Problem

The key is recognizing that both inradius and exradius formulas depend on area and semiperimeter—quantities you can express using the constraint that $P$ lies on the extension of $BA$. Since $\triangle ABC$ is equilateral with side $a$, and $P$ is positioned on line $BA$ extended beyond $A$, the triangles $\triangle PAC$ and $\triangle PBC$ share the vertex $C$ and have bases on the same line. Use $r = \frac{\text{Area}}{s}$ and $r_{\text{ex}} = \frac{\text{Area}}{s - \text{opposite side}}$ to express both radii, then look for how the geometry constrains their sum to depend only on $a$.

The key is recognizing that both inradius and exradius formulas depend on area and semiperimeter—quantities you can express using the constraint that $P$ lies on the extension of $BA$. Since $\triangle ABC$ is equilateral with side $a$, and $P$ is positioned on line $BA$ extended beyond $A$, the triangles $\triangle PAC$ and $\triangle PBC$ share the vertex $C$ and have bases on the same line. Use $r = \frac{\text{Area}}{s}$ and $r_{\text{ex}} = \frac{\text{Area}}{s - \text{opposite side}}$ to express both radii, then look for how the geometry constrains their sum to depend only on $a$.