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Question Number 141227 by ajfour last updated on 16/May/21

Commented by ajfour last updated on 16/May/21

$T h e c u b i c c u r v e i s y = x^{3} - x - c$ $F i n d r a d i i R a n d s w i t h o u t$ $h a v i n g t o f i n d α, β, γ .$
Commented by MJS_new last updated on 17/May/21

$just a thought : if we let$ $c = a (a^{2} - 1) \land a > 0$ $the equation gets$ $x^{3} - x - a (a^{2} - 1) = 0$ $with the solutions$ $α = a$ $β = - \frac{a}{2} + \frac{\sqrt{4 - 3 a^{2}}}{2}$ $γ = - \frac{a}{2} - \frac{\sqrt{4 - 3 a^{2}}}{2}$ $and the equation of the line connecting$ $(\begin{matrix} α \\ 0 \end{matrix}) and (\begin{matrix} 0 \\ - c \end{matrix}) is$ $(a^{2} - 1) x + y - a (a^{2} - 1) = 0$ $\Rightarrow$ $R = \frac{∣ (a^{2} - 1) (3 a + \sqrt{4 - 3 a^{2}}) ∣}{2 \sqrt{a^{4} - 2 a^{2} + 2}}$ $r = \frac{∣ (a^{2} - 1) (3 a - \sqrt{4 - 3 a^{2}}) ∣}{2 \sqrt{a^{4} - 2 a^{2} + 2}}$ $of course the original problem is not solved$ $because c = a (a^{2} - 1) \Leftrightarrow a^{3} - a - c = 0 so finding$ $a for a given c leads back to the start \infty$
	Answered by ajfour last updated on 17/May/21

	Commented by ajfour last updated on 17/May/21
	$α=(√(p^2 −c^2 )) β=α−q γ=α−r α+β+γ=0 αβ+βγ+γα=−1 αβ γ= c (q/r)=(s/R) (c/p)=(s/q)=(R/r) ⇒ c+s+R=0 let (q/r)=(s/R)=k ⇒ q=kr 2α=q+r−(√(p^2 −c^2 )) 2β= r−(√(p^2 −c^2 )) 2γ= q−(√(p^2 −c^2 )) ⇒ q+r=(3/2)(√(p^2 −c^2 )) ...(i) (q+r)^2 +2(p^2 −c^2 ) −3(√(p^2 −c^2 )) (q+r) +qr+(p^2 −c^2 ) −(√(p^2 −c^2 ))(q+r)=−4 ⇒ (q+r)^2 +3(p^2 −c^2 ) −4(√(p^2 −c^2 ))(q+r)+4+qr=0 .....(ii) (q+r−(√(p^2 −c^2 ))) ×{qr+(p^2 −c^2 )−(√(p^2 −c^2 ))(q+r)} =8c ....(iii) simplifying (ii) −(3/4)(p^2 −c^2 )+4+qr = 0 ..(I) now taking up (iii) −(5/2)(√(p^2 −c^2 ))(qr−((√(p^2 −c^2 ))/2)) =8c using (I) in here→ −(5/2)((2/( (√3))))(√(4+qr))(qr−((√(4+qr))/( (√3)))) = 8c let qr=4tan^2 θ ⇒ sec θ(tan^2 θ−(1/( 2(√3)))sec θ) +(((√3)c)/5) = 0 AB=−(((√3)c)/5) B=A^2 −1−(A/(2(√3))) ⇒ (((√3)c)/(5A))+(A/(2(√3)))=A(A−(1/A)) ⇒ A=(((A/(2(√3)))+(((√3)c)/(5A)))/(A−(1/A))) = (((A^2 /(2(√3)))+(((√3)c)/5))/(A^2 −1)) A^2 =(((A^4 /(12))+(A^2 /(10))+((3c^2 )/(25)))/(A^4 −2A^2 +1)) ⇒ 60A^2 (A^2 −1)^2 =5(A^2 +(3/5))^2 −(9/5)+((36c^2 )/5) let A^2 = t+b ⇒ 60(t+b)(t+b−1)^2 = 5(t+b+(3/5))^2 −(9/5)+((36c^2 )/5) 60{t^3 +b(4b−3)t^2 + [(b−1)^2 +2b(b−2)]t+b(b−1)^2 } = 5t^2 +10(b+(3/5))t +5(b+(3/5))^2 −(9/5)+((36c^2 )/5) ⇒ 60t^3 +5(48b^2 −36b−1)t^2 +2(90b^2 −185b+27)t +(60b^3 −125b^2 +54b+((36c^2 )/5)) = 0 .....(★) let 5(48b^2 −36b−1)^2 =72(90b^2 −185b+27) Nice eq. I saw the exact answer and ran away asap..! otherwise let→ 48b^2 −36b−1=0 ⇒ 3(4b)^2 −9(4b)−1=0 4b=(3/2)±(√((9/4)+(1/3))) b=((6±(√(62)))/(16)) Now P =90b^2 −185b+27 = 540b+15−185b+27 =355b+42 The positive b=((6+(√(62)))/(16)) should lead to the answer.. (3/4)α^2 =4+qr ∀ qr>−4 qr= 4(A^2 −1) =4(t+b−1) ⇒ t >−b and i checked that for chosen b, t>0>−b b is known, we get t from(★)$
	$α = \sqrt{p^{2} - c^{2}}$ $β = α - q$ $γ = α - r$ $α + β + γ = 0$ $α β + β γ + γ α = - 1$ $α β γ = c$ $\frac{q}{r} = \frac{s}{R}$ $\frac{c}{p} = \frac{s}{q} = \frac{R}{r}$ $\Rightarrow c + s + R = 0$ $l e t \frac{q}{r} = \frac{s}{R} = k$ $\Rightarrow q = k r$ $2 α = q + r - \sqrt{p^{2} - c^{2}}$ $2 β = r - \sqrt{p^{2} - c^{2}}$ $2 γ = q - \sqrt{p^{2} - c^{2}}$ $\Rightarrow q + r = \frac{3}{2} \sqrt{p^{2} - c^{2}} . . . (i)$ ${(q + r)}^{2} + 2 (p^{2} - c^{2})$ $- 3 \sqrt{p^{2} - c^{2}} (q + r)$ $+ q r + (p^{2} - c^{2})$ $- \sqrt{p^{2} - c^{2}} (q + r) = - 4$ $\Rightarrow$ ${(q + r)}^{2} + 3 (p^{2} - c^{2})$ $- 4 \sqrt{p^{2} - c^{2}} (q + r) + 4 + q r = 0$ $. . . . . (i i)$ $(q + r - \sqrt{p^{2} - c^{2}})$ $\times {q r + (p^{2} - c^{2}) - \sqrt{p^{2} - c^{2}} (q + r)}$ $= 8 c$ $. . . . (i i i)$ $s i m p l i f y i n g (i i)$ $- \frac{3}{4} (p^{2} - c^{2}) + 4 + q r = 0 . . (I)$ $n o w t a k i n g u p (i i i)$ $- \frac{5}{2} \sqrt{p^{2} - c^{2}} (q r - \frac{\sqrt{p^{2} - c^{2}}}{2}) = 8 c$ $u s i n g (I) i n h e r e \to$ $- \frac{5}{2} (\frac{2}{\sqrt{3}}) \sqrt{4 + q r} (q r - \frac{\sqrt{4 + q r}}{\sqrt{3}})$ $= 8 c$ $l e t q r = 4 \tan^{2} θ \Rightarrow$ $\sec θ (\tan^{2} θ - \frac{1}{2 \sqrt{3}} \sec θ)$ $+ \frac{\sqrt{3} c}{5} = 0$ $A B = - \frac{\sqrt{3} c}{5}$ $B = A^{2} - 1 - \frac{A}{2 \sqrt{3}}$ $\Rightarrow \frac{\sqrt{3} c}{5 A} + \frac{A}{2 \sqrt{3}} = A (A - \frac{1}{A})$ $\Rightarrow A = \frac{\frac{A}{2 \sqrt{3}} + \frac{\sqrt{3} c}{5 A}}{A - \frac{1}{A}} = \frac{\frac{A^{2}}{2 \sqrt{3}} + \frac{\sqrt{3} c}{5}}{A^{2} - 1}$ $A^{2} = \frac{\frac{A^{4}}{12} + \frac{A^{2}}{10} + \frac{3 c^{2}}{25}}{A^{4} - 2 A^{2} + 1}$ $\Rightarrow 60 A^{2} {(A^{2} - 1)}^{2}$ $= 5 {(A^{2} + \frac{3}{5})}^{2} - \frac{9}{5} + \frac{36 c^{2}}{5}$ $l e t A^{2} = t + b \Rightarrow$ $60 (t + b) {(t + b - 1)}^{2}$ $= 5 {(t + b + \frac{3}{5})}^{2} - \frac{9}{5} + \frac{36 c^{2}}{5}$ $60 {t^{3} + b (4 b - 3) t^{2} +$ $[{(b - 1)}^{2} + 2 b (b - 2)] t + b {(b - 1)}^{2}}$ $= 5 t^{2} + 10 (b + \frac{3}{5}) t$ $+ 5 {(b + \frac{3}{5})}^{2} - \frac{9}{5} + \frac{36 c^{2}}{5}$ $\Rightarrow$ $60 t^{3} + 5 (48 b^{2} - 36 b - 1) t^{2}$ $+ 2 (90 b^{2} - 185 b + 27) t$ $+ (60 b^{3} - 125 b^{2} + 54 b + \frac{36 c^{2}}{5})$ $= 0 . . . . . (★)$ $l e t 5 {(48 b^{2} - 36 b - 1)}^{2}$ $= 72 (90 b^{2} - 185 b + 27)$ $N i c e e q . I s a w t h e e x a c t a n s w e r$ $a n d r a n a w a y a s a p . .!$ $o t h e r w i s e l e t \to$ $48 b^{2} - 36 b - 1 = 0$ $\Rightarrow 3 {(4 b)}^{2} - 9 (4 b) - 1 = 0$ $4 b = \frac{3}{2} \pm \sqrt{\frac{9}{4} + \frac{1}{3}}$ $b = \frac{6 \pm \sqrt{62}}{16}$ $N o w$ $P = 90 b^{2} - 185 b + 27$ $= 540 b + 15 - 185 b + 27$ $= 355 b + 42$ $T h e p o s i t i v e b = \frac{6 + \sqrt{62}}{16}$ $s h o u l d l e a d t o t h e a n s w e r . .$ $\frac{3}{4} α^{2} = 4 + q r \forall q r > - 4$ $q r = 4 (A^{2} - 1)$ $= 4 (t + b - 1)$ $\Rightarrow t > - b$ $a n d i c h e c k e d t h a t f o r$ $c h o s e n b, t > 0 > - b$ $b i s k n o w n, w e g e t t f r o m (★)$
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