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Question Number 53910 by ajfour last updated on 27/Jan/19

Commented by ajfour last updated on 27/Jan/19

Find maximum area of quadrilateral  ABCD.

$${Find}\:{maximum}\:{area}\:{of}\:{quadrilateral} \\ $$$${ABCD}. \\ $$

Commented by ajfour last updated on 27/Jan/19

Commented by ajfour last updated on 27/Jan/19

At least find θ and φ in terms of  R and r such that area ABCDA is  maximum.

$${At}\:{least}\:{find}\:\theta\:{and}\:\phi\:{in}\:{terms}\:{of} \\ $$$${R}\:{and}\:{r}\:{such}\:{that}\:{area}\:{ABCDA}\:{is} \\ $$$${maximum}. \\ $$

Commented by ajfour last updated on 27/Jan/19

shall your result meet with what  i calculated, please confirm it, Sir..    tan θ = ((R+r(1+cos φ))/(rsin φ))    tan φ = ((r+R(1+cos θ))/(Rsin θ)) .

$${shall}\:{your}\:{result}\:{meet}\:{with}\:{what} \\ $$$${i}\:{calculated},\:{please}\:{confirm}\:{it},\:{Sir}.. \\ $$$$\:\:\mathrm{tan}\:\theta\:=\:\frac{{R}+{r}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{r}\mathrm{sin}\:\phi} \\ $$$$\:\:\mathrm{tan}\:\phi\:=\:\frac{{r}+{R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)}{{R}\mathrm{sin}\:\theta}\:. \\ $$

Commented by mr W last updated on 27/Jan/19

without calculation i guess the result  is when tangent at D is parallel to  diagonal AC and tangent at C is  parallel to diagonal BD.   i.e. CA⊥AD, BD⊥BC.

$${without}\:{calculation}\:{i}\:{guess}\:{the}\:{result} \\ $$$${is}\:{when}\:{tangent}\:{at}\:{D}\:{is}\:{parallel}\:{to} \\ $$$${diagonal}\:{AC}\:{and}\:{tangent}\:{at}\:{C}\:{is} \\ $$$${parallel}\:{to}\:{diagonal}\:{BD}.\: \\ $$$${i}.{e}.\:{CA}\bot{AD},\:{BD}\bot{BC}. \\ $$

Commented by ajfour last updated on 27/Jan/19

Brilliant professor Sir, but my  final cubic solution had error,  i am trying to see if Cardano′s fit,  please help there.

$${Brilliant}\:{professor}\:{Sir},\:{but}\:{my} \\ $$$${final}\:{cubic}\:{solution}\:{had}\:{error}, \\ $$$${i}\:{am}\:{trying}\:{to}\:{see}\:{if}\:{Cardano}'{s}\:{fit}, \\ $$$${please}\:{help}\:{there}. \\ $$

Commented by mr W last updated on 27/Jan/19

yes, i can confirm that your solution  matches that what i guess.

$${yes},\:{i}\:{can}\:{confirm}\:{that}\:{your}\:{solution} \\ $$$${matches}\:{that}\:{what}\:{i}\:{guess}. \\ $$

Commented by mr W last updated on 27/Jan/19

Commented by mr W last updated on 27/Jan/19

∠BAC=(π/2)−θ  [R+r(1+cos φ)]tan ((π/2)−θ)=r sin φ  ⇒tan θ=((R+r(1+cos φ))/(r sin φ))  similarly:  [r+R(1+cos θ)]tan ((π/2)−φ)=R sin θ  ⇒tan φ=((r+R(1+cos θ))/(R sin θ))

$$\angle{BAC}=\frac{\pi}{\mathrm{2}}−\theta \\ $$$$\left[{R}+{r}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)\right]\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)={r}\:\mathrm{sin}\:\phi \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{{R}+{r}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{r}\:\mathrm{sin}\:\phi} \\ $$$${similarly}: \\ $$$$\left[{r}+{R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\right]\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\phi\right)={R}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{tan}\:\phi=\frac{{r}+{R}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)}{{R}\:\mathrm{sin}\:\theta} \\ $$

Commented by mr W last updated on 27/Jan/19

this is how i guess the result:  if C is fixed on the second circle, i.e.  ΔACB is fixed, max. ΔACD is when  tangent at D is parallel to AC, for this  is the max. altitude over AC.  similarly if D is fixed on the first  circle, max. ΔBDC is when tangent  at C is parallel to BD.

$${this}\:{is}\:{how}\:{i}\:{guess}\:{the}\:{result}: \\ $$$${if}\:{C}\:{is}\:{fixed}\:{on}\:{the}\:{second}\:{circle},\:{i}.{e}. \\ $$$$\Delta{ACB}\:{is}\:{fixed},\:{max}.\:\Delta{ACD}\:{is}\:{when} \\ $$$${tangent}\:{at}\:{D}\:{is}\:{parallel}\:{to}\:{AC},\:{for}\:{this} \\ $$$${is}\:{the}\:{max}.\:{altitude}\:{over}\:{AC}. \\ $$$${similarly}\:{if}\:{D}\:{is}\:{fixed}\:{on}\:{the}\:{first} \\ $$$${circle},\:{max}.\:\Delta{BDC}\:{is}\:{when}\:{tangent} \\ $$$${at}\:{C}\:{is}\:{parallel}\:{to}\:{BD}. \\ $$

Commented by mr W last updated on 27/Jan/19

i think we got the correct exact solution  now.

$${i}\:{think}\:{we}\:{got}\:{the}\:{correct}\:{exact}\:{solution} \\ $$$${now}. \\ $$

Commented by ajfour last updated on 28/Jan/19

I too believed Cardano′s method  shall fetch all roots, whether real  or complex; MjS sir′s comment  once had let me doubt this belief  for some time; or may be i  misinterpreted it..thanks!

$${I}\:{too}\:{believed}\:{Cardano}'{s}\:{method} \\ $$$${shall}\:{fetch}\:{all}\:{roots},\:{whether}\:{real} \\ $$$${or}\:{complex};\:{MjS}\:{sir}'{s}\:{comment} \\ $$$${once}\:{had}\:{let}\:{me}\:{doubt}\:{this}\:{belief} \\ $$$${for}\:{some}\:{time};\:{or}\:{may}\:{be}\:{i} \\ $$$${misinterpreted}\:{it}..{thanks}! \\ $$

Answered by ajfour last updated on 27/Jan/19

Commented by ajfour last updated on 27/Jan/19

C(R+r+rcos φ, rsin φ)  A=Area(△ABC)+Area(△ACD)  A=(((R+r)/2))rsin φ+((Rh)/2)  Eq. of AD:   y=−xtan θ   h=[rsin φ+(R+r+rcos φ)tan θ]cos θ    =rsin φcos θ+(R+r+rcos φ)sin θ    = (R+r)sin θ+rsin (θ+φ)  2A=(R+r)rsin φ+R(R+r)sin θ                      +Rrsin (θ+φ)  let  μ=(R/r)   ⇒ 2A/r^2 =(μ+1)sin φ+μ(1+μ)sin θ                                 +μsin (θ+φ)  ((2A)/((μ+1)r^2 ))=sin φ+μsin θ+λsin (θ+φ)                         where λ=(μ/(μ+1)) =(R/(R+r))  ((2/((μ+1)r^2 )))((∂(A))/∂θ)=μcos θ+λcos (θ+φ)=0  ⇒  μcos θ = −λcos (θ+φ)    ...(i)  ((2/((μ+1)r^2 )))((∂(A))/∂φ)= cos φ+λcos (θ+φ)=0  ⇒  cos φ = −λcos (θ+φ)     ...(ii)  ⇒    cos φ = μcos θ      Now using (i) with μ/λ = b=μ+1     bcos θ+cos θcos φ−sin θsin φ=0  cos^2 θ(b+μcos θ)^2 =(1−cos^2 θ)(1−μ^2 cos^2 θ)  let  cos θ= z  ⇒ b^2 z^2 +2μbz^3 =1−μ^2 z^2 −z^2   but b=μ+1 , ⇒       2(μ^2 +μ+1)z^2 +2μ(μ+1)z^3 −1=0  ⇒  (1/z^3 )−2(μ^2 +μ+1)(1/z)−2μ(μ+1)=0  (1/z)=(1/(cos θ))=sec θ = t  ⇒ t^3 +pt+q =0  with  p=−2(μ^2 +μ+1) ; q=−2μ(μ+1)  (q^2 /4)+(p^3 /(27)) = μ^2 (μ+1)^2 −((8(μ^2 +μ+1)^3 )/(27))   D < 0 perhaps (as μ>1)  ....

$${C}\left({R}+{r}+{r}\mathrm{cos}\:\phi,\:{r}\mathrm{sin}\:\phi\right) \\ $$$${A}={Area}\left(\bigtriangleup{ABC}\right)+{Area}\left(\bigtriangleup{ACD}\right) \\ $$$${A}=\left(\frac{{R}+{r}}{\mathrm{2}}\right){r}\mathrm{sin}\:\phi+\frac{{Rh}}{\mathrm{2}} \\ $$$${Eq}.\:{of}\:{AD}:\:\:\:{y}=−{x}\mathrm{tan}\:\theta \\ $$$$\:{h}=\left[{r}\mathrm{sin}\:\phi+\left({R}+{r}+{r}\mathrm{cos}\:\phi\right)\mathrm{tan}\:\theta\right]\mathrm{cos}\:\theta \\ $$$$\:\:={r}\mathrm{sin}\:\phi\mathrm{cos}\:\theta+\left({R}+{r}+{r}\mathrm{cos}\:\phi\right)\mathrm{sin}\:\theta \\ $$$$\:\:=\:\left({R}+{r}\right)\mathrm{sin}\:\theta+{r}\mathrm{sin}\:\left(\theta+\phi\right) \\ $$$$\mathrm{2}{A}=\left({R}+{r}\right){r}\mathrm{sin}\:\phi+{R}\left({R}+{r}\right)\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{Rr}\mathrm{sin}\:\left(\theta+\phi\right) \\ $$$${let}\:\:\mu=\frac{{R}}{{r}}\: \\ $$$$\Rightarrow\:\mathrm{2}{A}/{r}^{\mathrm{2}} =\left(\mu+\mathrm{1}\right)\mathrm{sin}\:\phi+\mu\left(\mathrm{1}+\mu\right)\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mu\mathrm{sin}\:\left(\theta+\phi\right) \\ $$$$\frac{\mathrm{2}{A}}{\left(\mu+\mathrm{1}\right){r}^{\mathrm{2}} }=\mathrm{sin}\:\phi+\mu\mathrm{sin}\:\theta+\lambda\mathrm{sin}\:\left(\theta+\phi\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{where}\:\lambda=\frac{\mu}{\mu+\mathrm{1}}\:=\frac{{R}}{{R}+{r}} \\ $$$$\left(\frac{\mathrm{2}}{\left(\mu+\mathrm{1}\right){r}^{\mathrm{2}} }\right)\frac{\partial\left({A}\right)}{\partial\theta}=\mu\mathrm{cos}\:\theta+\lambda\mathrm{cos}\:\left(\theta+\phi\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mu\mathrm{cos}\:\theta\:=\:−\lambda\mathrm{cos}\:\left(\theta+\phi\right)\:\:\:\:...\left({i}\right) \\ $$$$\left(\frac{\mathrm{2}}{\left(\mu+\mathrm{1}\right){r}^{\mathrm{2}} }\right)\frac{\partial\left({A}\right)}{\partial\phi}=\:\mathrm{cos}\:\phi+\lambda\mathrm{cos}\:\left(\theta+\phi\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:\phi\:=\:−\lambda\mathrm{cos}\:\left(\theta+\phi\right)\:\:\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\:\mathrm{cos}\:\phi\:=\:\mu\mathrm{cos}\:\theta \\ $$$$\:\:\:\:{Now}\:{using}\:\left({i}\right)\:{with}\:\mu/\lambda\:=\:{b}=\mu+\mathrm{1} \\ $$$$\:\:\:{b}\mathrm{cos}\:\theta+\mathrm{cos}\:\theta\mathrm{cos}\:\phi−\mathrm{sin}\:\theta\mathrm{sin}\:\phi=\mathrm{0} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \theta\left({b}+\mu\mathrm{cos}\:\theta\right)^{\mathrm{2}} =\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta\right)\left(\mathrm{1}−\mu^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta\right) \\ $$$${let}\:\:\mathrm{cos}\:\theta=\:{z} \\ $$$$\Rightarrow\:{b}^{\mathrm{2}} {z}^{\mathrm{2}} +\mathrm{2}\mu{bz}^{\mathrm{3}} =\mathrm{1}−\mu^{\mathrm{2}} {z}^{\mathrm{2}} −{z}^{\mathrm{2}} \\ $$$${but}\:{b}=\mu+\mathrm{1}\:,\:\Rightarrow \\ $$$$\:\:\:\:\:\mathrm{2}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right){z}^{\mathrm{2}} +\mathrm{2}\mu\left(\mu+\mathrm{1}\right){z}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{{z}^{\mathrm{3}} }−\mathrm{2}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)\frac{\mathrm{1}}{{z}}−\mathrm{2}\mu\left(\mu+\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{\mathrm{cos}\:\theta}=\mathrm{sec}\:\theta\:=\:{t} \\ $$$$\Rightarrow\:\boldsymbol{{t}}^{\mathrm{3}} +\boldsymbol{{pt}}+\boldsymbol{{q}}\:=\mathrm{0} \\ $$$${with}\:\:{p}=−\mathrm{2}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)\:;\:{q}=−\mathrm{2}\mu\left(\mu+\mathrm{1}\right) \\ $$$$\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}\:=\:\mu^{\mathrm{2}} \left(\mu+\mathrm{1}\right)^{\mathrm{2}} −\frac{\mathrm{8}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{27}} \\ $$$$\:{D}\:<\:\mathrm{0}\:{perhaps}\:\left({as}\:\mu>\mathrm{1}\right) \\ $$$$.... \\ $$

Commented by ajfour last updated on 27/Jan/19

Thanks a lot Sir!

$${Thanks}\:{a}\:{lot}\:{Sir}! \\ $$

Commented by mr W last updated on 27/Jan/19

Commented by mr W last updated on 28/Jan/19

 t^3 −pt−q =0  with  p=2(μ^2 +μ+1) ; q=2μ(μ+1)  (u+v)^3 −3uv(u+v)−(u^3 +v^3 )=0  let t=u+v  u^3 v^3 =((p/3))^3 =(8/(27))(μ^2 +μ+1)^3   u^3 +v^3 =q=2μ(μ+1)  u^3  and v^3  are roots of   x^2 −2μ(μ+1)x+(8/(27))(μ^2 +μ+1)^3 =0  D=4[μ^2 (μ+1)^2 −(8/(27))(μ^2 +μ+1)^3 ]<0  u^3 (v^3 )=μ(μ+1)±i(√((8/(27))(μ^2 +μ+1)^3 −μ^2 (μ+1)^2 ))=A±iB  with A=μ(μ+1), B=(√((8/(27))(μ^2 +μ+1)^3 −μ^2 (μ+1)^2 ))  A+iB=(√(A^2 +B^2 ))((A/(√(A^2 +B^2 )))+i(B/(√(A^2 +B^2 ))))=(√(A^2 +B^2 ))(cos α+i sin α)  with α=tan^(−1) (B/A)=tan^(−1) (√(((8(μ^2 +μ+1)^3 )/(27μ^2 (μ+1)^2 ))−1))  A^2 +B^2 =(8/(27))(μ^2 +μ+1)^3   u^3 (v^3 )=(√(A^2 +B^2 ))(cos α±i sin α)  ⇒u=((A^2 +B^2 ))^(1/6) [cos (((α+2nπ)/3))+i sin (((α+2nπ)/3))]  ⇒v=((A^2 +B^2 ))^(1/6) [cos (((α+2nπ)/3))−i sin (((α+2nπ)/3))]  ⇒t=u+v=2((A^2 +B^2 ))^(1/6)  cos (((α+2nπ)/3)), n=0,1,2  ⇒t=sec θ=2(√((2(μ^2 +μ+1))/3)) cos ((1/3)tan^(−1) (√(((8(μ^2 +μ+1)^3 )/(27μ^2 (μ+1)^2 ))−1)))

$$\:\boldsymbol{{t}}^{\mathrm{3}} −\boldsymbol{{pt}}−\boldsymbol{{q}}\:=\mathrm{0} \\ $$$${with}\:\:{p}=\mathrm{2}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)\:;\:{q}=\mathrm{2}\mu\left(\mu+\mathrm{1}\right) \\ $$$$\left({u}+{v}\right)^{\mathrm{3}} −\mathrm{3}{uv}\left({u}+{v}\right)−\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$${let}\:{t}={u}+{v} \\ $$$${u}^{\mathrm{3}} {v}^{\mathrm{3}} =\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{8}}{\mathrm{27}}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)^{\mathrm{3}} \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} ={q}=\mathrm{2}\mu\left(\mu+\mathrm{1}\right) \\ $$$${u}^{\mathrm{3}} \:{and}\:{v}^{\mathrm{3}} \:{are}\:{roots}\:{of}\: \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\mu\left(\mu+\mathrm{1}\right){x}+\frac{\mathrm{8}}{\mathrm{27}}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$${D}=\mathrm{4}\left[\mu^{\mathrm{2}} \left(\mu+\mathrm{1}\right)^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{27}}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)^{\mathrm{3}} \right]<\mathrm{0} \\ $$$${u}^{\mathrm{3}} \left({v}^{\mathrm{3}} \right)=\mu\left(\mu+\mathrm{1}\right)\pm{i}\sqrt{\frac{\mathrm{8}}{\mathrm{27}}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)^{\mathrm{3}} −\mu^{\mathrm{2}} \left(\mu+\mathrm{1}\right)^{\mathrm{2}} }={A}\pm{iB} \\ $$$${with}\:{A}=\mu\left(\mu+\mathrm{1}\right),\:{B}=\sqrt{\frac{\mathrm{8}}{\mathrm{27}}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)^{\mathrm{3}} −\mu^{\mathrm{2}} \left(\mu+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${A}+{iB}=\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }\left(\frac{{A}}{\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }}+{i}\frac{{B}}{\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }}\right)=\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }\left(\mathrm{cos}\:\alpha+{i}\:\mathrm{sin}\:\alpha\right) \\ $$$${with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{{B}}{{A}}=\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{8}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{27}\mu^{\mathrm{2}} \left(\mu+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{1}} \\ $$$${A}^{\mathrm{2}} +{B}^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{27}}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)^{\mathrm{3}} \\ $$$${u}^{\mathrm{3}} \left({v}^{\mathrm{3}} \right)=\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }\left(\mathrm{cos}\:\alpha\pm{i}\:\mathrm{sin}\:\alpha\right) \\ $$$$\Rightarrow{u}=\sqrt[{\mathrm{6}}]{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }\left[\mathrm{cos}\:\left(\frac{\alpha+\mathrm{2}{n}\pi}{\mathrm{3}}\right)+{i}\:\mathrm{sin}\:\left(\frac{\alpha+\mathrm{2}{n}\pi}{\mathrm{3}}\right)\right] \\ $$$$\Rightarrow{v}=\sqrt[{\mathrm{6}}]{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }\left[\mathrm{cos}\:\left(\frac{\alpha+\mathrm{2}{n}\pi}{\mathrm{3}}\right)−{i}\:\mathrm{sin}\:\left(\frac{\alpha+\mathrm{2}{n}\pi}{\mathrm{3}}\right)\right] \\ $$$$\Rightarrow{t}={u}+{v}=\mathrm{2}\sqrt[{\mathrm{6}}]{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} }\:\mathrm{cos}\:\left(\frac{\alpha+\mathrm{2}{n}\pi}{\mathrm{3}}\right),\:{n}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$$\Rightarrow{t}=\mathrm{sec}\:\theta=\mathrm{2}\sqrt{\frac{\mathrm{2}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)}{\mathrm{3}}}\:\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{8}\left(\mu^{\mathrm{2}} +\mu+\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{27}\mu^{\mathrm{2}} \left(\mu+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{1}}\right) \\ $$

Commented by ajfour last updated on 27/Jan/19

EXCELLENT   Sir, but  wont any of  the other two solutions might fit?

$$\mathcal{EXCELLENT}\:\:\:{Sir},\:{but}\:\:{wont}\:{any}\:{of} \\ $$$${the}\:{other}\:{two}\:{solutions}\:{might}\:{fit}? \\ $$

Commented by mr W last updated on 27/Jan/19

no, because we expect that 0<θ<(π/2).

$${no},\:{because}\:{we}\:{expect}\:{that}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}. \\ $$

Commented by mr W last updated on 27/Jan/19

example:  μ=(R/r)=1, θ=φ=68.53°  check:  from tan θ=((R+r(1+cos φ))/(r sin φ))  ⇒tan θ=((2+cos θ)/(sin θ))  2cos^2  θ+2cos θ−1=0  cos θ=(((√3)−1)/2)  ⇒θ=cos^(−1) (((√3)−1)/2)=68.53°

$${example}: \\ $$$$\mu=\frac{{R}}{{r}}=\mathrm{1},\:\theta=\phi=\mathrm{68}.\mathrm{53}° \\ $$$${check}: \\ $$$${from}\:\mathrm{tan}\:\theta=\frac{{R}+{r}\left(\mathrm{1}+\mathrm{cos}\:\phi\right)}{{r}\:\mathrm{sin}\:\phi} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{2}+\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta} \\ $$$$\mathrm{2cos}^{\mathrm{2}} \:\theta+\mathrm{2cos}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{cos}\:\theta=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}=\mathrm{68}.\mathrm{53}°\: \\ $$

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