Question Number 201004 by Mingma last updated on 27/Nov/23
Answered by ajfour last updated on 27/Nov/23
Commented by Mingma last updated on 27/Nov/23
Very elegant, sir!
Answered by ajfour last updated on 28/Nov/23
Commented by ajfour last updated on 02/Dec/23
$${p}=\mathrm{2}\left(\frac{{a}}{\mathrm{2}}\right)\mathrm{cos}\:\theta={a}\mathrm{cos}\:{A} \\ $$$$\Sigma{a}^{\mathrm{2}} \left({b}_{{P}} +{c}_{{P}} \right)=\Sigma{a}^{\mathrm{2}} \left({r}+{q}\right) \\ $$$$=\:\:\:{a}^{\mathrm{2}} \left({b}\mathrm{cos}\:{B}+{c}\mathrm{cos}\:{C}\right)+ \\ $$$$\:\:\:\:\:\:\:{b}^{\mathrm{2}} \left({c}\mathrm{cos}\:{C}+{a}\mathrm{cos}\:{A}\right)+ \\ $$$$\:\:\:\:\:\:\:\:{c}^{\mathrm{2}} \left({a}\mathrm{cos}\:{A}+{b}\mathrm{cos}\:{B}\right) \\ $$$$=\Sigma{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\mathrm{cos}\:{A} \\ $$$$=\Sigma{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{abc}\Sigma\left\{\left(\frac{{b}}{{c}}+\frac{{c}}{{b}}\right)^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{{bc}}\left(\frac{{b}}{{c}}+\frac{{c}}{{b}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{abc}\Sigma\left\{\mathrm{2}+\left(\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)−\left(\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{abc}\Sigma\left\{\mathrm{2}−\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\right)−\left(\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }−\frac{{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{abc}\left(\mathrm{6}−\mathrm{0}−\mathrm{0}\right)=\mathrm{3}{abc} \\ $$