Derive Barycentric coordinate distance formula

Question

please pardon the poor formatting. (I'll work on learning it in time; I just started this account to see help with this question.)

I've recently started learning about affine geometry and Barycentric coordinates, and I have a question regarding the distance formula for Barycentric coordinates. The Wikipedia page on Barycentric coordinate system gives two versions of this formula, and while I have no trouble proving the first, (first I took the dot product of the displacement vector $PQ$ while setting $A$ to the origin, much as the author of the Mathematical Gazette, cited by Wikipedia, did; I also proved it by setting the origin to the circumcenter of triangle $ABC$. Also, I followed another citation in said article which should have lead to an answer-but alas, that article stated the result without even a "proof is obvious.")

my "proof" of the second relies on some (very simple) algebraic manipulation which lacks geometric intuition/motivation. Yes, it works, but there should be a better argument. (Both forms are written below.)

Essentially, my question is this: can anyone help me prove the second form, but without first proving the first form? (Presumably, such a proof would provide the geometric intuition I'm looking for.) I've been such on this for days and it's starting to get to me-I've tried many different approaches.

Setting: Triangle $ABC$ is positively oriented; $P, Q$ are vectors in the plane of $ABC$, with $P, Q$ having normalized/homogeneous Barycentric coordinates $P= [p_1, p_2, p_3], Q= [q_1,q_2,q_3].$ Thus, displacement vector $PQ= [q_1-p_1,q_2-p_2,q_3-p_3]=[x,y,z],$ with $x+y+z=0.$

Form $1$: (no problems here) $\textrm{dist}(P,Q)^2 = -yza^2-xzb^2-xyc^2.$

Form $2$: (subject of my question-and yes, I'm familiar with the polarization identity and its relation to the coefficients below-also familiar with the circumcenter's Barycentric coordinates and the similarity to those coefficients but I'm not sure how to relate the two in a proof.)

$\textrm{dist}(P,Q)^2 = \frac12\{(b^2+c^2-a^2)x^2 + (a^2+c^2-b^2)y^2 + (b^2+a^2-c^2)z^2\}.$

Thanks for any help/guidance-it's much appreciated. This one has me stumped.

Answer 1

This is a nice problem. The question states:

Setting: Triangle $ABC$ is positively oriented; $P, Q$ are vectors in the plane of $ABC$, with $P, Q$ having normalized/homogeneous Barycentric coordinates $P= [p_1, p_2, p_3], Q= [q_1,q_2,q_3].$ Thus, displacement vector $PQ= [q_1-p_1,q_2-p_2,q_3-p_3]=[x,y,z],$ with $x+y+z=0.$

Form $1$: (no problems here) $\textrm{dist}(P,Q)^2 = -yza^2-xzb^2-xyc^2.$

Form $2$: (subject of my question-and yes, I'm familiar with the polarization identity and its relation to the coefficients below-also familiar with the circumcenter's Barycentric coordinates and the similarity to those coefficients but I'm not sure how to relate the two in a proof.)

$\textrm{dist}(P,Q)^2 = \frac12\{(b^2+c^2-a^2)x^2 + (a^2+c^2-b^2)y^2 + (b^2+a^2-c^2)z^2\}.$

First, we assume we are in an affine plane over an inner product space such as the Euclidean plane. This means that the (inner) dot product defines a distance measure of line segments by $\;\textrm{dist}(P,Q)^2 := |PQ|^2 = (Q-P)\cdot(Q-P).\;$ Now given a triangle of reference $ABC$ with sides $\;a,b,c\;$ we have $\;a^2=|BC|^2,\;b^2=|AC|^2,\;c^2=|AB|^2.$

We want the length of a line segment $\;PQ=Q-P=xA+yB+zC,\;$ where $\;0 = x+y+z.\;$ Now $\;|PQ|^2 = (xA+yB+zC)\cdot(xA+yB+zC) = (x+y+z)(|A|^2x+|B|^2y+|C|^2z) + T,$ where $T = -yz|B-C|^2-xz|A-C|^2-xy|A-B|^2 = -yza^2-xzb^2-xyc^2.\;$ Since $\;0 = x+y+z,\;$ then $\;|PQ|^2=T\;$ which proves Form $1$.

The linear space of quadratics has a basis $(x^2,xy,y^2).$ Assuming that $\;0=x+y+z,\;$ it also has bases $(xy,xz,yz)\;$ and $\;(x^2,y^2,z^2).\;$ We used one of them for Form $1$. Using the other basis, we suppose that $\;|PQ|^2=ux^2+vy^2+wz^2.\;$ But $\;a^2=|BC|^2=v+w,\;$ $b^2=|AC|^2=u+w,\;$ $c^2=|AB|^2=u+v.\;$ Solving for $\;u,v,w\;$ proves Form $2$.