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Question-121793

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Question Number 121793 by ajfour last updated on 11/Nov/20
Commented by ajfour last updated on 13/Nov/20
If ODE^(⌢) is a sector, and △ABC is equilateral with minimum area has an area equal to △OEF , find θ.
$${If}\:{O}\overset{\frown} {{DE}}\:\:\:{is}\:{a}\:{sector},\:{and}\:\bigtriangleup{ABC}\: \\ $$$${is}\:{equilateral}\:{with}\:{minimum}\:{area} \\ $$$${has}\:{an}\:{area}\:{equal}\:{to}\:\bigtriangleup{OEF}\:,\:{find}\:\theta. \\ $$$$ \\ $$
Commented by ajfour last updated on 14/Nov/20
mrW Sir kindly help with this Q.
$${mrW}\:{Sir}\:{kindly}\:{help}\:{with}\:{this}\:{Q}. \\ $$
Answered by ajfour last updated on 13/Nov/20
let OD=OE=1, F be origin. OF=p=cos θ , FB=y, FC=x 2p(√(1−p^2 ))=(((√3)s^2 )/4) ⇒ 64((p^2 /s^2 ))((1/s^2 )−(p^2 /s^2 ))=3 ⇒ say ((p/s))^2 =t ⇒ t^2 −(t/s^2 )+(3/(64))=0 ⇒ t=(p^2 /s^2 )=(1/(2s^2 ))±(√((1/(4s^4 ))−(3/(64)))) ...(I) or p^2 =(1/2)±(√((1/4)−((3s^4 )/(64)))) BC^( 2) =s^2 =x^2 +y^2 let M be midpoint of BC. M((x/2),(y/2)) slope of AM m=(x/y) A(p+(x/2)+((s(√3))/2)(y/( s)) , (y/2)+((s(√3))/2)(x/s)) As OA=1 (p+(x/2)+((y(√3))/2))^2 +((y/2)+((x(√3))/2))^2 =1 ⇒ p^2 +p(x+y(√3))+x^2 +y^2 +(√3)xy=1 with s^2 =x^2 +y^2 a minimum let x=scos φ, y=ssin φ ⇒ p^2 +sp(cos φ+(√3)sin φ)+ s^2 +(√3)s^2 sin φcos φ=1 ⇒ (p^2 /s^2 )+(p/s)(cos φ+(√3)sin φ)+1 +(√3)sin φcos φ−(1/s^2 )=0 And from (I) we have (p^2 /s^2 )=(1/(2s^2 ))±(√((1/(4s^4 ))−(3/(64)))) ......
$${let}\:{OD}={OE}=\mathrm{1},\:{F}\:{be}\:{origin}. \\ $$$${OF}={p}=\mathrm{cos}\:\theta\:,\:{FB}={y},\:{FC}={x} \\ $$$$\mathrm{2}{p}\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }=\frac{\sqrt{\mathrm{3}}{s}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{64}\left(\frac{{p}^{\mathrm{2}} }{{s}^{\mathrm{2}} }\right)\left(\frac{\mathrm{1}}{{s}^{\mathrm{2}} }−\frac{{p}^{\mathrm{2}} }{{s}^{\mathrm{2}} }\right)=\mathrm{3} \\ $$$$\Rightarrow\:{say}\:\:\left(\frac{{p}}{{s}}\right)^{\mathrm{2}} ={t} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{2}} −\frac{{t}}{{s}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{64}}=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}=\frac{{p}^{\mathrm{2}} }{{s}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{s}^{\mathrm{2}} }\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}{s}^{\mathrm{4}} }−\frac{\mathrm{3}}{\mathrm{64}}}\:\:\:\:…\left({I}\right) \\ $$$${or}\:\:\:\:{p}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{3}{s}^{\mathrm{4}} }{\mathrm{64}}} \\ $$$${BC}^{\:\mathrm{2}} ={s}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${let}\:{M}\:{be}\:{midpoint}\:{of}\:{BC}. \\ $$$${M}\left(\frac{{x}}{\mathrm{2}},\frac{{y}}{\mathrm{2}}\right) \\ $$$${slope}\:{of}\:{AM}\:\:\:{m}=\frac{{x}}{{y}} \\ $$$${A}\left({p}+\frac{{x}}{\mathrm{2}}+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\frac{{y}}{\:{s}}\:,\:\:\frac{{y}}{\mathrm{2}}+\frac{{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\frac{{x}}{{s}}\right) \\ $$$${As}\:\:{OA}=\mathrm{1} \\ $$$$\left({p}+\frac{{x}}{\mathrm{2}}+\frac{{y}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{y}}{\mathrm{2}}+\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\:\:{p}^{\mathrm{2}} +{p}\left({x}+{y}\sqrt{\mathrm{3}}\right)+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\sqrt{\mathrm{3}}{xy}=\mathrm{1} \\ $$$${with}\:\:{s}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\:{a}\:{minimum} \\ $$$${let}\:\:\:{x}={s}\mathrm{cos}\:\phi,\:\:{y}={s}\mathrm{sin}\:\phi \\ $$$$\Rightarrow \\ $$$$\:{p}^{\mathrm{2}} +{sp}\left(\mathrm{cos}\:\phi+\sqrt{\mathrm{3}}\mathrm{sin}\:\phi\right)+ \\ $$$${s}^{\mathrm{2}} +\sqrt{\mathrm{3}}{s}^{\mathrm{2}} \mathrm{sin}\:\phi\mathrm{cos}\:\phi=\mathrm{1} \\ $$$$\Rightarrow\:\frac{{p}^{\mathrm{2}} }{{s}^{\mathrm{2}} }+\frac{{p}}{{s}}\left(\mathrm{cos}\:\phi+\sqrt{\mathrm{3}}\mathrm{sin}\:\phi\right)+\mathrm{1} \\ $$$$\:\:\:\:+\sqrt{\mathrm{3}}\mathrm{sin}\:\phi\mathrm{cos}\:\phi−\frac{\mathrm{1}}{{s}^{\mathrm{2}} }=\mathrm{0} \\ $$$${And}\:{from}\:\left({I}\right)\:{we}\:{have} \\ $$$$\:\:\frac{{p}^{\mathrm{2}} }{{s}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{s}^{\mathrm{2}} }\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}{s}^{\mathrm{4}} }−\frac{\mathrm{3}}{\mathrm{64}}}\: \\ $$$$…… \\ $$$$ \\ $$

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