Question Number 64686 by ajfour last updated on 20/Jul/19
Answered by ajfour last updated on 20/Jul/19
![Let B be origin and x axis along BC. x^2 +y^2 =c^2 (x−a)^2 +y^2 =b^2 ⇒ a(2x−a)=c^2 −b^2 x=((a^2 +c^2 −b^2 )/(2a)) y=(√(c^2 −(((a^2 +c^2 −b^2 )/(2a)))^2 )) △^2 =((a^2 y^2 )/4) = (a^2 /4)[(((2ac−a^2 −c^2 +b^2 )(2ac+a^2 +c^2 −b^2 )/(4a^2 ))] =(([b^2 −(a−c)^2 ][(a+c)^2 −b^2 ])/(16)) ⇒ 16△^2 =−(a^2 −c^2 )^2 −b^4 +2b^2 (a^2 +c^2 ) = −a^4 −b^4 −c^4 +2(a^2 c^2 +b^2 c^2 +c^2 a^2 ) =2Σa^2 b^2 −Σa^4 =4Σa^2 b^2 −(Σa^2 )^2 =4Σa^2 b^2 −[(a+b+c)^2 −2Σab]^2 =4Σa^2 b^2 −[16s^4 −16s^2 Σab+4(Σab)^2 ] =−16s^4 +16s^2 Σab−16(abc)s ⇒ △=(√(s^2 Σab−s^4 −(abc)s)) △=(√(s[s(ab+bc+ca)−s^3 −abc])) =(√(s[s(ab+bc+ca)+s^3 −s(a+b+c)−abc)) ⇒ __________________________ △=(√(s(s−a)(s−b)(s−c))) __________________________](Q64690.png)
$${Let}\:{B}\:{be}\:{origin}\:\:{and}\:{x}\:{axis}\:{along} \\ $$$${BC}. \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\left({x}−{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{a}\left(\mathrm{2}{x}−{a}\right)={c}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$${x}=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{a}} \\ $$$${y}=\sqrt{{c}^{\mathrm{2}} −\left(\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{a}}\right)^{\mathrm{2}} } \\ $$$$\bigtriangleup^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} {y}^{\mathrm{2}} }{\mathrm{4}}\:=\:\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\left[\frac{\left(\mathrm{2}{ac}−{a}^{\mathrm{2}} −{c}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{2}{ac}+{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right.}{\mathrm{4}{a}^{\mathrm{2}} }\right] \\ $$$$\:\:\:\:=\frac{\left[{b}^{\mathrm{2}} −\left({a}−{c}\right)^{\mathrm{2}} \right]\left[\left({a}+{c}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} \right]}{\mathrm{16}} \\ $$$$\Rightarrow\:\mathrm{16}\bigtriangleup^{\mathrm{2}} =−\left({a}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} −{b}^{\mathrm{4}} +\mathrm{2}{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:=\:−{a}^{\mathrm{4}} −{b}^{\mathrm{4}} −{c}^{\mathrm{4}} +\mathrm{2}\left({a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:=\mathrm{2}\Sigma{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\Sigma{a}^{\mathrm{4}} \\ $$$$\:\:\:\:\:=\mathrm{4}\Sigma{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left(\Sigma{a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{4}\Sigma{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left[\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\Sigma{ab}\right]^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{4}\Sigma{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left[\mathrm{16}{s}^{\mathrm{4}} −\mathrm{16}{s}^{\mathrm{2}} \Sigma{ab}+\mathrm{4}\left(\Sigma{ab}\right)^{\mathrm{2}} \right] \\ $$$$\:\:\:\:\:=−\mathrm{16}{s}^{\mathrm{4}} +\mathrm{16}{s}^{\mathrm{2}} \Sigma{ab}−\mathrm{16}\left({abc}\right){s} \\ $$$$\Rightarrow\:\bigtriangleup=\sqrt{{s}^{\mathrm{2}} \Sigma{ab}−{s}^{\mathrm{4}} −\left({abc}\right){s}} \\ $$$$\:\bigtriangleup=\sqrt{{s}\left[{s}\left({ab}+{bc}+{ca}\right)−{s}^{\mathrm{3}} −{abc}\right]} \\ $$$$\:\:\:\:\:=\sqrt{{s}\left[{s}\left({ab}+{bc}+{ca}\right)+{s}^{\mathrm{3}} −{s}\left({a}+{b}+{c}\right)−{abc}\right.} \\ $$$$\Rightarrow \\ $$$$\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\bigtriangleup=\sqrt{\boldsymbol{{s}}\left(\boldsymbol{{s}}−\boldsymbol{{a}}\right)\left(\boldsymbol{{s}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{s}}−\boldsymbol{{c}}\right)}\: \\ $$$$\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$
Commented by mr W last updated on 20/Jul/19
$${what}'{s}\:{the}\:{question}\:{sir}? \\ $$
Commented by ajfour last updated on 20/Jul/19
$${Nothing}\:{Sir},\:{just}\:{whiling}\:{away} \\ $$$${my}\:{time}.\:\:\left(^{\underset{\bullet} {\frown}} \underset{\smile} {\shortmid}^{\:\:\underset{\bullet} {\frown}} \partial\right. \\ $$