Question Number 206919 by BaliramKumar last updated on 30/Apr/24
Answered by A5T last updated on 30/Apr/24
$${MA}=\mathrm{2}{x};\:{MC}=\mathrm{3}{x}\Rightarrow{AC}=\mathrm{5}{x} \\ $$$$\Rightarrow{AB}+{BC}=\mathrm{4}+\mathrm{3}{x}+\mathrm{2}{x}=\mathrm{5}{x}+\mathrm{4} \\ $$$$\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{2}} +\left(\mathrm{2}+\mathrm{2}{x}\right)^{\mathrm{2}} =\mathrm{25}{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{2}=\mathrm{0}\Rightarrow{x}=\mathrm{2}\:{or}\:\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{AB}+{BC}=\mathrm{14} \\ $$
Answered by A5T last updated on 30/Apr/24
$${Let}\:{the}\:{perpendendicular}\:{from}\:{O}_{\mathrm{1}} \:{to}\:{AC}\:{meet} \\ $$$${it}\:{at}\:{D}. \\ $$$$\mathrm{1}−\frac{{AD}}{{AM}}=\frac{{AM}−{AD}}{{AM}}=\frac{{DM}}{\mathrm{4}}=\mathrm{1}−\frac{{r}_{{O}_{\mathrm{1}} } }{{r}_{{O}} } \\ $$$$\Rightarrow{DM}=\mathrm{4}−\mathrm{2}{r}_{{O}_{\mathrm{1}} } \\ $$$$\left({r}_{{O}_{\mathrm{1}} } +{r}_{{O}} \right)^{\mathrm{2}} =\left({r}_{{O}} −{r}_{{O}_{\mathrm{1}} } \right)^{\mathrm{2}} +\left({DM}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({r}_{{O}_{\mathrm{1}} } +\mathrm{2}\right)^{\mathrm{2}} =\left(\mathrm{2}−{r}_{{O}_{\mathrm{1}} } \right)^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{2}{r}_{{O}_{\mathrm{1}} } \right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}_{{O}_{\mathrm{1}} } =\mathrm{3}−\sqrt{\mathrm{5}} \\ $$$${Similarly},{we}\:{get}\:{for}\:{O}_{\mathrm{2}} \\ $$$$\left({r}_{{O}_{\mathrm{2}} } +\mathrm{2}\right)^{\mathrm{2}} =\left(\mathrm{2}−{r}_{{O}_{\mathrm{2}} } \right)^{\mathrm{2}} +\left(\mathrm{6}−\mathrm{3}{r}_{{O}_{\mathrm{2}} } \right) \\ $$$$\Rightarrow{r}_{{O}_{\mathrm{2}} } =\frac{\mathrm{22}−\mathrm{4}\sqrt{\mathrm{10}}}{\mathrm{9}} \\ $$