Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 164018 by cortano1 last updated on 13/Jan/22

Answered by mr W last updated on 13/Jan/22

Commented by cortano1 last updated on 13/Jan/22

superb solution

$${superb}\:{solution} \\ $$

Commented by mr W last updated on 13/Jan/22

b=2R a=(√((R+r)^2 −(R−r)^2 ))+(√((R+r)^2 −(2R−r)^2 )) a=2(√(Rr))+(√(3R(2r−R))) BC=2(√((R+r)^2 −r^2 ))=2(√(R(R+2r))) a=(√(BC^2 −R^2 ))=(√(4R(R+2r)−R^2 ))=(√(R(3R+8r))) 2(√(Rr))+(√(3R(2r−R)))=(√(R(3R+8r))) let λ=(R/r) 2+(√(3(2−λ)))=(√(3λ+8)) 3(2−λ)=3λ+8+4−4(√(3λ+8)) 3(λ+1)=2(√(3λ+8)) 9(λ^2 +2λ+1)=4(3λ+8) 9λ^2 +6λ−23=0 λ=((−1+2(√6))/3) ⇒(R/r)=((2(√6)−1)/3) ✓

$${b}=\mathrm{2}{R} \\ $$$${a}=\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }+\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left(\mathrm{2}{R}−{r}\right)^{\mathrm{2}} } \\ $$$${a}=\mathrm{2}\sqrt{{Rr}}+\sqrt{\mathrm{3}{R}\left(\mathrm{2}{r}−{R}\right)} \\ $$$${BC}=\mathrm{2}\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\mathrm{2}\sqrt{{R}\left({R}+\mathrm{2}{r}\right)} \\ $$$${a}=\sqrt{{BC}^{\mathrm{2}} −{R}^{\mathrm{2}} }=\sqrt{\mathrm{4}{R}\left({R}+\mathrm{2}{r}\right)−{R}^{\mathrm{2}} }=\sqrt{{R}\left(\mathrm{3}{R}+\mathrm{8}{r}\right)} \\ $$$$\mathrm{2}\sqrt{{Rr}}+\sqrt{\mathrm{3}{R}\left(\mathrm{2}{r}−{R}\right)}=\sqrt{{R}\left(\mathrm{3}{R}+\mathrm{8}{r}\right)} \\ $$$${let}\:\lambda=\frac{{R}}{{r}} \\ $$$$\mathrm{2}+\sqrt{\mathrm{3}\left(\mathrm{2}−\lambda\right)}=\sqrt{\mathrm{3}\lambda+\mathrm{8}} \\ $$$$\mathrm{3}\left(\mathrm{2}−\lambda\right)=\mathrm{3}\lambda+\mathrm{8}+\mathrm{4}−\mathrm{4}\sqrt{\mathrm{3}\lambda+\mathrm{8}} \\ $$$$\mathrm{3}\left(\lambda+\mathrm{1}\right)=\mathrm{2}\sqrt{\mathrm{3}\lambda+\mathrm{8}} \\ $$$$\mathrm{9}\left(\lambda^{\mathrm{2}} +\mathrm{2}\lambda+\mathrm{1}\right)=\mathrm{4}\left(\mathrm{3}\lambda+\mathrm{8}\right) \\ $$$$\mathrm{9}\lambda^{\mathrm{2}} +\mathrm{6}\lambda−\mathrm{23}=\mathrm{0} \\ $$$$\lambda=\frac{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{1}}{\mathrm{3}}\:\checkmark \\ $$

Commented by Tawa11 last updated on 13/Jan/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com