Question Number 218956 by Spillover last updated on 17/Apr/25
Answered by mr W last updated on 17/Apr/25
Commented by mr W last updated on 18/Apr/25
$${R}^{\mathrm{2}} =\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({c}+\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{5}{c}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\frac{{c}}{\:\sqrt{\mathrm{2}}}+{b}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\mathrm{5}{b}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}{bc}−\mathrm{2}{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{b}=\left(\frac{−\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{5}}\right){c} \\ $$$$\left(\frac{{c}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{2}}{a}+\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\left(\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\mathrm{5}{a}^{\mathrm{2}} +\mathrm{3}{ca}−\mathrm{2}{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\mathrm{2}{c}}{\mathrm{5}} \\ $$$$\frac{{blue}}{{pink}}=\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{2}}{\:\mathrm{2}\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\left(\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{7}\right)}{\mathrm{25}}\approx\mathrm{0}.\mathrm{952} \\ $$
Commented by Spillover last updated on 17/Apr/25
$${thank}\:{you} \\ $$
Answered by Spillover last updated on 18/Apr/25
