Gergonne's Go

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 $A\; (-1,0) \hspace{1cm} B\;(1,0) \hspace{1cm} C\; (\cos2\theta,\sin2\theta)$ The triangle $ABC$ has sides $a,b,c$ and semiperimeter $s$ where $a \; = \; 2\sin\theta \hspace{1cm} b \; = \; 2\cos\theta \hspace{1cm} c \; = \; 2 \hspace{1cm} s \; = \; 1 + \cos\theta + \sin\theta$ If $D,E,F$ are the touch points of the incircle with the sides of the triangle, then $AF=AE=s-a$ and $BD=s-b$ , and hence we have coordinates $\begin{array}{lll} D & \hspace{0.2cm} & \big(1 - (1 + \sin\theta - \cos\theta)\sin\theta,(1 + \sin\theta - \cos\theta)\cos\theta\big) \\ E & & \big(-1 + (1 + \cos\theta - \sin\theta)\cos\theta,(1 + \cos\theta - \sin\theta)\sin\theta\big) \\ F & & (\cos\theta - \sin\theta,0) \end{array}$ and the inradius of the triangle $ABC$ is $r = s-c = \cos\theta + \sin\theta - 1$ . Solving simultaneously the equations for the lines $AD$ and $BE$ , we find that the Gergonne point $Ge$ has coordinates $\big(X(\theta),Y(\theta)\big)$ , where $\begin{aligned} X(\theta) & = \; \cos2\theta - \frac{2(\cos\theta - \sin\theta)(\cos\theta + \sin\theta - 1)}{3 + \cos\theta + \sin\theta} \\ Y(\theta) & = \; \frac{(\cos\theta + \sin\theta - 1) (1 + \cos\theta + \sin\theta)^2}{3 + \cos\theta + \sin\theta} \end{aligned}$ By symmetry, the desired area is $\begin{aligned} \Delta & = \; 2\int_{\frac12\pi}^0 X'(\theta)Y(\theta)\,d\theta \\ & = \; 2\int_{\frac12\pi}^0 \frac{(-1 + cos\theta + \sin\theta) (1 + \cos\theta + \sin\theta)^3 (-6 +\cos\theta + \cos3\theta + \sin\theta - 8\sin2\theta - \sin3\theta)}{(3 + \cos\theta + \sin\theta)^3} \,d\theta \\ & = \; -32\int_0^1 \frac{(t - 1) t (1 + t)^3 (1 + 9 t + 8 t^2 - 6 t^3 + t^4 - 7 t^5 + 2 t^6)}{(1 + t^2)^5 (2 + t + t^2)^3} \,dt \end{aligned}$ using the $t = \tan\tfrac12\theta$ substitution. This last integrand can be expanded by partial fractions as $-32\left(\begin{array}{l} \displaystyle \frac{16}{(1 + t^2)^5} + \frac{4 (3t - 11)}{(1 + t^2)^4} - \frac{ 2 (14 t - 19)}{(1 + t^2)^3} + \frac{16 t - 5}{(1 + t^2)^2} + \frac{3}{( 1 + t^2)} \\[2ex] \displaystyle - \frac{14 (t - 1)}{(2 + t + t^2)^3} - \frac{19 t + 33}{(2 + t + t^2)^2} - \frac{3}{( 2 + t + t^2)} \end{array}\right)$ Since each of these terms can be integrated using either $t = \tan\varphi$ or $t + \tfrac12 = \tan\varphi$ as substitution, we deduce that $\begin{aligned} \Delta & = \; -43\pi + \frac{368}{7} +\frac{3200}{\sqrt{7}}\cos^{-1}\big(\tfrac58\sqrt{2}\big) \\ & = \; -43\pi + \frac{2^4 \times 23}{7} + \frac{2^7 \times 25}{\sqrt{7}}\cos^{-1}\big(\tfrac58\sqrt{2}\big) \end{aligned}$ so that $\alpha = 43$ , $\beta = 4$ , $\gamma = 23$ , $\delta = 25$ , $\varepsilon = 5$ , $\zeta = 8$ , $p = 2$ and $q=7$ , so that $\alpha +\beta + \gamma + \delta + \varepsilon + \zeta + p + q = \boxed{117}$ .