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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:IR2aparametriccurvedefinedastIα(t)=(t21t31,2tt31)Provethatfora,b,cIα(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=5c=914α(3)=(413313)α(5)=(631562)α(914)=(32240335282015)linethroughthe1stand2ndy=12192x423the3rdisnotonthislinewrong

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

iftheresatleastonecounterexamplethequestioniswrong

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,Imakeafewteststoensureitsworththeeffort

Answered by mind is power last updated on 21/Nov/19

false this is abc=1−a−b−c

falsethisisabc=1abc

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)=(a21a31,2aa31),α(b)=(b21b31,2bb31),α(c)=(c21c31,2cc31)α(a),α(b),α(c)insamelign2aa312bb31a21a31b21b31=2aa312cc31a21a312cc212a(b31)2b(a31)(b31)(a21)(a31)(b21)=2a(c21)2c(a21)(c31)(a21)(c21)(a31)..Hwewillworckwitheoneexpressiontheotheroneisjust(a,b)(a,c)2a(b31)2b(a31)=(ba)(2ab(b+a)+2)(b31)(a21)(a31)(b21)=(a1)(b1){(a+1)(b2+b+1)(b+1)(a2+a+1)}=(a1)(b1){ab2ba2+b2a2}=(a1)(b1)(ba){ab+b+a}Hab(b+a)+1(a1)(b1)(ab+a+b)=ac(c+a)+1(a1)(c1)(ac+a+c)ab(b+a)+1(b1)(ab+a+b)=ac(c+a)+1(c1)(ac+a+c)(b1)(ab+a+b)=ab2+b2ab(ab(a+b)+1)(ac2+c2ca)=(ca(c+a)+1)(ab2+b2ab){a2b+ab2+1}(ac2+c2ca)=a3bc2+a2bc2a2bca3b+a2b2c2+ac2b2acb2a2b2+ac2+c2ca(c2a+ca2+1)(ab2+b2ab)=a2b2c2+c2ab2c2a2bac2+ca3b2+ca2b2ca3cba2+ab2+b2ab(ab(a+b)+1)(ac2+c2ca)(ca(c+a)+1)(ab2+b2ab)=0a3bc(cb)+a3(cb)+acb(cb)+a2bc(cb)+a2(c2b2)+a(c2b2)+c2b2(cb)}=0(cb){a3bc+a3+acb+a2(c+b)+a2bc(cb)+a(c+b)+c+b1}=0a3bc+a3+acb+a2bc+a2c+a2b+ac+ab+c+b1=0a31+abc(a2+a+1)+c(a2+a+1)+b(a2+a+1)=0a31=(a1)(a2+a+1)(a2+a+1)(a1+abc+c+b)=0aRa2+a+1>0a1+abc+b+c=0abc=1abc

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