Determine the greatest possible value of $$\sum_{i=1}^{10}{\cos 3x_i}$$ for real numbers $x_1,x_2....x_{10}$ satisfying $$\sum_{i=0}^{10}{\cos x_i}=0$$

My attempt:

$$\sum \cos 3x = \sum 4\cos^3x -\sum3\cos x=4\sum \cos^3 x $$

So now we have to maximize sum of cubes of ten numbers when their sum is zero and each lie in interval $[-1,1]$. i often use AM GM inequalities but here are 10 numbers and they are not even positive. Need help to how to visualize and approach these kinds of questions.

  • $\begingroup$ Is your claim that $\cos(3x)=4\cos^3(x)+3\cos(x)$? that's not even true at $x=0$. $\endgroup$
    – lulu
    Sep 11, 2019 at 10:31
  • 3
    $\begingroup$ I agree with @lulu: it should be $4\cos^3x\color{blue}{-}3\cos x$. $\endgroup$
    – J.G.
    Sep 11, 2019 at 10:31
  • $\begingroup$ This question is question A3 from the 2018 Putnam. I would like to note that the Putnam competition is usually designed for the brightest undergraduate students, so this will not have a straightforward solution. $\endgroup$
    – Toby Mak
    Sep 11, 2019 at 10:32
  • $\begingroup$ For what it's worth, solutions to the 201 Putnam can be found here $\endgroup$
    – lulu
    Sep 11, 2019 at 10:33
  • 2
    $\begingroup$ Lagrange method yields the restrictions $$3\sin (3x_j) = \lambda \sin x_j ,\quad 1\leq j\leq 10 $$ $\endgroup$
    – AlvinL
    Sep 11, 2019 at 10:33

2 Answers 2


Visualising the solution

You have asked for help in visualising the solution. I think you will find it useful to have in mind the picture of $y=x^3$ for $-1\le x\le1$.

Now consider the arrangement of the 10 numbers in the maximum position. (We have a continuous function on a compact set and so the maximum is attained.)

First suppose that there is a number, $s$, smaller in magnitude than the least negative number $l$. Increasing $l$ whilst decreasing $s$ by the same amount would increase the sum of cubes and therefore cannot occur.

So, all the negative numbers are equal, to $l$ say, and all the positive numbers are greater than $|l|$.

Now suppose that a positive number was not $1$. Then increasing it to $1$ whilst reducing one of the $l$s would increase the sum of cubes and therefore cannot occur.

Hence we need only consider the case where we have $m$ $1$s and $10-m$ numbers equal to $-\frac{m}{10-m}$.


Hint: Use the fact that $$\cos a + \cos b=2\cos\left(\frac12(a+b)\right)\cos\left(\frac12(a-b)\right).$$

If you pair the summands and apply above transformation, then the sum becomes a product with $10$ cosine factors, and a scaling factor of $2^5.$ So now at least we have an estimate of the sum.

  • $\begingroup$ I think it won't work since the all entries in cos are different $\endgroup$
    – Rishi
    Sep 11, 2019 at 13:33

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