triangle 
(Heron) 
; 
; 
, where 
is the semiperimeter of the orthic triangle.
Therefore, 
.
For an acute triangle ,
, where 
a.s.o. 
.
I"ll use the identity 
in an acute . Thus, 
(Walker's inequality).
(Goldstone's inequality).
I"ll use the identity 
and
the inequality
From the sum of the simple inequalities 
a.s.o. obtain that 
Incircles and excircles. 
and 
If the incircle touches 
at 
, then 
.
; 
; 
are the roots of the equation
; 
.
are the roots of the equation 
are the roots of the equation 
si 
the exterior bisector of 
is right-angled 
; 
![$ \bullet\ [I_aI_bI_c]=2pR\ ,\ I_bI_c=4R\cos\frac A2$](//latex.artofproblemsolving.com/c/9/2/c92c05fa109851a8a302bd1beed9838425e4569f.png)
and the circumradius of 
has the length 
.
Remarkable distancies.
; 
and 
; 
: 
Power 
of the point w.r.t. the circumcircle 
For example, ![$ \boxed {\ \begin{array}{ccc}
M\in [BC] & \Longrightarrow & OM^2+MB\cdot MC=R^2\\\\
M\not\in [BC] & \Longrightarrow & OM^2-MB\cdot MC=R^2\end{array}\ }\ .$](//latex.artofproblemsolving.com/0/4/9/049e3a3aa5ebe9648c7631edad256dbd8527672d.png)
Arrow (dart) 
is the length of the arc 
which belongs to the interior of the angle 
and is 
; 
Characterization of the right angle in 
.
Euler's relation. 
is nonobtuse 
Theorem of Sines. 
; 
Stewart's relation. 
The length of a cevian. 
, 
Trigonometrical form of the Ceva's theorem. 
Observe that the first relation can write thus :
Symmetrical fundamental forms. 
- the roots of the equations 
.
, where 
. For example, 
and
Show similarly that 
![$ \bullet\ \sum a^2(b+c)=\sum bc(b+c)=\sum bc[(a+b+c)-a]=s_1s_2-3s_3$](//latex.artofproblemsolving.com/c/5/a/c5a3086466b3ff0f873c90bcb71e3f3fefa1bfb8.png)
![$ \sum bc\left(b^2+c^2\right)=\sum bc\left[\left(a^2+b^2+c^2\right)-a^2\right]=\sum bc\cdot\sum a^2-\sum a^2bc=$](//latex.artofproblemsolving.com/1/0/e/10e8f0f0ce3b3ba18141eb0943e332b7c653e839.png)
(Schur for 
) is equivalently with Hadwiger-Finsler. Indeed, 
Schur for 
Indeed, 
Here find the Schur's inequality and its Vornicu's extension.
Remark. The Blundon's inequality is in fact a simple consequence of the triangle inequality
Short proof. I'll use the well-known distances in a triangle
, where 
, 
and 
are the incenter , circumcenter and the Nagel's point
of the triangle . Now, one can easily show that Blundon's inequality reduces to
, which is just the triangle inequality applied in 
.
See the remarkable geometrical inequalities (1) and (2).
Are well-known (or prove easily) some relations in
w.r.t. the centroid 
, the orthocenter 
, the Nagel's point 
, the circumcircle 
, the incircle 
,the
-excircle 
, 
, 
, the Gergonne's point 
and the symmedian center 
(Lemoine's point). I"ll use the standard notations. For example, 
,
, 
, semiperimeter 
, where 
, area ![$ S= [ABC]$](http://latex.artofproblemsolving.com/d/c/f/dcfc48a99617713cb1e96f86c4d89f9bcd6444cf.png)
, 
- the length of 
-median, 
- length of -interior (exterior) bisector , 
- lengthof
-altitude, 
- length -symmedian, distance 
of the point 
to the line 
, the power 
of the point 
w.r.t. the circumcircle 
of a.s.o.
Linear identities.
Let and its interior point 
, where 
, 
, 
. Then 
, where 
is the cevian trangle of w.r.t. , i.e. (Van Aubel's relation) 
. Is evidently that 
.
![$\bullet\ \ \ (\forall )\ x_k\in (0,1]\ ,\ k\in \overline{1,n}$](http://latex.artofproblemsolving.com/2/4/9/24906ff7f51fc9bd6dc55bd9af9392bb2c1a0c4b.png)
there is the relation 
![$ \bullet\ \left\|\ \begin{array}{c}
\sum a^3=3abc+(a+b+c)\cdot \left[\ \left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\ \right]\\\\
\boxed{\ (a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)\ }\end{array}\ \right\|$](http://latex.artofproblemsolving.com/6/4/b/64bb53c093e397069c877bf740b29e9054ba33e7.png)
and 
.
; ![$ \sum a^2(p-b)(p-c)=p\left[abc-4\prod (p-a)\right]=4p^2r(R-r)\ .$](http://latex.artofproblemsolving.com/0/e/2/0e2a5d49f2e49ab7b537059e75d6e6359dc2be1a.png)
; 
; 
; 
![$ \bullet\ \boxed{\sum bc(s-b)(s-c)=r^2\left[s^2+(4R+r)^2\right]}\ ;\ \sum $](http://latex.artofproblemsolving.com/e/7/6/e7635aae57ba41705d0d1eb6a04b6fe369326be4.png)
![$ \bullet\ (p-b)(p-c)(b-c)^2\ +\ p(p-a)(b+c-2a)^2\ =\ 4\cdot\left[\ bcm_a^2\ -\ 2ap(p-a)^2\ \right]\ \Longrightarrow$](http://latex.artofproblemsolving.com/a/a/6/aa6f23abd9c5c8fa41e520ad07477fb6a7127a68.png)
Lagrange's identity 
Linear-angled identities/inequalities.
. Thus, 
, i.e. 
.
![$\implies\boxed{\begin{array}{c}
\sum a\cdot\cos A=2a\sin B\sin C=\frac {2S}{R}\le s\\\\
\sum a^2\cos A=\frac {r\left[3p^2-(2R+r)(4R+r)\right]}{R}\end{array}}$](http://latex.artofproblemsolving.com/a/f/f/aff6d10463779d3e1daa71a73742d907a0832321.png)
.
; 
; 
Prove easily that
.
and 
. If the triangle 
is acute, then there are the inequalities 
and 
. Area 
of the triangle 
(Heron) 
; 
; 
, where 
is the semiperimeter of the orthic triangle.
Therefore, 
.For an acute triangle ,
, where 
a.s.o. 
. I"ll use the identity 
in an acute . Thus, 
(Walker's inequality). 
(Goldstone's inequality). 
I"ll use the identity 
andthe inequality
From the sum of the simple inequalities 
a.s.o. obtain that 
Incircles and excircles. 
and 
If the incircle touches 
at 
, then 
.
; 
; 
are the roots of the equation
; 
.
are the roots of the equation 
are the roots of the equation 
si 
the exterior bisector of 
is right-angled 
; 
![$ \bullet\ [I_aI_bI_c]=2pR\ ,\ I_bI_c=4R\cos\frac A2$](http://latex.artofproblemsolving.com/c/9/2/c92c05fa109851a8a302bd1beed9838425e4569f.png)
and the circumradius of 
has the length 
. Remarkable distancies.
; 
and 
; 
: 
Power 
of the point w.r.t. the circumcircle 
For example, ![$ \boxed {\ \begin{array}{ccc}
M\in [BC] & \Longrightarrow & OM^2+MB\cdot MC=R^2\\\\
M\not\in [BC] & \Longrightarrow & OM^2-MB\cdot MC=R^2\end{array}\ }\ .$](http://latex.artofproblemsolving.com/0/4/9/049e3a3aa5ebe9648c7631edad256dbd8527672d.png)
Arrow (dart) 
is the length of the arc 
which belongs to the interior of the angle 
and is 
; 
Characterization of the right angle in 
.
Euler's relation. 
is nonobtuse 
Theorem of Sines. 
; 
Stewart's relation. 
The length of a cevian. 
, 
Trigonometrical form of the Ceva's theorem. 
Observe that the first relation can write thus :
Symmetrical fundamental forms. 
- the roots of the equations 
.
, where 
. For example, 
and
Show similarly that 
![$ \bullet\ \sum a^2(b+c)=\sum bc(b+c)=\sum bc[(a+b+c)-a]=s_1s_2-3s_3$](http://latex.artofproblemsolving.com/c/5/a/c5a3086466b3ff0f873c90bcb71e3f3fefa1bfb8.png)
![$ \sum bc\left(b^2+c^2\right)=\sum bc\left[\left(a^2+b^2+c^2\right)-a^2\right]=\sum bc\cdot\sum a^2-\sum a^2bc=$](http://latex.artofproblemsolving.com/1/0/e/10e8f0f0ce3b3ba18141eb0943e332b7c653e839.png)
(Schur for 
) is equivalently with Hadwiger-Finsler. Indeed, 
Schur for 
Indeed, 
Here find the Schur's inequality and its Vornicu's extension.
Remark. The Blundon's inequality is in fact a simple consequence of the triangle inequality 
Short proof. I'll use the well-known distances in a triangle
, where 
, 
and 
are the incenter , circumcenter and the Nagel's pointof the triangle . Now, one can easily show that Blundon's inequality reduces to
, which is just the triangle inequality applied in 
.See the remarkable geometrical inequalities (1) and (2).
This post has been edited 167 times. Last edited by Virgil Nicula, Nov 23, 2017, 9:38 AM
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