jmc

Let $AM=x$, then $CM=15-x$. Also let $BM=d$ Clearly, $\frac{[ABM]}{[CBM]}=\frac{x}{15-x}$. We can also express each area by the rs formula. Then $\frac{[ABM]}{[CBM]}=\frac{p(ABM)}{p(CBM)}=\frac{12+d+x}{28+d-x}$. Equating and cross-multiplying yields $25x+2dx=15d+180$ or $d=\frac{25x-180}{15-2x}.$ Note that for $d$ to be positive, we must have $7.2<x<7.5$. By Stewart's Theorem, we have $12^2(15-x)+13^{2}x=d^{2}15+15x(15-x)$ or $432=3d^2+40x-3x^2.$ Brute forcing by plugging in our previous result for $d$, we have $432=\frac{3(25x-180{)}^2}{(15-2x{)}^2}+40x-3x^2.$ Clearing the fraction and gathering like terms, we get $0=12x^4-340x^3+2928x^2-7920x.$ Aside: Since $x$ must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that $12x$ is an integer. The only such $x$ in the above-stated range is $\frac{22}{3}$. Legitimately solving that quartic, note that $x=0$ and $x=15$ should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get $0=4x(x-15)(3x^2-40x+132)=x(x-15)(x-6)(3x-22).$ The only solution in the desired range is thus $\frac{22}{3}$. Then $CM=\frac{23}{3}$, and our desired ratio $\frac{AM}{CM}=\frac{22}{23}$, giving us an answer of $\boxed{45}$.