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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=2((a/2))cos θ=acos A Σa^2 (b_P +c_P )=Σa^2 (r+q) = a^2 (bcos B+ccos C)+ b^2 (ccos C+acos A)+ c^2 (acos A+bcos B) =Σa(b^2 +c^2 )cos A =Σa(b^2 +c^2 )(((b^2 +c^2 −a^2 )/(2bc))) =(1/2)abcΣ{((b/c)+(c/b))^2 −(a^2 /(bc))((b/c)+(c/b))} =(1/2)abcΣ{2+((b^2 /c^2 )+(c^2 /b^2 ))−((a^2 /c^2 )+(a^2 /b^2 ))} =(1/2)abcΣ{2−((a^2 /b^2 )−(b^2 /c^2 ))−((a^2 /c^2 )−(c^2 /b^2 ))} =(1/2)abc(6−0−0)=3abc

$${p}=\mathrm{2}\left(\frac{{a}}{\mathrm{2}}\right)\mathrm{cos}\:\theta={a}\mathrm{cos}\:{A} \\ $$$$\Sigma{a}^{\mathrm{2}} \left({b}_{{P}} +{c}_{{P}} \right)=\Sigma{a}^{\mathrm{2}} \left({r}+{q}\right) \\ $$$$=\:\:\:{a}^{\mathrm{2}} \left({b}\mathrm{cos}\:{B}+{c}\mathrm{cos}\:{C}\right)+ \\ $$$$\:\:\:\:\:\:\:{b}^{\mathrm{2}} \left({c}\mathrm{cos}\:{C}+{a}\mathrm{cos}\:{A}\right)+ \\ $$$$\:\:\:\:\:\:\:\:{c}^{\mathrm{2}} \left({a}\mathrm{cos}\:{A}+{b}\mathrm{cos}\:{B}\right) \\ $$$$=\Sigma{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\mathrm{cos}\:{A} \\ $$$$=\Sigma{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{abc}\Sigma\left\{\left(\frac{{b}}{{c}}+\frac{{c}}{{b}}\right)^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{{bc}}\left(\frac{{b}}{{c}}+\frac{{c}}{{b}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{abc}\Sigma\left\{\mathrm{2}+\left(\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)−\left(\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{abc}\Sigma\left\{\mathrm{2}−\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }−\frac{{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\right)−\left(\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }−\frac{{c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{abc}\left(\mathrm{6}−\mathrm{0}−\mathrm{0}\right)=\mathrm{3}{abc} \\ $$

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