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Difference between revisions of "1979 IMO Problems/Problem 5"

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==Problem==
 
==Problem==
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Determine all real numbers a for which there exists positive reals <math>x_{1}, \ldots, x_{5}</math> which satisfy the relations <math> \sum_{k=1}^{5} kx_{k}=a,</math> <math> \sum_{k=1}^{5} k^{3}x_{k}=a^{2},</math> <math> \sum_{k=1}^{5} k^{5}x_{k}=a^{3}.</math>
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Determine all real numbers a for which there exists non-negative reals <math>x_{1}, \ldots, x_{5}</math> which satisfy the relations <math> \sum_{k=1}^{5} kx_{k}=a,</math> <math> \sum_{k=1}^{5} k^{3}x_{k}=a^{2},</math> <math> \sum_{k=1}^{5} k^{5}x_{k}=a^{3}.</math>
  
 
==Solution==
 
==Solution==

Revision as of 13:55, 30 November 2021

Problem

Determine all real numbers a for which there exists non-negative reals $x_{1}, \ldots, x_{5}$ which satisfy the relations $\sum_{k=1}^{5} kx_{k}=a,$ $\sum_{k=1}^{5} k^{3}x_{k}=a^{2},$ $\sum_{k=1}^{5} k^{5}x_{k}=a^{3}.$

Solution

Discussion thread can be found here: [1]

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See Also

1979 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions

Let $\Sigma_1= \sum_{k=1}^{5} kx_{k}$, $\Sigma_2=\sum_{k=1}^{5} k^{3}x_{k}$ and $\Sigma_3=\sum_{k=1}^{5} k^{5}x_{k}$. For all pairs $i,j\in \mathbb{Z}$, let \[\Sigma(i,j)=i^2j^2\Sigma_1-(i^2+j^2)\Sigma_2+\Sigma_3\] Then we have on one hand \[\Sigma(i,j)=i^2j^2\Sigma_1-(i^2+j^2)\Sigma_2+\Sigma_3=\sum_{k=1}^5(i^2j^2k-(i^2+j^2)k^3+k^5)x_k =\sum_{k=1}^5k(i^2j^2-(i^2+j^2)k^2+k^4)x_k\] Therefore \(1) \[\Sigma(i,j)=\sum_{k=1}^5k(k^2-i^2)(k^2-j^2)x_k\] and on the other hand \ (2) \[\Sigma(i,j)=i^2j^2a-(i^2+j^2)a^2+a^3=a(a-i^2)(a-j^2)\] Then from (1) we have\[\Sigma(0,5)=\sum_{k=1}^5k^3(k^2-5^2)x_k\leq 0\] and from (2) \[\Sigma(0,5)=a^2(a-25)\] so $a\in [0,25]$ Besides we also have from (1) \[\Sigma(0,1)=\sum_{k=1}^5k^3(k^2-1)x_k\geq 0\] and from (2) \[\Sigma(0,1)=a^2(a-1)\geq 0 \implies a\notin (0,1)\] and for $n=1,2,3,4$ \[\Sigma(n,n+1)=\sum_{k=1}^5k(k^2-n^2)(k^2-(n+1)^2)x_k\] where in the right hand we have that \[k<n \implies (k^2-n^2)<0, (k^2-(n+1)^2)<0\], so \[(k^2-n^2)(k^2-(n+1)^2)>0\], \[k=n,n+1 , \implies (k^2-n^2)(k^2-(n+1)^2)=0\] and \[k>n \implies (k^2-n^2)(k^2-(n+1)^2)>0\], so \[\Sigma(n,n+1)\geq 0\] for $n=1,2,3,4$ From the latter and (2) we also have \[\Sigma(n,n+1)=a(a-n^2)(a-(n+1)^2))\geq 0\implies a\notin (n^2,(n+1)^2)\] So we have that \[a\in [0,25]-\bigcup_{n=0}^4(n^2,(n+1^2))=\{0,1,4,9,16,25\}\]

If $a=k^2$, $k=0,1,2,3,4,5$ take $x_k=k$, $x_j=0$ for $j\neq k$. Then $\Sigma_1=k^2=a$, $\Sigma_2=k^3k=k^4=a^2$, and $\Sigma_3=k^5k=k^6=a^3$

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