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.

  • 1
    $\begingroup$ algebraic manipulation which lacks geometric intuition/motivation Maybe you should elaborate some more on what geometric intuition/motivation you see in the derivation of the first form to begin with. That would make it more clear what you expect as an answer here, since otherwise the difference between the two forms is indeed just a simple algebraic identity $-xy = \frac{1}{2}\left(x^2+y^2-z^2\right)$ which follows straight from $x+y+z=0$. $\endgroup$ – dxiv Dec 20 '17 at 4:36
  • $\begingroup$ Late response but yes, as I stated above, I know how to convert the one form to the other algebraically. I think I was as clear as possible in the explanation above regarding "intuition" but alas, such a notion is necessarily undefinable here. Thanks for the response however. $\endgroup$ – STK 299 Jun 22 '18 at 0:56

This is a nice problem. 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 with basis $(x^2,xy,y^2),$ assuming that $\;0=x+y+z,\;$ 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 $2$.

  • $\begingroup$ Late response but thank you Somos! It doesn't quite provide what I was looking for (moved on there; was looking for a more clear geometric interpretation/meaning) but it's a neat proof nonetheless. $\endgroup$ – STK 299 Jun 22 '18 at 1:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.