Question

Triangle $OAB$ has vertices $A (0,12), B (5,0)$ and $O (0,0)$. There exist line ' $l$ ' cutting $AB$ and $OA$ at $M$ and $N$ respectively, such that circles can be inscribed in $\triangle AMN$ and quadrilateral $O B M N$. Also these two circles are tangent to the line $l$ at the same point. If line $l$ pass through $(0,8)$, then the area of quadrilateral OBMN is $\frac{m}{n}$ where $m$ and $n$ are co-prime, then find the value of $\left(\frac{m}{10}-3 n+10\right)$.

Answer 1

$\text { Inradius of } \triangle OAB \text { ' } r \text { ' }=\frac{\Delta}{ S }=\frac{\frac{1}{2} \times 5 \times 12}{\frac{1}{2}(5+12+13)} $
$ r =\frac{60}{30}=2$

Hence, $I =(2,2)$
$m _{ AI }=\frac{12-2}{0-2}=-5$
Hence, slope of line $l=\frac{1}{5}$
Equation of line ' $T$ ' passing through $(0,8)$ is
$y-8=\frac{1}{5}(x-0) $
$5 y-40=x $
$x-5 y+40=0$
Point $M$ can be obtained by solving equation of line $A B$ and $x-5 y+40=0$
$\therefore M \left(\frac{20}{13}, \frac{108}{13}\right)$
Now, Area of $\triangle AMN =\frac{1}{2}\begin{vmatrix}0 & 12 & 1 \\ 0 & 8 & 1 \\ \frac{20}{13} & \frac{108}{13} & 1\end{vmatrix}=\frac{40}{13}$
Area of quadrilateral $O B M N=$ Area of $\triangle A O B-$ Area of $\triangle A M N=\frac{350}{13}=\frac{m}{n}$