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Find-the-cubic-equation-whose-roots-are-the-radius-of-three-escribed-circles-in-term-of-inradius-circumradius-and-semiperimeter-

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Question Number 19055 by Tinkutara last updated on 03/Aug/17
Find the cubic equation whose roots are the radius of three escribed circles in term of inradius, circumradius and semiperimeter.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{cubic}\:\mathrm{equation}\:\mathrm{whose}\:\mathrm{roots} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{three}\:\mathrm{escribed}\:\mathrm{circles} \\ $$$$\mathrm{in}\:\mathrm{term}\:\mathrm{of}\:\mathrm{inradius},\:\mathrm{circumradius}\:\mathrm{and} \\ $$$$\mathrm{semiperimeter}. \\ $$
Answered by behi.8.3.4.1.7@gmail.com last updated on 05/Aug/17
(x−r_a )(x−r_b )(x−r_c )=0 ⇒x^3 −(Σr_a )x^2 +(Σr_a r_b )x−Πr_a =0 1)A+B=π−C⇒tg(((A+B)/2))=tg((π/2)−(C/2)) ⇒((tg(A/2)+tg(B/2))/(1−tg(A/2)tg(B/2)))=(1/(tg(C/2)))⇒Σtg(A/2)tg(B/2)=1 2)Πtg(A/2)=(√((((p−b)(p−c))/(p(p−a))).(((p−a)(p−c))/(p(p−b))).(((p−a)(p−b))/(p(p−c)))))= =(√((p(p−a)(p−b)(p−c))/p^4 ))=(S/p^2 )=(r/p) 3)Σtg(A/2)=Σ(((p−b)(p−c))/S)=((p^2 −p(b+c)+bc)/S)+ +((p^2 −p(a+c)+ac)/S)+((p^2 −p(a+b)+ab)/S)= =((3p^2 −2p(a+b+c)+ab+bc+ca)/S)= =((ab+bc+ca−p^2 )/S)=((4Rr+r^2 )/S)=((4R+r)/p) 4)r_a =p.tg(A/2)⇒ { ((Σr_a =pΣtg(A/2)=p.((4R+r)/p)=4R+r)),((Σr_a r_b =p^2 Σtg(A/2)tg(B/2)=p^2 )) :} 5)Πr_a =p^3 .Πtg(A/2)=p^3 .(r/p)=r.p^2 ⇒x^3 −(4R+r)x^2 +p^2 .x−r.p^2 =0
$$\left({x}−{r}_{{a}} \right)\left({x}−{r}_{{b}} \right)\left({x}−{r}_{{c}} \right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\left(\Sigma{r}_{{a}} \right){x}^{\mathrm{2}} +\left(\Sigma{r}_{{a}} {r}_{{b}} \right){x}−\Pi{r}_{{a}} =\mathrm{0} \\ $$$$\left.\mathrm{1}\right){A}+{B}=\pi−{C}\Rightarrow{tg}\left(\frac{{A}+{B}}{\mathrm{2}}\right)={tg}\left(\frac{\pi}{\mathrm{2}}−\frac{{C}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\frac{{tg}\frac{{A}}{\mathrm{2}}+{tg}\frac{{B}}{\mathrm{2}}}{\mathrm{1}−{tg}\frac{{A}}{\mathrm{2}}{tg}\frac{{B}}{\mathrm{2}}}=\frac{\mathrm{1}}{{tg}\frac{{C}}{\mathrm{2}}}\Rightarrow\Sigma{tg}\frac{{A}}{\mathrm{2}}{tg}\frac{{B}}{\mathrm{2}}=\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\Pi{tg}\frac{{A}}{\mathrm{2}}=\sqrt{\frac{\left({p}−{b}\right)\left({p}−{c}\right)}{{p}\left({p}−{a}\right)}.\frac{\left({p}−{a}\right)\left({p}−{c}\right)}{{p}\left({p}−{b}\right)}.\frac{\left({p}−{a}\right)\left({p}−{b}\right)}{{p}\left({p}−{c}\right)}}= \\ $$$$=\sqrt{\frac{{p}\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)}{{p}^{\mathrm{4}} }}=\frac{{S}}{{p}^{\mathrm{2}} }=\frac{{r}}{{p}} \\ $$$$\left.\mathrm{3}\right)\Sigma{tg}\frac{{A}}{\mathrm{2}}=\Sigma\frac{\left({p}−{b}\right)\left({p}−{c}\right)}{{S}}=\frac{{p}^{\mathrm{2}} −{p}\left({b}+{c}\right)+{bc}}{{S}}+ \\ $$$$+\frac{{p}^{\mathrm{2}} −{p}\left({a}+{c}\right)+{ac}}{{S}}+\frac{{p}^{\mathrm{2}} −{p}\left({a}+{b}\right)+{ab}}{{S}}= \\ $$$$=\frac{\mathrm{3}{p}^{\mathrm{2}} −\mathrm{2}{p}\left({a}+{b}+{c}\right)+{ab}+{bc}+{ca}}{{S}}= \\ $$$$=\frac{{ab}+{bc}+{ca}−{p}^{\mathrm{2}} }{{S}}=\frac{\mathrm{4}{Rr}+{r}^{\mathrm{2}} }{{S}}=\frac{\mathrm{4}{R}+{r}}{{p}} \\ $$$$\left.\mathrm{4}\right){r}_{{a}} ={p}.{tg}\frac{{A}}{\mathrm{2}}\Rightarrow\begin{cases}{\Sigma{r}_{{a}} ={p}\Sigma{tg}\frac{{A}}{\mathrm{2}}={p}.\frac{\mathrm{4}{R}+{r}}{{p}}=\mathrm{4}{R}+{r}}\\{\Sigma{r}_{{a}} {r}_{{b}} ={p}^{\mathrm{2}} \Sigma{tg}\frac{{A}}{\mathrm{2}}{tg}\frac{{B}}{\mathrm{2}}={p}^{\mathrm{2}} }\end{cases} \\ $$$$\left.\mathrm{5}\right)\Pi{r}_{{a}} ={p}^{\mathrm{3}} .\Pi{tg}\frac{{A}}{\mathrm{2}}={p}^{\mathrm{3}} .\frac{{r}}{{p}}={r}.{p}^{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{{x}}^{\mathrm{3}} −\left(\mathrm{4}\boldsymbol{{R}}+\boldsymbol{{r}}\right)\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{p}}^{\mathrm{2}} .\boldsymbol{{x}}−\boldsymbol{{r}}.\boldsymbol{{p}}^{\mathrm{2}} =\mathrm{0} \\ $$
Commented by Tinkutara last updated on 05/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by behi.8.3.4.1.7@gmail.com last updated on 05/Aug/17
corrected!
$${corrected}! \\ $$

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