Difference between revisions of "1979 IMO Problems/Problem 5"
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==Problem== | ==Problem== | ||
− | Determine all real numbers a for which there exists | + | 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== | ||
− | + | 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> | |
+ | 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}} | == See Also == {{IMO box|year=1979|num-b=4|num-a=6}} | ||
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Latest revision as of 10:55, 30 September 2022
Problem
Determine all real numbers a for which there exists non-negative reals which satisfy the relations
Solution
Let ,
and
. For all pairs
, let
Then we have on one hand
Therefore \\(1)
and on the other hand \\ (2)
Then from (1) we have
and from (2)
so
Besides we also have from (1)
and from (2)
and for
where in the right hand we have that
, so
,
and
, so
for
From the latter and (2) we also have
So we have that
If
,
take
,
for
. Then
,
, and
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 |