Prove that all x_i are equal

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Prove that all x_i are equal

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Source: Baltic Way 2011

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WakeUp

#1 h Nov 6, 2011, 12:09 PM • 3 Y

Y by PhantomR, Adventure10, Mango247

The real numbers $x_1,\ldots ,x_{2011}$ satisfy
$x_1+x_2=2x_1',\ x_2+x_3=2x_2', \ \ldots, \ x_{2011}+x_1=2x_{2011}'$
where $x_1',x_2',\ldots,x_{2011}'$ is a permutation of $x_1,x_2,\ldots,x_{2011}$ . Prove that $x_1=x_2=\ldots =x_{2011}$ .

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pco

#2 h Nov 6, 2011, 1:51 PM • 5 Y

Y by PhantomR, Centralorbit, Adventure10, Mango247, and 1 other user

WakeUp wrote:

The real numbers $x_1,\ldots ,x_{2011}$ satisfy
$x_1+x_2=2x_1',\ x_2+x_3=2x_2', \ \ldots, \ x_{2011}+x_1=2x_{2011}'$
where $x_1',x_2',\ldots,x_{2011}'$ is a permutation of $x_1,x_2,\ldots,x_{2011}$ . Prove that $x_1=x_2=\ldots =x_{2011}$ .

Chose any $n<2011$ such equations : the number of $x_i$ involved in these equations is clearly $>n$

Let then $M=\max(x_i)$ and let $n$ be the number of $x_i$ (and also of $x_i'$ ) equal to $M$

In each equality $x_k+x_{k+1}=2M$ , we can deduce $x_k=x_{k+1}=M$

Suppose $n<2011$ : adding then the $n$ such equalities, we get that the number of $x_i=M$ is $>n$ , and so contradiction.
So $n=2011$

Hence the result.

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yair

#3 h Nov 6, 2011, 2:12 PM • 8 Y

Y by pco, PhantomR, az360, Centralorbit, Adventure10, Mango247, and 2 other users

Use $\sum_{i}x_{i}^{2}=\sum_{i}x_{i}^{'2}=\sum_{i}\left(\frac{x_{i}+x_{i+1}}{2}\right)^{2}$
which can also written as $\sum_{i}\left(\frac{x_{i}-x_{i+1}}{2}\right)^{2}=0$

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