Question-90499

Question Number 90499 by ajfour last updated on 24/Apr/20

Commented by ajfour last updated on 24/Apr/20

★Similar to a previous question. If parabola is y=x^2 , and △ABC is equilateral, find eq. of circle.

$$\bigstar{Similar}\:{to}\:{a}\:{previous}\:{question}. \\ $$$${If}\:{parabola}\:{is}\:{y}={x}^{\mathrm{2}} ,\:{and}\:\bigtriangleup{ABC} \\ $$$${is}\:{equilateral},\:{find}\:{eq}.\:{of}\:{circle}. \\ $$

Commented by mr W last updated on 24/Apr/20

$${for}\:{every}\:{side}\:{length}\:{s}\:{we}\:{get}\:{a}\:{circle}, \\ $$$${so}\:{no}\:{unique}\:{solution}. \\ $$

Commented by ajfour last updated on 24/Apr/20

$${then}\:{we}\:{can}\:{try}\:{choosing}\:{the} \\ $$$${circle}\:{that}\:{passes}\:{through}\:{the} \\ $$$${origin},\:{O}\:{even}. \\ $$$${Sir},\:{do}\:{you}\:{know}\:{the}\:{Q}.{no}\:{of}\:{the} \\ $$$${previous}\:{similar}\:{question}? \\ $$

Commented by mr W last updated on 24/Apr/20

$$\mathrm{89281},\:\mathrm{62938} \\ $$

Commented by ajfour last updated on 24/Apr/20

$${thanks},\:{sir}. \\ $$

Commented by mr W last updated on 25/Apr/20

your solution is 62938 is a milestone for solving all these questions!

$${your}\:{solution}\:{is}\:\mathrm{62938}\:{is}\:{a}\:{milestone} \\ $$$${for}\:{solving}\:{all}\:{these}\:{questions}! \\ $$

Answered by mr W last updated on 24/Apr/20

s=side length of ABC (h,k)=center of ABC h=(√((s^2 /(48))−(1/4))) k=((3s^2 )/(16))−(1/4) x_i =2(√(−p)) sin ((1/3) sin^(−1) (q/( (√(−p^3 ))))+((2iπ)/3))+h i=0,1,2 for C,A,B with p=((h^2 −k)/2) q=((h(5h^2 −3k))/4)+((3h^2 +3k^2 −s^2 )/(18h)) x_C =2(√(−p)) sin ((1/3) sin^(−1) (q/( (√(−p^3 )))))+h x_C ^2 (1+x_C ^2 )=s^2 ⇒s=8.0429

$${s}={side}\:{length}\:{of}\:{ABC} \\ $$$$\left({h},{k}\right)={center}\:{of}\:{ABC} \\ $$$${h}=\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{48}}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${k}=\frac{\mathrm{3}{s}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}_{{i}} =\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\:\sqrt{−{p}^{\mathrm{3}} }}+\frac{\mathrm{2}{i}\pi}{\mathrm{3}}\right)+{h} \\ $$$${i}=\mathrm{0},\mathrm{1},\mathrm{2}\:{for}\:{C},{A},{B} \\ $$$${with} \\ $$$${p}=\frac{{h}^{\mathrm{2}} −{k}}{\mathrm{2}} \\ $$$${q}=\frac{{h}\left(\mathrm{5}{h}^{\mathrm{2}} −\mathrm{3}{k}\right)}{\mathrm{4}}+\frac{\mathrm{3}{h}^{\mathrm{2}} +\mathrm{3}{k}^{\mathrm{2}} −{s}^{\mathrm{2}} }{\mathrm{18}{h}} \\ $$$${x}_{{C}} =\mathrm{2}\sqrt{−{p}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{{q}}{\:\sqrt{−{p}^{\mathrm{3}} }}\right)+{h} \\ $$$${x}_{{C}} ^{\mathrm{2}} \left(\mathrm{1}+{x}_{{C}} ^{\mathrm{2}} \right)={s}^{\mathrm{2}} \\ $$$$\Rightarrow{s}=\mathrm{8}.\mathrm{0429} \\ $$

Commented by mr W last updated on 24/Apr/20

Commented by ajfour last updated on 25/Apr/20

$$\mathcal{T}{hanks}\:\mathcal{S}{ir}. \\ $$

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