Points on Leg of Right Triangle with Distance One

We have $CM\cdot CB = CD_1\cdot CD_2$ , from which we get $CD_1=\frac{\sqrt{649}-1}{2}$ . If we take $O$ as the midpoint of $BM$ , we have $OD_1=\frac{9}{2},OC=\frac{27}{2}$ , so by applying the cosine theorem to the triangle $COD_1$ we have $\cos\widehat{ACB}=\frac{\sqrt{649}}{27}$ , hence: $\frac{m}{n}=\cot^2\widehat{ACB}=\frac{649}{80}.$

Let

$L$ - be the midpoint of $\overline{BM}$

$D$ - be the midpoint of $\overline{D_1D_2}$

Points $D_1$ & $D_2$ lies on the same circle with its center $L$ .

Therefore the radius $R$ is equal to:

$R = \frac{\overline{BM}}{2} = 4.5 = \overline{LD_1} = \overline{LD_2}$

$\overline{DL}$ perpendicular to $\overline{D_1D_2}$ .

$\overline{DL} = \sqrt{4.5^2 + 0.5^2} = \sqrt{20}$

$\frac{\overline{AC}}{\overline{AB}} = \frac{\overline{DC}}{\overline{DL}} = \frac{\sqrt{(9 + 4.5)^2 - (\sqrt{20})^2}}{\sqrt{20}} = \sqrt{\frac{649}{80}}$

$m + n = 649 + 80 = \boxed{729}$

The more elegant approach would be what Jack D'Aurizio posted, but anyways, I wanted to endorse the dark side: cartesian bash. Doesn't get very tedious... took me about five minutes.

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Set $A$ at the origin. Let $C= (a,0)$ and $B= (0,b),$ where $a^2 + b^2 = 18^2.$ Then, $M = \left( \dfrac{a}{2}, \dfrac{b}{2} \right).$ Note that $D_1, D_2$ lie on the circle whose diameter is $BM.$ The midpoint of $BM$ is $\left( \dfrac{a}{4}, \dfrac{3b}{4} \right),$ and the radius of the circle is $\dfrac{a^2+b^2}{2}.$ Thus, the equation of the circle is $\left( x - \dfrac{a}{4} \right) ^2 + \left ( y -\dfrac{3b}{4} \right) ^2 = \dfrac{a^2+b^2}{16}.$ At the points when it intersects the $x$ axis, we have $y=0.$ So, the $x$ coordinates of $D_1$ and $D_2$ satisfy the equation $\left( x - \dfrac{a}{4} \right) ^2 + \left( \dfrac{3b}{4} \right) ^2 = \dfrac{a^2+b^2}{16},$ the discriminant of which is $\dfrac{a^2 - 8b^2}{4}.$ The difference between the roots is equal to the square root of the discriminant, so we have $D_1D_2 = 1 = \dfrac{\sqrt{a^2-8b^2}}{2} \implies a^2 - 8b^2 = 4.$ We also have $a^2 + b^2 = 18^2.$ Solving this system of equations yields $a^2 = \dfrac{2596}{6}, b^2 = \dfrac{320}{9}, \dfrac{m}{n} = \dfrac{a^2}{b^2} = \dfrac{649}{80}.$ Hence, $m= 649, n= 90, m+n= \boxed{729}.$