It is indeed a Geometry Problem!

Let $x=\tan{\frac{A}{2}},y=\tan{\frac{B}{2}},z=\tan{\frac{C}{2}}$ for some $\triangle ABC$ with side $a,b,c$

$\begin{aligned} &\frac{6}{x+\frac{1}{x}}=3\sin{A}\\&\frac{8}{y+\frac{1}{y}}=4\sin{B}\\&\frac{10}{z+\frac{1}{z}}=5\sin{C} \end{aligned}$

so we have $\triangle ABC$ with $\sin{A}:\sin{B}:\sin{C}=a:b:c=20:15:12$ . From this we get

$\begin{aligned} &\cos{A}=\frac{15^2+12^2-20^2}{2\cdot 15\cdot 12}=\frac{-31}{360} \\ &\cos{B}=\frac{20^2+12^2-15^2}{2\cdot 20 \cdot 12}=\frac{319}{480}\\& \cos{C}= \frac{20^2+15^2-12^2}{2\cdot 20 \cdot 15}=\frac{581}{600} \\ &\text{and then we get} \\ & \cos^2{\frac{A}{2}}=\frac{\cos{A}+1}{2}=\frac{329}{720},\sec^2{\frac{A}{2}}=\frac{720}{329},\tan^2{\frac{A}{2}}=\frac{391}{329} \\ & \cos^2{\frac{B}{2}}=\frac{\cos{B}+1}{2}=\frac{799}{960},\sec^2{\frac{B}{2}}=\frac{960}{799},\tan^2{\frac{B}{2}}=\frac{161}{799}\\& \cos^2{\frac{C}{2}}=\frac{\cos{C}+1}{2}=\frac{1081}{1200},\sec^2{\frac{C}{2}}=\frac{1200}{1081},\tan^2{\frac{C}{2}}=\frac{119}{1081}\end{aligned}$

finally we get $x^2+y^2+z^2=\tan^2{\frac{A}{2}}+\tan^2{\frac{B}{2}}+\tan^2{\frac{C}{2}}=\frac{192963}{128639}$ and $m+n=321602$