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Question-124821

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Question Number 124821 by ajfour last updated on 06/Dec/20
Commented by ajfour last updated on 06/Dec/20
If the circles in red and blue have equal radii, find AB.
$${If}\:{the}\:{circles}\:{in}\:{red}\:{and}\:{blue}\:{have} \\ $$$${equal}\:{radii},\:{find}\:{AB}. \\ $$
Commented by ajfour last updated on 06/Dec/20
Answered by mr W last updated on 06/Dec/20
Commented by mr W last updated on 06/Dec/20
R=1=radius of semicircle let λ=(r/R) cos θ=((R−2r)/R)=1−2λ tan α=(r/(R+(√((R−r)^2 −r^2 ))))=(λ/(1+(√(1−2λ)))) 2α=(π/2)−θ tan 2α=(1/(tan θ)) ((2×(λ/(1+(√(1−2λ)))))/(1−((λ/(1+(√(1−2λ)))))^2 ))=((1−2λ)/( (√(1−(1−2λ)^2 )))) ((2λ(1+(√(1−2λ))))/((1+(√(1−2λ)))^2 −λ^2 ))=((1−2λ)/( (√(1−(1−2λ)^2 )))) ⇒λ≈0.3129 sin β=(r/(R−r))=(λ/(1−λ)) AB=2(R−r)sin ((π−θ−β)/2) ((AB)/R)=2(1−λ)cos ((θ+β)/2) =2(1−λ)cos (1/2)[cos^(−1) (1−2λ)+sin^(−1) ((λ/(1−λ)))] ≈0.9274
$${R}=\mathrm{1}={radius}\:{of}\:{semicircle} \\ $$$${let}\:\lambda=\frac{{r}}{{R}} \\ $$$$\mathrm{cos}\:\theta=\frac{{R}−\mathrm{2}{r}}{{R}}=\mathrm{1}−\mathrm{2}\lambda \\ $$$$\mathrm{tan}\:\alpha=\frac{{r}}{{R}+\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }}=\frac{\lambda}{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\lambda}} \\ $$$$\mathrm{2}\alpha=\frac{\pi}{\mathrm{2}}−\theta \\ $$$$\mathrm{tan}\:\mathrm{2}\alpha=\frac{\mathrm{1}}{\mathrm{tan}\:\theta} \\ $$$$\frac{\mathrm{2}×\frac{\lambda}{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\lambda}}}{\mathrm{1}−\left(\frac{\lambda}{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\lambda}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}−\mathrm{2}\lambda}{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}\lambda\right)^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{2}\lambda\left(\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\lambda}\right)}{\left(\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\lambda}\right)^{\mathrm{2}} −\lambda^{\mathrm{2}} }=\frac{\mathrm{1}−\mathrm{2}\lambda}{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}\lambda\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow\lambda\approx\mathrm{0}.\mathrm{3129} \\ $$$$\mathrm{sin}\:\beta=\frac{{r}}{{R}−{r}}=\frac{\lambda}{\mathrm{1}−\lambda} \\ $$$${AB}=\mathrm{2}\left({R}−{r}\right)\mathrm{sin}\:\frac{\pi−\theta−\beta}{\mathrm{2}} \\ $$$$\frac{{AB}}{{R}}=\mathrm{2}\left(\mathrm{1}−\lambda\right)\mathrm{cos}\:\frac{\theta+\beta}{\mathrm{2}} \\ $$$$=\mathrm{2}\left(\mathrm{1}−\lambda\right)\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{1}−\mathrm{2}\lambda\right)+\mathrm{sin}^{−\mathrm{1}} \left(\frac{\lambda}{\mathrm{1}−\lambda}\right)\right] \\ $$$$\approx\mathrm{0}.\mathrm{9274} \\ $$
Commented by MJS_new last updated on 06/Dec/20
btw. λ is the real solution of λ^3 −5λ^2 +((57)/4)λ−4=0 λ=(5/3)+(1/6)((−1133+120(√(114))))^(1/3) −(1/6)((1133+120(√(114))))^(1/3) but this doesn′t make anything better...
$$\mathrm{btw}.\:\lambda\:\mathrm{is}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{of} \\ $$$$\lambda^{\mathrm{3}} −\mathrm{5}\lambda^{\mathrm{2}} +\frac{\mathrm{57}}{\mathrm{4}}\lambda−\mathrm{4}=\mathrm{0} \\ $$$$\lambda=\frac{\mathrm{5}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}\sqrt[{\mathrm{3}}]{−\mathrm{1133}+\mathrm{120}\sqrt{\mathrm{114}}}−\frac{\mathrm{1}}{\mathrm{6}}\sqrt[{\mathrm{3}}]{\mathrm{1133}+\mathrm{120}\sqrt{\mathrm{114}}} \\ $$$$\mathrm{but}\:\mathrm{this}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{make}\:\mathrm{anything}\:\mathrm{better}… \\ $$
Commented by mr W last updated on 06/Dec/20
wow! i stoped to consider if the equation can be simplified further. you showed an exact solution exists. thanks!
$${wow}!\:{i}\:{stoped}\:{to}\:{consider}\:{if}\:{the} \\ $$$${equation}\:{can}\:{be}\:{simplified}\:{further}. \\ $$$${you}\:{showed}\:{an}\:{exact}\:{solution}\:{exists}. \\ $$$${thanks}! \\ $$
Commented by MJS_new last updated on 06/Dec/20
((2λ(1+(√(1−2λ))))/((1+(√(1−2λ)))^2 −λ^2 ))=((1−2λ)/( (√(1−(1−2λ)^2 )))) let t=(√(1−2x))∧t≥0 ⇔ x=((1−t^2 )/2) ((4(t−1))/((t−3)(t+1)))=(t^2 /( (√(1−t^4 )))) squaring and transforming (t+1)(t^2 −t+2)^2 (t^3 −3t^2 +8t−4)=0 as t≥0 t^3 −3t^2 +8t−4=0 and from here it′s easy
$$\frac{\mathrm{2}\lambda\left(\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\lambda}\right)}{\left(\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\lambda}\right)^{\mathrm{2}} −\lambda^{\mathrm{2}} }=\frac{\mathrm{1}−\mathrm{2}\lambda}{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}\lambda\right)^{\mathrm{2}} }} \\ $$$$\mathrm{let}\:{t}=\sqrt{\mathrm{1}−\mathrm{2}{x}}\wedge{t}\geqslant\mathrm{0}\:\Leftrightarrow\:{x}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{\mathrm{4}\left({t}−\mathrm{1}\right)}{\left({t}−\mathrm{3}\right)\left({t}+\mathrm{1}\right)}=\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }} \\ $$$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming} \\ $$$$\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{2}\right)^{\mathrm{2}} \left({t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}} +\mathrm{8}{t}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{as}\:{t}\geqslant\mathrm{0} \\ $$$${t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}} +\mathrm{8}{t}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{from}\:\mathrm{here}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy} \\ $$
Commented by ajfour last updated on 06/Dec/20
Thanks mrW Sir for solving, N MjS Sir for interfering!
$${Thanks}\:{mrW}\:{Sir}\:{for}\:{solving},\: \\ $$$${N}\:\:{MjS}\:{Sir}\:{for}\:{interfering}! \\ $$

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