Question Number 228451 by Math1 last updated on 15/Apr/26
Answered by fantastic2 last updated on 15/Apr/26
Commented by fantastic2 last updated on 15/Apr/26
$${O}_{\mathrm{2}} {Z}\bot{O}_{\mathrm{1}} {K} \\ $$$${O}_{\mathrm{1}} {Z}={R}−{r}=\mathrm{2}−\mathrm{1}=\mathrm{1} \\ $$$${O}_{\mathrm{1}} {O}_{\mathrm{2}} =\sqrt{\mathrm{2}}\left({R}+{r}\right)=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$$\because\angle{O}_{\mathrm{1}} {AL}=\mathrm{45}^{\mathrm{0}} ,\angle{O}_{\mathrm{2}} {AM}=\mathrm{45}^{\mathrm{0}} \&\angle{LAM}=\mathrm{90}^{\mathrm{0}} \\ $$$$\therefore\angle{O}_{\mathrm{1}} {AO}_{\mathrm{2}} =\mathrm{180}^{\mathrm{0}} \\ $$$$\therefore{O}_{\mathrm{1}} ,{A},{O}_{\mathrm{2}} \:{are}\:{co}\:{linear} \\ $$$$\therefore{ZO}_{\mathrm{2}} ^{\mathrm{2}} =\left(\sqrt{\mathrm{2}}\left({R}+{r}\right)\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} \\ $$$${ZO}_{\mathrm{2}} =\sqrt{\mathrm{18}−\mathrm{1}}=\sqrt{\mathrm{17}}\Rightarrow{KN}=\sqrt{\mathrm{17}} \\ $$$${let}\:{KB}={BL}={a}\:\&{CM}={CN}={b} \\ $$$${now}\:, \\ $$$${BN}={BX} \\ $$$$\Rightarrow{KN}−{KB}={BL}+{LA}+{AX} \\ $$$$\Rightarrow\sqrt{\mathrm{17}}−{a}={a}+\mathrm{2}+\mathrm{1} \\ $$$$\Rightarrow{a}=\frac{\sqrt{\mathrm{17}}−\mathrm{3}}{\mathrm{2}} \\ $$$${similarl}\bar {{y}} \\ $$$${CK}={CY} \\ $$$$\Rightarrow{KN}−{CN}={CM}+{MA}+{AY} \\ $$$$\Rightarrow\sqrt{\mathrm{17}}−{b}={b}+\mathrm{1}+\mathrm{2} \\ $$$$\Rightarrow{b}=\frac{\sqrt{\mathrm{17}}−\mathrm{3}}{\mathrm{2}} \\ $$$$\sigma=\frac{\mathrm{1}}{\mathrm{2}}×{BA}×{AC} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{17}}−\mathrm{3}}{\mathrm{2}}+\mathrm{2}\right)\left(\frac{\sqrt{\mathrm{17}}−\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\sqrt{\mathrm{17}}+\mathrm{1}\right)\left(\sqrt{\mathrm{17}}−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{16}}{\mathrm{8}}=\mathrm{2}\checkmark \\ $$