Dynamic Geometry: P105

This is probably the last of Dynamic Geometry: P80 Series by @Valentin Duringer . A lot of calculations in the earlier problems in the series are necessary before we can answer this problem. I have compiled the calculations into a note for better understanding and ease of reference so that my solution to this problem is shorter. I will update the solutions of other problems in the series.

Solution to Dynamic Geometry: P105

Let the center of the unit semicircle be $O(0,0)$ , the origin of the $xy$ -plane, its diameter base $AB$ , the center of the purple circle be $P$ , $\angle POB = \theta$ , and $t = \tan \frac \theta 2$ .

From the calculations in Dynamic Geometry: P80 Series , the radii of the purple, red, and orange circles, $r$ , $r_3$ , and $r_4$ respectively are:

$\begin{cases} r = \dfrac {2t}{(1+t)^2} \\ r_3 = \dfrac {2t}{9t^2+2t+1} \\ r_4 = \dfrac {2t}{t^2+2t+9} \end{cases}$

When the sum of the curvatures of these circles is $\dfrac {211}8$ , we have:

$\begin{aligned} \frac 1r + \frac 1{r_3} + \frac 1{r_4} & = \frac {211}8 \\ \frac {(1+t)^2}{2t} + \frac {9t^2+2t+1}{2t} + \frac {t^2+2t+9}{2t} & = \frac {211}8 \\ \implies t & = 4 \end{aligned}$

From the calculations in Dynamic Geometry: P80 Series again, the vertices of the two triangle are:

$\begin{cases} S_1 \left(\dfrac {1-13t^2}{13t^2+4t+1}, \dfrac {6t}{13t^2+4t+1} \right), & S_2 \left(\dfrac {1-5t^2}{5t^2+2t+1}, \dfrac {4t}{5t^2+2t+1} \right), & S_3 \left(\dfrac {1-9t^2}{9t^2+1}, \dfrac {6t}{9t^2+1} \right) \\ T_1 \left(\dfrac {13-t^2}{t^2+4t+13}, \dfrac {6t}{t^2+4t+13} \right), & T_2 \left(\dfrac {5-t^2}{t^2+2t+5}, \dfrac {4t}{t^2+2t+5} \right), & T_3 \left(\dfrac {9-t^2}{t^2+9}, \dfrac {6t}{t^2+9} \right) \end{cases}$

When $t=4$ ,

$\begin{cases} S_1 \left(-\dfrac {23}{25}, \dfrac 8{75} \right), & S_2 \left(-\dfrac {79}{89}, \dfrac {16}{89} \right), & S_3 \left(-\dfrac {143}{145}, \dfrac {24}{145} \right) \\ T_1 \left(-\dfrac 1{15}, \dfrac 8{15} \right), & T_2 \left(-\dfrac {11}{29}, \dfrac {16}{29} \right), & T_3 \left(-\dfrac 7{25}, \dfrac {24}{25} \right) \end{cases}$

We can find the areas of the two triangles using shoelace formula .

$\begin{cases} A_S = \dfrac 12 \left|\left(-\dfrac {23}{25} + \dfrac {143}{145} \right) \left(\dfrac {16}{89} - \dfrac 8{75} \right) - \left(-\dfrac {23}{25} + \dfrac {79}{89} \right) \left(\dfrac {24}{145} - \dfrac 8{75} \right) \right| & = \dfrac {1088}{322625} \\ A_T = \dfrac 12 \left|\left(-\dfrac 1{15} + \dfrac 7{25} \right) \left(\dfrac {16}{29} - \dfrac 8{15} \right) - \left(-\dfrac 1{15} + \dfrac {11}{29} \right) \left(\dfrac {24}{25} - \dfrac 8{15} \right) \right| & = \dfrac {704}{10875} \end{cases} \\ \implies \frac {A_S}{A_T} = \frac {1088}{322625} \cdot \frac {10875}{704} = \frac {51}{979}$

Therefore $p+q = 51+979 = \boxed{1030}$ .