0

I am trying to understand relationship between Median and barycentric coordinates.

here is a figure (fig_1) comes from wiki

enter image description here

that post gives a good explanation and I am trying to connect the notation in that post to fig_1.

Vertex of such small triangle at the first median has B.C.(baricentric coordinates) like (g,f,f), where g+f+f=1, at the second median (f,g,f), and (f,f,g) at the third one. We can see that for g=1/3 triangle size is zero (coefficient cf= 0), for g=1 triangle is equal to large triangle (cf = 0).

in this context, is it correct to consider point E as (g,f,f)?

Bernard
  • 175,478
JJJohn
  • 1,436
  • Every point on the line segment $AE$ has barycentric coordinates of the form $(g,f,f)$ where $g+2f=1$. In particular, $E$ corresponds to $g=0,f=1/2$. –  May 28 '19 at 10:59

1 Answers1

1

Yes, a way to understand the situation is as follows. Let $x,y,z$ be the normed barycentric coordinates of $O$, $x+y+z=1$. By definition: $$ O = xA+yB+zC\ . $$ (Understand this either using affix points $A,B,C$, or using a "testing point" $P$ and rewrite with vectors the relation as $PO =xPA+yPB+zPC$.)

These $x,y,z$ are barycentric coordinates, thus coordinates "in the plane", and the argument reduces them, their symmetry to one "on the line". Let us see how. Just rewrite: $$ \begin{aligned} O &= xA+(y+z)E\ ,\text{ where}\\ E &= \frac y{y+z}B+\frac z{y+z}C\ . \end{aligned} $$ (From the first relation, $E$ is on $OA$, from the second relation it is on $BC$, so it is indeed the point $E$ in the picture.)

Since $E$ is the middle of $BC$, we must have the corresponding coordinates "in the line" equal, so $y/(y+z)=z/(y+z)$, so $y=z$.

By symmetry to $y=z$ we also get $x=y$ (and $x=z$), so $x=y=z=1/3$.

Note that already knowing the porportion $OE:AE$ lets us know the $x$.

dan_fulea
  • 32,856
  • thanks for your answer! is $O = xA+yB+zC$ in Cartesian coordinates or in barycentric coordinates? – JJJohn May 28 '19 at 12:09
  • 2
    I will give an answer using examples. See also http://faculty.evansville.edu/ck6/encyclopedia/ETC.html for more on trilinear and barycentric coordinates. For the intersection of medians we have the (not normalized) coordinates $(1:1:1)$. To make them sum to one, we normalize to $(1/3,1/3,1/3)$. Then the relation $G=\frac 13A+\frac13 B+\frac 13C$ (using $G$ instead of $O$) holds in cartesian coordinates, for instance, if $A(1,2)$, $B(2,3)$, $C(3,-2)$, then we simply build $\frac 13(A+B+C)=\left(\frac 13(1+2+3),\ \frac 13(2+3-2)\right)$. – dan_fulea May 28 '19 at 14:15
  • 1
    A very good description is here: http://web.evanchen.cc/handouts/bary/bary-short.pdf – dan_fulea May 28 '19 at 14:15