Difference between revisions of "1979 IMO Problems/Problem 5"
Line 8: | Line 8: | ||
== See Also == {{IMO box|year=1979|num-b=4|num-a=6}} | == See Also == {{IMO box|year=1979|num-b=4|num-a=6}} | ||
+ | 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 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 later 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> |
Revision as of 20:17, 27 November 2021
Problem
Determine all real numbers a for which there exists positive reals which satisfy the relations
Solution
Discussion thread can be found here: [1]
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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 , and . For all pairs , let Then we have 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 later and (2) we also have So we have that
If , take , for . Then , , and