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Question-165879

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Question Number 165879 by ajfour last updated on 09/Feb/22
Answered by mr W last updated on 09/Feb/22
Commented by ajfour last updated on 09/Feb/22
Go ahead sir, i shall also try, n past two questions, extreme thanks to their solutions. Unbelievable presentation.I shall scrutinize them well, for sure.
Commented by mr W last updated on 10/Feb/22
tan 22.5°=(√2)−1 sin 22.5°=(((√2)−1)/( (√(4−2(√2))))) let t=(1/(tan 22.5°))=(√2)+1 let s=(1/(sin 22.5°))=((√2)+1)(√(2(2−(√2)))) OB=OC=OA=(1/(tan 22.5°)) OK=OH=DH+OD=1+(1/(sin 22.5°)) JB=(1/(tan 22.5°))−R ((1/(tan 22.5°))−R)^2 =(1+R)^2 −(1−R)^2 R^2 −2(t+2)R+t^2 =0 ⇒R=t+2−2(√(t+1)) CE=OK−OC−EK=1+(1/(sin 22.5°))−(1/(tan 22.5°))−r (r+1)^2 =1^2 +(1+(1/(sin 22.5°))−(1/(tan 22.5°))−r)^2 (1+s−t)^2 =2(2+s−t)r ⇒r=(((1+s−t)^2 )/(2(2+s−t))) (R/r)=((2(2+s−t)(2+t−2(√(t+1))))/((1+s−t)^2 ))≈2.1989
$$\mathrm{tan}\:\mathrm{22}.\mathrm{5}°=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\mathrm{sin}\:\mathrm{22}.\mathrm{5}°=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$$${let}\:{t}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$${let}\:{s}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{22}.\mathrm{5}°}=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\sqrt{\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)} \\ $$$${OB}={OC}={OA}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°} \\ $$$${OK}={OH}={DH}+{OD}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{22}.\mathrm{5}°} \\ $$$${JB}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°}−{R} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°}−{R}\right)^{\mathrm{2}} =\left(\mathrm{1}+{R}\right)^{\mathrm{2}} −\left(\mathrm{1}−{R}\right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\mathrm{2}\left({t}+\mathrm{2}\right){R}+{t}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{R}={t}+\mathrm{2}−\mathrm{2}\sqrt{{t}+\mathrm{1}} \\ $$$$ \\ $$$${CE}={OK}−{OC}−{EK}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{22}.\mathrm{5}°}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°}−{r} \\ $$$$\left({r}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{22}.\mathrm{5}°}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}°}−{r}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{s}−{t}\right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{2}+{s}−{t}\right){r} \\ $$$$\Rightarrow{r}=\frac{\left(\mathrm{1}+{s}−{t}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}+{s}−{t}\right)} \\ $$$$\frac{{R}}{{r}}=\frac{\mathrm{2}\left(\mathrm{2}+{s}−{t}\right)\left(\mathrm{2}+{t}−\mathrm{2}\sqrt{{t}+\mathrm{1}}\right)}{\left(\mathrm{1}+{s}−{t}\right)^{\mathrm{2}} }\approx\mathrm{2}.\mathrm{1989} \\ $$
Commented by mr W last updated on 09/Feb/22
Commented by MJS_new last updated on 10/Feb/22
great! the solution can be simplified: (R/r)=1−(√2)+(√(2(2+(√2))))
$$\mathrm{great}!\:\mathrm{the}\:\mathrm{solution}\:\mathrm{can}\:\mathrm{be}\:\mathrm{simplified}: \\ $$$$\frac{{R}}{{r}}=\mathrm{1}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)} \\ $$
Commented by ajfour last updated on 09/Feb/22
Awesome solution sir.
$${Awesome}\:{solution}\:{sir}. \\ $$
Commented by mr W last updated on 10/Feb/22
thanks sirs!
$${thanks}\:{sirs}! \\ $$
Commented by Tawa11 last updated on 10/Feb/22
Weldone sir.
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$
Answered by ajfour last updated on 09/Feb/22
Commented by ajfour last updated on 10/Feb/22
let quarter circle radius q. black circle radius=1 22.5°=α (q−1)sin α=1 (q−1)cos α=R+2(√R) ⇒ R+2(√R)=cot α=t (√R)=−1+(√(1+t)) R=t+2−2(√(1+t)) (q−1)cos α+r+(√((r+1)^2 −1))=q ⇒ r+1+(√((r+1)^2 −1)) =cosec α(1−cos α)+2 ⇒ (r+1)^2 −1={s−t+2−(r+1)}^2 ⇒ −1=(s−t+2)^2 −2(s−t+2)(r+1) r+1=(((s−t+2)^2 +1)/(2(s−t+2))) ⇒ r=(((s−t+1)^2 )/(2(s−t+2))) (R/r)=((2(s−t+2))/((s−t+1)^2 ))×{t+2−2(√(t+1))}
$${let}\:{quarter}\:{circle}\:{radius}\:{q}. \\ $$$${black}\:{circle}\:{radius}=\mathrm{1} \\ $$$$\mathrm{22}.\mathrm{5}°=\alpha \\ $$$$\left({q}−\mathrm{1}\right)\mathrm{sin}\:\alpha=\mathrm{1} \\ $$$$\left({q}−\mathrm{1}\right)\mathrm{cos}\:\alpha={R}+\mathrm{2}\sqrt{{R}} \\ $$$$\Rightarrow\:{R}+\mathrm{2}\sqrt{{R}}=\mathrm{cot}\:\alpha={t} \\ $$$$\sqrt{{R}}=−\mathrm{1}+\sqrt{\mathrm{1}+{t}} \\ $$$${R}={t}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+{t}} \\ $$$$\left({q}−\mathrm{1}\right)\mathrm{cos}\:\alpha+{r}+\sqrt{\left({r}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}={q} \\ $$$$\Rightarrow\:\:{r}+\mathrm{1}+\sqrt{\left({r}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{cosec}\:\alpha\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)+\mathrm{2} \\ $$$$\Rightarrow\:\:\left({r}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\left\{{s}−{t}+\mathrm{2}−\left({r}+\mathrm{1}\right)\right\}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\:−\mathrm{1}=\left({s}−{t}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({s}−{t}+\mathrm{2}\right)\left({r}+\mathrm{1}\right) \\ $$$${r}+\mathrm{1}=\frac{\left({s}−{t}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}\left({s}−{t}+\mathrm{2}\right)} \\ $$$$\Rightarrow\:\:{r}=\frac{\left({s}−{t}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\left({s}−{t}+\mathrm{2}\right)} \\ $$$$\frac{{R}}{{r}}=\frac{\mathrm{2}\left({s}−{t}+\mathrm{2}\right)}{\left({s}−{t}+\mathrm{1}\right)^{\mathrm{2}} }×\left\{{t}+\mathrm{2}−\mathrm{2}\sqrt{{t}+\mathrm{1}}\right\} \\ $$
Commented by Tawa11 last updated on 10/Feb/22
Weldone sir
$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

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