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)
 A: 
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.
A: Assume $\max\{PA, PB, PC\} = PA$. Then $$\max\{PA, PB, PC\} = \frac{PA + PB + PC}{2}$$ turns into $PA = PB + PC$. Let, for simplicity,  $AB = BC = CA = a$. Choose inside the segment $AP$ the unique point $Q$ with the property that $AQ \cdot AP = a^2 = CA^2 = AB^2$. Then the latter identities are equivalent to $$\frac{CA}{PA} = \frac{AQ}{CA} \,\,\, \text{ and } \,\,\, \frac{AB}{PA} = \frac{AQ}{AB}$$ These ratio identities imply that triangles $CAQ$ and $PAC$ are similar and triangles $BAQ$ and $PAB$ are also similar. Consequently, $\angle \, ACQ = \angle \, APC$ and $\angle \, ABQ = \angle \, APB$, as well as
$$\frac{CA}{PA} = \frac{AQ}{CA} = \frac{QC}{PC} \,\,\, \text{ and } \,\,\, \frac{AB}{PA} = \frac{AQ}{AB} = \frac{BQ}{PB}$$ From the last ratio identities derive 
\begin{align}
QC &= \frac{CA}{PA} \cdot PC = \frac{a}{PA} \cdot PC\\
BQ & = \frac{AB}{PA} \cdot PB = \frac{a}{PA} \cdot PB
\end{align}
And if we add them together we obtain
$$BQ + QC = \frac{a}{PA} \cdot PB + \frac{a}{PA} \cdot PC = \frac{a}{PA} \big(PB + PC\big) = \frac{a}{PA} \cdot PA = a = BC.$$ But $BQ + QC = BC$ is possible if and only if $Q$ lies on the segment $BC$. Hence $$60^{\circ} = \angle \, ACB = \angle \, ACQ = \angle \, APC.$$ Because $\angle \, APC = 60^{\circ} = \angle \, ABC$, quadrilateral $ABPC$ is inscribed in a circle: the circumcircle of triangle $ABC$. 
The converse is also true. Take $P$ to be on the circumcircle of $ABC$ so that $P$ is on the arc $BC$, that doesn't contain $A$.  Then you want to show that $PA = PB + PC$. Pick point $Q$ on $PA$ so that $PQ = PB$. Since $ABPC$ is inscribed in a circle, we have $\angle \, QPB = \angle \, APB =\angle \, ACB = 60^{\circ}$. Therefore triangle $PBQ$ is equilateral. Hence, triangles $ABQ$ and $CBP$ are congruent because $AB=BC$ and $BQ = BP$ and $\angle \, ABQ = \angle CBP$. Thus, $QA = PC$. Now we see that $PA = PQ + QA = PB + PC$.  
Much slicker proof. 
Assume $\max\{PA, PB, PC\} = PA$. Then $$\max\{PA, PB, PC\} = \frac{PA + PB + PC}{2}$$ turns into $PA = PB + PC$. Let, for simplicity,  $AB = BC = CA = a$. 
Statement. Point $P$ satisfies the property $PA = PB + PC$ if and only if $P$ lies on the circumcircle of the equilateral triangle $ABC$, on the arc between points $B$ and $C$ not containing $A$.  
Proof: Assume $P$ is a point on the plane so that $PA = \max\{PA, PB, PC\}$. Then perform a $60^{\circ}$ counter-clockwise rotation around the point $A$ and let $P^*$ be the image of $P$ under the rotation.  Then $PA=P^*A$ and $\angle \, PAP^* = 60^{\circ}$. Hence triangle $PAP^*$ is equilateral and $PA=P^*A = PP^*$. Moreover, since triangle $ABC$ is equilateral itself, point $C$ is the image of point $B$ under the counter-clockwise $60^{\circ}$ rotation. Consequently, triangle $ACP^*$ is the rotation image of $ABP$ and so these two triangles are congruent. Thus, $PB=P^*C$. 
Observe that triangle $PCP^*$ is a triangle with edges $PP^* = PA, \,\, P^*C = PB$ and $PC$. 
Assume first that $PA = PB + PC$. Then $PP^* = P^*C + PC$ which is possible if and only if $C$ lies on the edge $PP^*$. But then $\angle \, CPA = \angle \, P^*PA = 60^{\circ}$ because $P^*PA$ is equilateral. Consequently $\angle\, CPA = \angle \, CBA = 60^{\circ}$ which means that point $P$ lies on the circumcircle of triangle $ABC$.
Conversely, let $P$ be on the circumcircle of the equilateral triangle $ABC$. Then $\angle\, CPA = \angle \, CBA = 60^{\circ}$. On the other hand by construction triangle $P^*PA$ is equilateral so $60^{\circ} = \angle \, P^*PA = \angle CPA$ which is possible if and only if point $C$ lies on the segment $P^*P$ (here is where we use the condition that $P$ is located on the arc $BC$ disjoint form $A$ which guarantees that $C$ and $P^*$ are on the same side of line $AP$). But then, $PP^* = P^*C + PC$ and since it has already been established that $PP^* = PA$ and $P^*C = PB$, we conclude that $PA = PB + PC$.
