Nagel Angle Ankle II

Standard results give the distances between the outcentre $O$ of a triangle and its incentre $I$ and Nagel point $Na$ as: $OI \; = \; \sqrt{R(R-2r)} \hspace{2cm} ONa \; = \; \frac{4\Delta OI^2}{abc} \; = \; \frac{4R\Delta(R-2r)}{abc}$ where $R,r$ are the outradius and inradius, and $\Delta$ is the area. Since $R \; = \; \frac{abc}{4\Delta}$ we deduce that $ONa \; = \; R-2r$ In this case we deduce that $ONa = 8 - 2\times3=2$ .

For a fixed vertex $A$ and given vertex $B$ on a circle of radius $8$ , there are two possible triangles $ABC$ with inradius $3$ and outradius $8$ , one being the mirror image of the other in the perpendicular bisector of $AB$ . Thus there are two candidates for the Nagel point, and they are mirror images of the other in the same line. As $B$ varies, the two candidate Nagel points provide a double cover of the circle with centre $O$ and radius $2$ , which has perimeter $4\pi$ , making the answer $\boxed{4}$ .

Clearly we have $OI = 4$ in this case, so the locus of the incentre is a circle of radius $4$ , centre $O$ , and hence has perimeter $8\pi$ .