Another Property of Points on Incircle

WLOG, we'll assume $\omega\;$ is a unit circle $x^2+y^2=1,\;$ $B=(-u,-1),\;$ $C=(v,-1),\;$ $u,v\gt 0.\;$ As a consequence, $\displaystyle A=\left(\frac{u-v}{uv-1},\frac{uv+1}{uv-1}\right).\;$ Assume also that $P=(\cos t,\sin t).\;$

We have $a=u+v,\;$ $\displaystyle b=\frac{u(v^2+1)}{uv-1},\;$ $\displaystyle c=\frac{v(u^2+1)}{uv-1},\;$

$\displaystyle\begin{align} x^2&=\left(\frac{u-v}{uv-1}-\cos t\right)^2+\left(\frac{uv+1}{uv-1}-\sin t\right)^2,\\ y^2&=(u+\cos t)^2+(1+\sin t)^2,\\ z^2&=(v-\cos t)^2+(1+\sin t)^2. \end{align}$

Our goal is to show that the terms in the expression $ax^2+by^2+cz^2,\;$ involving $t,\;$ cancel out. Taking into account that $\cos^2t+\sin^2t=1,\;$ below are the remaining terms:

$\displaystyle\begin{align} \mbox{from}\;ax^2:\;&-2(u+v)\left(\frac{u-v}{uv-1}\right)\cos t-2(u+v)\left(\frac{uv+1}{uv-1}\right)\sin t\\ \mbox{from}\;by^2:\;&2\left(\frac{u(v^2+1)}{uv-1}\right)u\cos t+2\left(\frac{u(v^2+1)}{uv-1}\right)\sin t\\ \mbox{from}\;cz^2:\;&-2\left(\frac{v(u^2+1)}{uv-1}\right)v\cos t+2\left(\frac{v(u^2+1)}{uv-1}\right)\sin t, \end{align}$

and the sum of the six is indeed $0.$

The problem with the solution has been kindly communicated to me by Leo Giugiuc.

Note that, from the above proof, it follows that the statement holds for points on any circle concentric with $\omega.$ Thus, the result proved above can be formulated differently:

Circles concentric with the incircle are level curves for $f(P)=a\cdot AP^2+b\cdot BP^2+c\cdot CP^2.$

This result admits a very nice generalization that is proved using the barycentric coordinates.