Dynamic Geometry: P65

Let the center of the semicircle be $O$ , the horizontal distance between $O$ and the two centers of the circles be $a$ , and the radius of the two circles be $r$ . By Pythagorean theorem ,

$\begin{aligned} a^2 + (3r)^2 & = (1-r)^2 \\ a^2 & = 1 - 2r - 8r^2 \\ \implies a & = \sqrt{1-2r-8r^2} \end{aligned}$

The area of the triangle is given by

$\begin{aligned} A & = \frac 12 \cdot a \cdot 2r = ar = r\sqrt{1-2r-8r^2} \\ \frac {dA}{dr} & = \frac {2r-6r^2-32r^3}{2r\sqrt{1-2r-8r^2}} & \small \blue{\text{Putting }\frac {dA}{dr}=0} \\ 16r^2 + 3r - 1 & = 0 & \small \blue{\text{For }r \ne 0} \\ \implies r & = \frac {\sqrt{73}-3}{32} & \small \blue{\text{Since }r > 0} \end{aligned}$

Therefore $A$ is maximum, when $r = \dfrac {\sqrt{73}-3}{32}$ . Then

$\begin{aligned} \frac Ar & = \frac {ar}r = a = \sqrt{1-2r-8r^2} \\ & = \sqrt{(1-4r)(1+2r)} \\ & = \frac {\sqrt{(44-4\sqrt{73})(26+2\sqrt{73})}}{32} \\ & = \frac {\sqrt{2(11-\sqrt{73})(13+\sqrt{73})}}{16} \\ & = \frac {\sqrt{35-\sqrt{73}}}8 \end{aligned}$

Therefore $\sqrt{p+q-m} = \sqrt{35+73-8} = \boxed{10}$ .