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

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==Solution==
 
==Solution==
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Discussion thread can be found here: [https://aops.com/community/p367346]
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Let <math>\Sigma_1= \sum_{k=1}^{5} kx_{k}</math>, <math>\Sigma_2=\sum_{k=1}^{5} k^{3}x_{k}</math> and <math>\Sigma_3=\sum_{k=1}^{5} k^{5}x_{k}</math>. For all pairs <math>i,j\in \mathbb{Z}</math>, let<cmath>\Sigma(i,j)=i^2j^2\Sigma_1-(i^2+j^2)\Sigma_2+\Sigma_3</cmath>Then we have on one hand<cmath>\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</cmath>Therefore \\(1)<cmath>\Sigma(i,j)=\sum_{k=1}^5k(k^2-i^2)(k^2-j^2)x_k</cmath>and on the other hand \\ (2)<cmath>\Sigma(i,j)=i^2j^2a-(i^2+j^2)a^2+a^3=a(a-i^2)(a-j^2)</cmath>Then from (1) we have<cmath>\Sigma(0,5)=\sum_{k=1}^5k^3(k^2-5^2)x_k\leq 0</cmath>and from (2)<cmath>\Sigma(0,5)=a^2(a-25)</cmath>so <math>a\in [0,25]</math> Besides we also have from (1)<cmath>\Sigma(0,1)=\sum_{k=1}^5k^3(k^2-1)x_k\geq 0</cmath>and from (2)<cmath>\Sigma(0,1)=a^2(a-1)\geq 0 \implies a\notin (0,1)</cmath>and for <math>n=1,2,3,4</math><cmath>\Sigma(n,n+1)=\sum_{k=1}^5k(k^2-n^2)(k^2-(n+1)^2)x_k</cmath>where in the right hand we have that<cmath>k<n \implies (k^2-n^2)<0, (k^2-(n+1)^2)<0</cmath>, so<cmath>(k^2-n^2)(k^2-(n+1)^2)>0</cmath>,<cmath>k=n,n+1 , \implies (k^2-n^2)(k^2-(n+1)^2)=0</cmath>and<cmath>k>n \implies (k^2-n^2)(k^2-(n+1)^2)>0</cmath>, so<cmath>\Sigma(n,n+1)\geq 0</cmath>for <math>n=1,2,3,4</math> From the latter and (2) we also have<cmath>\Sigma(n,n+1)=a(a-n^2)(a-(n+1)^2))\geq 0\implies a\notin (n^2,(n+1)^2)</cmath>So we have that<cmath>a\in [0,25]-\bigcup_{n=0}^4(n^2,(n+1^2))=\{0,1,4,9,16,25\}</cmath>
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If <math>a=k^2</math>, <math>k=0,1,2,3,4,5</math> take <math>x_k=k</math>, <math>x_j=0</math> for <math>j\neq k</math>. Then <math>\Sigma_1=k^2=a</math>, <math>\Sigma_2=k^3k=k^4=a^2</math>, and <math>\Sigma_3=k^5k=k^6=a^3</math>
  
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{{solution}}
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==Solution II==
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Applying Cauchy-Schwarz on the three equations yields
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== See Also == {{IMO box|year=1979|num-b=4|num-a=6}}
 
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Let <math>\Sigma_1= \sum_{k=1}^{5} kx_{k}</math>, <math>\Sigma_2=\sum_{k=1}^{5} k^{3}x_{k}</math> and <math>\Sigma_3=\sum_{k=1}^{5} k^{5}x_{k}</math>. For all pairs <math>i,j\in \mathbb{Z}</math>, let <cmath>\Sigma(i,j)=i^2j^2\Sigma_1-(i^2+j^2)\Sigma_2+\Sigma_3</cmath>
 
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Then we have on one hand
 
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<cmath>
 
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\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
 
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=\sum_{k=1}^5k(i^2j^2-(i^2+j^2)k^2+k^4)x_k </cmath>
 
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Therefore \\(1)
 
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<cmath>
 
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\Sigma(i,j)=\sum_{k=1}^5k(k^2-i^2)(k^2-j^2)x_k
 
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</cmath>
 
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and on the other hand \\
 
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(2)
 
 
<cmath>
 
<cmath>
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\Sigma(i,j)=i^2j^2a-(i^2+j^2)a^2+a^3=a(a-i^2)(a-j^2)
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a\cdot a^3 = \left(\sum_{k=1}^5 kx_k\right)\left(\sum_{k=1}^5 k^5x_k\right)\geq \left(\sum_{k=1}^5 k^3x_k\right)^2=(a^2)^2
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</cmath>
 
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Then from (1) we have<cmath>
 
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\Sigma(0,5)=\sum_{k=1}^5k^3(k^2-5^2)x_k\leq 0
 
 
</cmath>
 
</cmath>
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and from (2)
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So this implies the following vectors are multiples of each other:
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<cmath>
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<cmath>(x_1,2x_2,3x_3,4x_4,5x_5) =\lambda (x_1,2^5x_2,3^5x_3,4^5x_4,5^5x_5)\quad \quad \lambda\in \mathbb{R}^+ </cmath>
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\Sigma(0,5)=a^2(a-25)
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This is only possible when at most one of the <math>x_i</math>'s are nonzero. Hence cycling through the cases where <math>x_i\neq 0</math>, we get the choices of <math>a=1,4,9,16,25</math> and the case of <math>a=0</math>.
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</cmath>
 
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so <math>a\in [0,25]</math>
 
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Besides we also have from (1)
 
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<cmath>
 
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\Sigma(0,1)=\sum_{k=1}^5k^3(k^2-1)x_k\geq 0
 
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</cmath>
 
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and from (2)
 
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<cmath>
 
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\Sigma(0,1)=a^2(a-1)\geq 0 \implies a\notin (0,1)
 
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</cmath>
 
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and for <math>n=1,2,3,4</math>
 
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<cmath>
 
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\Sigma(n,n+1)=\sum_{k=1}^5k(k^2-n^2)(k^2-(n+1)^2)x_k
 
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</cmath>
 
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where in  the right hand we have that  <cmath>k<n \implies (k^2-n^2)<0, (k^2-(n+1)^2)<0</cmath>, so
 
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<cmath>(k^2-n^2)(k^2-(n+1)^2)>0</cmath>,  <cmath>k=n,n+1 , \implies (k^2-n^2)(k^2-(n+1)^2)=0</cmath> and <cmath>k>n \implies (k^2-n^2)(k^2-(n+1)^2)>0</cmath>, so
 
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<cmath>
 
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\Sigma(n,n+1)\geq 0
 
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</cmath>
 
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for <math>n=1,2,3,4</math>
 
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From the latter and (2) we also have
 
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<cmath>
 
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\Sigma(n,n+1)=a(a-n^2)(a-(n+1)^2))\geq 0\implies a\notin (n^2,(n+1)^2)
 
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</cmath>
 
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So we have that <cmath>a\in [0,25]-\bigcup_{n=0}^4(n^2,(n+1^2))=\{0,1,4,9,16,25\}</cmath>
 
  
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If <math>a=k^2</math>, <math>k=0,1,2,3,4,5</math> take <math>x_k=k</math>, <math> x_j=0 </math> for <math>j\neq k</math>. Then <math>\Sigma_1=k^2=a</math>, <math>\Sigma_2=k^3k=k^4=a^2</math>, and <math>\Sigma_3=k^5k=k^6=a^3</math>
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== See Also == {{IMO box|year=1979|num-b=4|num-a=6}}

Latest revision as of 22:01, 18 March 2025

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

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 $\Sigma_3=k^5k=k^6=a^3$

Solution II

Applying Cauchy-Schwarz on the three equations yields \[a\cdot a^3 = \left(\sum_{k=1}^5 kx_k\right)\left(\sum_{k=1}^5 k^5x_k\right)\geq \left(\sum_{k=1}^5 k^3x_k\right)^2=(a^2)^2\] So this implies the following vectors are multiples of each other: \[(x_1,2x_2,3x_3,4x_4,5x_5) =\lambda (x_1,2^5x_2,3^5x_3,4^5x_4,5^5x_5)\quad \quad \lambda\in \mathbb{R}^+\] This is only possible when at most one of the $x_i$'s are nonzero. Hence cycling through the cases where $x_i\neq 0$, we get the choices of $a=1,4,9,16,25$ and the case of $a=0$.

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
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