Hard SUM

Algebra Level 4

Giving x 1 , x 2 , . . . . . . . . . x 2016 x_1,x_2,.........x_{2016} real numbers in the interval [ 1 , 1 ] [-1,1] such that x 1 3 + x 2 3 + . . . . . . . . . + x 2016 3 = 0 x_1^3+x_2^3+.........+x_{2016}^3=0 .

Which is the greatest value of x 1 + x 2 + . . . . . . . . . + x 2016 x_1+x_2+.........+x_{2016} ?


The answer is 672.

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2 solutions

汶良 林
Apr 30, 2015

Up voted. Can any one explain why 1/2 is chosen ? Thank you.

Niranjan Khanderia - 3 years, 4 months ago
Drop TheProblem
Apr 30, 2015

x i = s i n ( α i ) x_i=sin(\alpha_i) .

s i n ( 3 γ ) = 3 s i n ( γ ) 4 s i n ( γ ) 3 sin(3\gamma)=3sin(\gamma)-4sin(\gamma)^3

Summing all α i \alpha_i we get i = 1 n s i n ( 3 α i ) = 3 i = 1 n s i n ( α i ) 4 i = 1 n s i n ( α i ) 3 = 3 i = 1 n x i 4 i = 1 n x i 3 = 3 i = 1 n x i \displaystyle \sum_{i=1}^n sin(3\alpha_i)=3\displaystyle \sum_{i=1}^n sin(\alpha_i)-4\displaystyle \sum_{i=1}^n sin(\alpha_i)^3=3\displaystyle \sum_{i=1}^n x_i-4\displaystyle \sum_{i=1}^n x_i^3=3\displaystyle \sum_{i=1}^n x_i

s i n ( 3 γ ) 1 3 i = 1 n x i n sin(3\gamma)\leq1\Rightarrow 3\displaystyle \sum_{i=1}^n x_i\leq n and in this case, for n = 2016 n=2016 the greatest number is n 3 = 2016 3 = 672 \frac {n}{3}=\frac {2016}{3}=672

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