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 $△AMN$ and quadrilateral $OBMN$. 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 $\left(\frac{m}{10}-3n+10\right)$.

Answer 1

Answer: 6

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

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
$y-8=\frac{1}{5}(x-0)$
$5y-40=x$
$x-5y+40=0$
Point $M$ can be obtained by solving equation of line $AB$ and $x-5y+40=0$
$\therefore M\left(\frac{20}{13},\frac{108}{13}\right)$
Now, Area of $△AMN=\frac{1}{2}\left|\begin{array}{ccc}0& 12& 1\\ 0& 8& 1\\ \frac{20}{13}& \frac{108}{13}& 1\end{array}\right|=\frac{40}{13}$
Area of quadrilateral $OBMN=$ Area of $△AOB-$ Area of $△AMN=\frac{350}{13}=\frac{m}{n}$