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Question Number 74527 by Maclaurin Stickker last updated on 25/Nov/19
Commented by Maclaurin Stickker last updated on 25/Nov/19
In the figure determine the radius of the smallest circumference as a function of the radius R of the quadrant.
$${In}\:{the}\:{figure}\:{determine}\:{the}\:{radius} \\ $$$${of}\:{the}\:{smallest}\:{circumference}\:{as}\:{a} \\ $$$${function}\:{of}\:{the}\:{radius}\:\boldsymbol{\mathrm{R}}\:{of}\:{the}\:{quadrant}. \\ $$
Answered by ajfour last updated on 25/Nov/19
Commented by ajfour last updated on 25/Nov/19
let R=2 OQ=2−a (2−a)^2 −a^2 =(1+a)^2 −(1−a)^2 ⇒ 4−4a= 4a ⇒ a=(1/2) now sin α=(1/3) {(2−b)cos γ−a}^2 + {(2−a)cos α−(2−b)sin γ}^2 =(a+b)^2 .........(I) and (2−b)^2 sin^2 γ+{(2−b)cos γ−1}^2 = (1+b)^2 .....(II) ⇒ (2−b)^2 −(1+b)^2 +1= 2(2−b)cos γ ⇒ cos γ= ((2−3b)/(2−b)) now from (I) (2−a)^2 cos^2 α+a^2 +(2−b)^2 −(a+b)^2 = 2a(2−b)cos γ +2(2−a)(2−b)cos αsin γ ⇒ (9/4)×(8/9)+(1/4)+(2−b)^2 −(1/4)−b−b^2 = (2−3b)+2(√2)×(√((2−b)^2 −(2−3b)^2 )) (4−2b)^2 = 8(8b−8b^2 ) ⇒ 68b^2 −80b+16= 0 ⇒ 17b^2 −20b+4=0 ⇒ b= ((20−8(√2))/(34)) = ((10−4(√2))/(17)) (b/R) = ((5−2(√2))/(17)) = (1/(5+2(√2))) . as it was assumed R=2 .
$${let}\:{R}=\mathrm{2} \\ $$$${OQ}=\mathrm{2}−{a}\: \\ $$$$\left(\mathrm{2}−{a}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} =\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −\left(\mathrm{1}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{4}−\mathrm{4}{a}=\:\mathrm{4}{a}\:\:\: \\ $$$$\Rightarrow\:\:\:{a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:{now}\:\:\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\left\{\left(\mathrm{2}−{b}\right)\mathrm{cos}\:\gamma−{a}\right\}^{\mathrm{2}} + \\ $$$$\left\{\left(\mathrm{2}−{a}\right)\mathrm{cos}\:\alpha−\left(\mathrm{2}−{b}\right)\mathrm{sin}\:\gamma\right\}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:………\left({I}\right)\:\:\:\:{and} \\ $$$$\left(\mathrm{2}−{b}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \gamma+\left\{\left(\mathrm{2}−{b}\right)\mathrm{cos}\:\gamma−\mathrm{1}\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\:\left(\mathrm{1}+{b}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\:\:…..\left({II}\right) \\ $$$$\Rightarrow\:\: \\ $$$$\left(\mathrm{2}−{b}\right)^{\mathrm{2}} −\left(\mathrm{1}+{b}\right)^{\mathrm{2}} +\mathrm{1}=\:\mathrm{2}\left(\mathrm{2}−{b}\right)\mathrm{cos}\:\gamma \\ $$$$\Rightarrow\:\mathrm{cos}\:\gamma=\:\frac{\mathrm{2}−\mathrm{3}{b}}{\mathrm{2}−{b}} \\ $$$${now}\:{from}\:\left({I}\right) \\ $$$$\left(\mathrm{2}−{a}\right)^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \alpha+{a}^{\mathrm{2}} +\left(\mathrm{2}−{b}\right)^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\:\:\:=\:\mathrm{2}{a}\left(\mathrm{2}−{b}\right)\mathrm{cos}\:\gamma \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\left(\mathrm{2}−{a}\right)\left(\mathrm{2}−{b}\right)\mathrm{cos}\:\alpha\mathrm{sin}\:\gamma \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{9}}{\mathrm{4}}×\frac{\mathrm{8}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{4}}+\left(\mathrm{2}−{b}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}−{b}−{b}^{\mathrm{2}} \\ $$$$\:=\:\left(\mathrm{2}−\mathrm{3}{b}\right)+\mathrm{2}\sqrt{\mathrm{2}}×\sqrt{\left(\mathrm{2}−{b}\right)^{\mathrm{2}} −\left(\mathrm{2}−\mathrm{3}{b}\right)^{\mathrm{2}} } \\ $$$$\:\left(\mathrm{4}−\mathrm{2}{b}\right)^{\mathrm{2}} =\:\mathrm{8}\left(\mathrm{8}{b}−\mathrm{8}{b}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:\:\mathrm{68}{b}^{\mathrm{2}} −\mathrm{80}{b}+\mathrm{16}=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{17}{b}^{\mathrm{2}} −\mathrm{20}{b}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\:\:{b}=\:\frac{\mathrm{20}−\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{34}}\:=\:\frac{\mathrm{10}−\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{17}} \\ $$$$\:\:\:\:\:\:\frac{{b}}{{R}}\:=\:\frac{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{17}}\:=\:\frac{\mathrm{1}}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{2}}}\:\:. \\ $$$$\:\:{as}\:{it}\:{was}\:{assumed}\:{R}=\mathrm{2}\:. \\ $$
Commented by mr W last updated on 25/Nov/19
please check sir: (2−b)^2 −(1+b)^2 +1= 2(2−b)cos γ ⇒ cos γ= ((2−3b)/(2−b))
$${please}\:{check}\:{sir}: \\ $$$$\left(\mathrm{2}−{b}\right)^{\mathrm{2}} −\left(\mathrm{1}+{b}\right)^{\mathrm{2}} +\mathrm{1}=\:\mathrm{2}\left(\mathrm{2}−{b}\right)\mathrm{cos}\:\gamma \\ $$$$\Rightarrow\:\mathrm{cos}\:\gamma=\:\frac{\mathrm{2}−\mathrm{3}{b}}{\mathrm{2}−{b}} \\ $$
Commented by ajfour last updated on 25/Nov/19
many many thanks sir!
$${many}\:{many}\:{thanks}\:{sir}! \\ $$
Answered by mr W last updated on 25/Nov/19
Commented by mr W last updated on 25/Nov/19
OB=R−a OC=R−b BC=a+b DB=(R/2)+a DC=(R/2)+b sin α=(a/(R−a)) cos β=(((R−b)^2 +(R^2 /4)−((R/2)+b)^2 )/(2(R−b)(R/2)))=((R−3b)/(R−b)) cos γ=(((R−a)^2 +(R−b)^2 −(a+b)^2 )/(2(R−a)(R−b))) =((R^2 −R(a+b)−ab)/((R−a)(R−b))) cos (β+γ)=(((R−a)^2 +(R^2 /4)−((R/2)+a)^2 )/(2(R−a)(R/2)))=((R−3a)/(R−a)) cos (β+γ)=sin α ((R−3a)/(R−a))=(a/(R−a)) R^2 −5Ra+4a^2 =0 (R−4a)(R−a)=0 ⇒R=4a ⇒a=(R/4) cos (β+γ)=(1/3) ⇒γ=cos^(−1) (1/3)−β cos γ=(1/3)cos β+((2(√2))/3)sin β cos γ=((3R−5b)/(3(R−b))) cos β=((R−3b)/(R−b)) ⇒sin β=((2(√((R−2b)b)))/(R−b)) ((3R−5b)/(2(R−b)))=((R−3b)/(3(R−b)))+((4(√(2(R−2b)b)))/(3(R−b))) R−b=2(√(2b(R−2b))) R^2 −2Rb+b^2 =8b(R−2b) R^2 −10Rb+17b^2 =0 ⇒b=(R/(5+2(√2)))≈0.12774R
$${OB}={R}−{a} \\ $$$${OC}={R}−{b} \\ $$$${BC}={a}+{b} \\ $$$${DB}=\frac{{R}}{\mathrm{2}}+{a} \\ $$$${DC}=\frac{{R}}{\mathrm{2}}+{b} \\ $$$$\mathrm{sin}\:\alpha=\frac{{a}}{{R}−{a}} \\ $$$$\mathrm{cos}\:\beta=\frac{\left({R}−{b}\right)^{\mathrm{2}} +\frac{{R}^{\mathrm{2}} }{\mathrm{4}}−\left(\frac{{R}}{\mathrm{2}}+{b}\right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{b}\right)\frac{{R}}{\mathrm{2}}}=\frac{{R}−\mathrm{3}{b}}{{R}−{b}} \\ $$$$\mathrm{cos}\:\gamma=\frac{\left({R}−{a}\right)^{\mathrm{2}} +\left({R}−{b}\right)^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{a}\right)\left({R}−{b}\right)} \\ $$$$=\frac{{R}^{\mathrm{2}} −{R}\left({a}+{b}\right)−{ab}}{\left({R}−{a}\right)\left({R}−{b}\right)} \\ $$$$\mathrm{cos}\:\left(\beta+\gamma\right)=\frac{\left({R}−{a}\right)^{\mathrm{2}} +\frac{{R}^{\mathrm{2}} }{\mathrm{4}}−\left(\frac{{R}}{\mathrm{2}}+{a}\right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{a}\right)\frac{{R}}{\mathrm{2}}}=\frac{{R}−\mathrm{3}{a}}{{R}−{a}} \\ $$$$\mathrm{cos}\:\left(\beta+\gamma\right)=\mathrm{sin}\:\alpha \\ $$$$\frac{{R}−\mathrm{3}{a}}{{R}−{a}}=\frac{{a}}{{R}−{a}} \\ $$$${R}^{\mathrm{2}} −\mathrm{5}{Ra}+\mathrm{4}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({R}−\mathrm{4}{a}\right)\left({R}−{a}\right)=\mathrm{0} \\ $$$$\Rightarrow{R}=\mathrm{4}{a}\:\Rightarrow{a}=\frac{{R}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\left(\beta+\gamma\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\gamma=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}−\beta \\ $$$$\mathrm{cos}\:\gamma=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\:\beta+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\mathrm{sin}\:\beta \\ $$$$\mathrm{cos}\:\gamma=\frac{\mathrm{3}{R}−\mathrm{5}{b}}{\mathrm{3}\left({R}−{b}\right)} \\ $$$$\mathrm{cos}\:\beta=\frac{{R}−\mathrm{3}{b}}{{R}−{b}}\:\Rightarrow\mathrm{sin}\:\beta=\frac{\mathrm{2}\sqrt{\left({R}−\mathrm{2}{b}\right){b}}}{{R}−{b}} \\ $$$$\frac{\mathrm{3}{R}−\mathrm{5}{b}}{\mathrm{2}\left({R}−{b}\right)}=\frac{{R}−\mathrm{3}{b}}{\mathrm{3}\left({R}−{b}\right)}+\frac{\mathrm{4}\sqrt{\mathrm{2}\left({R}−\mathrm{2}{b}\right){b}}}{\mathrm{3}\left({R}−{b}\right)} \\ $$$${R}−{b}=\mathrm{2}\sqrt{\mathrm{2}{b}\left({R}−\mathrm{2}{b}\right)} \\ $$$${R}^{\mathrm{2}} −\mathrm{2}{Rb}+{b}^{\mathrm{2}} =\mathrm{8}{b}\left({R}−\mathrm{2}{b}\right) \\ $$$${R}^{\mathrm{2}} −\mathrm{10}{Rb}+\mathrm{17}{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{{R}}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{2}}}\approx\mathrm{0}.\mathrm{12774}{R} \\ $$
Commented by Maclaurin Stickker last updated on 25/Nov/19
Another perfect answer. Thank you.
$${Another}\:{perfect}\:{answer}.\:{Thank}\:{you}. \\ $$

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