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Question-103774




Question Number 103774 by dw last updated on 17/Jul/20
Answered by mr W last updated on 17/Jul/20
METHOD I  AB^(→) =a  BC^(→) =b  CA^(→) =c  a+b+c=0    AH=a+(1/9)b  BG=−a−(1/3)c  CP=−b−(5/6)a  BE=λBG=−λa−(λ/3)c  AE=μAH=μa+(μ/9)b=a−λa−(λ/3)c  μa+(μ/9)b=(1−λ)a+(λ/3)(a+b)  μa+(μ/9)b=(1−(2/3)λ)a+(λ/3)b  ⇒μ=3λ  ⇒μ=3λ=1−(2/3)λ  ⇒λ=(3/(11)) ⇒μ=(9/(11))    AF=ξAH=ξa+(ξ/9)b  CF=νCP=−νa−((5ν)/6)b=CA+AF  −νb−((5ν)/6)a=−a−b+ξa+(ξ/9)b  −((5ν)/6)b−νa=(ξ−1)a+((ξ/9)−1)b  ⇒−ν=(ξ/9)−1  ⇒−((5ν)/6)=ξ−1  ⇒(5/6)((ξ/9)−1)=ξ−1  ⇒ξ=(9/(49))  ⇒ν=((48)/(49))    BD=ηBG=−ηa+(η/3)(a+b)=−((2η)/3)a+(η/3)b  CD=εCP=−εb−((5ε)/6)a=CB+BD  −εb−((5ε)/6)a=−((2η)/3)a+((η/3)−1)b  ⇒−((5ε)/6)=−((2η)/3) ⇒5ε=4η  ⇒−ε=(η/3)−1  ⇒η=((15)/(17))  ⇒ε=((12)/(17))    ED=BD−BE=(η−λ)BG=  =(((15)/(17))−(3/(11)))BG=((114)/(187))BG  EG=BG−BE=(1−λ)BG=(8/(11))BG  ((ED)/(EG))=((114)/(187))×((11)/8)=((57)/(68))  EF=AE−AF=(μ−ξ)AH  =((9/(11))−(9/(49)))AH=((342)/(539))AH  ((EF)/(EA))=((342)/(539))×(1/μ)=((342)/(539))×((11)/9)=((38)/(49))    S_(DEF) =((EF)/(EA))×((ED)/(EG))×S_(AEG)   =((EF)/(EA))×((ED)/(EG))×((EG)/(BG))×S_(ABG)   =((EF)/(EA))×((ED)/(EG))×((EG)/(BG))×((AG)/(AC))×S_(ABC)   =((38)/(49))×((57)/(68))×(8/(11))×(1/3)×S_(ABC)   =((1444)/(9163))×S_(ABC)   ⇒(S_(DEF) /S_(ABC) )=((1444)/(9163))≈0.1576
$$\boldsymbol{{METHOD}}\:\boldsymbol{{I}} \\ $$$$\overset{\rightarrow} {{AB}}=\boldsymbol{{a}} \\ $$$$\overset{\rightarrow} {{BC}}=\boldsymbol{{b}} \\ $$$$\overset{\rightarrow} {{CA}}=\boldsymbol{{c}} \\ $$$$\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}=\mathrm{0} \\ $$$$ \\ $$$${AH}=\boldsymbol{{a}}+\frac{\mathrm{1}}{\mathrm{9}}\boldsymbol{{b}} \\ $$$${BG}=−\boldsymbol{{a}}−\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{c}} \\ $$$${CP}=−\boldsymbol{{b}}−\frac{\mathrm{5}}{\mathrm{6}}\boldsymbol{{a}} \\ $$$${BE}=\lambda{BG}=−\lambda\boldsymbol{{a}}−\frac{\lambda}{\mathrm{3}}\boldsymbol{{c}} \\ $$$${AE}=\mu{AH}=\mu\boldsymbol{{a}}+\frac{\mu}{\mathrm{9}}\boldsymbol{{b}}=\boldsymbol{{a}}−\lambda\boldsymbol{{a}}−\frac{\lambda}{\mathrm{3}}\boldsymbol{{c}} \\ $$$$\mu\boldsymbol{{a}}+\frac{\mu}{\mathrm{9}}\boldsymbol{{b}}=\left(\mathrm{1}−\lambda\right)\boldsymbol{{a}}+\frac{\lambda}{\mathrm{3}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right) \\ $$$$\mu\boldsymbol{{a}}+\frac{\mu}{\mathrm{9}}\boldsymbol{{b}}=\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\lambda\right)\boldsymbol{{a}}+\frac{\lambda}{\mathrm{3}}\boldsymbol{{b}} \\ $$$$\Rightarrow\mu=\mathrm{3}\lambda \\ $$$$\Rightarrow\mu=\mathrm{3}\lambda=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\lambda \\ $$$$\Rightarrow\lambda=\frac{\mathrm{3}}{\mathrm{11}}\:\Rightarrow\mu=\frac{\mathrm{9}}{\mathrm{11}} \\ $$$$ \\ $$$${AF}=\xi{AH}=\xi\boldsymbol{{a}}+\frac{\xi}{\mathrm{9}}\boldsymbol{{b}} \\ $$$${CF}=\nu{CP}=−\nu\boldsymbol{{a}}−\frac{\mathrm{5}\nu}{\mathrm{6}}\boldsymbol{{b}}={CA}+{AF} \\ $$$$−\nu\boldsymbol{{b}}−\frac{\mathrm{5}\nu}{\mathrm{6}}\boldsymbol{{a}}=−\boldsymbol{{a}}−\boldsymbol{{b}}+\xi\boldsymbol{{a}}+\frac{\xi}{\mathrm{9}}\boldsymbol{{b}} \\ $$$$−\frac{\mathrm{5}\nu}{\mathrm{6}}\boldsymbol{{b}}−\nu\boldsymbol{{a}}=\left(\xi−\mathrm{1}\right)\boldsymbol{{a}}+\left(\frac{\xi}{\mathrm{9}}−\mathrm{1}\right)\boldsymbol{{b}} \\ $$$$\Rightarrow−\nu=\frac{\xi}{\mathrm{9}}−\mathrm{1} \\ $$$$\Rightarrow−\frac{\mathrm{5}\nu}{\mathrm{6}}=\xi−\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{5}}{\mathrm{6}}\left(\frac{\xi}{\mathrm{9}}−\mathrm{1}\right)=\xi−\mathrm{1} \\ $$$$\Rightarrow\xi=\frac{\mathrm{9}}{\mathrm{49}} \\ $$$$\Rightarrow\nu=\frac{\mathrm{48}}{\mathrm{49}} \\ $$$$ \\ $$$${BD}=\eta{BG}=−\eta\boldsymbol{{a}}+\frac{\eta}{\mathrm{3}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)=−\frac{\mathrm{2}\eta}{\mathrm{3}}\boldsymbol{{a}}+\frac{\eta}{\mathrm{3}}\boldsymbol{{b}} \\ $$$${CD}=\epsilon{CP}=−\epsilon\boldsymbol{{b}}−\frac{\mathrm{5}\epsilon}{\mathrm{6}}\boldsymbol{{a}}={CB}+{BD} \\ $$$$−\epsilon\boldsymbol{{b}}−\frac{\mathrm{5}\epsilon}{\mathrm{6}}\boldsymbol{{a}}=−\frac{\mathrm{2}\eta}{\mathrm{3}}\boldsymbol{{a}}+\left(\frac{\eta}{\mathrm{3}}−\mathrm{1}\right)\boldsymbol{{b}} \\ $$$$\Rightarrow−\frac{\mathrm{5}\epsilon}{\mathrm{6}}=−\frac{\mathrm{2}\eta}{\mathrm{3}}\:\Rightarrow\mathrm{5}\epsilon=\mathrm{4}\eta \\ $$$$\Rightarrow−\epsilon=\frac{\eta}{\mathrm{3}}−\mathrm{1} \\ $$$$\Rightarrow\eta=\frac{\mathrm{15}}{\mathrm{17}} \\ $$$$\Rightarrow\epsilon=\frac{\mathrm{12}}{\mathrm{17}} \\ $$$$ \\ $$$${ED}={BD}−{BE}=\left(\eta−\lambda\right){BG}= \\ $$$$=\left(\frac{\mathrm{15}}{\mathrm{17}}−\frac{\mathrm{3}}{\mathrm{11}}\right){BG}=\frac{\mathrm{114}}{\mathrm{187}}{BG} \\ $$$${EG}={BG}−{BE}=\left(\mathrm{1}−\lambda\right){BG}=\frac{\mathrm{8}}{\mathrm{11}}{BG} \\ $$$$\frac{{ED}}{{EG}}=\frac{\mathrm{114}}{\mathrm{187}}×\frac{\mathrm{11}}{\mathrm{8}}=\frac{\mathrm{57}}{\mathrm{68}} \\ $$$${EF}={AE}−{AF}=\left(\mu−\xi\right){AH} \\ $$$$=\left(\frac{\mathrm{9}}{\mathrm{11}}−\frac{\mathrm{9}}{\mathrm{49}}\right){AH}=\frac{\mathrm{342}}{\mathrm{539}}{AH} \\ $$$$\frac{{EF}}{{EA}}=\frac{\mathrm{342}}{\mathrm{539}}×\frac{\mathrm{1}}{\mu}=\frac{\mathrm{342}}{\mathrm{539}}×\frac{\mathrm{11}}{\mathrm{9}}=\frac{\mathrm{38}}{\mathrm{49}} \\ $$$$ \\ $$$${S}_{{DEF}} =\frac{{EF}}{{EA}}×\frac{{ED}}{{EG}}×{S}_{{AEG}} \\ $$$$=\frac{{EF}}{{EA}}×\frac{{ED}}{{EG}}×\frac{{EG}}{{BG}}×{S}_{{ABG}} \\ $$$$=\frac{{EF}}{{EA}}×\frac{{ED}}{{EG}}×\frac{{EG}}{{BG}}×\frac{{AG}}{{AC}}×{S}_{{ABC}} \\ $$$$=\frac{\mathrm{38}}{\mathrm{49}}×\frac{\mathrm{57}}{\mathrm{68}}×\frac{\mathrm{8}}{\mathrm{11}}×\frac{\mathrm{1}}{\mathrm{3}}×{S}_{{ABC}} \\ $$$$=\frac{\mathrm{1444}}{\mathrm{9163}}×{S}_{{ABC}} \\ $$$$\Rightarrow\frac{{S}_{{DEF}} }{{S}_{{ABC}} }=\frac{\mathrm{1444}}{\mathrm{9163}}\approx\mathrm{0}.\mathrm{1576} \\ $$

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