One intersection P of the 3 Apollonius circles of the triangle $\triangle ABC$ (the 1st isodynamic point) always lies in the triangle interior, even if the triangle is obtuse. Invert the vertices A, B, C in a circle (P) centered at P and with arbitrary radius r. Then $A'B' = AB \cdot \frac{r^2}{PA \cdot PB},$ $B'C' = BC \cdot \frac{r^2}{PB \cdot PC},$ $C'A' = CA \cdot \frac{r^2}{PC \cdot PA}.$ Since P is on the B-Appolonius circle, $\frac{AB}{BC} = \frac{PA}{PC}$ and $\frac{A'B'}{B'C'} = \frac{AB}{BC} \cdot \frac{PB \cdot PC}{PA \cdot PB} = 1.$ Similarly, since P is on the C-Apollonius circle, $B'C' = C'A'.$ Thus the inverted triangle $\triangle A'B'C'$ is equilateral. But for any point P (not on the triangle circumcircle), the circumcevian triangle $\triangle XYZ$ of P, the pedal triangle $\triangle DEF$ with respect to P, and the triangle $\triangle A'B'C'$ inverted in an arbitrary circle centered at P are similar (for example, see http://www.mathlinks.ro/Forum/viewtopic.php?t=70444).

i think there's a little easier solution. since $\angle ABP+\angle ACP=\angle YXZ=60$, it follows that the isogonals of $AP$ and $BP$ form an angle of $120$. same for $AP$ and $CP$, and $CP$ and $BP$, then the isogonal conjugate of $P$ must be $F$ (the Fermat-Torricelli point). we conclude that $P$ is the isogonal conjugate of $F$. the converse is also easy to prove.

As $XYZ$ is equilateral it follows that $m\angle APC=60+A$. Let $P'$ be the isogonal conjugate of $P$ wrt $\triangle ABC$ we easily get that $m\angle APC+m\angle AP'C=180+A$ (look my lemma at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=109112 post 4)so $m\angle AP'C=120$. Similary for the other vertices we get that $P'$ is the fermat point of $\triangle ABC$ so that $P$ is the first isodynamic point of $\triangle ABC$.