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Question Number 35390 by ajfour last updated on 18/May/18

Commented by ajfour last updated on 18/May/18

$$Findmaximumareaof△ABC.$$$$Theellipseequationis\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$

Commented by MJS last updated on 18/May/18

$$\mathrm{lol}.\mathrm{I}\mathrm{solved}\mathrm{this}\mathrm{some}\mathrm{years}\mathrm{ago},\mathrm{not}\mathrm{knowing}$$$$\mathrm{that}\mathrm{Steiner}\mathrm{solved}\mathrm{it}\mathrm{long}\mathrm{before}\mathrm{me}.$$$$\mathrm{all}\mathrm{triangles}\mathrm{with}\mathrm{barycenter}\mathrm{in}\mathrm{the}\mathrm{center}$$$$\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{are}\mathrm{of}\mathrm{equal}\mathrm{area}.$$$$\mathrm{I}\mathrm{will}\mathrm{post}\mathrm{my}\mathrm{proof}\mathrm{but}\mathrm{it}\mathrm{might}\mathrm{take}\mathrm{some}$$$$\mathrm{time}.$$$$\mathrm{google}Steiner-ellipse\mathrm{or}\mathrm{look}\mathrm{it}\mathrm{up}\mathrm{on}$$$$Wikipedia$$

Commented by MJS last updated on 18/May/18

$$\mathrm{btw}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{still}\mathrm{engaged}\mathrm{with}\mathrm{that}\mathrm{other}\mathrm{problem}...$$

Commented by ajfour last updated on 18/May/18

$$sokindofyouSir!$$

Commented by MJS last updated on 18/May/18

$$\mathrm{start}\mathrm{with}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{radius}a$$$$x^2+{\left(y^{\ast }\right)}^2=a^2$$$$\mathrm{insert}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}{A}^{\ast }{B}^{\ast }{C}^{\ast }\mathrm{with}$$$$A^{\ast }=\left(\begin{array}{c}{x}_A\\ y_A^{\ast }=\frac{a}{b}{y}_A\end{array}\right);B^{\ast }=\left(\begin{array}{c}{x}_B\\ y_B^{\ast }=\frac{a}{b}{y}_B\end{array}\right);C^{\ast }=\left(\begin{array}{c}{x}_C\\ y_C^{\ast }=\frac{a}{b}{y}_C\end{array}\right)$$$$\mathrm{now}\mathrm{transform}y=\frac{b}{a}{y}^{\ast }$$$$\mathrm{the}\mathrm{circle}\mathrm{becomes}\mathrm{the}\mathrm{ellipse}$$$$\mathrm{with}{\mathrm{area}}_{\mathrm{ell}}=\frac{b}{a}\times {\mathrm{area}}_{\mathrm{circle}}=ab\pi$$$$\mathrm{the}\mathrm{equilateral}\mathrm{triangle}\mathrm{becomes}\mathrm{the}\mathrm{given}\mathrm{one}$$$$\mathrm{with}{\mathrm{area}}_{△}=\frac{b}{a}\times {\mathrm{area}}_{\mathrm{equi}△}$$$$\mathrm{it}\mathrm{should}\mathrm{be}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{barycenter}$$$$\mathrm{stays}\mathrm{in}\mathrm{place}\mathrm{and}\mathrm{all}\mathrm{possible}\mathrm{triangles}$$$$\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{area}.$$$$$$$$\mathrm{I}\mathrm{will}\mathrm{post}\mathrm{some}\mathrm{equations}\mathrm{for}\mathrm{calculating}$$$$\mathrm{both}\mathrm{directions}:\mathrm{find}\mathrm{smallest}\mathrm{ellipse}\mathrm{to}\mathrm{a}$$$$\mathrm{given}\mathrm{triangle}\mathrm{and}\mathrm{find}\mathrm{the}\mathrm{angle}\mathrm{and}\mathrm{ratio}$$$$\mathrm{to}\mathrm{transform}\mathrm{a}\mathrm{equilateral}\mathrm{triangle}\mathrm{into}$$$$\mathrm{any}\mathrm{triangle}...\mathrm{just}\mathrm{give}\mathrm{me}\mathrm{some}\mathrm{time}\mathrm{to}$$$$\mathrm{find}\mathrm{these}\mathrm{formulas}\mathrm{within}\mathrm{the}\mathrm{chaos}\mathrm{in}\mathrm{my}$$$$\mathrm{library}...$$

Commented by ajfour last updated on 18/May/18

$$A=\frac{1}{2}\times a\sqrt{3}\times \left(\frac{3a}{2}\right)\left(\frac{b}{a}\right)$$$$=\frac{3\sqrt{3}\mathit{a}\mathit{b}}{4}.$$$$\mathit{U}nderstood\mathit{S}ir,thankyou!$$

Answered by rahul 19 last updated on 18/May/18

$$\frac{3\sqrt{3}}{4}ab.$$

Commented by rahul 19 last updated on 19/May/18

$$Gotthesameanswerbyassuming$$$$triangletobeisosceles!:)$$

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