Question-39303

Question Number 39303 by behi83417@gmail.com last updated on 05/Jul/18

Commented by behi83417@gmail.com last updated on 05/Jul/18

equaliteral triangle with sides: BC=1,rotates about :B,at angle:15^• 1)find common area between rest and rotated case. 2)yes or no! common region is a cyclic shape.

$$\boldsymbol{{equaliteral}}\:\boldsymbol{{triangle}}\:\boldsymbol{{with}}\:\boldsymbol{{sides}}: \\ $$$$\boldsymbol{{BC}}=\mathrm{1},\boldsymbol{{rotates}}\:\boldsymbol{{about}}\::\boldsymbol{{B}},\boldsymbol{{at}}\:\boldsymbol{{angle}}:\mathrm{15}^{\bullet} \\ $$$$\left.\mathrm{1}\right)\boldsymbol{{find}}\:\boldsymbol{{common}}\:\boldsymbol{{area}}\:\boldsymbol{{between}} \\ $$$$\boldsymbol{{rest}}\:\boldsymbol{{and}}\:\boldsymbol{{rotated}}\:\boldsymbol{{case}}. \\ $$$$\left.\mathrm{2}\right)\boldsymbol{{yes}}\:\boldsymbol{{or}}\:\boldsymbol{{no}}! \\ $$$$\boldsymbol{{common}}\:\boldsymbol{{region}}\:\boldsymbol{{is}}\:\boldsymbol{{a}}\:\boldsymbol{{cyclic}}\:\boldsymbol{{shape}}. \\ $$

Answered by MrW3 last updated on 05/Jul/18

Commented by MrW3 last updated on 05/Jul/18

let a=side length =1 let θ=rotation angle=15° BD=(((√3)a)/(2 cos (θ/2))) α=30°−(θ/2) β=60°+θ ⇒α+β=90°+(θ/2) ((EB)/(sin (180°−α−β)))=((BD)/(sin β)) ⇒EB=(((√3)a sin (α+β))/(2 cos (θ/2) sin β))=(((√3)a)/(2 sin (60°+θ))) A_(ΔDBE) =(1/2)×DB×EB×sin α A_(common) =2A_(ΔDBE) =DB×EB×sin α =(((√3)a)/(2 cos (θ/2)))×(((√3)a)/(2 sin (60°+θ)))×sin (30°−(θ/2)) =((3a^2 sin (30°−(θ/2)))/(4 cos (θ/2) sin (60°+θ))) =A_0 (((√3) sin (30°−(θ/2)))/(cos (θ/2) sin (60°+θ))) the common area is not cyclic since 2β=120°+2θ is not always equal to 180°.

$${let}\:{a}={side}\:{length}\:=\mathrm{1} \\ $$$${let}\:\theta={rotation}\:{angle}=\mathrm{15}° \\ $$$$ \\ $$$${BD}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\alpha=\mathrm{30}°−\frac{\theta}{\mathrm{2}} \\ $$$$\beta=\mathrm{60}°+\theta \\ $$$$\Rightarrow\alpha+\beta=\mathrm{90}°+\frac{\theta}{\mathrm{2}} \\ $$$$\frac{{EB}}{\mathrm{sin}\:\left(\mathrm{180}°−\alpha−\beta\right)}=\frac{{BD}}{\mathrm{sin}\:\beta} \\ $$$$\Rightarrow{EB}=\frac{\sqrt{\mathrm{3}}{a}\:\mathrm{sin}\:\left(\alpha+\beta\right)}{\mathrm{2}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\:\mathrm{sin}\:\beta}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{60}°+\theta\right)} \\ $$$${A}_{\Delta{DBE}} =\frac{\mathrm{1}}{\mathrm{2}}×{DB}×{EB}×\mathrm{sin}\:\alpha \\ $$$${A}_{{common}} =\mathrm{2}{A}_{\Delta{DBE}} ={DB}×{EB}×\mathrm{sin}\:\alpha \\ $$$$=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}×\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{60}°+\theta\right)}×\mathrm{sin}\:\left(\mathrm{30}°−\frac{\theta}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{3}{a}^{\mathrm{2}} \:\mathrm{sin}\:\left(\mathrm{30}°−\frac{\theta}{\mathrm{2}}\right)}{\mathrm{4}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\:\mathrm{sin}\:\left(\mathrm{60}°+\theta\right)} \\ $$$$={A}_{\mathrm{0}} \frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\mathrm{30}°−\frac{\theta}{\mathrm{2}}\right)}{\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\:\mathrm{sin}\:\left(\mathrm{60}°+\theta\right)} \\ $$$$ \\ $$$${the}\:{common}\:{area}\:{is}\:{not}\:{cyclic}\:{since} \\ $$$$\mathrm{2}\beta=\mathrm{120}°+\mathrm{2}\theta\:{is}\:{not}\:{always}\:{equal}\:{to}\:\mathrm{180}°. \\ $$

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