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Question Number 199101 by cortano12 last updated on 28/Oct/23

Answered by ajfour last updated on 28/Oct/23

Commented by ajfour last updated on 28/Oct/23

Triangle side be 2(√3)s. Rcos 30°=s(√3) ⇒ R=2 Centre C(0,1) x^2 +(y−1)^2 =4 If M be bottom end of blue line, then M(−4cos θ, 3−4sin θ) p^2 +(3+q)^2 =20 −p=−4cos θ+2sin θ −q=3−4sin θ−2cos θ paeabola axis y+q=((sin θ)/(cos θ))(x+p) for y=0 , x_A =((qcos θ)/(sin θ))−p Line through (−p,−q) ⊥ to axis y+q+((cos θ)/(sin θ))(x+p)=0 for y=0 x_D =−p−((qsin θ)/(cos θ)) now (∓s(√3)−x_D )cos θ=(x_A ±s(√3))^2 sin^2 θ say ±s(√3) =t (t+p+((qsin θ)/(cos θ)))cos θ=(((qcos θ)/(sin θ))−p−t)^2 sin^2 θ sum of roots=0 ⇒ cos θ=2sin θ(psin θ−qcos θ) ⇒ cos θ=2sin θ{3cos θ−4sin θcos θ−2cos^2 θ +4sin θcos θ−2sin^2 θ} ⇒ cos θ=2sin θ(3cos θ−2) cos^2 θ=4(1−cos^2 θ)(3cos θ−2)^2 z^2 =4(1−z^2 )(3z−2)^2 .....

$${Triangle}\:{side}\:{be}\:\mathrm{2}\sqrt{\mathrm{3}}{s}. \\ $$$${R}\mathrm{cos}\:\mathrm{30}°={s}\sqrt{\mathrm{3}}\:\:\:\Rightarrow\:\:{R}=\mathrm{2} \\ $$$${Centre}\:{C}\left(\mathrm{0},\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${If}\:{M}\:{be}\:{bottom}\:{end}\:{of}\:{blue}\:\:{line}, \\ $$$${then}\:\:{M}\left(−\mathrm{4cos}\:\theta,\:\mathrm{3}−\mathrm{4sin}\:\theta\right) \\ $$$${p}^{\mathrm{2}} +\left(\mathrm{3}+{q}\right)^{\mathrm{2}} =\mathrm{20} \\ $$$$−{p}=−\mathrm{4cos}\:\theta+\mathrm{2sin}\:\theta \\ $$$$−{q}=\mathrm{3}−\mathrm{4sin}\:\theta−\mathrm{2cos}\:\theta \\ $$$${paeabola}\:{axis} \\ $$$${y}+{q}=\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\left({x}+{p}\right) \\ $$$${for}\:{y}=\mathrm{0}\:\:,\:\:\:{x}_{{A}} =\frac{{q}\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}−{p} \\ $$$${Line}\:{through}\:\left(−{p},−{q}\right)\:\bot\:{to}\:{axis} \\ $$$${y}+{q}+\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}\left({x}+{p}\right)=\mathrm{0}\:\:\: \\ $$$${for}\:{y}=\mathrm{0}\:\:\:{x}_{{D}} =−{p}−\frac{{q}\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta} \\ $$$${now}\:\:\:\left(\mp{s}\sqrt{\mathrm{3}}−{x}_{{D}} \right)\mathrm{cos}\:\theta=\left({x}_{{A}} \pm{s}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${say}\:\:\pm{s}\sqrt{\mathrm{3}}\:={t} \\ $$$$\left({t}+{p}+\frac{{q}\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right)\mathrm{cos}\:\theta=\left(\frac{{q}\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}−{p}−{t}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${sum}\:{of}\:{roots}=\mathrm{0}\:\:\Rightarrow \\ $$$$\mathrm{cos}\:\theta=\mathrm{2sin}\:\theta\left({p}\mathrm{sin}\:\theta−{q}\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow\:\mathrm{cos}\:\theta=\mathrm{2sin}\:\theta\left\{\mathrm{3cos}\:\theta−\mathrm{4sin}\:\theta\mathrm{cos}\:\theta−\mathrm{2cos}\:^{\mathrm{2}} \theta\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4sin}\:\theta\mathrm{cos}\:\theta−\mathrm{2sin}\:^{\mathrm{2}} \theta\right\} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:\theta=\mathrm{2sin}\:\theta\left(\mathrm{3cos}\:\theta−\mathrm{2}\right) \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \theta=\mathrm{4}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta\right)\left(\mathrm{3cos}\:\theta−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}−{z}^{\mathrm{2}} \right)\left(\mathrm{3}{z}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$..... \\ $$$$ \\ $$

Answered by mr W last updated on 28/Oct/23

y=x^2 line 1: y=4+(x−2)tan α line 2: y=4+(x−2)tan (α+(π/3)) with m_1 =tan α=k m_2 =tan (α+(π/3))=((k+(√3))/(1−(√3)k)) intersection line 1: y_1 =x_1 ^2 =4+(x_1 −2) m_1 ⇒x_1 ^2 −m_1 x_1 +2m_1 −4=0 ⇒x_1 =m_1 −2 intersection line 2: y_2 =x_2 ^2 =4+(x_2 −2) m_2 ⇒x_2 =m_2 −2 s^2 =(x_1 −2)^2 +(y_1 −4)^2 =(m_1 −4)^2 (m_1 ^2 +1) s^2 =(x_2 −2)^2 +(y_2 −4)^2 =(m_2 −4)^2 (m_2 ^2 +1) (m_1 −4)^2 (m_1 ^2 +1)=(m_2 −4)^2 (m_2 ^2 +1) we get 4 roots. they show 2 suitable solutions.

$${y}={x}^{\mathrm{2}} \\ $$$${line}\:\mathrm{1}:\:{y}=\mathrm{4}+\left({x}−\mathrm{2}\right)\mathrm{tan}\:\alpha \\ $$$${line}\:\mathrm{2}:\:{y}=\mathrm{4}+\left({x}−\mathrm{2}\right)\mathrm{tan}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right) \\ $$$${with}\:{m}_{\mathrm{1}} =\mathrm{tan}\:\alpha={k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{m}_{\mathrm{2}} =\mathrm{tan}\:\left(\alpha+\frac{\pi}{\mathrm{3}}\right)=\frac{{k}+\sqrt{\mathrm{3}}}{\mathrm{1}−\sqrt{\mathrm{3}}{k}} \\ $$$${intersection}\:{line}\:\mathrm{1}: \\ $$$${y}_{\mathrm{1}} ={x}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{4}+\left({x}_{\mathrm{1}} −\mathrm{2}\right)\:{m}_{\mathrm{1}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} ^{\mathrm{2}} −{m}_{\mathrm{1}} {x}_{\mathrm{1}} +\mathrm{2}{m}_{\mathrm{1}} −\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{1}} ={m}_{\mathrm{1}} −\mathrm{2} \\ $$$${intersection}\:{line}\:\mathrm{2}: \\ $$$${y}_{\mathrm{2}} ={x}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{4}+\left({x}_{\mathrm{2}} −\mathrm{2}\right)\:{m}_{\mathrm{2}} \\ $$$$\Rightarrow{x}_{\mathrm{2}} ={m}_{\mathrm{2}} −\mathrm{2} \\ $$$$ \\ $$$${s}^{\mathrm{2}} =\left({x}_{\mathrm{1}} −\mathrm{2}\right)^{\mathrm{2}} +\left({y}_{\mathrm{1}} −\mathrm{4}\right)^{\mathrm{2}} =\left({m}_{\mathrm{1}} −\mathrm{4}\right)^{\mathrm{2}} \left({m}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${s}^{\mathrm{2}} =\left({x}_{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} +\left({y}_{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} =\left({m}_{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} \left({m}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\left({m}_{\mathrm{1}} −\mathrm{4}\right)^{\mathrm{2}} \left({m}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}\right)=\left({m}_{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} \left({m}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${we}\:{get}\:\mathrm{4}\:{roots}.\:{they}\:{show}\:\mathrm{2}\:{suitable}\: \\ $$$${solutions}. \\ $$

Commented by mr W last updated on 28/Oct/23

Commented by cortano12 last updated on 28/Oct/23

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 28/Oct/23

Alternative method: acc. to the solution from ajfour sir in Q62938, for a given side length s of the equilateral triangle, its vertexes satisfy following equation: x^3 −3hx^2 +px+q=0 with h=((√(s^2 −12))/(4(√3))), k= ((3s^2 )/(16))−(1/4), p=((9h^2 −3k)/2), q=((h^2 +k^2 −s^2 /3)/(3h)) now it is given x=2, reversely we can get s.

$$\underline{{Alternative}\:{method}:} \\ $$$${acc}.\:{to}\:{the}\:{solution}\:{from}\:\boldsymbol{{ajfour}}\:{sir} \\ $$$${in}\:{Q}\mathrm{62938},\:{for}\:{a}\:{given}\:{side}\:{length}\:\boldsymbol{{s}}\: \\ $$$${of}\:{the}\:{equilateral}\:{triangle},\:{its}\: \\ $$$${vertexes}\:{satisfy}\:{following}\:{equation}: \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{hx}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\: \\ $$$${with}\:{h}=\frac{\sqrt{\boldsymbol{{s}}^{\mathrm{2}} −\mathrm{12}}}{\mathrm{4}\sqrt{\mathrm{3}}},\:{k}=\:\frac{\mathrm{3}\boldsymbol{{s}}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{4}}, \\ $$$${p}=\frac{\mathrm{9}{h}^{\mathrm{2}} −\mathrm{3}{k}}{\mathrm{2}},\:{q}=\frac{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} −\boldsymbol{{s}}^{\mathrm{2}} /\mathrm{3}}{\mathrm{3}{h}} \\ $$$${now}\:{it}\:{is}\:{given}\:{x}=\mathrm{2},\:{reversely}\:{we} \\ $$$${can}\:{get}\:{s}. \\ $$

Commented by mr W last updated on 28/Oct/23

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