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Question Number 40489 by ajfour last updated on 22/Jul/18

Commented by ajfour last updated on 22/Jul/18

In terms of radius of circle R, find a, and b.

$${In}\:{terms}\:{of}\:{radius}\:{of}\:{circle}\:{R}, \\ $$$${find}\:{a},\:{and}\:{b}. \\ $$

Commented by behi83417@gmail.com last updated on 23/Jul/18

BC.AF=AB.CD cosOAD=(a/b),OD=(√(b^2 −a^2 )) tgOED=((√(b^2 −a^2 ))/(OE))⇒OE=(√(b^2 −a^2 )).((√(b^2 −a^2 ))/a)=((b^2 −a^2 )/a) AE=a+((b^2 −a^2 )/a)=(b^2 /a),DE^2 =b^2 −a^2 +(((b^2 −a^2 )^2 )/a^2 )= =(((b^2 −a^2 )(a^2 +b^2 −a^2 ))/a^2 )⇒DE=(b/a)(√(b^2 −a^2 )) ((DE)/(EC))=((OE)/(EF))⇒EF=((b^2 −a^2 )/a).(b/((b/a)(√(b^2 −a^2 ))))=(√(b^2 −a^2 )) (a/b)=((b+BD)/(a+((b^2 −a^2 )/a)+(√(b^2 −a^2 ))))⇒BD=((b^2 +a(√(b^2 −a^2 )))/b)−b =(a/b)(√(b^2 −a^2 )) cosBDO=cos(90+EDO)=−sinEDO= =−((OE)/(ED))=−(((b^2 −a^2 )/a)/((b/a)(√(b^2 −a^2 ))))=−((√(b^2 −a^2 ))/b) cosBDO=−((√(b^2 −a^2 ))/b) R^2 =OD^2 +BD^2 −2OD.BD.cosBDO= =(b^2 −a^2 )+(a^2 /b^2 )(b^2 −a^2 )+2(√(b^2 −a^2 )).(a/b)(√(b^2 −a^2 )).((√(b^2 −a^2 ))/b)= =(b^2 −a^2 )(1+(a^2 /b^2 )+2(a/b^2 )(√(b^2 −a^2 )))= =((b^2 −a^2 )/b^2 )(b^2 +a^2 +2a(√(b^2 −a^2 ))) ⇒R=((√(b^2 +a^2 +2a(√(b^2 −a^2 ))))/b).(√(b^2 −a^2 )). BC.((b^2 +a(√(b^2 −a^2 )))/a)=((b^2 +a(√(b^2 −a^2 )))/b).(b+(b/a)(√(b^2 −a^2 ))) BC=a+(√(b^2 −a^2 )). OF.BC=R^2 .sinCOB⇒ sinCOB=(((a+(√(b^2 −a^2 )))(((b^2 −a^2 )/a)+(√(b^2 −a^2 ))))/R^2 )= =(((a+(√(b^2 −a^2 )))^2 .(√(b^2 −a^2 )))/(aR^2 ))

$${BC}.{AF}={AB}.{CD} \\ $$$${cosOAD}=\frac{{a}}{{b}},{OD}=\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${tgOED}=\frac{\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{OE}}\Rightarrow{OE}=\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }.\frac{\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}} \\ $$$${AE}={a}+\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}}=\frac{{b}^{\mathrm{2}} }{{a}},{DE}^{\mathrm{2}} ={b}^{\mathrm{2}} −{a}^{\mathrm{2}} +\frac{\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }= \\ $$$$=\frac{\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} }\Rightarrow{DE}=\frac{{b}}{{a}}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\frac{{DE}}{{EC}}=\frac{{OE}}{{EF}}\Rightarrow{EF}=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}}.\frac{{b}}{\frac{{b}}{{a}}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}=\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\frac{{a}}{{b}}=\frac{{b}+{BD}}{{a}+\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}}+\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\Rightarrow{BD}=\frac{{b}^{\mathrm{2}} +{a}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{b}}−{b} \\ $$$$=\frac{{a}}{{b}}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${cosBDO}={cos}\left(\mathrm{90}+{EDO}\right)=−{sinEDO}= \\ $$$$=−\frac{{OE}}{{ED}}=−\frac{\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}}}{\frac{{b}}{{a}}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}=−\frac{\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{b}} \\ $$$${cosBDO}=−\frac{\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{b}} \\ $$$${R}^{\mathrm{2}} ={OD}^{\mathrm{2}} +{BD}^{\mathrm{2}} −\mathrm{2}{OD}.{BD}.{cosBDO}= \\ $$$$=\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+\mathrm{2}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }.\frac{{a}}{{b}}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }.\frac{\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{b}}= \\ $$$$=\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\mathrm{2}\frac{{a}}{{b}^{\mathrm{2}} }\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)= \\ $$$$=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{a}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{R}=\frac{\sqrt{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{a}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}}{{b}}.\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }. \\ $$$${BC}.\frac{{b}^{\mathrm{2}} +{a}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}=\frac{{b}^{\mathrm{2}} +{a}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{b}}.\left({b}+\frac{{b}}{{a}}\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right) \\ $$$${BC}={a}+\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }. \\ $$$${OF}.{BC}={R}^{\mathrm{2}} .{sinCOB}\Rightarrow \\ $$$${sinCOB}=\frac{\left({a}+\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)\left(\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}}+\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)}{{R}^{\mathrm{2}} }= \\ $$$$=\frac{\left({a}+\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)^{\mathrm{2}} .\sqrt{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{aR}^{\mathrm{2}} } \\ $$

Commented by ajfour last updated on 23/Jul/18

Thanks Sir.

$${Thanks}\:{Sir}. \\ $$

Answered by ajfour last updated on 23/Jul/18

let ∠OAB=∠ECF =θ bcos θ = a ....(i) AE=bsec θ Also CF=BF=bcos θ =a EF=bsin θ =AF−AE ⇒ bsin θ=acot θ−bsec θ ..(ii) OF=(√(R^2 −BF^( 2) )) =(√(R^2 −a^2 )) OF=OE+EF ⇒ (√(R^2 −a^2 ))=bsin^2 θ+bsin θ ..(iii) using (i) in (ii): asec θsin θ=acot θ−asec^2 θ ⇒ tan θ=cot θ−1−tan^2 θ let tan θ=m ; then m^2 =1−m−m^3 or m^3 +m^2 +m=1 ⇒ m≈ 0.54369 using (i) in (iii) (√(R^2 −a^2 ))=asec θsin^2 θ+asec θsin θ =atan θ(1+sin θ) R^2 −a^2 =a^2 m^2 (1+(m/(√(1+m^2 ))))^2 ⇒ a^2 =(R^2 /(1+m^2 (1+(m/(√(1+m^2 ))))^2 )) ⇒ a ≈ 0.7743R b=a(√(1+m^2 )) ≈ 0.88R .

$${let}\:\angle{OAB}=\angle{ECF}\:=\theta \\ $$$${b}\mathrm{cos}\:\theta\:=\:{a}\:\:\:\:\:\:\:\:\:\:....\left({i}\right) \\ $$$${AE}={b}\mathrm{sec}\:\theta\:\:\:\:\:\:\:\: \\ $$$${Also}\:\:{CF}={BF}={b}\mathrm{cos}\:\theta\:={a} \\ $$$${EF}={b}\mathrm{sin}\:\theta\:={AF}−{AE} \\ $$$$\Rightarrow\:\:\:{b}\mathrm{sin}\:\theta={a}\mathrm{cot}\:\theta−{b}\mathrm{sec}\:\theta\:\:\:..\left({ii}\right) \\ $$$${OF}=\sqrt{{R}^{\mathrm{2}} −{BF}^{\:\mathrm{2}} }\:=\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${OF}={OE}+{EF} \\ $$$$\Rightarrow\:\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }={b}\mathrm{sin}\:^{\mathrm{2}} \theta+{b}\mathrm{sin}\:\theta\:\:\:..\left({iii}\right) \\ $$$${using}\:\left({i}\right)\:{in}\:\left({ii}\right): \\ $$$${a}\mathrm{sec}\:\theta\mathrm{sin}\:\theta={a}\mathrm{cot}\:\theta−{a}\mathrm{sec}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\theta=\mathrm{cot}\:\theta−\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta \\ $$$${let}\:\:\mathrm{tan}\:\theta={m}\:;\:{then} \\ $$$$\:\:{m}^{\mathrm{2}} =\mathrm{1}−{m}−{m}^{\mathrm{3}} \\ $$$${or}\:\:\:\:{m}^{\mathrm{3}} +{m}^{\mathrm{2}} +{m}=\mathrm{1} \\ $$$$\Rightarrow\:\:{m}\approx\:\mathrm{0}.\mathrm{54369} \\ $$$${using}\:\:\left({i}\right)\:{in}\:\left({iii}\right) \\ $$$$\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }={a}\mathrm{sec}\:\theta\mathrm{sin}\:^{\mathrm{2}} \theta+{a}\mathrm{sec}\:\theta\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}\mathrm{tan}\:\theta\left(\mathrm{1}+\mathrm{sin}\:\theta\right) \\ $$$${R}^{\mathrm{2}} −{a}^{\mathrm{2}} ={a}^{\mathrm{2}} {m}^{\mathrm{2}} \left(\mathrm{1}+\frac{{m}}{\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{a}^{\mathrm{2}} =\frac{{R}^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} \left(\mathrm{1}+\frac{{m}}{\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}\right)^{\mathrm{2}} } \\ $$$$\:\Rightarrow\:\:\:\:{a}\:\approx\:\mathrm{0}.\mathrm{7743}{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:{b}={a}\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }\:\approx\:\mathrm{0}.\mathrm{88}{R}\:. \\ $$

Commented by MJS last updated on 23/Jul/18

I get the same value for θ so either me or you made a mistake somewhere

$$\mathrm{I}\:\mathrm{get}\:\mathrm{the}\:\mathrm{same}\:\mathrm{value}\:\mathrm{for}\:\theta \\ $$$$\mathrm{so}\:\mathrm{either}\:\mathrm{me}\:\mathrm{or}\:\mathrm{you}\:\mathrm{made}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{somewhere} \\ $$

Answered by MJS last updated on 23/Jul/18

R=1 ∠DAO=∠ECF ⇒ ∣CF∣=∣BF∣=a ∣OF∣=(√(1−a^2 )) ∣OD∣=∣BF∣((∣AO∣)/(∣AF∣))=(a^2 /(a+(√(1−a^2 )))) AB^(⇀) =B−A= (((a+(√(1−a^2 )))),(a) ) CD^(⇀) =D−C= (((−(√(1−a^2 )))),((a((2a+(√(1−a^2 )))/(a+(√(1−a^2 )))))) ) AB^(⇀) ⊥CD^(⇀) (a+(√(1−a^2 )))(−(√(1−a^2 )))+a^2 ((2a+(√(1−a^2 )))/(a+(√(1−a^2 ))))=0 a^6 −((19)/(17))a^4 +(7/(17))a^2 −(1/(17))=0 a=(√x) x^3 −((19)/(17))x^2 +(7/(17))x−(1/(17))=0 x=z+((19)/(51)) z^3 −(4/(867))z−((1172)/(132651))=0 z≈.214167 x≈.586716 a≈.765974 ∣OD∣≈.416452 b≈.871865 a≈.765974R b≈.871865R

$${R}=\mathrm{1} \\ $$$$\angle{DAO}=\angle{ECF}\:\Rightarrow\:\mid{CF}\mid=\mid{BF}\mid={a} \\ $$$$\mid{OF}\mid=\sqrt{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$$\mid{OD}\mid=\mid{BF}\mid\frac{\mid{AO}\mid}{\mid{AF}\mid}=\frac{{a}^{\mathrm{2}} }{{a}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }} \\ $$$$\overset{\rightharpoonup} {{AB}}={B}−{A}=\begin{pmatrix}{{a}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\\{{a}}\end{pmatrix} \\ $$$$\overset{\rightharpoonup} {{CD}}={D}−{C}=\begin{pmatrix}{−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}\\{{a}\frac{\mathrm{2}{a}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{{a}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}}\end{pmatrix} \\ $$$$\overset{\rightharpoonup} {{AB}}\bot\overset{\rightharpoonup} {{CD}} \\ $$$$\left({a}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right)\left(−\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right)+{a}^{\mathrm{2}} \frac{\mathrm{2}{a}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{{a}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}=\mathrm{0} \\ $$$${a}^{\mathrm{6}} −\frac{\mathrm{19}}{\mathrm{17}}{a}^{\mathrm{4}} +\frac{\mathrm{7}}{\mathrm{17}}{a}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{17}}=\mathrm{0} \\ $$$${a}=\sqrt{{x}} \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{19}}{\mathrm{17}}{x}^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{17}}{x}−\frac{\mathrm{1}}{\mathrm{17}}=\mathrm{0} \\ $$$${x}={z}+\frac{\mathrm{19}}{\mathrm{51}} \\ $$$${z}^{\mathrm{3}} −\frac{\mathrm{4}}{\mathrm{867}}{z}−\frac{\mathrm{1172}}{\mathrm{132651}}=\mathrm{0} \\ $$$${z}\approx.\mathrm{214167} \\ $$$${x}\approx.\mathrm{586716} \\ $$$${a}\approx.\mathrm{765974} \\ $$$$\mid{OD}\mid\approx.\mathrm{416452} \\ $$$${b}\approx.\mathrm{871865} \\ $$$$ \\ $$$${a}\approx.\mathrm{765974}{R} \\ $$$${b}\approx.\mathrm{871865}{R} \\ $$

Commented by ajfour last updated on 23/Jul/18

matches close to my answer; thank you Sir.

$${matches}\:{close}\:{to}\:{my}\:{answer}; \\ $$$${thank}\:{you}\:{Sir}. \\ $$

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