An Arrowhead

Draw in the following segments:

$\triangle ABC$ is a $15$ - $20$ - $25$ right triangle combined with a $20$ - $21$ - $29$ right triangle, so that $AH = 15$ , $CH = 20$ , $AC = 25$ , $BH = 21$ , and $BC = 29$ .

Let $GC = x$ , so that $AG = AC - GC = 25 - x$ . Since $\triangle ADG \sim \triangle CFG \sim \triangle AHC$ by AA similarity, $FG = \cfrac{CH}{AC} \cdot GC = \cfrac{4}{5}x$ , $FC = \cfrac{AH}{AC} \cdot GC = \cfrac{3}{5}x$ , $GD = \cfrac{CH}{AC} \cdot AG = \cfrac{4}{5}(25 - x)$ , and $AD = \cfrac{AH}{AC} \cdot AG = \cfrac{3}{5}(25 - x)$ .

Since $\triangle EFC \sim \triangle CHB$ by AA similarity, $EF = \cfrac{CH}{HB} \cdot FC = \cfrac{4}{7}x$ and $EC = \cfrac{CB}{HB} \cdot FC = \cfrac{29}{35}x$ , which means $EG = EF + FG = \cfrac{4}{7}x + \cfrac{4}{5}x = \cfrac{48}{35}x$ .

The inradius of right $\triangle ADG$ is $r_1 = \cfrac{1}{2}(AD + GD - AG) = \cfrac{1}{2}\bigg(\cfrac{3}{5}(25 - x) + \cfrac{4}{5}(25 - x) - (25 - x)\bigg) = \cfrac{1}{5}(25 - x)$ .

And the inradius of $\triangle EGC$ is $r_2 = \cfrac{2 \cdot \frac{1}{2} \cdot EG \cdot FC}{EG + EC + GC} = \cfrac{2 \cdot \frac{1}{2} \cdot \frac{48}{35}x \cdot \frac{3}{5}x}{\frac{48}{35}x + \frac{29}{35}x + x} = \cfrac{9}{35}x$ .

Since the two small circles are congruent, $r_1 = r_2$ , so $\cfrac{1}{5}(25 - x) = \cfrac{9}{35}x$ , which solves to $x = \cfrac{175}{16}$ .

Then:

$DB = AB - AD = 36 - \cfrac{3}{5}(25 - x) = 36 - \cfrac{3}{5}\bigg(25 - \cfrac{175}{16}\bigg) = \cfrac{441}{16}$

$ED = EG + GD = \cfrac{48}{35}x + \cfrac{4}{5}(25 - x) = \cfrac{48}{35} \cdot \cfrac{175}{16} + \cfrac{4}{5}\bigg(25 - \cfrac{175}{16}\bigg) = \cfrac{105}{4}$

$EB = EC + CB = \cfrac{29}{35}x + 29 = \cfrac{29}{35} \cdot \cfrac{175}{16} + 29 = \cfrac{609}{16}$

That means the inradius of $\triangle EDB$ is $R = \cfrac{1}{2}(DB + ED - EB) = \cfrac{1}{2}\bigg(\cfrac{441}{16} + \cfrac{105}{4} - \cfrac{609}{16}\bigg) = \cfrac{63}{8}$ .

Therefore, $p = 63$ , $q = 8$ , and $p + q = \boxed{71}$ .

First we find that altitude $CN = 20$ , $AN=15$ , and $NB = 21$ . Then $\triangle ACN$ is a $3$ - $4$ - $5$ right triangle and $\triangle BCN$ , a $20$ - $21$ - $29$ right triangle.

Let $AD = x$ . Then $DF = \dfrac 43 x$ , where $F$ is the intersection point of $AC$ and $DE$ , and $AF = \frac 53 x$ . Since inradius of a triangle, $r = \dfrac As$ , where $A$ and $s$ are the area and semi-perimeter of the triangle, then the radius of the two congruent circles

$r = \frac {AD \cdot DF}{AD+DF+FA} = \frac {\frac 43 x^2}{x\left(1+\frac 43 + \frac 53 \right)} = \frac x3$

Let $CG$ be perpendicular to $DE$ . Then $CG = AN - AD = 15-x$ . Then $CE = \dfrac {29(15-x)}{21}$ , $EG = \dfrac {20(15-x)}{21}$ , $GF = \dfrac {4(15-x)}3$ , and $FC = \dfrac {5(15-x)}3$ . Then

$r = \frac {CG \cdot EF}{CE+EF+FC} = \frac {(15-x)\left(\frac {20}{21}+\frac 43\right)}{\frac {29}{21}+\frac {20}{21}+\frac 43 + \frac 53} = \frac {3(15-x)}7 \\ \begin{aligned} \frac x3 & = \frac {3(15-x)}7 \\ 7x & = 135 - 9x \\ \implies x & = \frac {135}{16} \end{aligned}$

Then the radius of the large circle,

$\begin{aligned} R & = \frac {DE \cdot BD}{BD+DE+EB} = \frac {(36-x) \cdot \frac {20}{21}}{1+\frac {20}{21}+\frac {29}{21}} = \frac {63}8 \end{aligned}$

Therefore $p+q = 63+8 = \boxed{71}$ .