Algebra and Analysis!
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ k = 1 ∑ 5 k x k = a k = 1 ∑ 5 k 3 x k = a 2 k = 1 ∑ 5 k 5 x k = a 3
Find the sum of all possible values of real number a for which there exists non-negative real numbers x 1 , x 2 , . . , x 5 satisfying the system of equations above.
The answer is 55.
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1 solution
Awesome problem by Priyanshu Mishra with an awesome solution by Mark Hennings! I thought the problem was extremely interesting, so I played with it for a couple hours before giving up. To help people see Mark's last step, where he determines a = k^2, the set of equations looks like: k(x k) = a, k^3(x k) = a^2, k^5(x_k) = a^3. Dividing consecutive equations gives the solution.
If we define u k = k x k and v k = k 2 u k for 1 ≤ k ≤ 5 , then we have ∥ u ∥ 2 = a u ⋅ v = a 2 ∥ v ∥ 2 = a 3 so that u ⋅ v = ∥ u ∥ × ∥ v ∥ . Using the Cauchy-Schwarz Inequality, we deduce that v = λ u for some λ . Thus ( λ − k 2 ) k x k = 0 1 ≤ k ≤ 5 If λ = k 2 for some 1 ≤ k ≤ 5 then it follows that x j = 0 for all j = k , and in this case a = k 2 . Otherwise we must have x k = 0 for all k , and hence a = 0 . Thus the answer is 0 + 1 2 + 2 2 + 3 2 + 4 2 + 5 2 = 5 5 .