Feuerbach's Bow-Tie

If we let $\angle BAC = \theta$ for $0 < \theta < \tfrac12\pi$ (so that $C$ is in the upper semicircle), then the vertices of the triangle $ABC$ are $A\; (-1,0) \hspace{1cm} B\;(1,0) \hspace{1cm} C\; (\cos2\theta,\sin2\theta)$ As has been shown in previous problems of this type, the incentre $I$ of $ABC$ has coordinates $I\; (\cos\theta - \sin\theta, \cos\theta + \sin\theta - 1)$ (note that $\cos\theta + \sin\theta - 1$ is the inradius). Since $ABC$ is right-angled, the orthocentre of $ABC$ is the vertex $C$ , and so the nine-point centre $N$ is the midpoint of $OC$ , and hence has coordinates $N\; \big(\tfrac12\cos2\theta,\tfrac12\sin2\theta\big)$ We calculate that $IN = \tfrac12(3 - 2\cos\theta - 2\sin\theta)$ , and so (since the radius of the nine-point circle is half that of the outcircle, namely $\tfrac12$ ) $\overrightarrow{OFe} \; = \; \overrightarrow{ON} + \frac{1}{3 - 2\cos\theta-2\sin\theta}\overrightarrow{NI}$ It is useful at this stage to introduce the coordinate $u = \theta - \tfrac14\pi$ , so that $Fe$ has coordinates $\big(X(u),Y(u)\big)$ , where $X(u) \; = \; \frac{\sqrt{2}\sin u(2\cos^2u - \sqrt{2}\cos u - 1)}{3 - 2\sqrt{2}\cos u} \hspace{2cm} Y(u) \; = \; \frac{2\sin^2u(\sqrt{2}\cos u - 1)}{3 - 2\sqrt{2}\cos u}$ We find that $X'(u)Y(u) \,=\, Z(\cos u)$ , where $\begin{aligned} Z(c) & = \; \frac{2(1-c^2)(\sqrt{2}c-1)(10 - 15 \sqrt{2} c - 4 c^2 + 22 \sqrt{2} c^3 - 16 c^4)}{(3 -2\sqrt{2}c)^3} \\ & = \; -\tfrac{15}{16} + \tfrac{1}{4}\sqrt{2}c + \tfrac{17}{8} c^2 - \tfrac{3}{4}\sqrt{2} c^3 - 2 c^4 + \frac{1}{64(3 - 2 \sqrt{2} c)^3} - \frac{5}{ 32 (3 - 2 \sqrt{2}c)^2} + \frac{41}{64 (3 - 2 \sqrt{2}c)} \end{aligned}$ Thus the desired area is $\Delta$ , where $\begin{aligned} -\tfrac12\Delta & = \; \int_{-\frac14\pi}^{\frac14\pi} Z(\cos u)\,du \\ & = \; \int_{-\frac14\pi}^{\frac14\pi}\left( -\tfrac{15}{16} + \tfrac{1}{4}\sqrt{2}\cos u + \tfrac{17}{8} \cos^2u - \tfrac{3}{4}\sqrt{2}\cos^3u - 2 \cos^4u \right)\,du + \tfrac{41}{64}Q_1 - \tfrac{5}{32}Q_2 + \tfrac{1}{64}Q_3 \\ & = \; -\tfrac{11}{16} - \tfrac{5}{16}\pi + \tfrac{41}{64}Q_1 - \tfrac{5}{32}Q_2 + \tfrac{1}{64}Q_3 \end{aligned}$ where $Q_j \; = \; \int_{-\frac14\pi}^{\frac14\pi} \frac{du}{(3 -2\sqrt{2}\cos u)^j} \hspace{2cm} j \in \mathbb{N} \cup \{0\}$ Now, integrating by parts, $\begin{aligned} Q_j & = \; 3Q_{j+1} - 2\sqrt{2}\int_{-\frac14\pi}^{\frac14\pi} \frac{\cos u}{(3 - 2\sqrt{2}\cos u)^{j+1}}\,du \\ & = \; 3Q_{j+1} - 2\sqrt{2}\left\{ \left[\frac{\sin u}{(3 - 2\sqrt{2}\cos u)^{j+1}}\right]_{-\frac14\pi}^{\frac14\pi} + 2\sqrt{2}(j+1) \int_{-\frac14\pi}^{\frac14\pi}\frac{\sin^2u}{(3 - 2\sqrt{2}\cos u)^{j+2}}\,du\right\} \\ & = \; 3Q_{j+1} - 4 - 8(j+1)Q_{j+2} + 8(j+1)\int_{-\frac14\pi}^{\frac14\pi} \frac{\cos^2u}{(3 - 2\sqrt{2}\cos u)^{j+2}}\,du \\ & = \; 3Q_{j+1} - 4 - 8(j+1)Q_{j+2} + (j+1)\int_{-\frac14\pi}^{\frac14\pi} \frac{9 - 6(3 - 2\sqrt{2}\cos u) + (3 - 2\sqrt{2}\cos u)^2}{(3 - 2\sqrt{2}\cos u)^{j+2}}\,du \\ & = \; 3Q_{j+1} - 4 - 8(j+1)Q_{j+2} + (j+1)\big(9Q_{j+2} - 6Q_{j+1} + Q_j\big) \end{aligned}$ so we obtain the recurrence relation $jQ_j - 3(2j+1)Q_{j+1} + (j+1)Q_{j+2} \; = \; 4 \hspace{2cm} j \ge 0$ Now $Q_0 = \tfrac12\pi$ and $Q_1 \; = \; 2\int_0^{\frac14\pi} \frac{du}{3 -2\sqrt{2}\cos u} \; = \; 4 \int_0^{\sqrt{2}-1} \frac{dt}{(3 - 2\sqrt{2}) + (3 + 2\sqrt{2})t^2} \; = \; 4(3 - 2\sqrt{2})\int_0^{\sqrt{2}-1} \frac{dt}{(3 - 2\sqrt{2})^2 + t^2} \; = \; 4\tan^{-1}\big(\sqrt{2}+1\big) \; = \; \tfrac32\pi$ using the substitution $t = \tan\tfrac12\theta$ . Thus we deduce that $Q_2 = 4 + \tfrac92\pi$ and $Q_3 = 20 + \tfrac{39}{2}\pi$ , and hence $\Delta \; = \; \tfrac{11}{8} + \tfrac58\pi - \tfrac{41}{32}Q_1 + \tfrac{5}{16}Q_2 - \tfrac{1}{32}Q_3\; = \; 2 - \tfrac12\pi$ making the answer $2 + 1 + 2 = \boxed{5}$ .