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Question Number 32184 by rahul 19 last updated on 21/Mar/18
If one vertex of the triangle having maximum area that can be inscribed in the circle ∣z−i∣=5 is 3−3i, then find other vertices of triangle.
$$\boldsymbol{{I}}{f}\:{one}\:{vertex}\:{of}\:{the}\:{triangle}\:{having} \\ $$$${maximum}\:{area}\:{that}\:{can}\:{be}\:{inscribed} \\ $$$${in}\:{the}\:{circle}\:\mid\boldsymbol{{z}}−\boldsymbol{{i}}\mid=\mathrm{5}\:{is}\:\mathrm{3}−\mathrm{3}\boldsymbol{{i}},\:{then} \\ $$$${find}\:{other}\:{vertices}\:{of}\:{triangle}. \\ $$
Answered by MJS last updated on 21/Mar/18
the triangle with maximum area is equilateral the given circle has it′s center at z=i, so we move everything to: ∣z∣=5 and the vertex to A′=3−4i=5e^(i(tan^(−1) (3/4)−(π/2))) the other vertices are B′=5e^(i(tan^(−1) (3/4)+(π/6))) =(2(√3)−(3/2))+(((3(√3))/2)+2)i C′=5e^(i(tan^(−1) (3/4)−((7π)/6))) =(−2(√3)−(3/2))+(−((3(√3))/3)+2)i now we move them back: A=A′+i=3−3i B=B′+i=(2(√3)−(3/2))+(((3(√3))/2)+3)i C=C′+i=(−2(√3)−(3/2))+(−((3(√3))/2)+3)i
$$\mathrm{the}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{maximum}\:\mathrm{area} \\ $$$$\mathrm{is}\:\mathrm{equilateral} \\ $$$$\mathrm{the}\:\mathrm{given}\:\mathrm{circle}\:\mathrm{has}\:\mathrm{it}'\mathrm{s}\:\mathrm{center} \\ $$$$\mathrm{at}\:{z}=\mathrm{i},\:\mathrm{so}\:\mathrm{we}\:\mathrm{move}\:\mathrm{everything} \\ $$$$\mathrm{to}:\:\mid{z}\mid=\mathrm{5}\:\mathrm{and}\:\mathrm{the}\:\mathrm{vertex}\:\mathrm{to} \\ $$$${A}'=\mathrm{3}−\mathrm{4i}=\mathrm{5e}^{\mathrm{i}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}−\frac{\pi}{\mathrm{2}}\right)} \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{vertices}\:\mathrm{are} \\ $$$${B}'=\mathrm{5e}^{\mathrm{i}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}+\frac{\pi}{\mathrm{6}}\right)} =\left(\mathrm{2}\sqrt{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{2}}\right)+\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{2}\right)\mathrm{i} \\ $$$${C}'=\mathrm{5e}^{\mathrm{i}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{7}\pi}{\mathrm{6}}\right)} =\left(−\mathrm{2}\sqrt{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{2}}\right)+\left(−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{3}}+\mathrm{2}\right)\mathrm{i} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{move}\:\mathrm{them}\:\mathrm{back}: \\ $$$${A}={A}'+\mathrm{i}=\mathrm{3}−\mathrm{3i} \\ $$$${B}={B}'+\mathrm{i}=\left(\mathrm{2}\sqrt{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{2}}\right)+\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{3}\right)\mathrm{i} \\ $$$${C}={C}'+\mathrm{i}=\left(−\mathrm{2}\sqrt{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{2}}\right)+\left(−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{3}\right)\mathrm{i} \\ $$
Commented by rahul 19 last updated on 21/Mar/18
thank u sir!
$${thank}\:{u}\:{sir}! \\ $$

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