Question-218607

Question Number 218607 by Spillover last updated on 13/Apr/25

Answered by mr W last updated on 13/Apr/25

Commented by mr W last updated on 13/Apr/25

bigger circle: (x−(s/2))^2 +(y−(s/2))^2 =(s^2 /2) smaller circle: (x−(s/2))^2 +(y−((3s)/8))^2 =(((5s)/8))^2 (−(s/2))^2 +(y_G −((3s)/8))^2 =(((5s)/8))^2 ⇒y_G =((3s)/4) eqn. of GH: y=s+(1/2)(x−(s/2))=(x/2)+((3s)/4) (x−(s/2))^2 +((x/2)+((3s)/4)−(s/2))^2 =(s^2 /2) 5x^2 −3sx−((3s^2 )/4)=0 x=(((3±2(√6))/(10)))s ⇒x_F =(((3−2(√6))s)/(10)) ⇒x_K =(((3+2(√6))s)/(10)) Δx_(GF) =x_G −x_F =(((2(√6)−3)s)/(10)) Δx_(HK) =x_K −x_H =(((3+2(√6))s)/(10))−(s/2)=(((2(√6)−2)s)/(10)) ((w−u)/v)=((Δx_(HK) −Δx_(GF) )/(Δx_(EG) )) =(((((2(√6)−2)s)/(10))−(((2(√6)−3)s)/(10)))/(s/2))=(1/5) ⇒v=5(w−u) ≡ y=5(z−x) in original question✓ eqn. of JC: y=s+((x−s)/( (√3)))=(x/( (√3)))+(1−(1/( (√3))))s eqn. of AE: y=(√3)x y_M =(√3)x_M =(x_M /( (√3)))+(1−(1/( (√3))))s ⇒x_M =((((√3)−1)s)/2) eqn. of BE: y=−(√3)(x−s) y_N =−(√3)(x_N −s)=(x_N /( (√3)))+(1−(1/( (√3))))s ⇒x_N =(((4−(√3))s)/4) Δx_(MJ) =x_M −x_J =((((√3)−1)s)/2) Δx_(CN) =x_C −x_N =s−(((4−(√3))s)/4)=(((√3)s)/4) Δx_(NM) =x_N −x_M =(((4−(√3))s)/4)−((((√3)−1)s)/2)=((3(2−(√3))s)/4) ((c−a)/b)=((Δx_(CN) −Δx_(MJ) )/(Δx_(MN) )) =(((((√3)s)/4)−((((√3)−1)s)/2))/((3(2−(√3))s)/4))=(1/3) ⇒b=3(c−a) ✓

$${bigger}\:{circle}: \\ $$$$\left({x}−\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{{s}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${smaller}\:{circle}: \\ $$$$\left({x}−\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{3}{s}}{\mathrm{8}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{5}{s}}{\mathrm{8}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left(−\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}_{{G}} −\frac{\mathrm{3}{s}}{\mathrm{8}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{5}{s}}{\mathrm{8}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{y}_{{G}} =\frac{\mathrm{3}{s}}{\mathrm{4}} \\ $$$${eqn}.\:{of}\:{GH}: \\ $$$${y}={s}+\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{{s}}{\mathrm{2}}\right)=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{3}{s}}{\mathrm{4}} \\ $$$$\left({x}−\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{3}{s}}{\mathrm{4}}−\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{{s}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} −\mathrm{3}{sx}−\frac{\mathrm{3}{s}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$${x}=\left(\frac{\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{10}}\right){s} \\ $$$$\Rightarrow{x}_{{F}} =\frac{\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{6}}\right){s}}{\mathrm{10}} \\ $$$$\Rightarrow{x}_{{K}} =\frac{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}\right){s}}{\mathrm{10}} \\ $$$$\Delta{x}_{{GF}} ={x}_{{G}} −{x}_{{F}} =\frac{\left(\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{3}\right){s}}{\mathrm{10}} \\ $$$$\Delta{x}_{{HK}} ={x}_{{K}} −{x}_{{H}} =\frac{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}\right){s}}{\mathrm{10}}−\frac{{s}}{\mathrm{2}}=\frac{\left(\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{2}\right){s}}{\mathrm{10}} \\ $$$$\frac{{w}−{u}}{{v}}=\frac{\Delta{x}_{{HK}} −\Delta{x}_{{GF}} }{\Delta{x}_{{EG}} } \\ $$$$\:\:\:=\frac{\frac{\left(\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{2}\right){s}}{\mathrm{10}}−\frac{\left(\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{3}\right){s}}{\mathrm{10}}}{\frac{{s}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\Rightarrow{v}=\mathrm{5}\left({w}−{u}\right)\:\equiv\:{y}=\mathrm{5}\left({z}−{x}\right)\:{in}\:{original}\:{question}\checkmark \\ $$$$ \\ $$$${eqn}.\:{of}\:{JC}: \\ $$$${y}={s}+\frac{{x}−{s}}{\:\sqrt{\mathrm{3}}}=\frac{{x}}{\:\sqrt{\mathrm{3}}}+\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right){s} \\ $$$${eqn}.\:{of}\:{AE}: \\ $$$${y}=\sqrt{\mathrm{3}}{x} \\ $$$${y}_{{M}} =\sqrt{\mathrm{3}}{x}_{{M}} =\frac{{x}_{{M}} }{\:\sqrt{\mathrm{3}}}+\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right){s} \\ $$$$\Rightarrow{x}_{{M}} =\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right){s}}{\mathrm{2}} \\ $$$${eqn}.\:{of}\:{BE}: \\ $$$${y}=−\sqrt{\mathrm{3}}\left({x}−{s}\right) \\ $$$${y}_{{N}} =−\sqrt{\mathrm{3}}\left({x}_{{N}} −{s}\right)=\frac{{x}_{{N}} }{\:\sqrt{\mathrm{3}}}+\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right){s} \\ $$$$\Rightarrow{x}_{{N}} =\frac{\left(\mathrm{4}−\sqrt{\mathrm{3}}\right){s}}{\mathrm{4}} \\ $$$$\Delta{x}_{{MJ}} ={x}_{{M}} −{x}_{{J}} =\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right){s}}{\mathrm{2}} \\ $$$$\Delta{x}_{{CN}} ={x}_{{C}} −{x}_{{N}} ={s}−\frac{\left(\mathrm{4}−\sqrt{\mathrm{3}}\right){s}}{\mathrm{4}}=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{4}} \\ $$$$\Delta{x}_{{NM}} ={x}_{{N}} −{x}_{{M}} =\frac{\left(\mathrm{4}−\sqrt{\mathrm{3}}\right){s}}{\mathrm{4}}−\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right){s}}{\mathrm{2}}=\frac{\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){s}}{\mathrm{4}} \\ $$$$\frac{{c}−{a}}{{b}}=\frac{\Delta{x}_{{CN}} −\Delta{x}_{{MJ}} }{\Delta{x}_{{MN}} } \\ $$$$\:\:\:\:\:\:\:=\frac{\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{4}}−\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right){s}}{\mathrm{2}}}{\frac{\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){s}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{b}=\mathrm{3}\left({c}−{a}\right)\:\checkmark \\ $$

Commented by Spillover last updated on 13/Apr/25