Question Number 201965 by ajfour last updated on 17/Dec/23
$${m}−{h}=\mathrm{2}{p} \\ $$$${p}\left({m}−{h}\right)={k}−{q} \\ $$$${mk}−{qh}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${k}−\mathrm{2}{q}=\mathrm{1}+{ph} \\ $$$${Assume}\:{one}\:{find}\:{the}\:{rest}! \\ $$$$\checkmark \\ $$
Answered by Frix last updated on 17/Dec/23
$$\mathrm{Solve}\:\left(\mathrm{1}\right)\:\mathrm{for}\:{h}\:\mathrm{then}\:\left(\mathrm{2}\right)\:\mathrm{for}\:{k}\:\mathrm{then}\:\left(\mathrm{3}\right)\:\mathrm{for}\:{m} \\ $$$$\Rightarrow \\ $$$${h}={m}−\mathrm{2}{p}=\frac{−\mathrm{12}{p}^{\mathrm{3}} −\mathrm{6}{pq}+\mathrm{1}}{\mathrm{6}{p}^{\mathrm{2}} } \\ $$$${k}=\mathrm{2}{p}^{\mathrm{2}} +{q} \\ $$$${m}=\frac{−\mathrm{6}{pq}+\mathrm{1}}{\mathrm{6}{p}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:{p}^{\mathrm{3}} −\frac{{p}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{24}}=\mathrm{0} \\ $$$$\Rightarrow\:{p}\in\left\{−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{9}},\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\pi}{\mathrm{9}},\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{cos}\:\frac{\pi}{\mathrm{18}}\right\} \\ $$$$\mathrm{We}\:\mathrm{are}\:\mathrm{free}\:\mathrm{to}\:\mathrm{choose}\:{q} \\ $$