386. Barycentrical coordinates.
by Virgil Nicula, Sep 1, 2013, 2:55 AM
Barycentrical coordinates in Olympiad Geometry.
Here are barycentrical coordinates w.r.t.
for: centroid
, orthocenter
, circumcenter
, incenter
,
-excenter
and the points Nagel's
, Gergogne's
and Lemoine's
.

Theorem 1. If
, where
, then
and
have
.
Remark. Observe that
and
.
Theorem 2. The power
of
w.r.t.
of 
.
Examples:





Remark.
.
Theorem 3.
and
for which
. For
obtain the second part of
.
Theorem 4. If
where
, then
and
is given by
, where
.
Remark. For
obtain that
where
.
Applications.
AP1 (Miguel Ochoa Sanchez). Let
be an interior point of an equilateral
with
so that
. Prove that
.
Proof 1. Oberve that
. From the relation
obtain that
. Observe that
. From the relation 
for
and
obtain that
. Apply again 
for
and
and obtain that
.
Proof 2 (elementary). Observe that
. Apply the Stewart's relation in the particular case
when
is equilateral to the cevians:
. Apply the Van Aubel relation:
.
Remark. Consider an
-isosceles triangle
and a point
. Denote the midpoint
od
and suppose w.l.o.g.
. Then 
.
Otherwise. Apply the Pythagoras's theorem:
.
AP2. Let
be the cevian triangle of the point
w.r.t.
and let
,
,
be the midpoints of
,
,
respectively. Denote the midpoints
,
, 
of
,
,
. Prove that
,
,
are concurrently. If
and
, then
.
Proof. Observe that
,
and
. Thus,
. Therefore,
the homogeneous system with
has at least solution, i.e.
what is evidently (the sum of its lines is zero).
The system
has the solution
,
i.e.
AP3. Let
be the cevian triangle of the point
w.r.t.
and let
,
,
be midpoints of
,
,
respectively. Denote midpoints
,
,
of
,
,
. Prove that
,
,
are concurrently. If
and
, then
.
Proof. Observe that
,
,
and
. Thus,

. Hence
the homogenous system with 
has at least solution, i.e.
what prove easily. The system
has the solution
, i.e.
AP4. Ascertain the area
, where
are the Lemoine's (symmedian) point, the incenter and the centroid of
respectively.
Proof.
.
AP5 (Miguel Ochoa Sanchez). Let
with the circumcircle
and a point
, where
. Prove that 
there is the identity
, where
is the power of the point
w.r.t. the circle
and
.
Proof.

Remark. Let
with the circumcircle
and two fixed points
and
, where
. Consider a variable circle
such that
and
belongs to a fixed line
. Prove that the sum
is constant
Here are barycentrical coordinates w.r.t.











Theorem 1. If





Remark. Observe that





Theorem 2. The power





Examples:







Remark.


Theorem 3.





Theorem 4. If


![$\boxed{M_1M_2^2=-\left[(\beta_1-\beta_2)(\gamma_1-\gamma_2)a^2+(\gamma_1-\gamma_2)(\alpha_1-\alpha_2)b^2+(\alpha_1-\alpha_2)(\beta_1-\beta_2)c^2\right]}\ (4^*)$](http://latex.artofproblemsolving.com/d/0/0/d003203c93b43b468c79317d80219ba4d2a1318b.png)


![$\boxed{\overrightarrow{OM_1}\cdot\overrightarrow{OM_2}=R^2-\frac 12\cdot\left[\left(\beta_1\gamma_2+\beta_2\gamma_1\right)a^2+\left(\gamma_1\alpha_2+\gamma_2\alpha_1\right)b^2+\left(\alpha_1\beta_2+\alpha_2\beta_1\right)c^2\right]}\ (5^*)$](http://latex.artofproblemsolving.com/1/7/9/179bec91ebd4b8d92e8890fa4580b53710f348ca.png)
Remark. For



Applications.
AP1 (Miguel Ochoa Sanchez). Let



![$\frac {[BPC]}2=\frac{[CPA]}4=\frac {[APB]}1$](http://latex.artofproblemsolving.com/5/a/d/5ad3bb66de0ada3e9bf1bfc65511888abe23388b.png)

Proof 1. Oberve that





for







for




Proof 2 (elementary). Observe that
![$\left\{\begin{array}{ccc}
\frac {DB}{DC}=\frac {[APB]}{[APC]}=\frac 14 & \implies & \frac{DB}1=\frac {DC}4=\frac 15\\\\
\frac {EC}{EA}=\frac {[BPC]}{[BPA]}=\frac 21 & \implies & \frac{EC}2=\frac {EA}1=\frac 13\\\\
\frac {FA}{FB}=\frac {[CPA]}{[CPB]}=\frac 42 & \implies & \frac{FA}4=\frac {BF}2=\frac 16\end{array}\right\|$](http://latex.artofproblemsolving.com/7/f/a/7fa0c61a8a27266f4080d882bb8a26ece493106c.png)






Remark. Consider an




![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)






Otherwise. Apply the Pythagoras's theorem:



AP2. Let






![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
![$[CA]$](http://latex.artofproblemsolving.com/4/5/c/45c1acd47628de406680d04c09fe6314c3847acf.png)
![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)



of
![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
![$[BE]$](http://latex.artofproblemsolving.com/f/b/0/fb061a8a7c5f9403b5f9261840de9dfea7cb68cf.png)
![$[CF]$](http://latex.artofproblemsolving.com/2/1/b/21bdd766ec1757878aeae83d55b7ab5917af4537.png)






Proof. Observe that









The system


i.e.

AP3. Let






![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
![$[CA]$](http://latex.artofproblemsolving.com/4/5/c/45c1acd47628de406680d04c09fe6314c3847acf.png)
![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)



![$[EF]$](http://latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
![$[DF]$](http://latex.artofproblemsolving.com/4/8/7/487608ba746e637d846b20401f23cc2b80336338.png)
![$[DE]$](http://latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)






Proof. Observe that



![$X\left[x(1+x), y(1-y) , z(1-z)\right]$](http://latex.artofproblemsolving.com/9/4/b/94b8553c8aa6f78fd1deb9e606cef0b55f345059.png)





has at least solution, i.e.




AP4. Ascertain the area
![$[KIG]=f(S,a,b,c)$](http://latex.artofproblemsolving.com/4/1/3/413eeada9168813be1989d97f2e327a825ea4b7a.png)


Proof.

![$(1+1+1)(a+b+c)\left(a^2+b^2+c^2\right)\cdot [KIG]=\mod\left|\begin{array}{ccc}
1 & 1 & 1\\\\
a & b & c\\\\
a^2 & b^2 & c^2\end{array}\right|\cdot [ABC]\implies$](http://latex.artofproblemsolving.com/4/f/c/4fc9b41a1899242cb84747d4f4878e8987f0869b.png)
![$\boxed{[KIG]=\frac {|(a-b)(b-c)(c-a)|S}{3(a+b+c)\left(a^2+b^2+c^2\right)}}$](http://latex.artofproblemsolving.com/7/6/6/766d7c5d303a63e98f3060878780a258f77d4dee.png)
AP5 (Miguel Ochoa Sanchez). Let





there is the identity





Proof.




Remark. Let










This post has been edited 90 times. Last edited by Virgil Nicula, Nov 24, 2016, 3:35 PM