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I'm trying to prove the fact that the incenter has the barycentric coordinates $(a,b,c)$. My reasoning goes like this; consider the triangle $\Delta ABC$ with barycentric coordinates $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ at $A$, $B$ and $C$, respectively. Let $|AB|=c$,$|AC|=b$ and $|BC|=a$. Let $D$ be the intersection between the internal angle bisector at $A$ and $BC$ and let $E$ the intersection between the internal angle bisector at $B$ and $AC$. According to the angle bisector theorem $\frac{BD}{CD}=\frac{c}{b}$ and $\frac{AE}{CE}=\frac{c}{a}$. If I'm correct that gives $D$ and $E$ the un-normalized barycentric coordinates $(0,b,c)$ and $(a,0,c)$, respectively. This is where I get stuck. How do I determine the intersection (ie the incenter of $AD$ and $BE$ from this. I am (as you can tell) very new to barycentric coordinates so please try to keep the answer simple. Thanks in advance!

Martin Sleziak
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nesHan
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3 Answers3

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For triangle $ABC$,

  • Let $a,b,c$ denote the lengths of sides $BC,CA,AB\;$, respectively.$\\[4pt]$
  • Let $h_a,h_b,h_c$ denote the lengths of the altitudes from vertices $A,B,C$, respectively.$\\[4pt]$
  • Let $r$ denote the inradius.$\\[4pt]$
  • Let $k$ denote the area.$\\[4pt]$

Since the distance from the incenter to each of the lines $BC,CA,AB\;$is $r$, it follows that

  • The "$A$" coordinate of the incenter is $\dfrac{r}{h_a}$.
  • The "$B$" coordinate of the incenter is $\dfrac{r}{h_b}$.
  • The "$C$" coordinate of the incenter is $\dfrac{r}{h_c}$.

From $$k = \frac{1}{2}\,a\,h_a = \frac{1}{2}\,b\,h_b = \frac{1}{2}\,c\,h_c$$ we get $$a = \frac{2k}{h_a},\;\;b = \frac{2k}{h_b},\;\;c = \frac{2k}{h_c}$$ hence $$ \frac{r}{h_a}:\frac{r}{h_b}:\frac{r}{h_c} = \frac{1}{h_a}:\frac{1}{h_b}:\frac{1}{h_c} = \frac{2k}{h_a}:\frac{2k}{h_b}:\frac{2k}{h_c} = a:b:c$$

It follows that the incenter has barycentric coordinates $(a,b,c)$, as was to be shown.

quasi
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Let $[XYZ]$ be the area of triangle $XYZ$. Then by the areal definition, $[BCI]:[CAI]:[ABI] = \frac{BC \cdot r}{2} : \frac{CA \cdot r}{2} : \frac{AB \cdot r}{2} = a:b:c$ where $r$ is the in radius.

littlecat
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Barycentric coordinates are a specific kind of homogeneous coordinates. In homogeneous coordinates you can join points and intersect lines both by computing the cross product. So you could compute $(A×D)×(B×E)$.

$$\left(\begin{pmatrix}1\\0\\0\end{pmatrix}\times \begin{pmatrix}0\\b\\c\end{pmatrix}\right)\times \left(\begin{pmatrix}0\\1\\0\end{pmatrix}\times \begin{pmatrix}a\\0\\c\end{pmatrix}\right)= \begin{pmatrix}0\\-c\\b\end{pmatrix}\times \begin{pmatrix}c\\0\\-a\end{pmatrix}= c\cdot\begin{pmatrix}a\\b\\c\end{pmatrix} $$

So if you have $c=0$, then the whole expression is the null vector, but otherwise it is just a representative of the coordinates $(a,b,c)$ as expected.

Community
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MvG
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  • So $(AxD)$ represents the line AD? – nesHan Jan 09 '17 at 13:49
  • @user405389: Yes. Better written as $(A\times B)$ though. That's because incidence in the projective plane corresponds to orthogonality in 3d, as described in more detail in the post I linked to. Essentially if you have points with barycentric coordinates $(x,y,z)$, i.e. $xA+yB+zC$, then the line $(0,b,c)$ would be the set of points satisfying $bx+cz=0$. – MvG Jan 09 '17 at 14:01
  • Wouldn't AxD give me (0,-c,b) such that the dot product of this with any point on plane containing line AD would be zero?. It looks like a representation of a plane to me. and Isn't (0,b,c) in your last line a point rather than line? – Vrisk Nov 06 '17 at 07:42
  • @Vrisk: Yes, a line in the projective plane is essentially a plane through the origin in the 3d embedding. So thinking of $A\times D$ as a plane makes sense for a while, although with practice people tend to care less about the 3d embedding. The $(0,b,c)$ as a line in my comment was meant as an example, but it was poorly chosen since it matches the point $D$. Should have used the example $A\times D=(0,-c,b)$ representing the line $${xA+yB+zC\mid 0x-cy+bz=0\wedge x+y+z=1}$$ – MvG Nov 06 '17 at 09:16
  • oh the -cy + bz and x + y + z =1 condition, this would give a line and not a plane I guess. But you didn't use that in when you took the cross product of the two planes- the cross product should have given me points on a line through the intersection. how did that cross product give you the intersection and not some random point in that line? – Vrisk Nov 06 '17 at 09:32
  • @Vrisk: The cross product did give me some random point on that line. Or some representative of the same homogeneous coordinates. That's the key concept behind homogeneous coordinates: multiples of the vector represent the same point in the plane. If you actually want to find the point, as a linear combination of corner points, then you dehomogenize, i.e. pick the multiple so that $x+y+z=1$ (in the case of barycentric coordinates). If yoou have further questions about using cross products for intersections in barycentric coordinates, I suggest you ask a separate question for that. – MvG Nov 06 '17 at 16:06
  • I think I got it, if (a, b, c) are solutions to system of linear equations then so are k(a, b, c) but in 3D the k results in different points on the line. Thanks, that's pretty neat. – Vrisk Nov 06 '17 at 16:18
  • @Vrisk: Exactly. – MvG Nov 06 '17 at 21:43