2018 AMC 12B Problem 12

Let $q=AC$ and $r=BD.$ The angle bisector theorem gives $\tfrac{q}{3}=\tfrac{10}{r},$ so $r=\tfrac{30}{q}.$ Applying the triangle inequalities to sides $q,$ $10,$ and $3+r$ and substituting $r=\tfrac{30}{q}$ yields $(q-15)(q+2)\lt0$ and $(q-3)(q+10)\gt0$ (the third inequality holds automatically). Together these force $3\lt q\lt15.$ So $(m,n)=(3,15)$ and $m+n=18.$ Thus, the correct answer is C.