Nagel Angle Ankle

If the points $A,B,C$ have coordinates $(-1,0)$ , $(1,0)$ and $(\cos2\theta,\sin2\theta)$ respectively (we are writing $\angle BAC = \theta$ ), then the Nagel points $Na$ is the intersection of the lines $AD$ and $BE$ , where $2 + AE = 2 + BD$ is equal to the semiperimeter $1+\cos\theta+\sin\theta$ of the triangle $ABC$ . This means that $\begin{aligned} D\;& \big(1 - (\sin\theta+\cos\theta-1)\sin\theta,(\sin\theta+\cos\theta-1)\cos\theta\big) \\ E\;&\big(-1 + (\sin\theta+\cos\theta-1)\cos\theta,(\sin\theta+\cos\theta-1)\sin\theta\big) \end{aligned}$ and hence the Nagel point has coordinates $Na\; \big(x(\theta),y(\theta)\big) \; = \; \big(\cos2\theta + 2\sin\theta - 2\cos\theta,(\sin\theta + \cos\theta - 1)^2\big)$ and so the upper half of the area enclosed by the locus of the Nagel point is $A \; = \; \int_0^{\frac12\pi} x'(\theta)y(\theta)\,d\theta \; = \; 8 - \tfrac52\pi$ Thus the desired proportion is $R \; = \; \tfrac{2A}{\pi} \; = \; \tfrac{16}{\pi} - 5 \; = \; 0.09295817894...$ and hence $\lfloor 10^6 R \rfloor \,=\, \boxed{92958}$ .