Question Number 186976 by mr W last updated on 12/Feb/23
Commented by mr W last updated on 12/Feb/23
$${the}\:{distances}\:{from}\:{a}\:{point}\:{to}\:{three} \\ $$$${vertexes}\:{of}\:{a}\:{regular}\:{hexagon}\:{are} \\ $$$${given}.\:{find}\:{the}\:{distances}\:{from}\:{this} \\ $$$${point}\:{to}\:{the}\:{other}\:{three}\:{vertexes}. \\ $$
Answered by mr W last updated on 12/Feb/23
$${applying}\:{the}\:{correlation}\:{between}\:{the} \\ $$$${distances}\:{which}\:{i}\:{found}\:{in}\:{Q}\mathrm{186924}. \\ $$$$\mathrm{2}\left({y}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \right)=\mathrm{3}×\mathrm{8}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\mathrm{121}\:\Rightarrow{y}=\sqrt{\mathrm{121}}=\mathrm{11} \\ $$$$\mathrm{2}\left({y}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} \right)=\mathrm{3}{z}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow{z}^{\mathrm{2}} =\frac{\mathrm{2}\left(\mathrm{11}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} \right)−\mathrm{5}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{139}\:\Rightarrow{z}=\sqrt{\mathrm{139}} \\ $$$$\mathrm{2}\left(\mathrm{5}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} \right)=\mathrm{3}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\frac{\mathrm{2}\left(\mathrm{5}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} \right)−\mathrm{11}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{43}\:\Rightarrow{x}=\sqrt{\mathrm{43}} \\ $$