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Question Number 107415 by ajfour last updated on 10/Aug/20

Commented by ajfour last updated on 10/Aug/20

Find  r/R.

$${Find}\:\:{r}/{R}. \\ $$

Answered by mr W last updated on 10/Aug/20

Commented by mr W last updated on 10/Aug/20

BA=2R−r  OD=R−2r  ((OD)/(BC))=((OA)/(BA))  ⇒((R−2r)/r)=(R/(2R−r))  ((R/r)−2)(2−(r/R))=1  let λ=(R/r)  (λ−2)(2λ−1)=λ  ⇒λ^2 −3λ+1=0  ⇒λ=((3+(√5))/2)≈2.618

$${BA}=\mathrm{2}{R}−{r} \\ $$$${OD}={R}−\mathrm{2}{r} \\ $$$$\frac{{OD}}{{BC}}=\frac{{OA}}{{BA}} \\ $$$$\Rightarrow\frac{{R}−\mathrm{2}{r}}{{r}}=\frac{{R}}{\mathrm{2}{R}−{r}} \\ $$$$\left(\frac{{R}}{{r}}−\mathrm{2}\right)\left(\mathrm{2}−\frac{{r}}{{R}}\right)=\mathrm{1} \\ $$$${let}\:\lambda=\frac{{R}}{{r}} \\ $$$$\left(\lambda−\mathrm{2}\right)\left(\mathrm{2}\lambda−\mathrm{1}\right)=\lambda \\ $$$$\Rightarrow\lambda^{\mathrm{2}} −\mathrm{3}\lambda+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\approx\mathrm{2}.\mathrm{618} \\ $$

Commented by ajfour last updated on 10/Aug/20

Very very nice solution Sir, thanks!

$${Very}\:{very}\:{nice}\:{solution}\:{Sir},\:{thanks}! \\ $$

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