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Question Number 214012 by ajfour last updated on 24/Nov/24

Commented by ajfour last updated on 24/Nov/24

Outer circle radius is R. Circle with center A has radius r=R/2. If △ABC is equilateral, find its edge length (say a).

$${Outer}\:{circle}\:{radius}\:{is}\:{R}.\:{Circle}\:{with} \\ $$$${center}\:{A}\:{has}\:{radius}\:{r}={R}/\mathrm{2}. \\ $$$${If}\:\bigtriangleup{ABC}\:{is}\:{equilateral},\:{find}\:{its} \\ $$$${edge}\:{length}\:\left({say}\:{a}\right). \\ $$

Answered by mr W last updated on 24/Nov/24

Commented by mr W last updated on 24/Nov/24

c=s−(R/2) C(s sin α, −(R/2)+s cos α) B(s sin (α+(π/3)), −(R/2)+s cos (α+(π/3))) s^2 sin^2 α+(−(R/2)+s cos α)^2 =(R−s+(R/2))^2 s^2 sin^2 (α+(π/3))+ [−(R/2)+s cos (α+(π/3))]^2 =R^2 let λ=(s/R) λ^2 sin^2 α+(−(1/2)+λ cos α)^2 =((3/2)−λ)^2 (1/4)−λ cos α=(9/4)−3λ ⇒λ cos α=3λ−2 ⇒λ sin α=2(√((1−λ)(2λ−1))) λ^2 sin^2 (α+(π/3))+ [−(1/2)+λ cos (α+(π/3))]^2 =1 λ^2 −λ cos (α+(π/3))=(3/4) λ^2 −(λ/2)(cos α−(√3) sin α)=(3/4) 2λ^2 −3λ+(1/2)+2(√(3(1−λ)(2λ−1)))=0 ⇒λ≈0.52391863, 0.97608137

$${c}={s}−\frac{{R}}{\mathrm{2}} \\ $$$${C}\left({s}\:\mathrm{sin}\:\alpha,\:−\frac{{R}}{\mathrm{2}}+{s}\:\mathrm{cos}\:\alpha\right) \\ $$$${B}\left({s}\:\mathrm{sin}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right),\:−\frac{{R}}{\mathrm{2}}+{s}\:\mathrm{cos}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right)\right) \\ $$$${s}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha+\left(−\frac{{R}}{\mathrm{2}}+{s}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} =\left({R}−{s}+\frac{{R}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\left(\alpha+\frac{\pi}{\mathrm{3}}\right)+\:\left[−\frac{{R}}{\mathrm{2}}+{s}\:\mathrm{cos}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right)\right]^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${let}\:\lambda=\frac{{s}}{{R}} \\ $$$$\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\lambda\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} =\left(\frac{\mathrm{3}}{\mathrm{2}}−\lambda\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}−\lambda\:\mathrm{cos}\:\alpha=\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{3}\lambda \\ $$$$\Rightarrow\lambda\:\mathrm{cos}\:\alpha=\mathrm{3}\lambda−\mathrm{2}\:\Rightarrow\lambda\:\mathrm{sin}\:\alpha=\mathrm{2}\sqrt{\left(\mathrm{1}−\lambda\right)\left(\mathrm{2}\lambda−\mathrm{1}\right)} \\ $$$$\lambda^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\left(\alpha+\frac{\pi}{\mathrm{3}}\right)+\:\left[−\frac{\mathrm{1}}{\mathrm{2}}+\lambda\:\mathrm{cos}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right)\right]^{\mathrm{2}} =\mathrm{1} \\ $$$$\lambda^{\mathrm{2}} −\lambda\:\mathrm{cos}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\lambda^{\mathrm{2}} −\frac{\lambda}{\mathrm{2}}\left(\mathrm{cos}\:\alpha−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\alpha\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{2}\lambda^{\mathrm{2}} −\mathrm{3}\lambda+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}\left(\mathrm{1}−\lambda\right)\left(\mathrm{2}\lambda−\mathrm{1}\right)}=\mathrm{0} \\ $$$$\Rightarrow\lambda\approx\mathrm{0}.\mathrm{52391863},\:\mathrm{0}.\mathrm{97608137} \\ $$

Commented by mr W last updated on 24/Nov/24

Commented by mr W last updated on 24/Nov/24

Commented by ajfour last updated on 24/Nov/24

Yes Sir . I took △ side =2s (8s^2 −6s+(1/2))^2 +12(8s^2 −6s+(1/2))+6=0 ⇒ 8s^2 +6s+(1/2)=−6−(√(30)) 2s=((3±(√(8(√(30))−43)))/4) ≈ 0.97608 , 0.52392

$${Yes}\:{Sir}\:.\:{I}\:{took}\:\bigtriangleup\:{side}\:=\mathrm{2}{s} \\ $$$$\left(\mathrm{8}{s}^{\mathrm{2}} −\mathrm{6}{s}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{12}\left(\mathrm{8}{s}^{\mathrm{2}} −\mathrm{6}{s}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{8}{s}^{\mathrm{2}} +\mathrm{6}{s}+\frac{\mathrm{1}}{\mathrm{2}}=−\mathrm{6}−\sqrt{\mathrm{30}} \\ $$$$\:\:\mathrm{2}{s}=\frac{\mathrm{3}\pm\sqrt{\mathrm{8}\sqrt{\mathrm{30}}−\mathrm{43}}}{\mathrm{4}} \\ $$$$\:\:\approx\:\:\mathrm{0}.\mathrm{97608}\:\:,\:\:\mathrm{0}.\mathrm{52392} \\ $$

Commented by ajfour last updated on 25/Nov/24

Let △ side be 2s. 2s=r+c And mid point of AC be M(ssin α, −r+scos α) BM=s(√3) B(ssin α+s(√3)cos α, −r+scos α−s(√3)sin α) OB^( 2) =R^2 s^2 sin^2 α+3s^2 cos^2 α+2(√3)s^2 sin αcos α +r^2 +s^2 cos^2 α+3s^2 sin^2 α−2(√3)s^2 sin αcos α −2rscos α+2(√3)rssin α=R^2 ⇒ 4s^2 −2rscos α+2(√3)rssin α=R^2 −r^2 say (s/R)=t, and (r/R)=(1/2) (given) 4t^2 −tcos α+(√3)tsin α=1−(1/4) .......(i) Now C(2ssin α, −r+2scos α) OC^( 2) =(R−c)^2 =(R+r−2s)^2 4s^2 −4rscos α=(R+2r−2s)(R−2s) 2t^2 −tcos α=(1−t)(1−2t) .....(ii) ⇒ tcos α=3t−1 from ..(i) (√3)tsin α=1−(1/4)−4t^2 +3t−1 ⇒ 3(3t−1)^2 +(3t−4t^2 −(1/4))^2 =3t^2 ⇒ (3t−4t^2 −(1/4))^2 +24t^2 −18t+3=0 (3t−4t^2 −(1/4))^2 −6(3t−4t^2 −(1/4))+(3/2)=0 ⇒ 3t−4t^2 −(1/4)=3−(√((15)/2)) ⇒ 16t^2 −12t+13−(√(120))=0 (s/R)=t=(3/8)±(√((9/(64))−((52)/(64))+((8(√(30)))/(64)))) 2s = side a=(((3±(√(8(√(30))−43)))/4))R ✓ ★

$${Let}\:\bigtriangleup\:{side}\:{be}\:\mathrm{2}{s}. \\ $$$$\mathrm{2}{s}={r}+{c} \\ $$$${And}\:{mid}\:{point}\:{of}\:{AC}\:{be} \\ $$$${M}\left({s}\mathrm{sin}\:\alpha,\:−{r}+{s}\mathrm{cos}\:\alpha\right) \\ $$$${BM}={s}\sqrt{\mathrm{3}} \\ $$$${B}\left({s}\mathrm{sin}\:\alpha+{s}\sqrt{\mathrm{3}}\mathrm{cos}\:\alpha,\:−{r}+{s}\mathrm{cos}\:\alpha−{s}\sqrt{\mathrm{3}}\mathrm{sin}\:\alpha\right) \\ $$$${OB}^{\:\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \alpha+\mathrm{3}{s}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \alpha+\cancel{\mathrm{2}\sqrt{\mathrm{3}}{s}^{\mathrm{2}} \mathrm{sin}\:\alpha\mathrm{cos}\:\alpha} \\ $$$$\:\:+{r}^{\mathrm{2}} +{s}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \alpha+\mathrm{3}{s}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \alpha−\cancel{\mathrm{2}\sqrt{\mathrm{3}}{s}^{\mathrm{2}} \mathrm{sin}\:\alpha\mathrm{cos}\:\alpha} \\ $$$$−\mathrm{2}{rs}\mathrm{cos}\:\alpha+\mathrm{2}\sqrt{\mathrm{3}}{rs}\mathrm{sin}\:\alpha={R}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{4}{s}^{\mathrm{2}} −\mathrm{2}{rs}\mathrm{cos}\:\alpha+\mathrm{2}\sqrt{\mathrm{3}}{rs}\mathrm{sin}\:\alpha={R}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$${say}\:\:\frac{{s}}{{R}}={t},\:\:{and}\:\:\frac{{r}}{{R}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\left({given}\right) \\ $$$$\mathrm{4}{t}^{\mathrm{2}} −{t}\mathrm{cos}\:\alpha+\sqrt{\mathrm{3}}{t}\mathrm{sin}\:\alpha=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\:\:.......\left({i}\right) \\ $$$${Now} \\ $$$${C}\left(\mathrm{2}{s}\mathrm{sin}\:\alpha,\:−{r}+\mathrm{2}{s}\mathrm{cos}\:\alpha\right) \\ $$$${OC}^{\:\mathrm{2}} =\left({R}−{c}\right)^{\mathrm{2}} =\left({R}+{r}−\mathrm{2}{s}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{s}^{\mathrm{2}} −\mathrm{4}{rs}\mathrm{cos}\:\alpha=\left({R}+\mathrm{2}{r}−\mathrm{2}{s}\right)\left({R}−\mathrm{2}{s}\right) \\ $$$$\mathrm{2}{t}^{\mathrm{2}} −{t}\mathrm{cos}\:\alpha=\left(\mathrm{1}−{t}\right)\left(\mathrm{1}−\mathrm{2}{t}\right)\:\:\:\:.....\left({ii}\right) \\ $$$$\Rightarrow\:\:{t}\mathrm{cos}\:\alpha=\mathrm{3}{t}−\mathrm{1}\:\:\: \\ $$$${from}\:\:..\left({i}\right) \\ $$$$\sqrt{\mathrm{3}}{t}\mathrm{sin}\:\alpha=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}{t}−\mathrm{1} \\ $$$$\Rightarrow\:\:\mathrm{3}\left(\mathrm{3}{t}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{3}{t}−\mathrm{4}{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{3}{t}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left(\mathrm{3}{t}−\mathrm{4}{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\mathrm{24}{t}^{\mathrm{2}} −\mathrm{18}{t}+\mathrm{3}=\mathrm{0} \\ $$$$\left(\mathrm{3}{t}−\mathrm{4}{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\mathrm{6}\left(\mathrm{3}{t}−\mathrm{4}{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)+\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{3}{t}−\mathrm{4}{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{3}−\sqrt{\frac{\mathrm{15}}{\mathrm{2}}} \\ $$$$\Rightarrow\:\:\mathrm{16}{t}^{\mathrm{2}} −\mathrm{12}{t}+\mathrm{13}−\sqrt{\mathrm{120}}=\mathrm{0} \\ $$$$\frac{{s}}{{R}}={t}=\frac{\mathrm{3}}{\mathrm{8}}\pm\sqrt{\frac{\mathrm{9}}{\mathrm{64}}−\frac{\mathrm{52}}{\mathrm{64}}+\frac{\mathrm{8}\sqrt{\mathrm{30}}}{\mathrm{64}}} \\ $$$$\mathrm{2}{s}\:=\:{side}\:{a}=\left(\frac{\mathrm{3}\pm\sqrt{\mathrm{8}\sqrt{\mathrm{30}}−\mathrm{43}}}{\mathrm{4}}\right){R}\:\checkmark \\ $$$$\bigstar \\ $$

Commented by mr W last updated on 25/Nov/24

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