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Question Number 77504 by Maclaurin Stickker last updated on 07/Jan/20

Answered by mr W last updated on 07/Jan/20

say the radius of large outer circle=R

$${say}\:{the}\:{radius}\:{of}\:{large}\:{outer}\:{circle}={R} \\ $$

Commented by mr W last updated on 07/Jan/20

Commented by mr W last updated on 07/Jan/20

(a/(R−a))=sin 30°=(1/2) ⇒a=(R/3) (x/(R−2a−x))=sin 30°=(1/2) 3x=R−2a=R−((2R)/3)=(R/3) ⇒x=(R/9)

$$\frac{{a}}{{R}−{a}}=\mathrm{sin}\:\mathrm{30}°=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{{R}}{\mathrm{3}} \\ $$$$\frac{{x}}{{R}−\mathrm{2}{a}−{x}}=\mathrm{sin}\:\mathrm{30}°=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{3}{x}={R}−\mathrm{2}{a}={R}−\frac{\mathrm{2}{R}}{\mathrm{3}}=\frac{{R}}{\mathrm{3}} \\ $$$$\Rightarrow{x}=\frac{{R}}{\mathrm{9}} \\ $$

Commented by mr W last updated on 07/Jan/20

Commented by mr W last updated on 07/Jan/20

OT=R cos 30°=(((√3)R)/2) 2b=R−(((√3)R)/2)=(((2−(√3))R)/2) ⇒b=(((2−(√3))R)/4) cos β=((OP^2 +OQ^2 −PQ^2 )/(2×OP×OQ))=(((R−r)^2 +(R−b)^2 −(r+b)^2 )/(2(R−r)(R−b))) cos β=((R^2 −(R+b)r−Rb)/((R−r)(R−b))) cos β=((R−(15−8(√3))r)/((R−r))) ST^2 =(b+r)^2 −(b−r)^2 =4br ⇒ST=2(√(br))=(√((2−(√3))Rr)) ST=PS tan β+OT tan β=(r+(((√3)R)/2))tan β ST=(r+(((√3)R)/2))((√((R−r)^2 −[R−(15−8(√3))r]^2 ))/(R−(15−8(√3))r)) ST=((√3)R+2r)(2−(√3))((√([R−4(2−(√3))r]r))/(R−(15−8(√3))r)) (√((2−(√3))Rr))=((√3)R+2r)(2−(√3))((√([R−4(2−(√3))r]r))/(R−(15−8(√3))r)) (√((2+(√3))R))=((√3)R+2r)((√(R−4(2−(√3))r))/(R−(15−8(√3))r)) ((1−(15−8(√3))(r/R))/((√3)+2(r/R)))=(√((2−(√3))[1−4(2−(√3))(r/R)])) ⇒((1−(15−8(√3))λ)/((√3)+2λ))=(√((2−(√3))[1−4(2−(√3))λ])) ⇒λ=(1/(16)) ⇒r=(R/(16)) ⇒(x/r)=((R/9)/(R/(16)))=((16)/9)

$${OT}={R}\:\mathrm{cos}\:\mathrm{30}°=\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}} \\ $$$$\mathrm{2}{b}={R}−\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}}=\frac{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){R}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\frac{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){R}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\beta=\frac{{OP}^{\mathrm{2}} +{OQ}^{\mathrm{2}} −{PQ}^{\mathrm{2}} }{\mathrm{2}×{OP}×{OQ}}=\frac{\left({R}−{r}\right)^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} −\left({r}+{b}\right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{r}\right)\left({R}−{b}\right)} \\ $$$$\mathrm{cos}\:\beta=\frac{{R}^{\mathrm{2}} −\left({R}+{b}\right){r}−{Rb}}{\left({R}−{r}\right)\left({R}−{b}\right)} \\ $$$$\mathrm{cos}\:\beta=\frac{{R}−\left(\mathrm{15}−\mathrm{8}\sqrt{\mathrm{3}}\right){r}}{\left({R}−{r}\right)} \\ $$$${ST}^{\mathrm{2}} =\left({b}+{r}\right)^{\mathrm{2}} −\left({b}−{r}\right)^{\mathrm{2}} =\mathrm{4}{br} \\ $$$$\Rightarrow{ST}=\mathrm{2}\sqrt{{br}}=\sqrt{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){Rr}} \\ $$$${ST}={PS}\:\mathrm{tan}\:\beta+{OT}\:\mathrm{tan}\:\beta=\left({r}+\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}}\right)\mathrm{tan}\:\beta \\ $$$${ST}=\left({r}+\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}}\right)\frac{\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −\left[{R}−\left(\mathrm{15}−\mathrm{8}\sqrt{\mathrm{3}}\right){r}\right]^{\mathrm{2}} }}{{R}−\left(\mathrm{15}−\mathrm{8}\sqrt{\mathrm{3}}\right){r}} \\ $$$${ST}=\left(\sqrt{\mathrm{3}}{R}+\mathrm{2}{r}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\frac{\sqrt{\left[{R}−\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){r}\right]{r}}}{{R}−\left(\mathrm{15}−\mathrm{8}\sqrt{\mathrm{3}}\right){r}} \\ $$$$\sqrt{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){Rr}}=\left(\sqrt{\mathrm{3}}{R}+\mathrm{2}{r}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\frac{\sqrt{\left[{R}−\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){r}\right]{r}}}{{R}−\left(\mathrm{15}−\mathrm{8}\sqrt{\mathrm{3}}\right){r}} \\ $$$$\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){R}}=\left(\sqrt{\mathrm{3}}{R}+\mathrm{2}{r}\right)\frac{\sqrt{{R}−\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){r}}}{{R}−\left(\mathrm{15}−\mathrm{8}\sqrt{\mathrm{3}}\right){r}} \\ $$$$\frac{\mathrm{1}−\left(\mathrm{15}−\mathrm{8}\sqrt{\mathrm{3}}\right)\frac{{r}}{{R}}}{\sqrt{\mathrm{3}}+\mathrm{2}\frac{{r}}{{R}}}=\sqrt{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left[\mathrm{1}−\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\frac{{r}}{{R}}\right]} \\ $$$$\Rightarrow\frac{\mathrm{1}−\left(\mathrm{15}−\mathrm{8}\sqrt{\mathrm{3}}\right)\lambda}{\sqrt{\mathrm{3}}+\mathrm{2}\lambda}=\sqrt{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left[\mathrm{1}−\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\lambda\right]} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\Rightarrow{r}=\frac{{R}}{\mathrm{16}} \\ $$$$\Rightarrow\frac{{x}}{{r}}=\frac{\frac{{R}}{\mathrm{9}}}{\frac{{R}}{\mathrm{16}}}=\frac{\mathrm{16}}{\mathrm{9}} \\ $$

Commented by Maclaurin Stickker last updated on 07/Jan/20

Amazing

$${Amazing} \\ $$

Commented by TawaTawa last updated on 07/Jan/20

God bless you sir. Thanks for your time.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$

Commented by mr W last updated on 08/Jan/20

alternative: ST=(√((2−(√3))Rr))=OP×sin β (2−(√3))Rr=OP^2 ×(1−cos^2 β) (2−(√3))Rr=(R−r)^2 (1−[((R−(15−8(√3))r)/((R−r)))]^2 ) (2−(√3))Rr=(R−r)^2 −[R−(15−8(√3))r]^2 (2−(√3))Rr=4[R−4(2−(√3))r](2−(√3))^2 r R=4[R−4(2−(√3))r](2−(√3)) ⇒r=(R/(4(2−(√3))))[1−(1/(4(2−(√3))))] ⇒r=(R/(16))[((7−4(√3))/((2−(√3))^2 ))] ⇒r=(R/(16))[(((2−(√3))^2 )/((2−(√3))^2 ))] ⇒r=(R/(16))

$${alternative}: \\ $$$${ST}=\sqrt{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){Rr}}={OP}×\mathrm{sin}\:\beta \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){Rr}={OP}^{\mathrm{2}} ×\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\beta\right) \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){Rr}=\left({R}−{r}\right)^{\mathrm{2}} \left(\mathrm{1}−\left[\frac{{R}−\left(\mathrm{15}−\mathrm{8}\sqrt{\mathrm{3}}\right){r}}{\left({R}−{r}\right)}\right]^{\mathrm{2}} \right) \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){Rr}=\left({R}−{r}\right)^{\mathrm{2}} −\left[{R}−\left(\mathrm{15}−\mathrm{8}\sqrt{\mathrm{3}}\right){r}\right]^{\mathrm{2}} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){Rr}=\mathrm{4}\left[{R}−\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){r}\right]\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} {r} \\ $$$${R}=\mathrm{4}\left[{R}−\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){r}\right]\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow{r}=\frac{{R}}{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}\right] \\ $$$$\Rightarrow{r}=\frac{{R}}{\mathrm{16}}\left[\frac{\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\right] \\ $$$$\Rightarrow{r}=\frac{{R}}{\mathrm{16}}\left[\frac{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\right] \\ $$$$\Rightarrow{r}=\frac{{R}}{\mathrm{16}} \\ $$

Commented by TawaTawa last updated on 08/Jan/20

Wow, great sir. God bless you sir.

$$\mathrm{Wow},\:\mathrm{great}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 29/Dec/21

This too. Hahaha

$$\mathrm{This}\:\mathrm{too}.\:\mathrm{Hahaha} \\ $$

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