Prove that x1=x2=....=x2025

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Prove that x1=x2=....=x2025

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#1 h Apr 9, 2025, 5:22 AM • 1 Y

Y by cubres

The real numbers $x_1,x_2,\cdots,x_{2025}$ satisfy,
$x_1+x_2=2\bar{x_1}, x_2+x_3=2\bar{x_2},\cdots, x_{2025}+x_1=2\bar{x_{2025}}$ Where { $\bar{x_1},\cdots,\bar{x_{2025}}$ } is a permutation of $x_1,x_2,\cdots,x_{2025}$ . Prove that $x_1=x_2=\cdots=x_{2025}$

This post has been edited 1 time. Last edited by Rohit-2006, Apr 9, 2025, 5:23 AM

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#2 h Apr 9, 2025, 8:10 AM

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Proof:
$Re(x_k+x_{k+1})=2Re(x_k)\implies Re(x_k)=Re(x_{k+1})\implies Re(x_i)=Re(x_j)$
$Im(x_k+x_{k+1})=-2Im(x_k)\implies Im(x_{k+1})=3Im(x_{k})\implies Im(x_k)=Im(x_{k+2025})=3^{2025}Im(x_k)\implies Im(x_k)=0$
Done.

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#3 h Apr 9, 2025, 11:04 AM

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Right?

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#4 h Apr 9, 2025, 11:11 AM

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flower417477 wrote:

Proof:
$Re(x_k+x_{k+1})=2Re(x_k)\implies Re(x_k)=Re(x_{k+1})\implies Re(x_i)=Re(x_j)$
$Im(x_k+x_{k+1})=-2Im(x_k)\implies Im(x_{k+1})=3Im(x_{k})\implies Im(x_k)=Im(x_{k+2025})=3^{2025}Im(x_k)\implies Im(x_k)=0$
Done.

@2above All the nos are real :skull: . $\overline{x_i} = \sigma(x_i)$ which is the image of the element after permuting.

KevinKV01 wrote:

Right?

@above you cannot assume them to be increasing because the condition is cyclic and not symmetric.

This post has been edited 3 times. Last edited by kamatadu, Apr 9, 2025, 11:13 AM

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#5 h Apr 9, 2025, 5:11 PM • 1 Y

Y by alexanderhamilton124

Restated Problem wrote:

The real numbers $x_1,x_2,\ldots,x_{2025}$ satisfy, $x_1+x_2=2x_{\sigma(1)}, x_2+x_3=2{x_{\sigma(2)}}, \ldots, x_{2025}+x_1=2{x_{\sigma(2025)}} .$ Where $(\sigma(1), \sigma(2),\ldots,\sigma(2025))$ is a permutation of $(1,2,\ldots,2025)$ . Prove that $x_1=x_2=\cdots=x_{2025}$ .

Here is my solution using arrows.

Draw a graph with the $x_i$ as the vertices and draw a directed edge from $x_i\to x_{\sigma(i)}$ . It is well known that this graph can be decomposed into disjoint cycles (including $1$ -cycles).

Claim: Let $M=\max\left\{ x_1,x_2,\ldots,x_{2025} \right\}$ . Then all the elements in a cycle containing $M$ must be equal to $M$ .

Proof: FTSOC assume not. This means that there is another element $x_{i}$ in the same cycle as that of $M$ such that $x_{i}<M$ .

Firstly note that $2x_{\sigma(i)}=x_{i}+x_{i+1} <M+x_{i+1}\le 2M\implies x_{\sigma(i)}<M$ . Similarly, $x_{\sigma(i)}<M\implies x_{\sigma(\sigma(i))}<M$ .

We can continue this pattern till we get back to the element with value $M$ which implies $M<M$ giving our desired contradiction. $\square$

Now pick a cycle containing $M$ . Therefore all the elements in this cycle are equal to $M$ .

Let the elements of this cycle be denoted as $\left\{ x_{i_1},x_{i_2},\ldots,x_{i_k} \right\}$ . Here $\sigma(i_1)=i_2$ , $\sigma(i_2)=i_3$ and so on up to $\sigma(i_k)=i_1$ .

Now, $x_{i_1-1}+x_{i_1}=2x_{i_2}\implies x_{i_1-1}=M$ . Similarly, we can also derive that $x_{i_2-1}=x_{i_3-1}=\cdots=x_{i_k-1}=M$ .

Note that from the indices $\left\{ i_1,i_2,\ldots,i_k \right\}$ , we just traversed to $\left\{ i_1-1, i_2-1,\ldots,i_k-1\right\}$ . We can continue doing so till we cover all the indices modulo $2025$ .

This shows that all the elements $x_i=M$ and we are done.

This post has been edited 5 times. Last edited by kamatadu, Apr 9, 2025, 5:19 PM

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#6 h Apr 10, 2025, 3:38 AM

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Alternative proof but same intuition to take max... :ninja:

:ninja:

Let $M$ be the maximum element of the set say occurs at $x_i$ . So,
$x_{i-1}+x_i\leq2M\Rightarrow \bar{x_{i-1}}\leq M$ Similarly we can say for other elements that all $x_j$ 's are less than or equal to $M$ . If one of the $x_j$ is strictly less than $M$ then there exist another $x_k$ such that $x_k>M$ , a contradiction!!! So all $x_i$ 's are equal.

We can do the same thing using minimalist...

Avigyan 5 bar edit korar por ki sundor akta solution dilo....love you baby

This post has been edited 2 times. Last edited by Rohit-2006, Apr 10, 2025, 3:50 AM

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#7 h Apr 10, 2025, 8:50 AM

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Rohit-2006 wrote:

Alternative proof but same intuition to take max... :ninja:

:ninja:

Let $M$ be the maximum element of the set say occurs at $x_i$ . So,
$x_{i-1}+x_i\leq2M\Rightarrow \bar{x_{i-1}}\leq M$ Similarly we can say for other elements that all $x_j$ 's are less than or equal to $M$ . If one of the $x_j$ is strictly less than $M$ then there exist another $x_k$ such that $x_k>M$ , a contradiction!!! So all $x_i$ 's are equal.

We can do the same thing using minimalist...

Avigyan 5 bar edit korar por ki sundor akta solution dilo....love you baby

Quote:

there exist another $x_{k}$ such that $x_{k} > M$

i don't understand how u are concluding this can u please explain

i understand it as if $x_{j} < M$ then $x_{j}+x_{j-1}<2M$ $\Rightarrow$ $2\bar{x_{j-1}}<2M$ $\Rightarrow$ $\bar{x_{j-1}}<M$ which is valid because
$x_{j-1}$ is not the maximal element

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#8 h Apr 10, 2025, 8:51 AM

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@2above. Couldn't understand your solution

This post has been edited 1 time. Last edited by Fibonacci_math, Apr 10, 2025, 8:51 AM

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Project_Donkey_into_M4

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#10 h Apr 15, 2025, 6:25 PM

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Rohit-2006 wrote:

The real numbers $x_1,x_2,\cdots,x_{2025}$ satisfy,
$x_1+x_2=2\bar{x_1}, x_2+x_3=2\bar{x_2},\cdots, x_{2025}+x_1=2\bar{x_{2025}}$ Where { $\bar{x_1},\cdots,\bar{x_{2025}}$ } is a permutation of $x_1,x_2,\cdots,x_{2025}$ . Prove that $x_1=x_2=\cdots=x_{2025}$

Why are the above sols so complicated ?

Claim : There are more than $1$ minimum element and consecutive minimum elements exist.

Proof : Consider one minimal element. That gives that there are more than $1$ minimum elements and atleast $2$ of them must be consecutive.
(As $x_i+x_j \geq 2x_{\text{min}}$ ).

Now say there are $k$ consecutive minimum elements $\{x_1,x_2,\cdots x_k\}$ . Now note that this generates $k-1$ minimal sums , $\{x_1+x_2,x_2+x_3,\cdots ,x_{k-1}+x_k\}$ .All of them are equal to $2x_{\text{min}}$ . But there are atleast $k$ minimum elements , so one of them will generate either $2$ consecutive minimal elements or will yield that $x_{k+1}$ is also minimum (this happens if $\{\bar{x_1},\bar{x_2}\cdots \bar{x_{k-1}}\}\in \{x_1,x_2,\cdots x_{n-1}\}$ ) or will give that there are more consecutive minimum elements (this happens if $\{\bar{x_1},\bar{x_2}\cdots \bar{x_{k-1}}\} \not \in \{x_1,x_2,\cdots x_{n-1}\}$ ).In this way we can generate an infinite number of minimum elements, but that can't be true.Hence all the elements are equal $\blacksquare$

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