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Question Number 187808 by LowLevelLump last updated on 22/Feb/23
Answered by a.lgnaoui last updated on 22/Feb/23
△ABC′ { ((AB=AC=BC=a)),((S_1 =((a^2 (√3))/2))) :} △ACB′ { ((AC=AB′=CB′=b)),((S_2 =((b^2 (√3))/2))) :} △BCC′ { ((BC=BC′=CC^′ =c)),((S_3 =((c^2 (√3))/2))) :} sin B=((AH)/(AB))=(1/3)⇒AH=asin B S_1 −S_2 +S_3 =(((a^2 −b^2 +c^2 )(√3))/2) =((√3)/2)(1) S(△ABC)=((c×((AH)/(AB)))/2)=((ac)/6)= (2) { ((a^2 −b^2 +c^2 =1)),((c=acos B+(√(b^2 −a^2 sin^2 B)))) :} c=((2a(√2))/3)+(1/3)(√( 9b^2 −a^2 )) { ((c^2 =b^2 −a^2 +1 (3))),((c=((2a(√2) +(√(9b^2 −a^2 )))/3) (4))) :} b^2 = (((9−16a^2 )^2 )/(288a^2 ))+(a^2 /9) (3)⇒c^2 =1+(((9−16a^2 )^2 )/(288a^2 ))−((8a^2 )/9) c^2 =((81)/(288a^2 )) ⇒ c=((9(√2))/(24a)) (5) a) AreaABC=((ac)/6)= ((3(√2))/(48)) Area(ABC)=((√2)/(16)) b)If sin Asin C=((√2)/3) ((sin A)/c)=((sin B)/b)=((sin C)/a)⇒asin A=csin C ((a(√2))/(3sin C))=csin C ⇒3csin^2 C=a(√2) sin^2 c=((a(√2))/(3c)); sin c=((AH)/b)=((asin B)/b)=(a/(3b))(sin B=(1/3)) =(a^2 /(9b^2 )) =((a(√2))/(3c)) b^2 =((ac(√2))/6)= (5) ⇒b^2 =((9(√2))/(24)) ×((√2)/6) ⇒ b=((√2)/4)
$$\bigtriangleup{ABC}'\:\:\begin{cases}{{AB}={AC}={BC}={a}}\\{{S}_{\mathrm{1}} =\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$$$\bigtriangleup{ACB}'\:\:\:\begin{cases}{{AC}={AB}'={CB}'={b}}\\{{S}_{\mathrm{2}} =\frac{{b}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$$$\bigtriangleup{BCC}'\:\:\:\begin{cases}{{BC}={BC}'={CC}^{'} ={c}}\\{{S}_{\mathrm{3}} =\frac{{c}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{sin}\:{B}=\frac{{AH}}{{AB}}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{AH}={a}\mathrm{sin}\:{B} \\ $$$${S}_{\mathrm{1}} −{S}_{\mathrm{2}} +{S}_{\mathrm{3}} =\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{1}\right) \\ $$$${S}\left(\bigtriangleup{ABC}\right)=\frac{{c}×\frac{{AH}}{{AB}}}{\mathrm{2}}=\frac{{ac}}{\mathrm{6}}=\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\begin{cases}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}}\\{{c}={a}\mathrm{cos}\:{B}+\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {B}}}\end{cases} \\ $$$${c}=\frac{\mathrm{2}{a}\sqrt{\mathrm{2}}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\:\mathrm{9}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\begin{cases}{{c}^{\mathrm{2}} ={b}^{\mathrm{2}} −{a}^{\mathrm{2}} +\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{3}\right)}\\{{c}=\frac{\mathrm{2}{a}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{9}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\mathrm{3}}\:\:\:\:\:\:\:\:\left(\mathrm{4}\right)}\end{cases} \\ $$$${b}^{\mathrm{2}} =\:\frac{\left(\mathrm{9}−\mathrm{16}{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{288}{a}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{\mathrm{9}} \\ $$$$\left(\mathrm{3}\right)\Rightarrow{c}^{\mathrm{2}} =\mathrm{1}+\frac{\left(\mathrm{9}−\mathrm{16}{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{288}{a}^{\mathrm{2}} }−\frac{\mathrm{8}{a}^{\mathrm{2}} }{\mathrm{9}} \\ $$$${c}^{\mathrm{2}} =\frac{\mathrm{81}}{\mathrm{288}{a}^{\mathrm{2}} }\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\:{c}=\frac{\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{24}{a}}\:\:\:\:\:\left(\mathrm{5}\right)\:\:\:\:\:\: \\ $$$$\left.{a}\right)\:\:\:\:{AreaABC}=\frac{{ac}}{\mathrm{6}}=\:\:\:\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{48}} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{Area}}\left(\boldsymbol{{ABC}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{16}} \\ $$$$\left.{b}\right)\boldsymbol{{If}}\:\:\mathrm{sin}\:\boldsymbol{{A}}\mathrm{sin}\:\boldsymbol{{C}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\:\:\: \\ $$$$\:\:\:\:\:\frac{\mathrm{sin}\:{A}}{{c}}=\frac{\mathrm{sin}\:{B}}{{b}}=\frac{\mathrm{sin}\:{C}}{{a}}\Rightarrow{a}\mathrm{sin}\:{A}={c}\mathrm{sin}\:{C} \\ $$$$\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{3sin}\:{C}}={c}\mathrm{sin}\:{C}\:\:\:\Rightarrow\mathrm{3}{c}\mathrm{sin}^{\mathrm{2}} \:{C}={a}\sqrt{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{2}} {c}=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{3}{c}};\:\mathrm{sin}\:{c}=\frac{{AH}}{{b}}=\frac{{a}\mathrm{sin}\:{B}}{{b}}=\frac{{a}}{\mathrm{3}{b}}\left(\mathrm{sin}\:{B}=\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{9}{b}^{\mathrm{2}} }\:=\frac{{a}\sqrt{\mathrm{2}}}{\mathrm{3}{c}}\:\:\:\:\:{b}^{\mathrm{2}} =\frac{{ac}\sqrt{\mathrm{2}}}{\mathrm{6}}= \\ $$$$\left(\mathrm{5}\right)\:\Rightarrow{b}^{\mathrm{2}} =\frac{\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{24}}\:×\frac{\sqrt{\mathrm{2}}}{\mathrm{6}}\:\:\:\:\:\Rightarrow\:\:\boldsymbol{{b}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:\: \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 22/Feb/23

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