jmc

Let $O$ be center of the circle and $P$,$Q$ be the two points of tangent such that $P$ is on $BI$ and $Q$ is on $AI$. We know that $AD:CD=CD:BD=12:35$. Since the ratios between corresponding lengths of two similar diagrams are equal, we can let $AD=144,CD=420$ and $BD=1225$. Hence $AQ=144,BP=1225,AB=1369$ and the radius $r=OD=210$. Since we have $\tan OAB=\frac{35}{24}$ and $\tan OBA=\frac{6}{35}$ , we have $\sin (OAB+OBA)=\frac{1369}{\sqrt{(1801\ast 1261)}},$$\cos (OAB+OBA)=\frac{630}{\sqrt{(1801\ast 1261)}}$. Hence $\sin I=\sin (2OAB+2OBA)=\frac{2\ast 1369\ast 630}{1801\ast 1261}$. let $IP=IQ=x$ , then we have Area$(IBC)$ = $(2x+1225\ast 2+144\ast 2)\ast \frac{210}{2}$ = $(x+144)(x+1225)\ast \sin \frac{I}{2}$. Then we get $x+1369=\frac{3\ast 1369\ast (x+144)(x+1225)}{1801\ast 1261}$. Now the equation looks very complex but we can take a guess here. Assume that $x$ is a rational number (If it's not then the answer to the problem would be irrational which can't be in the form of $\frac{m}{n}$) that can be expressed as $\frac{a}{b}$ such that $(a,b)=1$. Look at both sides; we can know that $a$ has to be a multiple of $1369$ and not of $3$ and it's reasonable to think that $b$ is divisible by $3$ so that we can cancel out the $3$ on the right side of the equation. Let's see if $x=\frac{1369}{3}$ fits. Since $\frac{1369}{3}+1369=\frac{4\ast 1369}{3}$, and $\frac{3\ast 1369\ast (x+144)(x+1225)}{1801\ast 1261}=\frac{3\ast 1369\ast \frac{1801}{3}\ast \frac{1261\ast 4}{3}}{1801\ast 1261}=\frac{4\ast 1369}{3}$. Amazingly it fits! Since we know that $3\ast 1369\ast 144\ast 1225-1369\ast 1801\ast 1261<0$, the other solution of this equation is negative which can be ignored. Hence $x=1369/3$. Hence the perimeter is $1225\ast 2+144\ast 2+\frac{1369}{3}\ast 2=1369\ast \frac{8}{3}$, and $BC$ is $1369$. Hence $\frac{m}{n}=\frac{8}{3}$, $m+n=\boxed{11}$.