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Question Number 30455 by ajfour last updated on 22/Feb/18

Commented by mrW2 last updated on 25/Feb/18

p=((ac(a^2 −b^2 +c^2 ))/(a(a^2 −b^2 +c^2 )+c(a^2 +b^2 −c^2 ))) q=((c^2 (a^2 +b^2 −c^2 ))/(a(a^2 −b^2 +c^2 )+c(a^2 +b^2 −c^2 ))) (p/q)=((a(a^2 −b^2 +c^2 ))/(c(a^2 +b^2 −c^2 ))) please confirm.

$${p}=\frac{{ac}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)} \\ $$$${q}=\frac{{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)} \\ $$$$\frac{{p}}{{q}}=\frac{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)} \\ $$$${please}\:{confirm}. \\ $$

Commented by ajfour last updated on 25/Feb/18

yes sir, with some effort i obtained the same.

$${yes}\:{sir},\:{with}\:{some}\:{effort}\:{i}\:{obtained} \\ $$$${the}\:{same}. \\ $$

Commented by ajfour last updated on 25/Feb/18

Commented by ajfour last updated on 25/Feb/18

λa^� +μ(c^� −λa^� )=η(pc^� −a^� )+a^� if BA is divided in the ratio p/(1−p). comparing coefficient of c^� , ⇒ μ = ηp ...(i) further since ∠B is bisected, μ = ((λa)/(c+λa)) , η = (a/(a+pc)) ..(ii) and 𝛌 = ((ccos B)/a) ....(iii) from (i) and (ii): p = (μ/η) = ((λ(a+pc))/(c+λa)) ⇒ p(c+λa)=λa+λcp p=((λa)/(λa+c−λc)) now, using (iii) p = ((ccos B)/(ccos B+c−((c^2 cos B)/a))) = ((accos B)/(accos B+ac−c^2 cos B)) =(((a^2 +c^2 −b^2 ))/((a^2 +c^2 −b^2 )+2ac−(c/a)(a^2 +c^2 −b^2 ))) pc =((ac(a^2 +c^2 −b^2 ))/(a(a^2 +c^2 −b^2 )+c(a^2 +b^2 −c^2 ))) qc = c−pc qc = ((c^2 (a^2 +b^2 −c^2 ))/(a(a^2 +c^2 −b^2 )+c(a^2 +b^2 −c^2 ))) .

$$\lambda\bar {{a}}+\mu\left(\bar {{c}}−\lambda\bar {{a}}\right)=\eta\left({p}\bar {{c}}−\bar {{a}}\right)+\bar {{a}} \\ $$$${if}\:{BA}\:{is}\:{divided}\:{in}\:{the}\:{ratio} \\ $$$${p}/\left(\mathrm{1}−{p}\right). \\ $$$${comparing}\:{coefficient}\:{of}\:\bar {{c}}\:, \\ $$$$\:\:\:\:\Rightarrow\:\:\mu\:=\:\eta{p}\:\:\:\:...\left({i}\right) \\ $$$${further}\:{since}\:\angle{B}\:{is}\:{bisected}, \\ $$$$\:\:\:\:\mu\:=\:\frac{\lambda{a}}{{c}+\lambda{a}}\:\:\:,\:\:\:\eta\:=\:\frac{{a}}{{a}+{pc}}\:\:..\left({ii}\right) \\ $$$${and}\:\:\boldsymbol{\lambda}\:=\:\frac{\boldsymbol{{c}}\mathrm{cos}\:\boldsymbol{{B}}}{\boldsymbol{{a}}}\:\:\:\:....\left({iii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\:\:\:\:\:{p}\:=\:\frac{\mu}{\eta}\:=\:\frac{\lambda\left({a}+{pc}\right)}{{c}+\lambda{a}} \\ $$$$\Rightarrow\:\:\:{p}\left({c}+\lambda{a}\right)=\lambda{a}+\lambda{cp} \\ $$$$\:\:\:\:\:\:{p}=\frac{\lambda{a}}{\lambda{a}+{c}−\lambda{c}}\:\:\:\:{now},\:{using}\:\left({iii}\right) \\ $$$$\:\:\:\boldsymbol{{p}}\:=\:\frac{\boldsymbol{{c}}\mathrm{cos}\:\boldsymbol{{B}}}{\boldsymbol{{c}}\mathrm{cos}\:\boldsymbol{{B}}+\boldsymbol{{c}}−\frac{\boldsymbol{{c}}^{\mathrm{2}} \mathrm{cos}\:\boldsymbol{{B}}}{\boldsymbol{{a}}}} \\ $$$$\:\:\:\:\:\:=\:\frac{\boldsymbol{{ac}}\mathrm{cos}\:\boldsymbol{{B}}}{\boldsymbol{{ac}}\mathrm{cos}\:\boldsymbol{{B}}+\boldsymbol{{ac}}−\boldsymbol{{c}}^{\mathrm{2}} \mathrm{cos}\:\boldsymbol{{B}}} \\ $$$$\:\:=\frac{\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+\mathrm{2}{ac}−\frac{{c}}{{a}}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)} \\ $$$${pc}\:=\frac{{ac}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)} \\ $$$${qc}\:=\:{c}−{pc} \\ $$$${qc}\:=\:\frac{{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}\:. \\ $$

Commented by mrW2 last updated on 25/Feb/18

GREAT !

$$\mathbb{GREAT}\:! \\ $$

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