Bored with congruent circles, yet?

Consider a general right triangle with side lengths $a$ , $b$ , and $c$ . Let the radius of the four congruent circles be $r$ and $\angle C = \theta$ . Also $DE$ be the line passing through the centers right two circles, which is perpendicular to $AC$ . Then $DE = r\sec \theta + 3r$ and $EC = b-3r$ . Now we have:

$\begin{aligned} \frac {DE}{EC} & = \tan \theta \\ \frac {r\sec \theta + 3r}{b-3r} & = \tan \theta & \small \blue{\text{Note that }\sec \theta = \frac ab \text{ and }\tan \theta = \frac cb} \\ \frac {\frac ab r + 3r}{b-3r} & = \frac cb \\ ar +3br & = bc - 3cr \\ \implies r & = \frac {bc}{a+3(b+c)} \end{aligned}$

For $a=5$ , $b=4$ , and $c=3$ , $r = \dfrac {4 \times 3}{5+3(4+3)} = \dfrac {12}{26} = \dfrac 6{13}$ . Then $a+b = 6+13 = \boxed{19}$ .

No trigonometry is needed. The area of $\Delta ABC$ can be written as $\Delta BGC + \Delta BGD + \Delta GFC + \$ square $ADGF$ , which can be expressed all in terms of $r$ :

$\frac{1}{2} \left(5r + (3-3r)(3r) + (4-3r)(3r) \right) + (3r)^2 = \frac{12}{2}$ $5r + 21r - 18r^2 + 18r^2 =12$ $r = \frac{12}{5 + 21} = \frac{6}{13}$

Hence $p + q = 6 + 13 = \boxed{19}$ .

Labelling the triangle's sides the same as Chew-Seong does, we get that $ar + 3cr + 3br = ab \Rightarrow r = \frac{bc}{a+3(b+c)}$ in general.

There are two solutions r= 6/13 and r=3/4 to the equation ✓[(3-3r)^2+(3r)^2-r^2]+ ✓[(4-3r)^2+(3r)^2-r^2]=5.

Answer = 19