Dynamic Geometry: P134

Let the radius of the cyan and green semicircles be $r_1$ and $r_2$ respectively. From the compilation of calculations in Dynamic Geometry: P32 Series , we find that $r_1+r_2 = 1$ and the radii of the two circles are always the same and it is $r = r_1r_2$ . If the center $O$ of the unit semicircle is the origin $(0,0)$ of the $xy$ -plane, the coordinates of the centers and the two circles and the vertices of the triangles they inscribe are:

$\begin{cases} S(r_1^2-r_2, 2r_1\sqrt{r_2}), & T(r_1-r_2^2, 2r_2\sqrt{r_1}) \\ S_1 \left(\dfrac {r_1-2r_2}{1+r_2}, \dfrac {2r_1\sqrt{r_2}}{1+r_2}\right), & T_1 \left(\dfrac {2r_1-r_2}{1+r_1}, \dfrac {2r_2\sqrt{r_1}}{1+r_1} \right) \\ S_2 \left(r_1 - r_2, 2r_1\sqrt{r_2} \right), & T_2 \left(r_1 - r_2, 2r_2\sqrt{r_1} \right) \\ S_3 \left(\dfrac {r_1^2-r_2}{1-r_1r_2}, \dfrac {2r_1\sqrt{r_2}}{1-r_1r_2} \right), & T_3 \left(\dfrac {r_1-r_2^2}{1-r_1r_2}, \dfrac {2r_2\sqrt{r_1}}{1-r_1r_2} \right) \end{cases}$

Then the centroids of the two triangles are:

$\begin{cases} G_s = \left(\dfrac{1}{3}\left(\dfrac{r_{1}-2r_{2}}{1+r_{2}}+r_{1}-r_{2}+\dfrac{r_{1}^{2}-r_{2}}{1-r_{1}r_{2}}\right), \dfrac{2r_{1}\sqrt{r_{2}}}{3}\left(\dfrac{1}{1+r_{2}}+1+\dfrac{1}{1-r_{1}r_{2}}\right) \right) \\ G_t = \left(\dfrac{1}{3}\left(\dfrac{2r_{1}-r_{2}}{1+r_{1}}+r_{1}-r_{2}+\dfrac{r_{1}-r_{2}^{2}}{1-r_{1}r_{2}}\right) , \dfrac{2r_{2}\sqrt{r_{1}}}{3}\left(\dfrac{1}{1+r_{1}}+1+\dfrac{1}{1-r_{1}r_{2}}\right) \right) \end{cases}$

As Euler line passes through the centroid and circumcenter, the equations of the two Euler lines are:

$\begin{cases} SG_s: & \dfrac {y-2r_1\sqrt{r_2}}{x-r_1^2+r_2} = \dfrac {\frac{2r_{1}\sqrt{r_{2}}}{3}\left(\frac{1}{1+r_{2}}+1+\frac{1}{1-r_{1}r_{2}}\right)-2r_1\sqrt{r_2}}{\frac{1}{3}\left(\frac{r_{1}-2r_{2}}{1+r_{2}}+r_{1}-r_{2}+\frac{r_{1}^{2}-r_{2}}{1-r_{1}r_{2}}\right)-r_1^2+r_2} \\ SG_t: & \dfrac {y-2r_2\sqrt{r_1}}{x-r_1+r_2^2} = \dfrac {\frac{2r_2\sqrt{r_1}}3\left(\frac{1}{1+r_{1}}+1+\frac{1}{1-r_{1}r_{2}}\right)-2r_2\sqrt{r_1}}{\frac{1}{3}\left(\frac{2r_{1}-r_{2}}{1+r_1}+r_{1}-r_{2}+\frac{r_1-r_2^2}{1-r_{1}r_{2}}\right)-r_1+r_2^2} \end{cases} \\ \implies \begin{cases} SG_s: & \dfrac {y-2r_1\sqrt{r_2}}{x-r_1^2+r_2} = \dfrac{2r_{1}\sqrt{r_{2}}\left(\frac{1}{1+r_{2}}-2+\frac{1}{1-r_{1}r_{2}}\right)}{\frac{r_{1}-2r_{2}}{1+r_{2}}+r_{1}+2r_{2}-3r_{1}^{2}+\frac{r_{1}^{2}-r_{2}}{1-r_{1}r_{2}}} \\ SG_t: & \dfrac {y-2r_2\sqrt{r_1}}{x-r_1+r_2^2} = \dfrac{2r_{2}\sqrt{r_{1}}\left(\frac{1}{1+r_{1}}-2+\frac{1}{1-r_{1}r_{2}}\right)}{\frac{2r_{1}-r_{2}}{1+r_{1}}-2r_{1}-r_{2}+3r_{2}^{2}+\frac{r_{1}-r_{2}^{2}}{1-r_{1}r_{2}}} \end{cases}$

Note that the right-hand sides of the equations are the gradient of the respectively Euler lines. When the Euler lines are perpendicular, we have:

$\frac{2r_{1}\sqrt{r_{2}}\left(\frac{1}{1+r_{2}}-2+\frac{1}{1-r_{1}r_{2}}\right)}{\frac{r_{1}-2r_{2}}{1+r_{2}}+r_{1}+2r_{2}-3r_{1}^{2}+\frac{r_{1}^{2}-r_{2}}{1-r_{1}r_{2}}} \cdot \frac{2r_{2}\sqrt{r_{1}}\left(\frac{1}{1+r_{1}}-2+\frac{1}{1-r_{1}r_{2}}\right)}{\frac{2r_{1}-r_{2}}{1+r_{1}}-2r_{1}-r_{2}+3r_{2}^{2}+\frac{r_{1}-r_{2}^{2}}{1-r_{1}r_{2}}} = -1$

With the intuition that the equation above can be written to involve only $r$ , coupled with determination and hard work, it beautifully reduced to:

$\begin{aligned} \frac{4\sqrt r (4r-1)}{1-4r-9r^2} & = -1 \\ 4\sqrt r (4r-1) = 9r^2 + 4r - 1 \\ 256 r^3 - 128 r^2 + 16 r & = 81 r^4 + 72 r^3 - 2 r^2 -8 r + 1 \\ 81r^4 - 184r^3 + 126r^2 - 24 r + 1 & = 0 \\ (r-1)^2(81r^2-22r + 1) & = 0 & \small \blue{\text{Since } r < 1 \text{ and } \frac {11+2\sqrt{10}}{81} \text{ is not}} \\ \implies r & = \frac {11-2\sqrt{10}}{81} & \small \blue{\text{a solution of }\frac{4\sqrt r (4r-1)}{1-4r-9r^2} = -1.} \end{aligned}$

Therefore $p+q+m+l = 11+2+10+81 = \boxed{104}$ .