Heptaparaparshinokh

Let $EB$ and $DG$ intersects at $H$ . We note that the four triangles $EDH$ , $DBH$ , $HBG$ , and $EHG$ are congruent and $EG \parallel DB$ . Also that $\triangle EDC$ , $\triangle AGE$ , and $\triangle ABC$ are similar.

Let $DB = EG = x$ . Then $\dfrac {EG}{AG} = \dfrac {BC}{AB}$ , $\implies \dfrac x{14-x} = \dfrac {15}{14} \implies x = \dfrac {14 \times 15}{29}$ . Let the radius of the large circle be $r$ and the inradius of $\triangle ABC$ be $R$ . Then

$\begin{aligned} \frac rR & = \frac {CD}{BC} = \frac {15-x}{15} = \frac {15}{29} \\ \implies r & = \frac {15}{29} R = \frac {15}{29} \times \frac As = \frac {15}{29} \times \frac {\frac 12 \times 14 \times 12}{\frac 12 (13+14+15)} = \frac {60}{29} \end{aligned}$

Therefore $p+q = 60+29 = \boxed{89}$ .

The congruence of the smaller circles hinted at the rhombic overall shape of the right triangles containing them, and we can use the property of rhombus' angle bisection here.

3/5 = 2(cos² €) – 1
cos € = 2/√5

Let's have the small RTs with the little incircles having legs x and 2x and hypothenuse √5x. With that, the triangle containing the larger circle would have 2 of its edges as √5x and (15 – √5x) that are relatively comparable to the original triangle's edges of 14 and 15 respectively.

√5x / (15 – √5x) = 14 / 15
x = 42√5 / 29

Ratio of triangles (smaller / original)
= [ (42√5 / 29) × √5 ] / 14
= 15 / 29

Original triangle's inradius
= (Double area) / Perimeter
= 14 × 12 / (13 + 14 + 15)
= 4

Inradius
= (Ratio) × (Original's)
= 60 / 29

Answer = 60 + 29 = 89