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Question Number 202967 by sonukgindia last updated on 06/Jan/24
Answered by mr W last updated on 06/Jan/24
Commented by mr W last updated on 07/Jan/24
side length of hexagon=1 (1/(sin (60°−θ)))=(c/(sin θ)) ⇒c=((2 sin θ)/( (√3) cos θ−sin θ)) ⇒d=2−c=2−((2 sin θ)/( (√3) cos θ−sin θ)) ((1−x+1)/(sin (30°+θ)))=(d/(sin (90°+θ))) ((2(2−x))/(cos θ+(√3) sin θ))=(1/(cos θ))×(2−((2 sin θ)/( (√3) cos θ−sin θ))) x=2−(((1+(√3) tan θ)((√3)−2 tan θ))/( (√3)−tan θ)) let t=tan θ x=3−((2(√3)(1−t^2 ))/( (√3)−t)) (dx/dt)=0 ⇒((−2t)/( (√3)−t))+((1−t^2 )/(((√3)−t)^2 ))=0 ⇒t^2 −2(√3)t+1=0 ⇒t=(√3)−(√2) ⇒θ=tan^(−1) ((√3)−(√2))≈17.63° x_(min) =4(√6)−9≈0.798 (((yellow)/(red)))_(min) =(x_(min) /(1−x_(min) ))=((4(√6)−9)/(10−4(√6))) =(3/2)+(√6)≈3.9495
$${side}\:{length}\:{of}\:{hexagon}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{60}°−\theta\right)}=\frac{{c}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{c}=\frac{\mathrm{2}\:\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{d}=\mathrm{2}−{c}=\mathrm{2}−\frac{\mathrm{2}\:\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta} \\ $$$$\frac{\mathrm{1}−{x}+\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{30}°+\theta\right)}=\frac{{d}}{\mathrm{sin}\:\left(\mathrm{90}°+\theta\right)} \\ $$$$\frac{\mathrm{2}\left(\mathrm{2}−{x}\right)}{\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta}=\frac{\mathrm{1}}{\mathrm{cos}\:\theta}×\left(\mathrm{2}−\frac{\mathrm{2}\:\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}\right) \\ $$$${x}=\mathrm{2}−\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta\right)\left(\sqrt{\mathrm{3}}−\mathrm{2}\:\mathrm{tan}\:\theta\right)}{\:\sqrt{\mathrm{3}}−\mathrm{tan}\:\theta} \\ $$$${let}\:{t}=\mathrm{tan}\:\theta \\ $$$${x}=\mathrm{3}−\frac{\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{3}}−{t}} \\ $$$$\frac{{dx}}{{dt}}=\mathrm{0} \\ $$$$\Rightarrow\frac{−\mathrm{2}{t}}{\:\sqrt{\mathrm{3}}−{t}}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\sqrt{\mathrm{3}}−{t}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\:\Rightarrow\theta=\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)\approx\mathrm{17}.\mathrm{63}° \\ $$$${x}_{{min}} =\mathrm{4}\sqrt{\mathrm{6}}−\mathrm{9}\approx\mathrm{0}.\mathrm{798} \\ $$$$\left(\frac{{yellow}}{{red}}\right)_{{min}} =\frac{{x}_{{min}} }{\mathrm{1}−{x}_{{min}} }=\frac{\mathrm{4}\sqrt{\mathrm{6}}−\mathrm{9}}{\mathrm{10}−\mathrm{4}\sqrt{\mathrm{6}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}+\sqrt{\mathrm{6}}\approx\mathrm{3}.\mathrm{9495} \\ $$
Commented by mr W last updated on 06/Jan/24

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