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Question Number 32184 by rahul 19 last updated on 21/Mar/18

$$\mathit{I}fonevertexofthetrianglehaving$$$$maximumareathatcanbeinscribed$$$$inthecircle\mid \mathit{z}-\mathit{i}\mid =5is3-3\mathit{i},then$$$$findotherverticesoftriangle.$$

Answered by MJS last updated on 21/Mar/18

$$\mathrm{the}\mathrm{triangle}\mathrm{with}\mathrm{maximum}\mathrm{area}$$$$\mathrm{is}\mathrm{equilateral}$$$$\mathrm{the}\mathrm{given}\mathrm{circle}\mathrm{has}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{center}$$$$\mathrm{at}z=\mathrm{i},\mathrm{so}\mathrm{we}\mathrm{move}\mathrm{everything}$$$$\mathrm{to}:\mid z\mid =5\mathrm{and}\mathrm{the}\mathrm{vertex}\mathrm{to}$$$$A^{\prime }=3-4\mathrm{i}={5\mathrm{e}}^{\mathrm{i}({\tan }^{-1}\frac{3}{4}-\frac{\pi }{2})}$$$$\mathrm{the}\mathrm{other}\mathrm{vertices}\mathrm{are}$$$$B^{\prime }={5\mathrm{e}}^{\mathrm{i}({\tan }^{-1}\frac{3}{4}+\frac{\pi }{6})}=(2\sqrt{3}-\frac{3}{2})+(\frac{3\sqrt{3}}{2}+2)\mathrm{i}$$$$C^{\prime }={5\mathrm{e}}^{\mathrm{i}({\tan }^{-1}\frac{3}{4}-\frac{7\pi }{6})}=(-2\sqrt{3}-\frac{3}{2})+(-\frac{3\sqrt{3}}{3}+2)\mathrm{i}$$$$\mathrm{now}\mathrm{we}\mathrm{move}\mathrm{them}\mathrm{back}:$$$$A=A^{\prime }+\mathrm{i}=3-3\mathrm{i}$$$$B=B^{\prime }+\mathrm{i}=(2\sqrt{3}-\frac{3}{2})+(\frac{3\sqrt{3}}{2}+3)\mathrm{i}$$$$C=C^{\prime }+\mathrm{i}=(-2\sqrt{3}-\frac{3}{2})+(-\frac{3\sqrt{3}}{2}+3)\mathrm{i}$$

Commented by rahul 19 last updated on 21/Mar/18

$$thankusir!$$

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