A geometric locus in a equilateral triangle Find the locus of the points $P$ in the plane of an equilateral triangle $ABC$ that satisfy :
$$\max\{PA,PB,PC\} = \frac{PA+PB+PC}{2}.$$
I have never dealt with locus problems like these. So any help would be appreciated. (And please mention the intuition behind the answer too, if possible)


Tricky question. I will give you just a substantial hint. Let we consider a point $P$ on the minor $BC$-arc of the circumcircle of $ABC$. By applying Ptolemy's theorem to the cyclic quadrilateral $PBAC$ we get that $PA=PB+PC$, from which
$$ PA = \frac{PA+PB+PC}{2}.$$
Can you guess now what the wanted locus is? Consider that $\max\{PA,PB,PC\}$ equals $PA$ iff $P$ lies in the $\widehat{BOC}$ angle, where $O$ is the circumcircle of $ABC$. To finish the proof, prove that along a ray emanating from $O$ in the $\widehat{BOC}$ angle, the function $PA-(PB+PC)$ has a unique zero.
