Question Number 27771 by ajfour last updated on 14/Jan/18
Commented by ajfour last updated on 14/Jan/18
$${If}\:{an}\:{equilateral}\:{triangle}\:{of} \\ $$$${edge}\:{length}\:{a}\:{is}\:{rotated}\:{about} \\ $$$${about}\:{its}\:{vertex}\:{A}\:{by}\:{an}\:{angle}\:\theta \\ $$$${find}\:{the}\:{area}\:{common}\:{in}\:{its} \\ $$$${new}\:{and}\:{previous}\:{positions}. \\ $$$$\left({the}\:{pink}\:{area}\right). \\ $$
Answered by mrW2 last updated on 14/Jan/18
Commented by mrW2 last updated on 14/Jan/18
![2α+θ=(π/3) ⇒α=(π/6)−(θ/2) ((BP)/(sin θ))=((AB)/(sin (π−(π/3)−θ)))=(a/(sin ((π/3)+θ))) ⇒BP=((sin θ)/(sin ((π/3)+θ)))×a similarly ((CM)/(sin α))=(a/(sin ((π/3)+α))) ⇒CM=((sin α)/(sin ((π/3)+α)))×a=((sin ((π/6)−(θ/2)))/(sin ((π/2)−(θ/2))))×a=((sin ((π/6)−(θ/2)))/(cos (θ/2)))×a PM=a−((sin θ)/(sin ((π/3)+θ)))×a−((sin ((π/6)−(θ/2)))/(cos (θ/2)))×a =a[1−((sin θ)/(((√3)/2) cos θ+(1/2) sin θ))−(((1/2) cos (θ/2)−((√3)/2) sin (θ/2))/(cos (θ/2)))] =a[(1/2)−((2 tan θ)/( (√3)+tan θ))+((√3)/2) tan (θ/2)] A_(ΔAPM) =(1/2)×PM×(((√3) a)/2)=(((√3) a^2 )/4)[(1/2)−((2 tan θ)/( (√3)+tan θ))+((√3)/2) tan (θ/2)] ⇒A_(APMQ) =2A_(ΔAPM) =(((√3) a^2 )/4)[1−((4 tan θ)/( (√3)+tan θ))+(√3) tan (θ/2)]](https://www.tinkutara.com/question/Q27777.png)
$$\mathrm{2}\alpha+\theta=\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow\alpha=\frac{\pi}{\mathrm{6}}−\frac{\theta}{\mathrm{2}} \\ $$$$\frac{{BP}}{\mathrm{sin}\:\theta}=\frac{{AB}}{\mathrm{sin}\:\left(\pi−\frac{\pi}{\mathrm{3}}−\theta\right)}=\frac{{a}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)} \\ $$$$\Rightarrow{BP}=\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)}×{a} \\ $$$${similarly} \\ $$$$\frac{{CM}}{\mathrm{sin}\:\alpha}=\frac{{a}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\alpha\right)} \\ $$$$\Rightarrow{CM}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\alpha\right)}×{a}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}−\frac{\theta}{\mathrm{2}}\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right)}×{a}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}−\frac{\theta}{\mathrm{2}}\right)}{\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}×{a} \\ $$$$ \\ $$$${PM}={a}−\frac{\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\theta\right)}×{a}−\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}−\frac{\theta}{\mathrm{2}}\right)}{\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}×{a} \\ $$$$={a}\left[\mathrm{1}−\frac{\mathrm{sin}\:\theta}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{sin}\:\theta}−\frac{\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}{\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}\right] \\ $$$$={a}\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}\:\mathrm{tan}\:\theta}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\theta}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right] \\ $$$$ \\ $$$${A}_{\Delta{APM}} =\frac{\mathrm{1}}{\mathrm{2}}×{PM}×\frac{\sqrt{\mathrm{3}}\:{a}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} }{\mathrm{4}}\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}\:\mathrm{tan}\:\theta}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\theta}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right] \\ $$$$ \\ $$$$\Rightarrow{A}_{{APMQ}} =\mathrm{2}{A}_{\Delta{APM}} =\frac{\sqrt{\mathrm{3}}\:{a}^{\mathrm{2}} }{\mathrm{4}}\left[\mathrm{1}−\frac{\mathrm{4}\:\mathrm{tan}\:\theta}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\theta}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right] \\ $$
Commented by ajfour last updated on 15/Jan/18
$${thank}\:{you}\:{sir}.\:{please}\:{view}\:{my} \\ $$$${way}\:{if}\:{you}\:{haven}'{t}. \\ $$
Commented by mrW2 last updated on 16/Jan/18
$${Your}\:{way}\:{is}\:{very}\:{nice}\:{too}.\:{I}\:{didn}'{t}\:{come} \\ $$$${to}\:{it}. \\ $$
Answered by ajfour last updated on 14/Jan/18
Commented by ajfour last updated on 14/Jan/18
![eq. of BP : y=−(√3)(x−a) eq. of AP : y=xtan θ eq. of AM : y=xtan (θ+α) =xtan ((π/6)+(θ/2)) y_P =−(√3)(x_P −a)=x_P tan θ ⇒ y_P =((a(√3)tan θ)/( (√3)+tan θ)) similarly y_M =((a(√3)tan ((π/6)+(θ/2)))/( (√3)+tan ((π/6)+(θ/2)))) =((a(√3))/( (√3)(((1−((tan (θ/2))/( (√3))))/((1/( (√3)))+tan (θ/2))))+1)) =((a(√3))/( (√3)[(((√3)−tan (θ/2))/(1+(√3)tan (θ/2)))]+1)) =((a(√3)[1+(√3)tan (θ/2)])/4) reqd. area=2×(a/2)(y_M −y_P ) =((a^2 (√3))/4)[1+(√3)tan (θ/2)−((4tan θ)/( (√3)+tan θ))] .](https://www.tinkutara.com/question/Q27783.png)
$${eq}.\:{of}\:{BP}\:: \\ $$$${y}=−\sqrt{\mathrm{3}}\left({x}−{a}\right) \\ $$$${eq}.\:{of}\:{AP}\::\:\:\:{y}={x}\mathrm{tan}\:\theta \\ $$$${eq}.\:{of}\:{AM}\::\:\:{y}={x}\mathrm{tan}\:\left(\theta+\alpha\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}\mathrm{tan}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\theta}{\mathrm{2}}\right) \\ $$$${y}_{{P}} =−\sqrt{\mathrm{3}}\left({x}_{{P}} −{a}\right)={x}_{{P}} \mathrm{tan}\:\theta \\ $$$$\Rightarrow\:{y}_{{P}} =\frac{{a}\sqrt{\mathrm{3}}\mathrm{tan}\:\theta}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\theta} \\ $$$${similarly}\:{y}_{{M}} =\frac{{a}\sqrt{\mathrm{3}}\mathrm{tan}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\theta}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\theta}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:=\frac{{a}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}−\frac{\mathrm{tan}\:\left(\theta/\mathrm{2}\right)}{\:\sqrt{\mathrm{3}}}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right)+\mathrm{1}} \\ $$$$=\frac{{a}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}\left[\frac{\sqrt{\mathrm{3}}−\mathrm{tan}\:\left(\theta/\mathrm{2}\right)}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\:\left(\theta/\mathrm{2}\right)}\right]+\mathrm{1}} \\ $$$$=\frac{{a}\sqrt{\mathrm{3}}\left[\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\:\left(\theta/\mathrm{2}\right)\right]}{\mathrm{4}} \\ $$$${reqd}.\:{area}=\mathrm{2}×\frac{{a}}{\mathrm{2}}\left({y}_{{M}} −{y}_{{P}} \right) \\ $$$$=\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}}\left[\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\:\frac{\theta}{\mathrm{2}}−\frac{\mathrm{4tan}\:\theta}{\:\sqrt{\mathrm{3}}+\mathrm{tan}\:\theta}\right]\:. \\ $$