Question-125491

Question Number 125491 by ajfour last updated on 11/Dec/20

Commented by ajfour last updated on 11/Dec/20

$${If}\:{outermost}\:{triangle}\:{is}\:{equilateral} \\ $$$${and}\:{radius}\:{of}\:{circle}\:{is}\:{unity},\:{find} \\ $$$${maximum}\:{brown}\:{shaded}\:{area}. \\ $$

Answered by mr W last updated on 11/Dec/20

Commented by mr W last updated on 11/Dec/20

BC=2 cos θ AB=(√7) tan β=((2 sin^2 θ )/( (√3)+2 sin θ cos θ)) DE=(√3) tan β=((2(√3) sin^2 θ )/( (√3)+2 sin θ cos θ)) BD=2−((2(√3) sin^2 θ )/( (√3)+2 sin θ cos θ)) A=Δ_(BCD) =(1/2)×2 cos θ×(2−((2(√3) sin^2 θ )/( (√3)+2 sin θ cos θ))) sin θ ⇒A=sin 2θ (1−(((√3) sin^2 θ )/( (√3)+sin 2θ))) A_(max) =0.7393 at θ=35.2354°

$${BC}=\mathrm{2}\:\mathrm{cos}\:\theta \\ $$$${AB}=\sqrt{\mathrm{7}} \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:}{\:\sqrt{\mathrm{3}}+\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta} \\ $$$${DE}=\sqrt{\mathrm{3}}\:\mathrm{tan}\:\beta=\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:}{\:\sqrt{\mathrm{3}}+\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta} \\ $$$${BD}=\mathrm{2}−\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:}{\:\sqrt{\mathrm{3}}+\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta} \\ $$$${A}=\Delta_{{BCD}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\:\mathrm{cos}\:\theta×\left(\mathrm{2}−\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:}{\:\sqrt{\mathrm{3}}+\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}\right)\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{A}=\mathrm{sin}\:\mathrm{2}\theta\:\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:}{\:\sqrt{\mathrm{3}}+\mathrm{sin}\:\mathrm{2}\theta}\right) \\ $$$$ \\ $$$${A}_{{max}} =\mathrm{0}.\mathrm{7393}\:{at}\:\theta=\mathrm{35}.\mathrm{2354}° \\ $$

Commented by ajfour last updated on 11/Dec/20

Commented by mr W last updated on 11/Dec/20

Commented by ajfour last updated on 11/Dec/20

tan α=((√3)/2) Let bottommost point of circle F. ∠ADF=φ , center of circle O. let OD=y ((sin 2θ)/(CD))=((sin φ)/1)=((sin (φ+2θ))/y) ⇒ y=cos 2θ+((sin 2θ)/(tan φ)) tan φ=((√3)/(1−OD))=((√3)/(1−y)) ⇒ y=cos 2θ+(((1−y)sin 2θ)/( (√3))) ⇒ y=(((√3)cos 2θ+sin 2θ)/( (√3)+sin 2θ)) Area s=2△=(1+y)sin 2θ s=(1+y)sin 2θ =(1+(((√3)cos 2θ+sin 2θ)/( (√3)+sin 2θ)))sin 2θ let tan θ=m s=sin 2θ(2−((2(√3)sin^2 θ)/( (√3)+2sin θcos θ))) =((4m)/((1+m^2 )))[1−(((√3)m^2 )/( (√3)(1+m^2 )+2m))] ⇒ △_(max) =(s_(max) /2)≈ 0.7393 θ≈ tan^(−1) (0.7063)≈35.234°

$$\mathrm{tan}\:\alpha=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${Let}\:{bottommost}\:{point}\:{of}\:{circle}\:{F}. \\ $$$$\angle{ADF}=\phi\:\:\:\:,\:\:{center}\:{of}\:{circle}\:{O}. \\ $$$${let}\:\:{OD}={y} \\ $$$$\frac{\mathrm{sin}\:\mathrm{2}\theta}{{CD}}=\frac{\mathrm{sin}\:\phi}{\mathrm{1}}=\frac{\mathrm{sin}\:\left(\phi+\mathrm{2}\theta\right)}{{y}} \\ $$$$\Rightarrow\:\:{y}=\mathrm{cos}\:\mathrm{2}\theta+\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{tan}\:\phi} \\ $$$$\mathrm{tan}\:\phi=\frac{\sqrt{\mathrm{3}}}{\mathrm{1}−{OD}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{1}−{y}} \\ $$$$\Rightarrow\:\:{y}=\mathrm{cos}\:\mathrm{2}\theta+\frac{\left(\mathrm{1}−{y}\right)\mathrm{sin}\:\mathrm{2}\theta}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:\:{y}=\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{2}\theta+\mathrm{sin}\:\mathrm{2}\theta}{\:\sqrt{\mathrm{3}}+\mathrm{sin}\:\mathrm{2}\theta} \\ $$$${Area}\:\:{s}=\mathrm{2}\bigtriangleup=\left(\mathrm{1}+{y}\right)\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\:\:\:{s}=\left(\mathrm{1}+{y}\right)\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\:\:\:\:\:\:=\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{2}\theta+\mathrm{sin}\:\mathrm{2}\theta}{\:\sqrt{\mathrm{3}}+\mathrm{sin}\:\mathrm{2}\theta}\right)\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\:\:\:\:{let}\:\:\mathrm{tan}\:\theta={m} \\ $$$$\:\:{s}=\mathrm{sin}\:\mathrm{2}\theta\left(\mathrm{2}−\frac{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:^{\mathrm{2}} \theta}{\:\sqrt{\mathrm{3}}+\mathrm{2sin}\:\theta\mathrm{cos}\:\theta}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{4}{m}}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)}\left[\mathrm{1}−\frac{\sqrt{\mathrm{3}}{m}^{\mathrm{2}} }{\:\sqrt{\mathrm{3}}\left(\mathrm{1}+{m}^{\mathrm{2}} \right)+\mathrm{2}{m}}\right] \\ $$$$\:\Rightarrow\:\bigtriangleup_{{max}} =\frac{{s}_{{max}} }{\mathrm{2}}\approx\:\mathrm{0}.\mathrm{7393} \\ $$$$\:\:\:\theta\approx\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{7063}\right)\approx\mathrm{35}.\mathrm{234}° \\ $$

Commented by ajfour last updated on 11/Dec/20

$${s}={f}\left({m}\right) \\ $$

Commented by ajfour last updated on 11/Dec/20

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

Commented by mr W last updated on 12/Dec/20

s=sin 2θ(2−((2(√3)sin^2 θ)/( (√3)+2sin θcos θ))) ⇒ s=2 sin 2θ(1−(((√3)sin^2 θ)/( (√3)+sin 2θ))) we got the same equation in terms of θ.

$$\:\:{s}=\mathrm{sin}\:\mathrm{2}\theta\left(\mathrm{2}−\frac{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sin}\:^{\mathrm{2}} \theta}{\:\sqrt{\mathrm{3}}+\mathrm{2sin}\:\theta\mathrm{cos}\:\theta}\right) \\ $$$$\Rightarrow\:{s}=\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}\mathrm{sin}\:^{\mathrm{2}} \theta}{\:\sqrt{\mathrm{3}}+\mathrm{sin}\:\mathrm{2}\theta}\right) \\ $$$${we}\:{got}\:{the}\:{same}\:{equation}\:{in}\:{terms} \\ $$$${of}\:\theta. \\ $$

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