Solution:

By Heron's formula with $s = 27,$ the area of $\triangle AB_0C_0$ is $\sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90.$ In the similarity $\triangle B_0C_1C_0 \sim \triangle AB_0C_0,$ side $B_0C_0$ corresponds to $AC_0,$ so the ratio is $r = \frac{17}{25},$ and $C_1C_0$ (corresponding to $B_0C_0$) equals $17r.$ Hence $$\frac{AC_1}{AC_0} = \frac{25 - 17r}{25} = 1 - r^2,$$ which is the similarity ratio of $\triangle AB_1C_1$ to $\triangle AB_0C_0.$

Segments $\overline{B_1C_1}$ and $\overline{B_0C_1}$ split $\triangle AB_0C_0$ into the three pieces, so $$[B_0C_1B_1] = 90\left(1 - r^2 - (1 - r^2)^2\right) = 90\,r^2(1 - r^2).$$ Each successive stage repeats the construction inside $\triangle AB_nC_n,$ scaling all areas by $(1 - r^2)^2,$ and the triangles $B_{n-1}C_nB_n$ have disjoint interiors.

The union's area is the geometric series $$\frac{90\,r^2(1 - r^2)}{1 - (1 - r^2)^2} = \frac{90(1 - r^2)}{2 - r^2} = 90 \cdot \frac{336/625}{961/625} = \frac{90 \cdot 336}{961}.$$ Since $961 = 31^2$ shares no factor with $90 \cdot 336 = 30240,$ the answer is $q = 961.$