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Let-consider-I-R-2-a-parametric-curve-defined-as-t-I-t-t-2-1-t-3-1-2t-t-3-1-Prove-that-for-a-b-c-I-a-b-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
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
Letconsiderα:I→R2aparametriccurvedefinedas∀t∈Iα(t)=(t2−1t3−1,2tt3−1)Provethatfora,b,c∈Iα(a),α(b),α(c)areonthesameligniffabc=a+b+c+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
testa=3b=5⇒c=914α(3)=(413313)α(5)=(631562)α(914)=(322403−35282015)linethroughthe1stand2ndy=12192x−423the3rdisnotonthisline⇒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
pleasesiricorrectlycopiedtheexercise.Whatonthattestprovethatthequeztionizwrong
Commented by MJS last updated on 21/Nov/19
if there′s at least one counter example the question is wrong
ifthere′satleastonecounterexamplethequestioniswrong
Commented by ~blr237~ last updated on 21/Nov/19
thanks you sir , i have seen the problem
thanksyousir,ihaveseentheproblem
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
wheneveranexamplelikethisonelookscomplicatedtotransform,Imakeafewteststoensureit′sworththeeffort
Answered by mind is power last updated on 21/Nov/19
false this is abc=1−a−b−c
falsethisisabc=1−a−b−c
Commented by ~blr237~ last updated on 21/Nov/19
Okay sir have you proved it?
Okaysirhaveyouprovedit?
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
letα(a)=(a2−1a3−1,2aa3−1),α(b)=(b2−1b3−1,2bb3−1),α(c)=(c2−1c3−1,2cc3−1)α(a),α(b),α(c)insamelign⇔2aa3−1−2bb3−1a2−1a3−1−b2−1b3−1=2aa3−1−2cc3−1a2−1a3−1−2cc2−1⇔2a(b3−1)−2b(a3−1)(b3−1)(a2−1)−(a3−1)(b2−1)=2a(c2−1)−2c(a2−1)(c3−1)(a2−1)−(c2−1)(a3−1)..Hwewillworckwitheoneexpressiontheotheroneisjust(a,b)→(a,c)2a(b3−1)−2b(a3−1)=(b−a)(2ab(b+a)+2)(b3−1)(a2−1)−(a3−1)(b2−1)=(a−1)(b−1){(a+1)(b2+b+1)−(b+1)(a2+a+1)}=(a−1)(b−1){ab2−ba2+b2−a2}=(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)=ab2+b2−a−b⇔(ab(a+b)+1)(ac2+c2−c−a)=(ca(c+a)+1)(ab2+b2−a−b){a2b+ab2+1}(ac2+c2−c−a)=a3bc2+a2bc2−a2bc−a3b+a2b2c2+ac2b2−acb2−a2b2+ac2+c2−c−a(c2a+ca2+1)(ab2+b2−a−b)=a2b2c2+c2ab2−c2a2−bac2+ca3b2+ca2b2−ca3−cba2+ab2+b2−a−b(ab(a+b)+1)(ac2+c2−c−a)−(ca(c+a)+1)(ab2+b2−a−b)=0⇔a3bc(c−b)+a3(c−b)+acb(c−b)+a2bc(c−b)+a2(c2−b2)+a(c2−b2)+c2−b2−(c−b)}=0⇔(c−b){a3bc+a3+acb+a2(c+b)+a2bc(c−b)+a(c+b)+c+b−1}=0⇔a3bc+a3+acb+a2bc+a2c+a2b+ac+ab+c+b−1=0⇔a3−1+abc(a2+a+1)+c(a2+a+1)+b(a2+a+1)=0a3−1=(a−1)(a2+a+1)⇒(a2+a+1)(a−1+abc+c+b)=0∀a∈Ra2+a+1>0⇒a−1+abc+b+c=0⇔abc=1−a−b−c

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