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Question Number 111988 by ajfour last updated on 05/Sep/20

Commented by ajfour last updated on 05/Sep/20

If △ABC is equilateral with side s, find x-coordinate of point A in terms of s.

$${If}\:\bigtriangleup{ABC}\:{is}\:{equilateral}\:{with}\:{side}\:\boldsymbol{{s}}, \\ $$$${find}\:{x}-{coordinate}\:{of}\:{point}\:{A}\:{in}\: \\ $$$${terms}\:{of}\:{s}. \\ $$

Answered by mr W last updated on 06/Sep/20

center of equilateral D(h,k) A(a,0) R=(2/3)×(((√3)s)/2)=(s/( (√3))) (a−h)^2 +k^2 =R^2 =(s^2 /3) ⇒a=h+(√((s^2 /3)−k^2 )) x_B =h+(a−h)cos 120°−(0−k)sin 120° =h−(1/2)(a−h)+((√3)/2)k =(1/2)(3h+(√3)k−a) y_B =k+(a−h)sin 120°+(0−k)cos 120° =k+((√3)/2)(a−h)+(1/2)k =(1/2)(−(√3)h+3k+(√3)a) y_B =x_B ^2 (1/2)(−(√3)h+3k+(√3)a)=(1/4)(3h+(√3)k−a)^2 ⇒2(−(√3)h+3k+(√3)a)=(3h+(√3)k−a)^2 ⇒2(3k+(√(s^2 −3k^2 )))=(2h+(√3)k−(√((s^2 /3)−k^2 )))^2 ...(i) x_C =h+(a−h)cos 240°−(0−k)sin 240° =h−(1/2)(a−h)+((√3)/2)(0−k) =(1/2)(3h−(√3)k−a) y_C =k+(a−h)sin 240°+(0−k)cos 240° =k−((√3)/2)(a−h)−(1/2)(0−k) =(1/2)((√3)h+3k−(√3)a) y_C =x_C ^2 (1/2)((√3)h+3k−(√3)a)=(1/4)(3h−(√3)k−a)^2 ⇒2((√3)h+3k−(√3)a)=(3h−(√3)k−a)^2 ⇒2(3k−(√(s^2 −3k^2 )))=(2h−(√3)k−(√((s^2 /3)−k^2 )))^2 ..(ii) we get h and k from (i) and (ii). from (i): 2h=(√(2(3k+(√(s^2 −3k^2 )))))−(√3)k+(√((s^2 /3)−k^2 )) from (ii): 2h=(√(2(3k−(√(s^2 −3k^2 )))))+(√3)k+(√((s^2 /3)−k^2 )) (√(2(3k+(√(s^2 −3k^2 )))))−(√3)k=(√(2(3k−(√(s^2 −3k^2 )))))+(√3)k ⇒(√(2(3k+(√(s^2 −3k^2 )))))−(√(2(3k−(√(s^2 −3k^2 )))))=2(√3)k solution possible for s≤≈4.75

$${center}\:{of}\:{equilateral}\:{D}\left({h},{k}\right) \\ $$$${A}\left({a},\mathrm{0}\right) \\ $$$${R}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}}=\frac{{s}}{\:\sqrt{\mathrm{3}}} \\ $$$$\left({a}−{h}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} ={R}^{\mathrm{2}} =\frac{{s}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow{a}={h}+\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{3}}−{k}^{\mathrm{2}} } \\ $$$${x}_{{B}} ={h}+\left({a}−{h}\right)\mathrm{cos}\:\mathrm{120}°−\left(\mathrm{0}−{k}\right)\mathrm{sin}\:\mathrm{120}° \\ $$$$={h}−\frac{\mathrm{1}}{\mathrm{2}}\left({a}−{h}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{k} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}{h}+\sqrt{\mathrm{3}}{k}−{a}\right) \\ $$$${y}_{{B}} ={k}+\left({a}−{h}\right)\mathrm{sin}\:\mathrm{120}°+\left(\mathrm{0}−{k}\right)\mathrm{cos}\:\mathrm{120}° \\ $$$$={k}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({a}−{h}\right)+\frac{\mathrm{1}}{\mathrm{2}}{k} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(−\sqrt{\mathrm{3}}{h}+\mathrm{3}{k}+\sqrt{\mathrm{3}}{a}\right) \\ $$$${y}_{{B}} ={x}_{{B}} ^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(−\sqrt{\mathrm{3}}{h}+\mathrm{3}{k}+\sqrt{\mathrm{3}}{a}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}{h}+\sqrt{\mathrm{3}}{k}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\left(−\sqrt{\mathrm{3}}{h}+\mathrm{3}{k}+\sqrt{\mathrm{3}}{a}\right)=\left(\mathrm{3}{h}+\sqrt{\mathrm{3}}{k}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{3}{k}+\sqrt{{s}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} }\right)=\left(\mathrm{2}{h}+\sqrt{\mathrm{3}}{k}−\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{3}}−{k}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$$ \\ $$$${x}_{{C}} ={h}+\left({a}−{h}\right)\mathrm{cos}\:\mathrm{240}°−\left(\mathrm{0}−{k}\right)\mathrm{sin}\:\mathrm{240}° \\ $$$$={h}−\frac{\mathrm{1}}{\mathrm{2}}\left({a}−{h}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{0}−{k}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}{h}−\sqrt{\mathrm{3}}{k}−{a}\right) \\ $$$${y}_{{C}} ={k}+\left({a}−{h}\right)\mathrm{sin}\:\mathrm{240}°+\left(\mathrm{0}−{k}\right)\mathrm{cos}\:\mathrm{240}° \\ $$$$={k}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({a}−{h}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{0}−{k}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}{h}+\mathrm{3}{k}−\sqrt{\mathrm{3}}{a}\right) \\ $$$${y}_{{C}} ={x}_{{C}} ^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}{h}+\mathrm{3}{k}−\sqrt{\mathrm{3}}{a}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}{h}−\sqrt{\mathrm{3}}{k}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\left(\sqrt{\mathrm{3}}{h}+\mathrm{3}{k}−\sqrt{\mathrm{3}}{a}\right)=\left(\mathrm{3}{h}−\sqrt{\mathrm{3}}{k}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{3}{k}−\sqrt{{s}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} }\right)=\left(\mathrm{2}{h}−\sqrt{\mathrm{3}}{k}−\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{3}}−{k}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\:\:..\left({ii}\right) \\ $$$${we}\:{get}\:{h}\:{and}\:{k}\:{from}\:\left({i}\right)\:{and}\:\left({ii}\right). \\ $$$$ \\ $$$${from}\:\left({i}\right): \\ $$$$\mathrm{2}{h}=\sqrt{\mathrm{2}\left(\mathrm{3}{k}+\sqrt{{s}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} }\right)}−\sqrt{\mathrm{3}}{k}+\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{3}}−{k}^{\mathrm{2}} } \\ $$$${from}\:\left({ii}\right): \\ $$$$\mathrm{2}{h}=\sqrt{\mathrm{2}\left(\mathrm{3}{k}−\sqrt{{s}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} }\right)}+\sqrt{\mathrm{3}}{k}+\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{3}}−{k}^{\mathrm{2}} } \\ $$$$\sqrt{\mathrm{2}\left(\mathrm{3}{k}+\sqrt{{s}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} }\right)}−\sqrt{\mathrm{3}}{k}=\sqrt{\mathrm{2}\left(\mathrm{3}{k}−\sqrt{{s}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} }\right)}+\sqrt{\mathrm{3}}{k} \\ $$$$\Rightarrow\sqrt{\mathrm{2}\left(\mathrm{3}{k}+\sqrt{{s}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} }\right)}−\sqrt{\mathrm{2}\left(\mathrm{3}{k}−\sqrt{{s}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} }\right)}=\mathrm{2}\sqrt{\mathrm{3}}{k} \\ $$$$ \\ $$$${solution}\:{possible}\:{for}\:{s}\leqslant\approx\mathrm{4}.\mathrm{75} \\ $$

Commented by mr W last updated on 06/Sep/20

Commented by mr W last updated on 06/Sep/20

Commented by mr W last updated on 06/Sep/20

Answered by ajfour last updated on 06/Sep/20

Let s=2p(√3) A(a, 0) (a/p)=3cos θ{1+tan^2 θ−tan θ(√(tan^2 θ−(1/3))) } and θ is obtained from tan θ+(√(tan^2 θ−(1/3)))=2psin θtan θ or (1/(sin^2 θ))+3(2psin θ−1)^2 =4 ★

$${Let}\:\:\:{s}=\mathrm{2}{p}\sqrt{\mathrm{3}} \\ $$$${A}\left({a},\:\mathrm{0}\right) \\ $$$$\frac{{a}}{{p}}=\mathrm{3cos}\:\theta\left\{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta−\mathrm{tan}\:\theta\sqrt{\mathrm{tan}\:^{\mathrm{2}} \theta−\frac{\mathrm{1}}{\mathrm{3}}}\:\right\} \\ $$$${and}\:\theta\:{is}\:{obtained}\:{from} \\ $$$$\mathrm{tan}\:\theta+\sqrt{\mathrm{tan}\:^{\mathrm{2}} \theta−\frac{\mathrm{1}}{\mathrm{3}}}=\mathrm{2}{p}\mathrm{sin}\:\theta\mathrm{tan}\:\theta \\ $$$${or} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \theta}+\mathrm{3}\left(\mathrm{2}{p}\mathrm{sin}\:\theta−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\:\:\:\:\:\:\:\bigstar \\ $$

Commented by ajfour last updated on 06/Sep/20

Commented by ajfour last updated on 06/Sep/20

Commented by ajfour last updated on 06/Sep/20

p=1, s=2(√3) , a=2(√3) , θ=30°

$${p}=\mathrm{1},\:{s}=\mathrm{2}\sqrt{\mathrm{3}}\:,\:{a}=\mathrm{2}\sqrt{\mathrm{3}}\:,\:\theta=\mathrm{30}° \\ $$

Commented by mr W last updated on 06/Sep/20

nice solution sir!

$${nice}\:{solution}\:{sir}! \\ $$

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