Question Number 175637 by mnjuly1970 last updated on 04/Sep/22
Answered by behi834171 last updated on 04/Sep/22
![I. r_a =p.tg(A/2)=p.((√(((p−b)(p−c))/(bc)))/( (√((p(p−a))/(bc)))))=(√((p(p−b)(p−c))/((p−a))))= =(S/(p−a)) and so on.... ⇒Σr_a =S.Σ((1/(p−a)))=S.((Σ(p−b)(p−c))/(Π(p−a)))= =(p/(4S)).[2ab+2bc+2ca−a^2 −b^2 −c^2 ]= =((4(r^2 +4R.r))/(4r))=r+4R .■ [(p−b)(p−c)=(1/4)(a+c−b)(a+b−c)= =(1/4)[a^2 +ab−ac+ac+bc−c^2 −ab−b^2 +bc]= =(1/4)[a^2 −(b−c)^2 ] (and so on) +(1/4)[c^2 −(b−a)^2 ]+ +(1/4)[b^2 −(a−c)^2 ]= =(1/4)[Σa^2 −(2Σa^2 −2Σab)]=(1/4)[2Σab−Σa^2 ]] II. d^2 =R^2 −2R.r ,d≥0⇒R≥2r . [d=distance from center of: incircle to center of outcircle]](https://www.tinkutara.com/question/Q175648.png)
$$\boldsymbol{{I}}. \\ $$$$\boldsymbol{{r}}_{\boldsymbol{{a}}} =\boldsymbol{{p}}.\boldsymbol{{tg}}\frac{\boldsymbol{{A}}}{\mathrm{2}}=\boldsymbol{{p}}.\frac{\sqrt{\frac{\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\boldsymbol{{bc}}}}}{\:\sqrt{\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}{\boldsymbol{{bc}}}}}=\sqrt{\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}}= \\ $$$$=\frac{\boldsymbol{{S}}}{\boldsymbol{{p}}−\boldsymbol{{a}}}\:\:\:\:\boldsymbol{{and}}\:\boldsymbol{{so}}\:\boldsymbol{{on}}…. \\ $$$$\Rightarrow\Sigma\boldsymbol{{r}}_{\boldsymbol{{a}}} =\boldsymbol{{S}}.\Sigma\left(\frac{\mathrm{1}}{\boldsymbol{{p}}−\boldsymbol{{a}}}\right)=\boldsymbol{{S}}.\frac{\Sigma\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\Pi\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}= \\ $$$$=\frac{\boldsymbol{{p}}}{\mathrm{4}\boldsymbol{{S}}}.\left[\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\mathrm{2}\boldsymbol{{ca}}−\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right]= \\ $$$$=\frac{\mathrm{4}\left(\boldsymbol{{r}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{R}}.\boldsymbol{{r}}\right)}{\mathrm{4}\boldsymbol{{r}}}=\boldsymbol{{r}}+\mathrm{4}\boldsymbol{{R}}\:\:\:.\blacksquare \\ $$$$\left[\left({p}−{b}\right)\left({p}−{c}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left({a}+{c}−{b}\right)\left({a}+{b}−{c}\right)=\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[{a}^{\mathrm{2}} +{ab}−{ac}+{ac}+{bc}−{c}^{\mathrm{2}} −{ab}−{b}^{\mathrm{2}} +{bc}\right]= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[{a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} \right]\:\:\:\:\left({and}\:{so}\:{on}\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{4}}\left[{c}^{\mathrm{2}} −\left({b}−{a}\right)^{\mathrm{2}} \right]+ \\ $$$$+\frac{\mathrm{1}}{\mathrm{4}}\left[{b}^{\mathrm{2}} −\left({a}−{c}\right)^{\mathrm{2}} \right]= \\ $$$$\left.=\frac{\mathrm{1}}{\mathrm{4}}\left[\Sigma\boldsymbol{{a}}^{\mathrm{2}} −\left(\mathrm{2}\Sigma\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{2}\Sigma\boldsymbol{{ab}}\right)\right]=\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{2}\Sigma\boldsymbol{{ab}}−\Sigma\boldsymbol{{a}}^{\mathrm{2}} \right]\right] \\ $$$$ \\ $$$$\boldsymbol{{II}}. \\ $$$$\boldsymbol{{d}}^{\mathrm{2}} =\boldsymbol{{R}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{R}}.\boldsymbol{{r}}\:\:\:\:,\boldsymbol{{d}}\geqslant\mathrm{0}\Rightarrow\boldsymbol{{R}}\geqslant\mathrm{2}\boldsymbol{{r}}\:\:\:. \\ $$$$\left[\boldsymbol{{d}}=\boldsymbol{{distance}}\:\boldsymbol{{from}}\:\:\boldsymbol{{center}}\:\boldsymbol{{of}}:\right. \\ $$$$\left.\:\boldsymbol{{incircle}}\:\boldsymbol{{to}}\:\boldsymbol{{center}}\:\boldsymbol{{of}}\:\:\boldsymbol{{outcircle}}\right] \\ $$