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

Commented by ZiYangLee last updated on 05/Oct/20

$$A_1{B}_1{C}_1\mathrm{is}\mathrm{a}\mathrm{regular}\mathrm{triangle}.$$$$A_2{B}_2{C}_2\mathrm{are}\mathrm{points}\mathrm{on}{A}_1{B}_1,A_2{B}_2,C_1{A}_1$$$$\mathrm{respectively}\mathrm{such}\mathrm{that}{A}_1{A}_2=B_1{B}_2=C_1{C}_2=\frac{1}{5}{A}_1{B}_1.$$$$A_3{B}_3{C}_3\mathrm{are}\mathrm{points}\mathrm{on}{A}_2{B}_2,B_2{C}_2,C_2{A}_2$$$$\mathrm{respectively}\mathrm{such}\mathrm{that}{A}_2{A}_3=B_2{B}_3=C_2{C}_3=\frac{1}{5}{A}_2{A}_3$$$$\mathrm{Repeat}\mathrm{this}\mathrm{process}\mathrm{to}\mathrm{constract}\mathrm{infinitely}$$$$\mathrm{many}\mathrm{triangles}△A_1{B}_1{C}_1,△A_2{B}_2{C}_2,△A_3{B}_3{C}_3,\dots$$$$\mathrm{If}{S}_n\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}△A_n{B}_n{C}_n,$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{\underset{n=1}{\sum ^{\mathrm{\infty }}}{S}_n}{S_1}=\frac{S_1+S_2+S_3+\dots }{S_1}$$

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

$$A_2{B}_2=A_1{B}_1\sqrt{{\left(\frac{4}{5}\right)}^2+{\left(\frac{1}{5}\right)}^2-2\times \frac{4}{5}\times \frac{1}{5}\times \frac{1}{2}}$$$$\frac{A_2{B}_2}{A_1{B}_1}=\frac{\sqrt{13}}{5}$$$$\frac{S_2}{S_1}={\left(\frac{\sqrt{13}}{5}\right)}^2=\frac{13}{25}$$$$\frac{\underset{n=1}{\sum ^{\mathrm{\infty }}}{S}_n}{S_1}=\frac{S_1+\frac{13}{25}{S}_1+{\left(\frac{13}{25}\right)}^2{S}_1+{\left(\frac{13}{25}\right)}^3{S}_1+...}{S_1}$$$$=\frac{1}{1-\frac{13}{25}}=\frac{25}{12}$$

Commented by ZiYangLee last updated on 06/Oct/20

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

Answered by Olaf last updated on 05/Oct/20

$$c_n={\mathrm{A}}_n{\mathrm{B}}_n\mathrm{and}{c}_{n+1}={\mathrm{A}}_{n+1}{\mathrm{B}}_{n+1}$$$${\mathrm{A}}_{n+1}{\mathrm{B}}_{n+1}^2={\mathrm{A}}_{n+1}{\mathrm{B}}_n^2+{\mathrm{B}}_n{\mathrm{B}}_{n+1}^2-{2\mathrm{A}}_{n+1}{\mathrm{B}}_n.{\mathrm{B}}_n{\mathrm{B}}_{n+1}\cos 60°$$$$(\mathrm{Al}-\mathrm{Kashi})$$$$c_{n+1}^2={\left(\frac{4}{5}{c}_n\right)}^2+{\left(\frac{1}{5}{c}_n\right)}^2-2.\frac{4}{5}{c}_n.\frac{1}{5}{c}_n\frac{1}{2}$$$$c_{n+1}^2=\frac{13}{25}{c}_n^2$$$$$$$${\mathrm{S}}_{n+1}=\frac{\sqrt{3}}{4}{c}_{n+1}^2=\frac{13}{25}\left(\frac{\sqrt{3}}{4}{c}_n^2\right)=\frac{13}{25}{\mathrm{S}}_n$$$${\mathrm{S}}_{n+1}={\left(\frac{13}{25}\right)}^n{\mathrm{S}}_1$$$$\frac{\underset{n=1}{\sum ^{\mathrm{\infty }}}{\mathrm{S}}_n}{{\mathrm{S}}_1}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{13}{25}\right)}^n=\frac{1}{1-\frac{13}{25}}=\frac{25}{12}$$

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