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Prove that they are equal
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Rushil
Navier-Stokes Equations
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#1
Prove that they are equal
INMO 2000 Problem 3
If a,b,c,x are real numbers such that abc \not= 0 and \frac{xb + (1-x)c}{a} = \frac{xc + (1-x)a}{b} = \frac{xa + (1-x) b }{c}, then prove that a = b = c.

PostPosted: Sun Oct 09, 2005 8:49 pm  Back to top 
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darij grinberg
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#2
Re: prove that they are equal
INMO 2000 Problem 3
Rushil wrote:
If a,b,c,x are real numbers such that abc \not= 0 and \frac{xb + (1-x)c}{a} = \frac{xc + (1-x)a}{b} = \frac{xa + (1-x) b }{c}, then prove that a = b = c.


The problem is wrong as stated: Take any three reals a, b, c such that a + b + c = 0, and take x=\frac12. Then,

\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c}=-\frac12.

Here is a possible corrected version of the problem:

Problem. If four real numbers a, b, c, x satisfy abc\neq 0 and

\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c},

then either a = b = c, or a + b + c = 0 and x=\frac12.

Solution. Since abc\neq 0, the numbers a, b, c are all nonzero, so we can divide by them without remorses.

Denote

\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c}=k.

Then,

xb+\left(1-x\right)c=ka;
xc+\left(1-x\right)a=kb;
xa+\left(1-x\right)b=kc;

adding these three equations together, we arrive at

\left(xb+\left(1-x\right)c\right)+\left(xc+\left(1-x\right)a\right)+\left(xa+\left(1-x\right)b\right)=ka+kb+kc,

what simplifies to a+b+c=k\left(a+b+c\right). Hence, either k = 1, or a + b + c = 0.

First consider the case when a + b + c = 0.

In this case, a = - (b + c), so that xb+\left(1-x\right)c=ka becomes xb+\left(1-x\right)c=k\left(-\left(b+c\right)\right)=-kb-kc. Addition of kb-\left(1-x\right)c yields xb+kb=-kc-\left(1-x\right)c; in other words, \left(x+k\right)b=\left(x-k-1\right)c.

Now, if x+k\neq 0, then this equation becomes \frac{b}{c}=\frac{x-k-1}{x+k}. Similarly, \frac{c}{a}=\frac{x-k-1}{x+k} and \frac{a}{b}=\frac{x-k-1}{x+k}. Hence,

\left(\frac{x-k-1}{x+k}\right)^3=\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}=1,

and thus \frac{x-k-1}{x+k}=1. Thus, \frac{b}{c}=\frac{x-k-1}{x+k}=1, so that b = c. Similarly, c = a. Thus, a = b = c, and we are done.

If x + k = 0, then the equation \left(x+k\right)b=\left(x-k-1\right)c yields x - k - 1 = 0 (since c is nonzero). Hence, x + k = x - k - 1 = 0, so that k = - k - 1, and thus 2k = -1; hence, k=-\frac12, so that x + k = 0 yields x=-k=-\left(-\frac12\right)=\frac12. Hence, we have a + b + c = 0 and x=\frac12. So in this case we are done as well.

Now, we consider the case when k = 1.

In this case, the equation xb+\left(1-x\right)c=ka becomes xb+\left(1-x\right)c=a, what is equivalent to xb+c-xc=a, i. e. to xb-xc=a-c, i. e. to x\left(b-c\right)=-\left(c-a\right). Similarly, x\left(c-a\right)=-\left(a-b\right) and x\left(a-b\right)=-\left(b-c\right). Mulitplying these three equations, we get

x\left(b-c\right)\cdot x\left(c-a\right)\cdot x\left(a-b\right)=\left(-\left(c-a\right)\right)\cdot\left(-\left(a-b\right)\ri...,

i. e.

x^3\left(b-c\right)\left(c-a\right)\left(a-b\right)=-\left(b-c\right)\left(c-a\right)\left(a-b\right).

Now, if one of the terms b - c, c - a, a - b equals 0, then we are quickly done (in fact, assuming that b - c = 0, we see that x\left(b-c\right)=-\left(c-a\right) yields c - a = 0, so that a = b = c). If not, then we can divide this equation by \left(b-c\right)\left(c-a\right)\left(a-b\right), and thus we get x^3=-1, so that x = -1. Hence, the equation xb+\left(1-x\right)c=a becomes -1b+\left(1-\left(-1\right)\right)c=a, what simplifies to -b + 2c = a. In other words, c - (b - c) = a, so that b - c = c - a. Similarly, c - a = a - b. So the numbers b - c, c - a, a - b are all equal to each other; but their sum is 0, and thus they are all equal to \frac03=0. Hence, a = b = c. Thus, the problem is solved in all cases.

darij
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PostPosted: Tue Oct 11, 2005 4:28 am  Back to top 
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rem
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#3
Re: Prove that they are equal
INMO 2000 Problem 3
Rushil wrote:
If a,b,c,x are real numbers such that abc \not= 0 and \frac{xb + (1-x)c}{a} = \frac{xc + (1-x)a}{b} = \frac{xa + (1-x) b }{c}, then prove that a = b = c.

A very simple solution:
First assume a<b<c. Then:
\frac{1}{a} > \frac{1}{c}
xb > xa
(1-x)c > (1-x)b, so:
\frac{xb + (1-x)c}{a} > \frac{xa + (1-x) b }{c}, contradiction.
So two of a,b,c are equal. wolog a=b. Have
\frac{xb + (1-x)c}{a} = \frac{xc + (1-x)a}{b}
xb+(1-x)c=xc+(1-x)a. Subst b with a:
xa+c-xc=xc+a-xa
(c-a)+2x(a-c)=0
(a-c)(2x-1)=0
Have a=c or x=1/2 which also yields a=c.
Hence a=b=c, qed
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PostPosted: Mon Jun 26, 2006 8:39 am  Back to top 
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FieryHydra
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#4
Re: Prove that they are equal
INMO 2000 Problem 3
rem wrote:
Rushil wrote:
If a,b,c,x are real numbers such that abc \not= 0 and \frac{xb+(1-x)c}{a}= \frac{xc+(1-x)a}{b}= \frac{xa+(1-x) b }{c}, then prove that a = b = c.

A very simple solution:
First assume a<b<c. Then:
\frac{1}{a}> \frac{1}{c}


This isn't necessarily true. Take a=-1 and c=1.
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PostPosted: Sun Jul 09, 2006 4:08 am  Back to top 
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euvtl
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#5
Re: Prove that they are equal
INMO 2000 Problem 3
Rushil wrote:
If a,b,c,x are real numbers such that abc \not= 0 and
\frac{xb+(1-x)c}{a}= \frac{xc+(1-x)a}{b}= \frac{xa+(1-x) b }{c},
then prove that a = b = c.


Cross-multiplying the first and third terms we get
xbc+(1-x)c^{2}= xa^{2}+(1-x)ab
\Longrightarrow x(c^{2}+a^{2}-ab-ca) = c^{2}-ab
And similarly,
x(a^{2}+b^{2}-bc-ab) = a^{2}-bc
x(b^{2}+c^{2}-ca-bc) = b^{2}-ca

Adding the three equations we get
2x(a^{2}+b^{2}+c^{2}-bc-ab-ca) = (a^{2}+b^{2}+c^{2}-bc-ab-ca)
(2x-1)(a^{2}+b^{2}+c^{2}-bc-ab-ca) = 0

Case 1:
(a^{2}+b^{2}+c^{2}-bc-ab-ca) = 0
\Longrightarrow \frac{1}{2}((a-b)^{2}+(b-c)^{2}+(c-a)^{2}) = 0
\Longrightarrow a=b=c

Case 2:
x=\frac{1}{2}
In the three equations we get
\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}
Add 1 to all three equations
\frac{a+b+c}{c}=\frac{b+c+a}{a}=\frac{c+a+b}{b}
So we again get a=b=c or a+b+c=0

PostPosted: Fri Jun 29, 2007 2:01 pm  Back to top 
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tim26270746
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#6
Re: Prove that they are equal
INMO 2000 Problem 3
Rushil wrote:
If a,b,c,x are real numbers such that abc \not = 0 and
\frac {xb + (1 - x)c}{a} = \frac {xc + (1 - x)a}{b} = \frac {xa + (1 - x) b }{c},
then prove that a = b = c.

um...
i thought if we added the top and the bottom together respectively we could obtain
xb+c(1-x)=a
xc+a(1-x)=b
xa+b(1-x)=c

OR a+b+c=0

PostPosted: Sat Dec 06, 2008 7:41 pm  Back to top 
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murgi
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#7
INMO in India is rare in mistake............... Shocked
Probably they wanted to make a,b,c non-negative real no.
I think in that case the sum holds good. Smile
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PostPosted: Tue Feb 17, 2009 9:58 am  Back to top 
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dgreenb801
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#8
Since the three fractions are equal, they equal the sum of the numerators over the sum of the denominators, which is 1, unless a+b+c=0. If two of a,b,c are equal, then all three are. So we find x = \frac {a - c}{b - c} = \frac {b - a}{c - a} = \frac {c - b}{a - b} if no two are equal and a+b+c is not 0. But if we have let's say a>b>c or a>c>b we get a contradiction because one of the fractions will be positive and the other negative, thus they are all equal.
If a+b+c=0 we can just substitute, we have
\frac{xb+(1-x)c}{-b-c}=\frac{xc+(1-x)(-b-c)}{b}, which is equivalent to
2x(b^2+bc+c^2)=b^2+bc+c^2, in which case x=1/2 or a=b=c=0.

PostPosted: Sat May 09, 2009 3:15 pm  Back to top 
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phymax
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#9
its too simple a problem to be hyped up so much..!
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PostPosted: Tue May 12, 2009 3:22 am  Back to top 
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kushal.sharma09
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#10
INMO problem
Possible solution
Put x=1 and solve taking RHS as k or else take two sets of eqns and solve for x separately and then compare them

PostPosted: Tue Oct 06, 2009 8:42 am  Back to top 
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b555
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#11
murgi wrote:
INMO in India is rare in mistake............... Shocked
Probably they wanted to make a,b,c non-negative real no.
I think in that case the sum holds good. Smile


not rare but i have seen it getting pointed as wrong quite a good number of times

PostPosted: Tue Oct 06, 2009 8:46 am  Back to top 
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manosij
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#12
they really should check these problems more thoroughly

PostPosted: Mon Oct 12, 2009 10:20 pm  Back to top 
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