Question Number 189517 by mr W last updated on 18/Mar/23
Commented by mr W last updated on 18/Mar/23
$${an}\:{unsolved}\:{old}\:{question} \\ $$
Answered by mr W last updated on 18/Mar/23
Commented by mr W last updated on 18/Mar/23
$${r}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{6}} \\ $$$${OA}={OB}={OC}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{3}}={d},\:{say} \\ $$$${x}^{\mathrm{2}} ={r}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{rd}\:\mathrm{cos}\:\theta \\ $$$${y}^{\mathrm{2}} ={r}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{rd}\:\mathrm{cos}\:\left(\mathrm{120}°−\theta\right) \\ $$$$\:\:\:\:={r}^{\mathrm{2}} +{d}^{\mathrm{2}} −{rd}\:\left(−\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right) \\ $$$${z}^{\mathrm{2}} ={r}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{rd}\:\mathrm{cos}\:\left(\mathrm{120}°+\theta\right) \\ $$$$\:\:\:\:={r}^{\mathrm{2}} +{d}^{\mathrm{2}} −{rd}\:\left(−\mathrm{cos}\:\theta−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{3}\left({r}^{\mathrm{2}} +{d}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{36}}+\frac{\mathrm{3}}{\mathrm{9}}\right){a}^{\mathrm{2}} =\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{4}}\:\checkmark \\ $$