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Question-215094

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Question Number 215094 by ajfour last updated on 28/Dec/24
Commented by ajfour last updated on 28/Dec/24
A question discussed Statics+Geometry https://youtu.be/vktsHDGgBlw?si=z6XC7xw6G8m6XjrS
Commented by mr W last updated on 28/Dec/24
Commented by mr W last updated on 28/Dec/24
((sin ((π/2)−α−θ))/(sin α))=((sin (α+θ))/(sin ((π/2)−α))) ⇒tan (α+θ)=(1/(tan α))=tan ((π/2)−α) ⇒α+θ=(π/2)−α ⇒θ=(π/2)−2α
$$\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\alpha−\theta\right)}{\mathrm{sin}\:\alpha}=\frac{\mathrm{sin}\:\left(\alpha+\theta\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\alpha\right)} \\ $$$$\Rightarrow\mathrm{tan}\:\left(\alpha+\theta\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\alpha\right) \\ $$$$\Rightarrow\alpha+\theta=\frac{\pi}{\mathrm{2}}−\alpha\:\Rightarrow\theta=\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha \\ $$
Commented by Frix last updated on 29/Dec/24
(R^2 /r^2 )=12+4(√3) (Simply had forgotten to (√(...)))
$$\frac{{R}^{\mathrm{2}} }{{r}^{\mathrm{2}} }=\mathrm{12}+\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{Simply}\:\mathrm{had}\:\mathrm{forgotten}\:\mathrm{to}\:\sqrt{…}\right) \\ $$
Commented by ajfour last updated on 28/Dec/24
Explain me the first line itself sir.
Commented by mr W last updated on 28/Dec/24
((sin ((π/2)−α−θ))/(sin α))=((CD)/(AC))=((CD)/(BC))=((sin (α+θ))/(sin ((π/2)−α)))
$$\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\alpha−\theta\right)}{\mathrm{sin}\:\alpha}=\frac{{CD}}{{AC}}=\frac{{CD}}{{BC}}=\frac{\mathrm{sin}\:\left(\alpha+\theta\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\alpha\right)} \\ $$
Commented by ajfour last updated on 28/Dec/24
Yeah θ_(ajfour) +θ_(mrw) =0 we have just marked differently. we have same situation. for example check for α=60° .
$${Yeah} \\ $$$$\:\theta_{{ajfour}} +\theta_{{mrw}} =\mathrm{0}\:\:\:\: \\ $$$${we}\:{have}\:{just}\:{marked}\:{differently}. \\ $$$${we}\:{have}\:{same}\:{situation}.\:{for}\:{example} \\ $$$${check}\:{for}\:\alpha=\mathrm{60}°\:. \\ $$
Commented by ajfour last updated on 28/Dec/24
Thanks sir Frix, you came back after really some time.
$${Thanks}\:{sir}\:{Frix},\:{you}\:{came}\:{back}\: \\ $$$${after}\:{really}\:{some}\:{time}. \\ $$
Commented by mr W last updated on 29/Dec/24
in this case we can even “see” the result obviously:
$${in}\:{this}\:{case}\:{we}\:{can}\:{even}\:“{see}''\:{the}\: \\ $$$${result}\:{obviously}: \\ $$
Commented by mr W last updated on 29/Dec/24
Commented by ajfour last updated on 29/Dec/24
https://youtu.be/m_z86aXCFIo?si=vpI3_hoVBncvRfE_
Answered by mr W last updated on 29/Dec/24
Commented by mr W last updated on 29/Dec/24
sin θ=(r/R)=(1/k) AB=2(√(R^2 −r^2 )) OB=(√(R^2 −r^2 ))+(√(((√2)R)^2 −r^2 ))=(√(R^2 −r^2 ))+(√(2R^2 −r^2 )) cos 2θ=((4(R^2 −r^2 )+R^2 −r^2 +2R^2 −r^2 +2(√((R^2 −r^2 )(2R^2 −r^2 )))−R^2 )/(4(√(R^2 −r^2 ))((√(R^2 −r^2 ))+(√(2R^2 −r^2 ))))) =((3k^2 −3+(√((k^2 −1)(2k^2 −1))))/( 2(√(k^2 −1))((√(k^2 −1))+(√(2k^2 −1))))) ((3k^2 −3+(√((k^2 −1)(2k^2 −1))))/( 2(√(k^2 −1))((√(k^2 −1))+(√(2k^2 −1)))))=1−(2/k^2 ) ⇒((k^2 +4)/(k^2 −4))=(√((2k^2 −1)/(k^2 −1))) ⇒k^4 −24k^2 +32=0 ⇒k^2 =4(3+(√7)) (R/r)=k=2(√(3+(√7)))≈4.752158
$$\mathrm{sin}\:\theta=\frac{{r}}{{R}}=\frac{\mathrm{1}}{{k}} \\ $$$${AB}=\mathrm{2}\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${OB}=\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }+\sqrt{\left(\sqrt{\mathrm{2}}{R}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }+\sqrt{\mathrm{2}{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{4}\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)+{R}^{\mathrm{2}} −{r}^{\mathrm{2}} +\mathrm{2}{R}^{\mathrm{2}} −{r}^{\mathrm{2}} +\mathrm{2}\sqrt{\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)\left(\mathrm{2}{R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)}−{R}^{\mathrm{2}} }{\mathrm{4}\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }\left(\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }+\sqrt{\mathrm{2}{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }\right)} \\ $$$$\:\:\:\:\:=\frac{\mathrm{3}{k}^{\mathrm{2}} −\mathrm{3}+\sqrt{\left({k}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{2}{k}^{\mathrm{2}} −\mathrm{1}\right)}}{\:\mathrm{2}\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\left(\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}+\sqrt{\mathrm{2}{k}^{\mathrm{2}} −\mathrm{1}}\right)} \\ $$$$\frac{\mathrm{3}{k}^{\mathrm{2}} −\mathrm{3}+\sqrt{\left({k}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{2}{k}^{\mathrm{2}} −\mathrm{1}\right)}}{\:\mathrm{2}\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\left(\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}+\sqrt{\mathrm{2}{k}^{\mathrm{2}} −\mathrm{1}}\right)}=\mathrm{1}−\frac{\mathrm{2}}{{k}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{k}^{\mathrm{2}} +\mathrm{4}}{{k}^{\mathrm{2}} −\mathrm{4}}=\sqrt{\frac{\mathrm{2}{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\Rightarrow{k}^{\mathrm{4}} −\mathrm{24}{k}^{\mathrm{2}} +\mathrm{32}=\mathrm{0} \\ $$$$\Rightarrow{k}^{\mathrm{2}} =\mathrm{4}\left(\mathrm{3}+\sqrt{\mathrm{7}}\right) \\ $$$$\frac{{R}}{{r}}={k}=\mathrm{2}\sqrt{\mathrm{3}+\sqrt{\mathrm{7}}}\approx\mathrm{4}.\mathrm{752158} \\ $$
Commented by mr W last updated on 29/Dec/24
Commented by Frix last updated on 29/Dec/24
Let R=1 ⇒ C_1 : (x−1)^2 +(y−1)^2 =1 C_2 : (x−1)^2 +(y−1)^2 =r^2 L_1 : y=((1+(√(2p−p^2 )))/p)x L_2 : y=((√(2p−p^2 ))/p)x+1 C_2 ∩L_2 ={T_2 } ⇒ p=2−2r^2 C_2 ∩L_1 ={T_1 } ⇒ r=2(√(3+(√7)))=4.75215...
$$\mathrm{Let}\:{R}=\mathrm{1} \\ $$$$\Rightarrow \\ $$$${C}_{\mathrm{1}} :\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${C}_{\mathrm{2}} :\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${L}_{\mathrm{1}} :\:{y}=\frac{\mathrm{1}+\sqrt{\mathrm{2}{p}−{p}^{\mathrm{2}} }}{{p}}{x} \\ $$$${L}_{\mathrm{2}} :\:{y}=\frac{\sqrt{\mathrm{2}{p}−{p}^{\mathrm{2}} }}{{p}}{x}+\mathrm{1} \\ $$$${C}_{\mathrm{2}} \cap{L}_{\mathrm{2}} =\left\{{T}_{\mathrm{2}} \right\}\:\Rightarrow\:{p}=\mathrm{2}−\mathrm{2}{r}^{\mathrm{2}} \\ $$$${C}_{\mathrm{2}} \cap{L}_{\mathrm{1}} =\left\{{T}_{\mathrm{1}} \right\}\:\Rightarrow\:{r}=\mathrm{2}\sqrt{\mathrm{3}+\sqrt{\mathrm{7}}}=\mathrm{4}.\mathrm{75215}… \\ $$
Commented by mr W last updated on 29/Dec/24
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