First, note that $AB=37$; let the tangents from $I$ to $\omega$ have length $x$. Then the perimeter of $\triangle ABI$ is equal to \[2(x+AD+DB)=2(x+37).\] It remains to compute $\dfrac{2(x+37)}{37}=2+\dfrac{2}{37}x$.
Observe $CD=\dfrac{12\cdot 35}{37}=\dfrac{420}{37}$, so the radius of $\omega$ is $\dfrac{210}{37}$. We may also compute $AD=\dfrac{12^{2}}{37}$ and $DB=\dfrac{35^{2}}{37}$ by similar triangles. Let $O$ be the center of $\omega$; notice that \[\tan(\angle DAO)=\dfrac{DO}{AD}=\dfrac{210/37}{144/37}=\dfrac{35}{24}\] so it follows \[\sin(\angle DAO)=\dfrac{35}{\sqrt{35^{2}+24^{2}}}=\dfrac{35}{\sqrt{1801}}\] while $\cos(\angle DAO)=\dfrac{24}{\sqrt{1801}}$. By the double-angle formula $\sin(2\theta)=2\sin\theta\cos\theta$, it turns out that \[\sin(\angle BAI)=\sin(2\angle DAO)=\dfrac{2\cdot 35\cdot 24}{1801}=\dfrac{1680}{1801}\]
Using the area formula $\dfrac{1}{2}ab\sin(C)$ in $\triangle ABI$, \[[ABI]=\left(\dfrac{1}{2}\right)\left(\dfrac{144}{37}+x\right)(37)\left(\dfrac{1680}{1801}\right)=\left(\dfrac{840}{1801}\right)(144+37x).\] But also, using $rs$, \[[ABI]=\left(\dfrac{210}{37}\right)(37+x).\] Now we can get \[\dfrac{[ABI]}{210}=\dfrac{4(144+37x)}{1801}=\dfrac{37+x}{37}\] so multiplying everything by $37\cdot 1801=66637$ lets us solve for $x$: \[21312+5476x=66637+1801x.\] We have $x=\dfrac{66637-21312}{5476-1801}=\dfrac{45325}{3675}=\dfrac{37}{3}$, and now \[2+\dfrac{2}{37}x=2+\dfrac{2}{3}=\dfrac{8}{3}\] giving the answer, $\boxed{011}$.
[AIME 2009-I12 Geogebra Diagram.png](https://artofproblemsolving.com/wiki/index.php/File:AIME_2009-I12_Geogebra_Diagram.png)
In right $\triangle ABC$ with hypotenuse $\overline{AB}$, $AC = 12$, $BC = 35$, and $\overline{CD}$ is the altitude to $\overline{AB}$. Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$. The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.