Question Number 145246 by gsk2684 last updated on 03/Jul/21
$$\mathrm{prove}\:\mathrm{that}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{inscribed}\: \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{r}\:\mathrm{having}\:\mathrm{maximum} \\ $$$$\mathrm{area}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{with} \\ $$$$\mathrm{side}\:\sqrt{\mathrm{3}}\mathrm{r}. \\ $$
Answered by Olaf_Thorendsen last updated on 03/Jul/21
$$\mathrm{Necessarily},\:\mathrm{by}\:\mathrm{symmetry}\:{a}\:=\:{b}\:=\:{c} \\ $$$$\mathrm{AB}^{\mathrm{2}} \:=\:\mathrm{OA}^{\mathrm{2}} +\mathrm{OB}^{\mathrm{2}} −\mathrm{2OA}.\mathrm{OB}.\mathrm{cos120}° \\ $$$${a}^{\mathrm{2}} \:=\:{r}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{r}.{r}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${a}^{\mathrm{2}} \:=\:\mathrm{3}{r}^{\mathrm{2}} \Rightarrow\:{a}\:=\:\sqrt{\mathrm{3}}{r} \\ $$$$\mathrm{The}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{equilateral}\:\mathrm{and}\:: \\ $$$$\mathrm{S}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} \:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left(\mathrm{3}{r}^{\mathrm{2}} \right)\:=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}{r}^{\mathrm{2}} \\ $$
Commented by gsk2684 last updated on 03/Jul/21
$$\mathrm{thank}\:\mathrm{you}. \\ $$$$\mathrm{Will}\:\mathrm{you}\:\mathrm{please}\:\mathrm{explain}\:\mathrm{how}\:\mathrm{to}\:\mathrm{prove} \\ $$$$\mathrm{to}\:\mathrm{prove}\:\mathrm{required}\:\mathrm{triangle}\:\mathrm{is}\: \\ $$$$\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$
Commented by ajfour last updated on 03/Jul/21
$${circle}\:{is}\:{circular}. \\ $$
Commented by peter frank last updated on 04/Jul/21
$${thank}\:{you}.{can}\:{you}\:{give}\:{a} \\ $$$${diagram}\: \\ $$