Question Number 38849 by behi83417@gmail.com last updated on 30/Jun/18
Commented by behi83417@gmail.com last updated on 30/Jun/18
$$\:\boldsymbol{{raw}}\:\boldsymbol{{idea}}\:\boldsymbol{{to}}\:\boldsymbol{{mr}}\:\boldsymbol{{Ajfour}}! \\ $$$$\boldsymbol{{a}}^{\mathrm{2}} =\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{t}}^{\mathrm{2}} \:\:,\boldsymbol{{z}}=\boldsymbol{{c}}.\boldsymbol{{cos}\theta} \\ $$$$\boldsymbol{{b}}^{\mathrm{2}} =\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} \:\:\:,\boldsymbol{{x}}=\mathrm{1}β\boldsymbol{{c}}.\boldsymbol{{cos}\theta} \\ $$$$\boldsymbol{{c}}^{\mathrm{2}} =\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} \:\:\:\:,\boldsymbol{{t}}=\mathrm{1}β\boldsymbol{{c}}.\boldsymbol{{sin}\theta} \\ $$$$\boldsymbol{{d}}^{\mathrm{2}} =\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{t}}^{\mathrm{2}} \:\:\:\:\:,\boldsymbol{{y}}=\boldsymbol{{c}}.\boldsymbol{{sin}\theta} \\ $$$$βββββββββββββββ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} =\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{d}}^{\mathrm{2}} \\ $$
Answered by ajfour last updated on 30/Jun/18
$${ultimately}\:{all}\:{of}\:{these}\:{leads}\:{to} \\ $$$$\left({if}\:\:\boldsymbol{{c}}\mathrm{cot}\:\theta\:={x}\:\right) \\ $$$$\:\:\:\:\boldsymbol{{x}}^{\mathrm{2}} \left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} \right)=\left(\boldsymbol{{x}}+\boldsymbol{{c}}\right)^{\mathrm{2}} \:. \\ $$