Circles in Barycentric Coordinates

Problem Proof

We start with the formula for the distance between two points in barycentric coordinates (Theorem 7, Barycentric Coordinates in Olympiad Geometry, Max Schindler and Evan Chen):

The distance between two points $P(m,n,p)\;$ and $Q(q,s,t),\;$ where the barycentric coordinates relative to $\Delta ABC\;$ satisfy $m+n+p=1\;$ and $q+s+t=1\;$ is defined by

$PQ^2=-a^2(n-s)(p-t)-b^2(p-t)(m-q)-c^2(m-q)(n-s).$

Using that we are able to express distances from $P(x,y,z),\;$ $x+y+z=1\;$ of point $P\;$ on a circle of radius $r\;$ around $M(u,v,w),\;$ $u+v+w=1,\;$ to $A(1,0,0),\;$ $B(0,1,0),\;$ $C(0,0,1),\;$ and $M:$

\begin{align} PA^2&=-a^2yz-b^2z(x-1)-c^2(x-1)y\\ PB^2&=-a^2(y-1)z-b^2zx-c^2x(y-1)\\ PC^2&=-a^2y(z-1)z-b^2(z-1)x-c^2xy\\ PM^2&=-a^2(y-v)(z-w)-b^2(z-w)(x-u)-c^2(x-u)(y-v). \end{align}

Multiply the first of these by $u,\;$ the second by $v,\;$ and the third by $w:$

\begin{align} uPA^2&+vPB^2+wPC^2=\\ &-a^2[(u+v+w)yz-vz-wx]\\ &-b^2[(u+v+w)zx-uz-wx]\\ &-c^2[(u+v+w)xy-uy-vx]\\ &=-a^2[yz-vz-wx]-b^2[zx-uz-wx]-c^2[xy-uy-vx]\\ &=[-a^2(y-v)(z-w)-b^2(z-w)(x-u)-c^2(x-u)(y-v)]\\ &\;\;\;\;+(a^2vw+b^2wu+c^2uv)\\ &=r^2+(a^2vw+b^2wu+c^2uv). \end{align}

Corollary

The level curves of the function $f(P)=uPA^2+vPB^2+wPC^2\;$ are the circles centered at $u:v:w.$

In particular, for the incenter $I(a:b:c)\;$ of triangle, the values of the function $f(P)=aPA^2+bPB^2+cPC^2\;$ are independent of $P\;$ on circles centered at $I,\;$ which is exactly the previous result concerning the circles concentric with the incircle.

Acknowledgment

The statement was brought to my attention on google+ by the member under the user name Snail Erato in a comment to my post concerning Another Property of Points on Incircle.

Applications

The simplest application is to the circles centered at the centroid $G=(1:1:1)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right).$ If $P\;$ lies on such a circle with radius $r,\;$ then the circle is described by

$\displaystyle \frac{1}{3}(PA^2+PB^2+PC^2)=r^2+\frac{1}{9}(GA^2+GB^2+GC^2).$

This can be rewritten as

$\displaystyle \frac{1}{3}(PA^2+PB^2+PC^2)=r^2+\frac{4}{81}(m_a^2+m_b^2+m_c^2),$

where $m_a,m_b,m_c\;$ are the three medians in $\Delta ABC.\;$ However, say, $\displaystyle m_a^2=\frac{-a^2+2b^2+2c^2}{4},\;$ from which we get

$\displaystyle m_a^2+m_b^2+m_c^2=\frac{3}{4}(a^2+b^2+c^2).$

Such that

$\displaystyle PA^2+PB^2+PC^2=3r^2+\frac{1}{9}(a^2+b^2+c^2),$

Here are additional examples:

• Another Property of Points on Incircle
• Incircle in Equilateral Triangle
• Points on Incircle: Another Look 