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Question Number 116674 by ZiYangLee last updated on 05/Oct/20

Commented by ZiYangLee last updated on 05/Oct/20

A_1 B_1 C_1 is a regular triangle. A_2 B_2 C_2 are points on A_1 B_1 ,A_2 B_2 ,C_1 A_1 respectively such that A_1 A_2 =B_1 B_2 =C_1 C_2 =(1/5)A_1 B_1 . A_3 B_3 C_3 are points on A_2 B_2 ,B_2 C_2 ,C_2 A_2 respectively such that A_2 A_3 =B_2 B_3 =C_2 C_3 =(1/5)A_2 A_3 Repeat this process to constract infinitely many triangles △A_1 B_1 C_1 ,△A_2 B_2 C_2 ,△A_3 B_3 C_3 ,… If S_n is the area of △A_n B_n C_n , find the value of ((Σ_(n=1) ^∞ S_n )/S_1 )=((S_1 +S_2 +S_3 +…)/S_1 )

$${A}_{\mathrm{1}} {B}_{\mathrm{1}} {C}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{triangle}. \\ $$$${A}_{\mathrm{2}} {B}_{\mathrm{2}} {C}_{\mathrm{2}} \mathrm{are}\:\mathrm{points}\:\mathrm{on}\:{A}_{\mathrm{1}} {B}_{\mathrm{1}} ,{A}_{\mathrm{2}} {B}_{\mathrm{2}} ,{C}_{\mathrm{1}} {A}_{\mathrm{1}} \: \\ $$$$\mathrm{respectively}\:\mathrm{such}\:\mathrm{that}\:{A}_{\mathrm{1}} {A}_{\mathrm{2}} ={B}_{\mathrm{1}} {B}_{\mathrm{2}} ={C}_{\mathrm{1}} {C}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{5}}{A}_{\mathrm{1}} {B}_{\mathrm{1}} . \\ $$$${A}_{\mathrm{3}} {B}_{\mathrm{3}} {C}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:{A}_{\mathrm{2}} {B}_{\mathrm{2}} ,{B}_{\mathrm{2}} {C}_{\mathrm{2}} ,{C}_{\mathrm{2}} {A}_{\mathrm{2}} \\ $$$$\mathrm{respectively}\:\mathrm{such}\:\mathrm{that}\:{A}_{\mathrm{2}} {A}_{\mathrm{3}} ={B}_{\mathrm{2}} {B}_{\mathrm{3}} ={C}_{\mathrm{2}} {C}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{5}}{A}_{\mathrm{2}} {A}_{\mathrm{3}} \\ $$$$\mathrm{Repeat}\:\mathrm{this}\:\mathrm{process}\:\mathrm{to}\:\mathrm{constract}\:\mathrm{infinitely} \\ $$$$\mathrm{many}\:\mathrm{triangles}\:\bigtriangleup{A}_{\mathrm{1}} {B}_{\mathrm{1}} {C}_{\mathrm{1}} ,\bigtriangleup{A}_{\mathrm{2}} {B}_{\mathrm{2}} {C}_{\mathrm{2}} ,\bigtriangleup{A}_{\mathrm{3}} {B}_{\mathrm{3}} {C}_{\mathrm{3}} ,\ldots \\ $$$$\mathrm{If}\:{S}_{{n}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\bigtriangleup{A}_{{n}} {B}_{{n}} {C}_{{n}} , \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{S}_{{n}} }{{S}_{\mathrm{1}} }=\frac{{S}_{\mathrm{1}} +{S}_{\mathrm{2}} +{S}_{\mathrm{3}} +\ldots}{{S}_{\mathrm{1}} } \\ $$

Answered by mr W last updated on 05/Oct/20

A_2 B_2 =A_1 B_1 (√(((4/5))^2 +((1/5))^2 −2×(4/5)×(1/5)×(1/2))) ((A_2 B_2 )/(A_1 B_1 ))=((√(13))/5) (S_2 /S_1 )=(((√(13))/5))^2 =((13)/(25)) ((Σ_(n=1) ^∞ S_n )/S_1 )=((S_1 +((13)/(25))S_1 +(((13)/(25)))^2 S_1 +(((13)/(25)))^3 S_1 +...)/S_1 ) =(1/(1−((13)/(25))))=((25)/(12))

$${A}_{\mathrm{2}} {B}_{\mathrm{2}} ={A}_{\mathrm{1}} {B}_{\mathrm{1}} \sqrt{\left(\frac{\mathrm{4}}{\mathrm{5}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{2}×\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{{A}_{\mathrm{2}} {B}_{\mathrm{2}} }{{A}_{\mathrm{1}} {B}_{\mathrm{1}} }=\frac{\sqrt{\mathrm{13}}}{\mathrm{5}} \\ $$$$\frac{{S}_{\mathrm{2}} }{{S}_{\mathrm{1}} }=\left(\frac{\sqrt{\mathrm{13}}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{13}}{\mathrm{25}} \\ $$$$\frac{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{S}_{{n}} }{{S}_{\mathrm{1}} }=\frac{{S}_{\mathrm{1}} +\frac{\mathrm{13}}{\mathrm{25}}{S}_{\mathrm{1}} +\left(\frac{\mathrm{13}}{\mathrm{25}}\right)^{\mathrm{2}} {S}_{\mathrm{1}} +\left(\frac{\mathrm{13}}{\mathrm{25}}\right)^{\mathrm{3}} {S}_{\mathrm{1}} +...}{{S}_{\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{13}}{\mathrm{25}}}=\frac{\mathrm{25}}{\mathrm{12}} \\ $$

Commented by ZiYangLee last updated on 06/Oct/20

Finally...This morning my Tinkutara had a trouble that it couldnt load for the Forum

$$\mathrm{Finally}...\mathrm{This}\:\mathrm{morning}\:\mathrm{my}\:\mathrm{Tinkutara}\:\mathrm{had}\:\mathrm{a}\: \\ $$$$\mathrm{trouble}\:\mathrm{that}\:\mathrm{it}\:\mathrm{couldnt}\:\mathrm{load}\:\mathrm{for}\:\mathrm{the}\:\mathrm{Forum} \\ $$

Answered by Olaf last updated on 05/Oct/20

c_n = A_n B_n and c_(n+1) = A_(n+1) B_(n+1) A_(n+1) B_(n+1) ^2 = A_(n+1) B_n ^2 +B_n B_(n+1) ^2 −2A_(n+1) B_n .B_n B_(n+1) cos60° (Al−Kashi) c_(n+1) ^2 = ((4/5)c_n )^2 +((1/5)c_n )^2 −2.(4/5)c_n .(1/5)c_n (1/2) c_(n+1) ^2 = ((13)/(25))c_n ^2 S_(n+1) = ((√3)/4)c_(n+1) ^2 = ((13)/(25))(((√3)/4)c_n ^2 ) = ((13)/(25))S_n S_(n+1) = (((13)/(25)))^n S_1 ((Σ_(n=1) ^∞ S_n )/S_1 ) = Σ_(n=0) ^∞ (((13)/(25)))^n = (1/(1−((13)/(25)))) = ((25)/(12))

$${c}_{{n}} \:=\:\mathrm{A}_{{n}} \mathrm{B}_{{n}} \:\mathrm{and}\:{c}_{{n}+\mathrm{1}} \:=\:\mathrm{A}_{{n}+\mathrm{1}} \mathrm{B}_{{n}+\mathrm{1}} \\ $$$$\mathrm{A}_{{n}+\mathrm{1}} \mathrm{B}_{{n}+\mathrm{1}} ^{\mathrm{2}} \:=\:\mathrm{A}_{{n}+\mathrm{1}} \mathrm{B}_{{n}} ^{\mathrm{2}} +\mathrm{B}_{{n}} \mathrm{B}_{{n}+\mathrm{1}} ^{\mathrm{2}} −\mathrm{2A}_{{n}+\mathrm{1}} \mathrm{B}_{{n}} .\mathrm{B}_{{n}} \mathrm{B}_{{n}+\mathrm{1}} \mathrm{cos60}° \\ $$$$\left(\mathrm{Al}−\mathrm{Kashi}\right) \\ $$$${c}_{{n}+\mathrm{1}} ^{\mathrm{2}} \:=\:\left(\frac{\mathrm{4}}{\mathrm{5}}{c}_{{n}} \right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{5}}{c}_{{n}} \right)^{\mathrm{2}} −\mathrm{2}.\frac{\mathrm{4}}{\mathrm{5}}{c}_{{n}} .\frac{\mathrm{1}}{\mathrm{5}}{c}_{{n}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$$${c}_{{n}+\mathrm{1}} ^{\mathrm{2}} \:=\:\frac{\mathrm{13}}{\mathrm{25}}{c}_{{n}} ^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{S}_{{n}+\mathrm{1}} \:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{c}_{{n}+\mathrm{1}} ^{\mathrm{2}} \:=\:\frac{\mathrm{13}}{\mathrm{25}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{c}_{{n}} ^{\mathrm{2}} \right)\:=\:\frac{\mathrm{13}}{\mathrm{25}}\mathrm{S}_{{n}} \\ $$$$\mathrm{S}_{{n}+\mathrm{1}} \:=\:\left(\frac{\mathrm{13}}{\mathrm{25}}\right)^{{n}} \mathrm{S}_{\mathrm{1}} \\ $$$$\frac{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{S}_{{n}} }{\mathrm{S}_{\mathrm{1}} }\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{13}}{\mathrm{25}}\right)^{{n}} =\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{13}}{\mathrm{25}}}\:=\:\frac{\mathrm{25}}{\mathrm{12}} \\ $$

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