Parabolas and Tangents

There is a lot of hardcore geometry in this question, and I shall not attempt to supply all the details of the theory of tangent parabolae and Simson lines, but shall content myself by stating the results.

Using the Cosine Rule, $\cos A \,=\, \tfrac15$ , and hence $\sin A \,=\, \tfrac{2}{5}\sqrt{6}$ . Thus the radius of the circumcircle of $ABC$ is $R$ , where $2R \,=\, \tfrac{a}{\sin A} \,=\, \tfrac{70}{\sqrt{6}}$ , and hence $R = \tfrac{35}{\sqrt{6}}$ . The circumcircle of the triangle formed by three tangents to a parabola always passes through the focus of the parabola, and hence both $F_B$ and $F_C$ lie on the circumcircle of $ABC$ . Since $F_BF_C = R\sqrt{2}$ , we deduce that $\angle F_BOF_C = \tfrac12\pi$ , where $O$ is the circumcentre.

The directrix for the parabola $P_B$ can be found by reflecting the focus $F_B$ in each of the lines $BC$ , $AC$ and $AB$ . The resulting three points $X^B_{A},\,X^B_B,\,X^B_C$ are collinear, and lie on the directrix of $P_B$ (which, incidentally, always passes through the orthocentre of $ABC$ ). The midpoints of the lines $F_BX^B_A$ , $F_BX^B_B$ and $F_BX^B_C$ are thus also collinear, lying on a line parallel to the directrix of $P_B$ . But the midpoints of these lines are the feet of the perpendiculars from $F_B$ to the lines $BC$ , $AC$ and $AB$ , and hence they lie on the Simson line of $F_B$ . Thus the directrix of $P_B$ is parallel to the Simson line of $F_B$ .

The angle between the Simson lines of $F_B$ and $F_C$ is equal to half the angle $\angle F_BOF_C$ , and hence is $\tfrac14\pi$ . Thus the angle between the directrices of $P_B$ and $P_C$ is $\tfrac14\pi$ . Since the principal axes of the two parabolae are each perpendicular to their directrices, the two principal axes are at an angle of $\tfrac14\pi$ to each other. Thus $\angle F_CXF_B$ is either $\tfrac14\pi$ or $\tfrac34\pi$ . If $\angle F_CXF_B = \tfrac14\pi$ , then $X$ either lies on the major arc $F_BF_C$ of the circumcircle, or else would lie on the reflection of that major arc in the line $F_BF_C$ (remember that $\angle F_BOF_C = \tfrac12\pi$ ). In either case, $X$ would not lie inside the circumcircle. Thus $\angle F_CXF_B = \tfrac34\pi$ .

The point $X$ must lie on the reflection of the minor arc $F_BF_C$ of the circumcircle of $ABC$ in the line $F_BF_C$ .