Let consider α : I→R^2 a parametric curve defined as ∀ t∈I α(t)=(((t^2 −1)/(t^3 −1)) ,((2t)/(t^3 −1))) Prove that for a,b,c∈I α(a),α(b),α(c) are on the same lign iff abc=a+b+c+1


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Question Number 74280 by ~blr237~ last updated on 21/Nov/19

$L e t c o n s i d e r α : I \to R^{2} a p a r a m e t r i c c u r v e d e f i n e d a s$ $\forall t \in I α (t) = (\frac{t^{2} - 1}{t^{3} - 1}, \frac{2 t}{t^{3} - 1})$ $P r o v e t h a t f o r a, b, c \in I$ $α (a), α (b), α (c) a r e o n t h e s a m e l i g n i f f a b c = a + b + c + 1$
Commented by MJS last updated on 21/Nov/19

$test$ $a = 3 b = 5 \Rightarrow c = \frac{9}{14}$ $α (3) = (\begin{matrix} \frac{4}{13} \\ \frac{3}{13} \end{matrix}) α (5) = (\begin{matrix} \frac{6}{31} \\ \frac{5}{62} \end{matrix}) α (\frac{9}{14}) = (\begin{matrix} \frac{322}{403} \\ - \frac{3528}{2015} \end{matrix})$ $line through the 1^{st} and 2^{nd}$ $y = \frac{121}{92} x - \frac{4}{23}$ $the 3^{rd} is not on this line$ $\Rightarrow wrong$
Commented by ~blr237~ last updated on 21/Nov/19

$p l e a s e s i r i c o r r e c t l y c o p i e d t h e e x e r c i s e . W h a t o n t h a t t e s t p r o v e t h a t t h e q u e z t i o n i z w r o n g$
Commented by MJS last updated on 21/Nov/19

$if {there}^{'} s at least one counter example the$ $question is wrong$
Commented by ~blr237~ last updated on 21/Nov/19

$t h a n k s y o u s i r, i h a v e s e e n t h e p r o b l e m$
Commented by MJS last updated on 21/Nov/19

$whenever an example like this one looks$ $complicated to transform, I make a few$ $tests to ensure {it}^{'} s worth the effort$
	Answered by mind is power last updated on 21/Nov/19

	$f a l s e t h i s i s a b c = 1 - a - b - c$
	Commented by ~blr237~ last updated on 21/Nov/19

	$O k a y s i r h a v e y o u p r o v e d i t ?$
	Answered by mind is power last updated on 21/Nov/19
	$let α(a)=(((a^2 −1)/(a^3 −1)),((2a)/(a^3 −1))),α(b)=(((b^2 −1)/(b^3 −1)),((2b)/(b^3 −1))),α(c)=(((c^2 −1)/(c^3 −1)),((2c)/(c^3 −1))) α(a),α(b),α(c) in same lign ⇔ ((((2a)/(a^3 −1))−((2b)/(b^3 −1)))/(((a^2 −1)/(a^3 −1))−((b^2 −1)/(b^3 −1))))=((((2a)/(a^3 −1))−((2c)/(c^3 −1)))/(((a^2 −1)/(a^3 −1))−((2c)/(c^2 −1)))) ⇔((2a(b^3 −1)−2b(a^3 −1))/((b^3 −1)(a^2 −1)−(a^3 −1)(b^2 −1)))=((2a(c^2 −1)−2c(a^2 −1))/((c^3 −1)(a^2 −1)−(c^2 −1)(a^3 −1)))..H we will worck withe one expression the other one is just (a,b)→(a,c) 2a(b^3 −1)−2b(a^3 −1)=(b−a)(2ab(b+a)+2) (b^3 −1)(a^2 −1)−(a^3 −1)(b^2 −1)=(a−1)(b−1){(a+1)(b^2 +b+1)−(b+1)(a^2 +a+1)} =(a−1)(b−1){ab^2 −ba^2 +b^2 −a^2 }=(a−1)(b−1)(b−a){ab+b+a} H ⇔((ab(b+a)+1)/((a−1)(b−1)(ab+a+b)))=((ac(c+a)+1)/((a−1)(c−1)(ac+a+c))) ⇔((ab(b+a)+1)/((b−1)(ab+a+b)))=((ac(c+a)+1)/((c−1)(ac+a+c))) (b−1)(ab+a+b)=ab^2 +b^2 −a−b ⇔(ab(a+b)+1)(ac^2 +c^2 −c−a)=(ca(c+a)+1)(ab^2 +b^2 −a−b) {a^2 b+ab^2 +1}(ac^2 +c^2 −c−a) =a^3 bc^2 +a^2 bc^2 −a^2 bc−a^3 b+a^2 b^2 c^2 +ac^2 b^2 −acb^2 −a^2 b^2 +ac^2 +c^2 −c−a (c^2 a+ca^2 +1)(ab^2 +b^2 −a−b)=a^2 b^2 c^2 +c^2 ab^2 −c^2 a^2 −bac^2 +ca^3 b^2 +ca^2 b^2 −ca^3 −cba^2 +ab^2 +b^2 −a−b (ab(a+b)+1)(ac^2 +c^2 −c−a)−(ca(c+a)+1)(ab^2 +b^2 −a−b)=0 ⇔ a^3 bc(c−b) +a^3 (c−b)+acb(c−b)+a^2 bc(c−b)+a^2 (c^2 −b^2 )+a(c^2 −b^2 )+c^2 −b^2 −(c−b)}=0 ⇔(c−b){a^3 bc+a^3 +acb+a^2 (c+b)+a^2 bc(c−b)+a(c+b)+c+b−1}=0 ⇔a^3 bc+a^3 +acb+a^2 bc+a^2 c+a^2 b+ac+ab+c+b−1=0 ⇔a^3 −1+abc(a^2 +a+1)+c(a^2 +a+1)+b(a^2 +a+1)=0 a^3 −1=(a−1)(a^2 +a+1) ⇒(a^2 +a+1)(a−1+abc+c+b)=0 ∀a∈R a^2 +a+1>0 ⇒a−1+abc+b+c=0 ⇔abc=1−a−b−c$
	$l e t α (a) = (\frac{a^{2} - 1}{a^{3} - 1}, \frac{2 a}{a^{3} - 1}), α (b) = (\frac{b^{2} - 1}{b^{3} - 1}, \frac{2 b}{b^{3} - 1}), α (c) = (\frac{c^{2} - 1}{c^{3} - 1}, \frac{2 c}{c^{3} - 1})$ $α (a), α (b), α (c) i n s a m e l i g n$ $\Leftrightarrow$ $\frac{\frac{2 a}{a^{3} - 1} - \frac{2 b}{b^{3} - 1}}{\frac{a^{2} - 1}{a^{3} - 1} - \frac{b^{2} - 1}{b^{3} - 1}} = \frac{\frac{2 a}{a^{3} - 1} - \frac{2 c}{c^{3} - 1}}{\frac{a^{2} - 1}{a^{3} - 1} - \frac{2 c}{c^{2} - 1}}$ $\Leftrightarrow \frac{2 a (b^{3} - 1) - 2 b (a^{3} - 1)}{(b^{3} - 1) (a^{2} - 1) - (a^{3} - 1) (b^{2} - 1)} = \frac{2 a (c^{2} - 1) - 2 c (a^{2} - 1)}{(c^{3} - 1) (a^{2} - 1) - (c^{2} - 1) (a^{3} - 1)} . . H$ $w e w i l l w o r c k w i t h e o n e e x p r e s s i o n t h e o t h e r o n e i s j u s t$ $(a, b) \to (a, c)$ $2 a (b^{3} - 1) - 2 b (a^{3} - 1) = (b - a) (2 a b (b + a) + 2)$ $(b^{3} - 1) (a^{2} - 1) - (a^{3} - 1) (b^{2} - 1) = (a - 1) (b - 1) {(a + 1) (b^{2} + b + 1) - (b + 1) (a^{2} + a + 1)}$ $= (a - 1) (b - 1) {{a b}^{2} - {b a}^{2} + b^{2} - a^{2}} = (a - 1) (b - 1) (b - a) {a b + b + a}$ $H$ $\Leftrightarrow \frac{a b (b + a) + 1}{(a - 1) (b - 1) (a b + a + b)} = \frac{a c (c + a) + 1}{(a - 1) (c - 1) (a c + a + c)}$ $\Leftrightarrow \frac{a b (b + a) + 1}{(b - 1) (a b + a + b)} = \frac{a c (c + a) + 1}{(c - 1) (a c + a + c)}$ $(b - 1) (a b + a + b) = {a b}^{2} + b^{2} - a - b$ $\Leftrightarrow (a b (a + b) + 1) ({a c}^{2} + c^{2} - c - a) = (c a (c + a) + 1) ({a b}^{2} + b^{2} - a - b)$ ${a^{2} b + {a b}^{2} + 1} ({a c}^{2} + c^{2} - c - a)$ $= a^{3} {b c}^{2} + a^{2} {b c}^{2} - a^{2} b c - a^{3} b + a^{2} b^{2} c^{2} + {a c}^{2} b^{2} - {a c b}^{2} - a^{2} b^{2} + {a c}^{2} + c^{2} - c - a$ $(c^{2} a + {c a}^{2} + 1) ({a b}^{2} + b^{2} - a - b) = a^{2} b^{2} c^{2} + c^{2} {a b}^{2} - c^{2} a^{2} - {b a c}^{2} + {c a}^{3} b^{2} + {c a}^{2} b^{2}$ $- {c a}^{3} - {c b a}^{2} + {a b}^{2} + b^{2} - a - b$ $(a b (a + b) + 1) ({a c}^{2} + c^{2} - c - a) - (c a (c + a) + 1) ({a b}^{2} + b^{2} - a - b) = 0$ $\Leftrightarrow$ $a^{3} b c (c - b) + a^{3} (c - b) + a c b (c - b) + a^{2} b c (c - b) + a^{2} (c^{2} - b^{2}) + a (c^{2} - b^{2}) + c^{2} - b^{2} - (c - b)} = 0$ $\Leftrightarrow (c - b) {a^{3} b c + a^{3} + a c b + a^{2} (c + b) + a^{2} b c (c - b) + a (c + b) + c + b - 1} = 0$ $\Leftrightarrow a^{3} b c + a^{3} + a c b + a^{2} b c + a^{2} c + a^{2} b + a c + a b + c + b - 1 = 0$ $\Leftrightarrow a^{3} - 1 + a b c (a^{2} + a + 1) + c (a^{2} + a + 1) + b (a^{2} + a + 1) = 0$ $a^{3} - 1 = (a - 1) (a^{2} + a + 1)$ $\Rightarrow (a^{2} + a + 1) (a - 1 + a b c + c + b) = 0$ $\forall a \in R a^{2} + a + 1 > 0$ $\Rightarrow a - 1 + a b c + b + c = 0$ $\Leftrightarrow a b c = 1 - a - b - c$
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