MHB Barycentric coordinates in a triangle - proof

[Samwise1]

I want to prove that the barycentric coordinates of a point $P$ inside the triangle with vertices in $(1,0,0), (0,1,0), (0,0,1)$ are distances from $P$ to the sides of the triangle.

Let's denote the triangle by $ABC, \ A = (1,0,0), B=(0,1,0), C= (0,0,1)$.

We consider triangles $ABP, \ BCP, \ CAP$.

The barycentric coordinates of $P$ will then be $(h_1, h_2, h_3)$ where $h_1$ is the height of $ABP$, $h_2 \rightarrow BCP$, $h_3 \rightarrow CAP$

I know that $h_1 = \frac{S_{ABP}}{S_{ABC}}$ and similarly for $h_2, \ h_3$

My problem is that I don't know how to prove that if $P= (p_1, p_2, p_3)$

then $(h_1 + h_2 + h_3)P = h_1 (1,0,0) + h_2 (0,1,0) + h_3 (0,0,1)$

Could you tell me what to do about it?

Thank you!

I have one more question. Similarly, barycentric coordinates of a point $p$ in a regular tetrahedron $\Delta_3$ are distances of $p$ from its faces. How can we prove that?

 

[Evgeny.Makarov]

I want to prove that the barycentric coordinates of a point $P$ inside the triangle with vertices in $(1,0,0), (0,1,0), (0,0,1)$ are distances from $P$ to the sides of the triangle.

...

The barycentric coordinates of $P$ will then be $(h_1, h_2, h_3)$ where $h_1$ is the height of $ABP$, $h_2 \rightarrow BCP$, $h_3 \rightarrow CAP$

Isn't it what you need to prove? Because the height of $ABP$ is the distance from $P$ to $AB$.

I know that $h_1 = \frac{S_{ABP}}{S_{ABC}}$ and similarly for $h_2, \ h_3$

My problem is that I don't know how to prove that if $P= (p_1, p_2, p_3)$

then $(h_1 + h_2 + h_3)p = h_1 (1,0,0) + h_2 (0,1,0) + h_3 (0,0,1)$

What is $p$ in the last equation?

 

[Samwise1]

Isn't it what you need to prove? Because the height of $ABP$ is the distance from $P$ to $AB$.

What is $p$ in the last equation?

I see. I didn't express myself very well.

It was supposed to be a reformulation of the statement which I want to prove, that is that the barycentric coordinates of a point $p$ inside a triangle are distances of that point from the sides of the triangle = heights of respective smaller triangles.

So far I only know that $$1 = \frac{\frac{1}{2}(|AB|h_1 + |BC|h_2 + |CA|h_3)}{S_{ABC}} = \frac{S_{ABP}}{S} + \frac{S_{BCP}}{S} + \frac{S_{CAP}}{S}$$, but I don't know how to show that the condition :

$$(h_1 + h_2 + h_3)P = h_1 (1,0,0) + h_2 (0,1,0) + h_3 (0,0,1)$$

is satisfied.

And the $p$ in the last equation was a typo, I guess I didn't press "Shift" correctly :)

 

[Evgeny.Makarov]

First, the length of each side in $\triangle ABC$ where $A=(1,0,0)$, $B=(0,1,0)$ and $C=(0,0,1)$ is $\sqrt{2}$, and $S_{\triangle ABC}=\dfrac{\sqrt{3}}{2}$. Therefore, if $h_1$ is the height of $ABP$ and $h=\sqrt{\dfrac{3}{2}}$ is the height of $ABC$, then $h_1$ is indeed the distance from $P$ to $AB$, but $\dfrac{S_{\triangle ABP}}{S_{\triangle ABC}}=\dfrac{h_1}{h}\ne h_1$.

I don't know how to show that the condition :

$$(h_1 + h_2 + h_3)P = h_1 (1,0,0) + h_2 (0,1,0) + h_3 (0,0,1)$$

is satisfied.

The general fact is that
\[
P=\dfrac{S_{PBC}}{S_{ABC}}A+\dfrac{S_{PAC}}{S_{ABC}}B+\dfrac{S_{PAB}}{S_{ABC}}C\qquad(*)
\]
and it is trivial to show that $\dfrac{S_{PAB}}{S_{ABC}}=\dfrac{h_1}{h}$ and similarly for other fractions. Do I understand right that you need a proof of (*)?

 

[Samwise1]

Thank you.

I've come up with a proof of $(*)$.

Here it is:

Let $h_i : \Delta_3 \rightarrow \mathbb{R}$ be functions assigning $P \in \Delta_3$ its $i$-th barycentric coordinate. These maps are affine.

Then we consider a plane, an affine subspace of $T$ containing the triangle and extend the above mappings to $T$.

Next we define $H(P) = h_1(P) + h_2(P) + h_3(P)$

$T$ is $2$-dimensional. We take three non-collinear points - the vertices of our triangle and see that $H(A)=H(B)=H(C) = \frac{\sqrt{3}}{2}$

So $H$ is constant.

Next we set $f(P) = \frac{1}{H} (h_1(P) , h_2(P) ,h_3(P))$

Here $f(A)=A, f(B)=B, f(C)=C$, so $f$ is the identity map.

Is that right?

 

[Samwise1]

The general fact is that
\[
P=\dfrac{S_{PBC}}{S_{ABC}}A+\dfrac{S_{PAC}}{S_{ABC}}B+\dfrac{S_{PAB}}{S_{ABC}}C\qquad(*)
\]
and it is trivial to show that $\dfrac{S_{PAB}}{S_{ABC}}=\dfrac{h_1}{h}$ and similarly for other fractions. Do I understand right that you need a proof of (*)?

Yes, as a matter of fact, I do need the proof of that fact.

My "proof" doesn't seem to prove much.

Could you tell me why the equality $(*)$ is true?