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Question-77425




Question Number 77425 by mr W last updated on 06/Jan/20
Commented by mr W last updated on 06/Jan/20
Given: r_a , r_b , r_c   Find: r=?    (see also Q77229)
$${Given}:\:{r}_{{a}} ,\:{r}_{{b}} ,\:{r}_{{c}} \\ $$$${Find}:\:{r}=? \\ $$$$ \\ $$$$\left({see}\:{also}\:{Q}\mathrm{77229}\right) \\ $$
Commented by jagoll last updated on 06/Jan/20
r = (√(r_a ×r_b )) +(√(r_a ×r_c )) +(√(r_b ×r_c ))
$${r}\:=\:\sqrt{{r}_{{a}} ×{r}_{{b}} }\:+\sqrt{{r}_{{a}} ×{r}_{{c}} }\:+\sqrt{{r}_{{b}} ×{r}_{{c}} } \\ $$
Answered by mr W last updated on 06/Jan/20
Commented by mr W last updated on 06/Jan/20
Found an other way than in Q77229:  cos α=((r−r_a )/(r+r_a ))  ⇒ sin α=((2(√(rr_a )))/(r+r_a ))  cos β=((r−r_b )/(r+r_b ))  ⇒ sin β=((2(√(rr_b )))/(r+r_b ))  cos γ=((r−r_c )/(r+r_c ))  2α+2β+2γ=2π  ⇒α+β+γ=π  cos (α+β)=cos (π−γ)=−cos γ  ((r−r_a )/(r+r_a ))×((r−r_b )/(r+r_b ))−((2(√(rr_a )))/(r+r_a ))×((2(√(rr_b )))/(r+r_b ))=−((r−r_c )/(r+r_c ))  (((r−r_a )(r−r_b )−4r(√(r_a r_b )))/((r+r_a )(r+r_b )))=−((r−r_c )/(r+r_c ))  ((r^2 −(r_a +r_b +4(√(r_a r_b )))r+r_a r_b )/(r^2 +(r_a +r_b )r+r_a r_b ))=−((r−r_c )/(r+r_c ))  r^3 −(r_a +r_b +4(√(r_a r_b )))r^2 +r_a r_b r+r_c r^2 −(r_c r_a +r_b r_c +4r_c (√(r_a r_b )))r+r_a r_b r_c       +r^3 +(r_a +r_b )r^2 +r_a r_b r−r_c r^2 −(r_c r_a +r_b r_c )r−r_a r_b r_c =0  r^3 −2(√(r_a r_b ))r^2 +r_a r_b r−r_c (r_a +r_b +2(√(r_a r_b )))r=0  r^2 −2(√(r_a r_b ))r+r_a r_b −r_c ((√r_a )+(√r_b ))^2 =0  (r−(√(r_a r_b )))^2 −((√(r_c r_a ))+(√(r_b r_c )))^2 =0  r−(√(r_a r_b ))=±((√(r_c r_a ))+(√(r_b r_c )))  r=(√(r_a r_b ))±((√(r_c r_a ))+(√(r_b r_c )))  since r>0,>r_a ,>r_b ,>r_c   ⇒r=(√(r_a r_b ))+(√(r_b r_c ))+(√(r_c r_a ))
$${Found}\:{an}\:{other}\:{way}\:{than}\:{in}\:{Q}\mathrm{77229}: \\ $$$$\mathrm{cos}\:\alpha=\frac{{r}−{r}_{{a}} }{{r}+{r}_{{a}} }\:\:\Rightarrow\:\mathrm{sin}\:\alpha=\frac{\mathrm{2}\sqrt{{rr}_{{a}} }}{{r}+{r}_{{a}} } \\ $$$$\mathrm{cos}\:\beta=\frac{{r}−{r}_{{b}} }{{r}+{r}_{{b}} }\:\:\Rightarrow\:\mathrm{sin}\:\beta=\frac{\mathrm{2}\sqrt{{rr}_{{b}} }}{{r}+{r}_{{b}} } \\ $$$$\mathrm{cos}\:\gamma=\frac{{r}−{r}_{{c}} }{{r}+{r}_{{c}} } \\ $$$$\mathrm{2}\alpha+\mathrm{2}\beta+\mathrm{2}\gamma=\mathrm{2}\pi \\ $$$$\Rightarrow\alpha+\beta+\gamma=\pi \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\mathrm{cos}\:\left(\pi−\gamma\right)=−\mathrm{cos}\:\gamma \\ $$$$\frac{{r}−{r}_{{a}} }{{r}+{r}_{{a}} }×\frac{{r}−{r}_{{b}} }{{r}+{r}_{{b}} }−\frac{\mathrm{2}\sqrt{{rr}_{{a}} }}{{r}+{r}_{{a}} }×\frac{\mathrm{2}\sqrt{{rr}_{{b}} }}{{r}+{r}_{{b}} }=−\frac{{r}−{r}_{{c}} }{{r}+{r}_{{c}} } \\ $$$$\frac{\left({r}−{r}_{{a}} \right)\left({r}−{r}_{{b}} \right)−\mathrm{4}{r}\sqrt{{r}_{{a}} {r}_{{b}} }}{\left({r}+{r}_{{a}} \right)\left({r}+{r}_{{b}} \right)}=−\frac{{r}−{r}_{{c}} }{{r}+{r}_{{c}} } \\ $$$$\frac{{r}^{\mathrm{2}} −\left({r}_{{a}} +{r}_{{b}} +\mathrm{4}\sqrt{{r}_{{a}} {r}_{{b}} }\right){r}+{r}_{{a}} {r}_{{b}} }{{r}^{\mathrm{2}} +\left({r}_{{a}} +{r}_{{b}} \right){r}+{r}_{{a}} {r}_{{b}} }=−\frac{{r}−{r}_{{c}} }{{r}+{r}_{{c}} } \\ $$$${r}^{\mathrm{3}} −\left({r}_{{a}} +{r}_{{b}} +\mathrm{4}\sqrt{{r}_{{a}} {r}_{{b}} }\right){r}^{\mathrm{2}} +{r}_{{a}} {r}_{{b}} {r}+{r}_{{c}} {r}^{\mathrm{2}} −\left({r}_{{c}} {r}_{{a}} +{r}_{{b}} {r}_{{c}} +\mathrm{4}{r}_{{c}} \sqrt{{r}_{{a}} {r}_{{b}} }\right){r}+{r}_{{a}} {r}_{{b}} {r}_{{c}} \\ $$$$\:\:\:\:+{r}^{\mathrm{3}} +\left({r}_{{a}} +{r}_{{b}} \right){r}^{\mathrm{2}} +{r}_{{a}} {r}_{{b}} {r}−{r}_{{c}} {r}^{\mathrm{2}} −\left({r}_{{c}} {r}_{{a}} +{r}_{{b}} {r}_{{c}} \right){r}−{r}_{{a}} {r}_{{b}} {r}_{{c}} =\mathrm{0} \\ $$$${r}^{\mathrm{3}} −\mathrm{2}\sqrt{{r}_{{a}} {r}_{{b}} }{r}^{\mathrm{2}} +{r}_{{a}} {r}_{{b}} {r}−{r}_{{c}} \left({r}_{{a}} +{r}_{{b}} +\mathrm{2}\sqrt{{r}_{{a}} {r}_{{b}} }\right){r}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}\sqrt{{r}_{{a}} {r}_{{b}} }{r}+{r}_{{a}} {r}_{{b}} −{r}_{{c}} \left(\sqrt{{r}_{{a}} }+\sqrt{{r}_{{b}} }\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({r}−\sqrt{{r}_{{a}} {r}_{{b}} }\right)^{\mathrm{2}} −\left(\sqrt{{r}_{{c}} {r}_{{a}} }+\sqrt{{r}_{{b}} {r}_{{c}} }\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${r}−\sqrt{{r}_{{a}} {r}_{{b}} }=\pm\left(\sqrt{{r}_{{c}} {r}_{{a}} }+\sqrt{{r}_{{b}} {r}_{{c}} }\right) \\ $$$${r}=\sqrt{{r}_{{a}} {r}_{{b}} }\pm\left(\sqrt{{r}_{{c}} {r}_{{a}} }+\sqrt{{r}_{{b}} {r}_{{c}} }\right) \\ $$$${since}\:{r}>\mathrm{0},>{r}_{{a}} ,>{r}_{{b}} ,>{r}_{{c}} \\ $$$$\Rightarrow{r}=\sqrt{{r}_{{a}} {r}_{{b}} }+\sqrt{{r}_{{b}} {r}_{{c}} }+\sqrt{{r}_{{c}} {r}_{{a}} } \\ $$
Commented by behi83417@gmail.com last updated on 06/Jan/20
AmAzInG proph: mrW!  thanks in advance sir.
$$\mathrm{AmAzInG}\:\mathrm{proph}:\:\mathrm{mrW}! \\ $$$$\mathrm{thanks}\:\mathrm{in}\:\mathrm{advance}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 06/Jan/20
thank you sir!  i remember that you once posted  this question, but i didn′t take  much notice.
$${thank}\:{you}\:{sir}! \\ $$$${i}\:{remember}\:{that}\:{you}\:{once}\:{posted} \\ $$$${this}\:{question},\:{but}\:{i}\:{didn}'{t}\:{take} \\ $$$${much}\:{notice}. \\ $$
Commented by Maclaurin Stickker last updated on 06/Jan/20
Great method, sir. Thanks for your work.
$${Great}\:{method},\:{sir}.\:{Thanks}\:{for}\:{your}\:{work}. \\ $$
Commented by Tawa11 last updated on 29/Dec/21
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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