Question Number 187808 by LowLevelLump last updated on 22/Feb/23
Answered by a.lgnaoui last updated on 22/Feb/23
$$\bigtriangleup{ABC}'\:\:\begin{cases}{{AB}={AC}={BC}={a}}\\{{S}_{\mathrm{1}} =\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$$$\bigtriangleup{ACB}'\:\:\:\begin{cases}{{AC}={AB}'={CB}'={b}}\\{{S}_{\mathrm{2}} =\frac{{b}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$$$\bigtriangleup{BCC}'\:\:\:\begin{cases}{{BC}={BC}'={CC}^{'} ={c}}\\{{S}_{\mathrm{3}} =\frac{{c}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{sin}\:{B}=\frac{{AH}}{{AB}}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{AH}={a}\mathrm{sin}\:{B} \\ $$$${S}_{\mathrm{1}} −{S}_{\mathrm{2}} +{S}_{\mathrm{3}} =\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{1}\right) \\ $$$${S}\left(\bigtriangleup{ABC}\right)=\frac{{c}×\frac{{AH}}{{AB}}}{\mathrm{2}}=\frac{{ac}}{\mathrm{6}}=\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\begin{cases}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}}\\{{c}={a}\mathrm{cos}\:{B}+\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {B}}}\end{cases} \\ $$$${c}=\frac{\mathrm{2}{a}\sqrt{\mathrm{2}}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\:\mathrm{9}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\begin{cases}{{c}^{\mathrm{2}} ={b}^{\mathrm{2}} −{a}^{\mathrm{2}} +\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{3}\right)}\\{{c}=\frac{\mathrm{2}{a}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{9}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\mathrm{3}}\:\:\:\:\:\:\:\:\left(\mathrm{4}\right)}\end{cases} \\ $$$${b}^{\mathrm{2}} =\:\frac{\left(\mathrm{9}−\mathrm{16}{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{288}{a}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{\mathrm{9}} \\ $$$$\left(\mathrm{3}\right)\Rightarrow{c}^{\mathrm{2}} =\mathrm{1}+\frac{\left(\mathrm{9}−\mathrm{16}{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{288}{a}^{\mathrm{2}} }−\frac{\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{9}} \\ $$$${c}^{\mathrm{2}} =\frac{\mathrm{81}}{\mathrm{288}{a}^{\mathrm{2}} }\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\:{c}=\frac{\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{24}{a}}\:\:\:\:\:\left(\mathrm{5}\right)\:\:\:\:\:\: \\ $$$$\left.{a}\right)\:\:\:\:{AreaABC}=\frac{{ac}}{\mathrm{6}}=\:\:\:\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{48}} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{Area}}\left(\boldsymbol{{ABC}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{16}} \\ $$$$\left.{b}\right)\boldsymbol{{If}}\:\:\mathrm{sin}\:\boldsymbol{{A}}\mathrm{sin}\:\boldsymbol{{C}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\:\:\: \\ $$$$\:\:\:\:\:\frac{\mathrm{sin}\:{A}}{{c}}=\frac{\mathrm{sin}\:{B}}{{b}}=\frac{\mathrm{sin}\:{C}}{{a}}\Rightarrow{a}\mathrm{sin}\:{A}={c}\mathrm{sin}\:{C} \\ $$$$\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{3sin}\:{C}}={c}\mathrm{sin}\:{C}\:\:\:\Rightarrow\mathrm{3}{c}\mathrm{sin}^{\mathrm{2}} \:{C}={a}\sqrt{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{2}} {c}=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{3}{c}};\:\mathrm{sin}\:{c}=\frac{{AH}}{{b}}=\frac{{a}\mathrm{sin}\:{B}}{{b}}=\frac{{a}}{\mathrm{3}{b}}\left(\mathrm{sin}\:{B}=\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{9}{b}^{\mathrm{2}} }\:=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{3}{c}}\:\:\:\:\:{b}^{\mathrm{2}} =\frac{{ac}\sqrt{\mathrm{2}}}{\mathrm{6}}= \\ $$$$\left(\mathrm{5}\right)\:\Rightarrow{b}^{\mathrm{2}} =\frac{\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{24}}\:×\frac{\sqrt{\mathrm{2}}}{\mathrm{6}}\:\:\:\:\:\Rightarrow\:\:\boldsymbol{{b}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:\: \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 22/Feb/23