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Question Number 74280 by ~blr237~ last updated on 21/Nov/19
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
$${Let}\:\:{consider}\:\alpha\::\:{I}\rightarrow\mathbb{R}^{\mathrm{2}} \:\:{a}\:{parametric}\:{curve}\:{defined}\:{as} \\ $$$$\forall\:{t}\in{I}\:\:\:\alpha\left({t}\right)=\left(\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{3}} −\mathrm{1}}\:,\frac{\mathrm{2}{t}}{{t}^{\mathrm{3}} −\mathrm{1}}\right)\: \\ $$$${Prove}\:{that}\:{for}\:{a},{b},{c}\in{I}\:\:\: \\ $$$$\:\:\alpha\left({a}\right),\alpha\left({b}\right),\alpha\left({c}\right)\:{are}\:{on}\:{the}\:{same}\:{lign}\:{iff}\:\:{abc}={a}+{b}+{c}+\mathrm{1} \\ $$
Commented by MJS last updated on 21/Nov/19
test a=3 b=5 ⇒ c=(9/(14)) α(3)= (((4/(13))),((3/(13))) ) α(5)= (((6/(31))),((5/(62))) ) α((9/(14)))= ((((322)/(403))),((−((3528)/(2015)))) ) line through the 1^(st) and 2^(nd) y=((121)/(92))x−(4/(23)) the 3^(rd) is not on this line ⇒ wrong
$$\mathrm{test} \\ $$$${a}=\mathrm{3}\:\:{b}=\mathrm{5}\:\:\Rightarrow\:{c}=\frac{\mathrm{9}}{\mathrm{14}} \\ $$$$\alpha\left(\mathrm{3}\right)=\begin{pmatrix}{\frac{\mathrm{4}}{\mathrm{13}}}\\{\frac{\mathrm{3}}{\mathrm{13}}}\end{pmatrix}\:\:\alpha\left(\mathrm{5}\right)=\begin{pmatrix}{\frac{\mathrm{6}}{\mathrm{31}}}\\{\frac{\mathrm{5}}{\mathrm{62}}}\end{pmatrix}\:\:\alpha\left(\frac{\mathrm{9}}{\mathrm{14}}\right)=\begin{pmatrix}{\frac{\mathrm{322}}{\mathrm{403}}}\\{−\frac{\mathrm{3528}}{\mathrm{2015}}}\end{pmatrix} \\ $$$$\mathrm{line}\:\mathrm{through}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{and}\:\mathrm{2}^{\mathrm{nd}} \\ $$$${y}=\frac{\mathrm{121}}{\mathrm{92}}{x}−\frac{\mathrm{4}}{\mathrm{23}} \\ $$$$\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{is}\:\mathrm{not}\:\mathrm{on}\:\mathrm{this}\:\mathrm{line} \\ $$$$\Rightarrow\:\mathrm{wrong} \\ $$
Commented by ~blr237~ last updated on 21/Nov/19
please sir i correctly copied the exercise.What on that test prove that the queztion iz wrong
$${please}\:{sir}\:{i}\:{correctly}\:{copied}\:{the}\:{exercise}.{What}\:{on}\:{that}\:{test}\:{prove}\:{that}\:{the}\:{queztion}\:{iz}\:{wrong} \\ $$
Commented by MJS last updated on 21/Nov/19
if there′s at least one counter example the question is wrong
$$\mathrm{if}\:\mathrm{there}'\mathrm{s}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{counter}\:\mathrm{example}\:\mathrm{the} \\ $$$$\mathrm{question}\:\mathrm{is}\:\mathrm{wrong} \\ $$
Commented by ~blr237~ last updated on 21/Nov/19
thanks you sir , i have seen the problem
$${thanks}\:{you}\:{sir}\:,\:{i}\:{have}\:{seen}\:{the}\:{problem} \\ $$
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
$$\mathrm{whenever}\:\mathrm{an}\:\mathrm{example}\:\mathrm{like}\:\mathrm{this}\:\mathrm{one}\:\mathrm{looks} \\ $$$$\mathrm{complicated}\:\mathrm{to}\:\mathrm{transform},\:\mathrm{I}\:\mathrm{make}\:\mathrm{a}\:\mathrm{few} \\ $$$$\mathrm{tests}\:\mathrm{to}\:\mathrm{ensure}\:\mathrm{it}'\mathrm{s}\:\mathrm{worth}\:\mathrm{the}\:\mathrm{effort} \\ $$
Answered by mind is power last updated on 21/Nov/19
false this is abc=1−a−b−c
$${false}\:{this}\:{is}\:{abc}=\mathrm{1}−{a}−{b}−{c} \\ $$$$ \\ $$
Commented by ~blr237~ last updated on 21/Nov/19
Okay sir have you proved it?
$${Okay}\:{sir}\:\:{have}\:{you}\:{proved}\:{it}? \\ $$
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
$${let}\:\alpha\left({a}\right)=\left(\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}^{\mathrm{3}} −\mathrm{1}},\frac{\mathrm{2}{a}}{{a}^{\mathrm{3}} −\mathrm{1}}\right),\alpha\left({b}\right)=\left(\frac{{b}^{\mathrm{2}} −\mathrm{1}}{{b}^{\mathrm{3}} −\mathrm{1}},\frac{\mathrm{2}{b}}{{b}^{\mathrm{3}} −\mathrm{1}}\right),\alpha\left({c}\right)=\left(\frac{{c}^{\mathrm{2}} −\mathrm{1}}{{c}^{\mathrm{3}} −\mathrm{1}},\frac{\mathrm{2}{c}}{{c}^{\mathrm{3}} −\mathrm{1}}\right) \\ $$$$\alpha\left({a}\right),\alpha\left({b}\right),\alpha\left({c}\right)\:{in}\:{same}\:{lign} \\ $$$$\Leftrightarrow \\ $$$$\frac{\frac{\mathrm{2}{a}}{{a}^{\mathrm{3}} −\mathrm{1}}−\frac{\mathrm{2}{b}}{{b}^{\mathrm{3}} −\mathrm{1}}}{\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}^{\mathrm{3}} −\mathrm{1}}−\frac{{b}^{\mathrm{2}} −\mathrm{1}}{{b}^{\mathrm{3}} −\mathrm{1}}}=\frac{\frac{\mathrm{2}{a}}{{a}^{\mathrm{3}} −\mathrm{1}}−\frac{\mathrm{2}{c}}{{c}^{\mathrm{3}} −\mathrm{1}}}{\frac{{a}^{\mathrm{2}} −\mathrm{1}}{{a}^{\mathrm{3}} −\mathrm{1}}−\frac{\mathrm{2}{c}}{{c}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\Leftrightarrow\frac{\mathrm{2}{a}\left({b}^{\mathrm{3}} −\mathrm{1}\right)−\mathrm{2}{b}\left({a}^{\mathrm{3}} −\mathrm{1}\right)}{\left({b}^{\mathrm{3}} −\mathrm{1}\right)\left({a}^{\mathrm{2}} −\mathrm{1}\right)−\left({a}^{\mathrm{3}} −\mathrm{1}\right)\left({b}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{\mathrm{2}{a}\left({c}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{2}{c}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}{\left({c}^{\mathrm{3}} −\mathrm{1}\right)\left({a}^{\mathrm{2}} −\mathrm{1}\right)−\left({c}^{\mathrm{2}} −\mathrm{1}\right)\left({a}^{\mathrm{3}} −\mathrm{1}\right)}..{H} \\ $$$${we}\:{will}\:{worck}\:{withe}\:{one}\:{expression}\:{the}\:{other}\:{one}\:{is}\:{just} \\ $$$$\left({a},{b}\right)\rightarrow\left({a},{c}\right) \\ $$$$\mathrm{2}{a}\left({b}^{\mathrm{3}} −\mathrm{1}\right)−\mathrm{2}{b}\left({a}^{\mathrm{3}} −\mathrm{1}\right)=\left({b}−{a}\right)\left(\mathrm{2}{ab}\left({b}+{a}\right)+\mathrm{2}\right) \\ $$$$\left({b}^{\mathrm{3}} −\mathrm{1}\right)\left({a}^{\mathrm{2}} −\mathrm{1}\right)−\left({a}^{\mathrm{3}} −\mathrm{1}\right)\left({b}^{\mathrm{2}} −\mathrm{1}\right)=\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left\{\left({a}+\mathrm{1}\right)\left({b}^{\mathrm{2}} +{b}+\mathrm{1}\right)−\left({b}+\mathrm{1}\right)\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)\right\} \\ $$$$=\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left\{{ab}^{\mathrm{2}} −{ba}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right\}=\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left({b}−{a}\right)\left\{{ab}+{b}+{a}\right\} \\ $$$${H} \\ $$$$\Leftrightarrow\frac{{ab}\left({b}+{a}\right)+\mathrm{1}}{\left({a}−\mathrm{1}\right)\left({b}−\mathrm{1}\right)\left({ab}+{a}+{b}\right)}=\frac{{ac}\left({c}+{a}\right)+\mathrm{1}}{\left({a}−\mathrm{1}\right)\left({c}−\mathrm{1}\right)\left({ac}+{a}+{c}\right)} \\ $$$$\Leftrightarrow\frac{{ab}\left({b}+{a}\right)+\mathrm{1}}{\left({b}−\mathrm{1}\right)\left({ab}+{a}+{b}\right)}=\frac{{ac}\left({c}+{a}\right)+\mathrm{1}}{\left({c}−\mathrm{1}\right)\left({ac}+{a}+{c}\right)} \\ $$$$\left({b}−\mathrm{1}\right)\left({ab}+{a}+{b}\right)={ab}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}−{b} \\ $$$$\Leftrightarrow\left({ab}\left({a}+{b}\right)+\mathrm{1}\right)\left({ac}^{\mathrm{2}} +{c}^{\mathrm{2}} −{c}−{a}\right)=\left({ca}\left({c}+{a}\right)+\mathrm{1}\right)\left({ab}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}−{b}\right) \\ $$$$\left\{{a}^{\mathrm{2}} {b}+{ab}^{\mathrm{2}} +\mathrm{1}\right\}\left({ac}^{\mathrm{2}} +{c}^{\mathrm{2}} −{c}−{a}\right) \\ $$$$={a}^{\mathrm{3}} {bc}^{\mathrm{2}} +{a}^{\mathrm{2}} {bc}^{\mathrm{2}} −{a}^{\mathrm{2}} {bc}−{a}^{\mathrm{3}} {b}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +{ac}^{\mathrm{2}} {b}^{\mathrm{2}} −{acb}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{ac}^{\mathrm{2}} +{c}^{\mathrm{2}} −{c}−{a} \\ $$$$\left({c}^{\mathrm{2}} {a}+{ca}^{\mathrm{2}} +\mathrm{1}\right)\left({ab}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}−{b}\right)={a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {ab}^{\mathrm{2}} −{c}^{\mathrm{2}} {a}^{\mathrm{2}} −{bac}^{\mathrm{2}} +{ca}^{\mathrm{3}} {b}^{\mathrm{2}} +{ca}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$−{ca}^{\mathrm{3}} −{cba}^{\mathrm{2}} +{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}−{b} \\ $$$$\left({ab}\left({a}+{b}\right)+\mathrm{1}\right)\left({ac}^{\mathrm{2}} +{c}^{\mathrm{2}} −{c}−{a}\right)−\left({ca}\left({c}+{a}\right)+\mathrm{1}\right)\left({ab}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}−{b}\right)=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$$\left.{a}^{\mathrm{3}} {bc}\left({c}−{b}\right)\:+{a}^{\mathrm{3}} \left({c}−{b}\right)+{acb}\left({c}−{b}\right)+{a}^{\mathrm{2}} {bc}\left({c}−{b}\right)+{a}^{\mathrm{2}} \left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+{a}\left({c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} −{b}^{\mathrm{2}} −\left({c}−{b}\right)\right\}=\mathrm{0} \\ $$$$\Leftrightarrow\left({c}−{b}\right)\left\{{a}^{\mathrm{3}} {bc}+{a}^{\mathrm{3}} +{acb}+{a}^{\mathrm{2}} \left({c}+{b}\right)+{a}^{\mathrm{2}} {bc}\left({c}−{b}\right)+{a}\left({c}+{b}\right)+{c}+{b}−\mathrm{1}\right\}=\mathrm{0} \\ $$$$\Leftrightarrow{a}^{\mathrm{3}} {bc}+{a}^{\mathrm{3}} +{acb}+{a}^{\mathrm{2}} {bc}+{a}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}+{ac}+{ab}+{c}+{b}−\mathrm{1}=\mathrm{0} \\ $$$$\Leftrightarrow{a}^{\mathrm{3}} −\mathrm{1}+{abc}\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)+{c}\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)+{b}\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)=\mathrm{0} \\ $$$${a}^{\mathrm{3}} −\mathrm{1}=\left({a}−\mathrm{1}\right)\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right) \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)\left({a}−\mathrm{1}+{abc}+{c}+{b}\right)=\mathrm{0} \\ $$$$\forall{a}\in\mathbb{R}\:\:{a}^{\mathrm{2}} +{a}+\mathrm{1}>\mathrm{0} \\ $$$$\Rightarrow{a}−\mathrm{1}+{abc}+{b}+{c}=\mathrm{0} \\ $$$$\Leftrightarrow{abc}=\mathrm{1}−{a}−{b}−{c} \\ $$$$ \\ $$

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