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Question Number 206053 by mr W last updated on 05/Apr/24

Commented by mr W last updated on 05/Apr/24

an unsolved old question (Q182165)

$${an}\:{unsolved}\:{old}\:{question}\:\left({Q}\mathrm{182165}\right) \\ $$

Answered by mr W last updated on 05/Apr/24

Commented by mr W last updated on 06/Apr/24

((AE)/(BE))=((CF)/(DF))=(1/k), say ((ΔBCF)/(ΔBDF))=((CF)/(DF))=(1/k) ⇒k(S_1 +S_3 )=ΔDBF ...(i) ((ΔDAE)/(ΔDBE))=((AE)/(BE))=(1/k) ⇒k(S_2 +S_6 )=ΔDBE ...(ii) (i)+(ii): k(S_1 +S_3 +S_2 +S_6 )=ΔDBF+ΔDBE ⇒k(S_1 +S_3 +S_2 +S_6 )=S+S_4 +S_5 ...(iii) similarly ⇒S_1 +S_5 +S_2 +S_4 =k(S+S_3 +S_6 ) ...(iv) (iii)+(iv): k(S_1 +S_3 +S_2 +S_6 )+S_1 +S_5 +S_2 +S_4 =S+S_4 +S_5 +k(S+S_3 +S_6 ) (k+1)(S_1 +S_2 )=(k+1)S ⇒S=S_1 +S_2 ✓

$$\frac{{AE}}{{BE}}=\frac{{CF}}{{DF}}=\frac{\mathrm{1}}{{k}},\:{say} \\ $$$$\frac{\Delta{BCF}}{\Delta{BDF}}=\frac{{CF}}{{DF}}=\frac{\mathrm{1}}{{k}} \\ $$$$\Rightarrow{k}\left({S}_{\mathrm{1}} +{S}_{\mathrm{3}} \right)=\Delta{DBF}\:\:\:...\left({i}\right) \\ $$$$\frac{\Delta{DAE}}{\Delta{DBE}}=\frac{{AE}}{{BE}}=\frac{\mathrm{1}}{{k}} \\ $$$$\Rightarrow{k}\left({S}_{\mathrm{2}} +{S}_{\mathrm{6}} \right)=\Delta{DBE}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$${k}\left({S}_{\mathrm{1}} +{S}_{\mathrm{3}} +{S}_{\mathrm{2}} +{S}_{\mathrm{6}} \right)=\Delta{DBF}+\Delta{DBE} \\ $$$$\Rightarrow{k}\left({S}_{\mathrm{1}} +{S}_{\mathrm{3}} +{S}_{\mathrm{2}} +{S}_{\mathrm{6}} \right)={S}+{S}_{\mathrm{4}} +{S}_{\mathrm{5}} \:\:\:...\left({iii}\right) \\ $$$${similarly} \\ $$$$\Rightarrow{S}_{\mathrm{1}} +{S}_{\mathrm{5}} +{S}_{\mathrm{2}} +{S}_{\mathrm{4}} ={k}\left({S}+{S}_{\mathrm{3}} +{S}_{\mathrm{6}} \right)\:\:\:...\left({iv}\right) \\ $$$$\left({iii}\right)+\left({iv}\right): \\ $$$${k}\left({S}_{\mathrm{1}} +{S}_{\mathrm{3}} +{S}_{\mathrm{2}} +{S}_{\mathrm{6}} \right)+{S}_{\mathrm{1}} +{S}_{\mathrm{5}} +{S}_{\mathrm{2}} +{S}_{\mathrm{4}} ={S}+{S}_{\mathrm{4}} +{S}_{\mathrm{5}} +{k}\left({S}+{S}_{\mathrm{3}} +{S}_{\mathrm{6}} \right) \\ $$$$\left({k}+\mathrm{1}\right)\left({S}_{\mathrm{1}} +{S}_{\mathrm{2}} \right)=\left({k}+\mathrm{1}\right){S} \\ $$$$\Rightarrow{S}={S}_{\mathrm{1}} +{S}_{\mathrm{2}} \:\checkmark \\ $$

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