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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
	$p=2((a/2))cos θ=acos A Σa^2 (b_P +c_P )=Σa^2 (r+q) = a^2 (bcos B+ccos C)+ b^2 (ccos C+acos A)+ c^2 (acos A+bcos B) =Σa(b^2 +c^2 )cos A =Σa(b^2 +c^2 )(((b^2 +c^2 −a^2 )/(2bc))) =(1/2)abcΣ{((b/c)+(c/b))^2 −(a^2 /(bc))((b/c)+(c/b))} =(1/2)abcΣ{2+((b^2 /c^2 )+(c^2 /b^2 ))−((a^2 /c^2 )+(a^2 /b^2 ))} =(1/2)abcΣ{2−((a^2 /b^2 )−(b^2 /c^2 ))−((a^2 /c^2 )−(c^2 /b^2 ))} =(1/2)abc(6−0−0)=3abc$
	$p = 2 (\frac{a}{2}) \cos θ = a \cos A$ $Σ a^{2} (b_{P} + c_{P}) = Σ a^{2} (r + q)$ $= a^{2} (b \cos B + c \cos C) +$ $b^{2} (c \cos C + a \cos A) +$ $c^{2} (a \cos A + b \cos B)$ $= Σ a (b^{2} + c^{2}) \cos A$ $= Σ a (b^{2} + c^{2}) (\frac{b^{2} + c^{2} - a^{2}}{2 b c})$ $= \frac{1}{2} a b c Σ {{(\frac{b}{c} + \frac{c}{b})}^{2} - \frac{a^{2}}{b c} (\frac{b}{c} + \frac{c}{b})}$ $= \frac{1}{2} a b c Σ {2 + (\frac{b^{2}}{c^{2}} + \frac{c^{2}}{b^{2}}) - (\frac{a^{2}}{c^{2}} + \frac{a^{2}}{b^{2}})}$ $= \frac{1}{2} a b c Σ {2 - (\frac{a^{2}}{b^{2}} - \frac{b^{2}}{c^{2}}) - (\frac{a^{2}}{c^{2}} - \frac{c^{2}}{b^{2}})}$ $= \frac{1}{2} a b c (6 - 0 - 0) = 3 a b c$
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