# A last year Putnam question maximum $\sum{\cos 3x}$

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.

• Is your claim that $\cos(3x)=4\cos^3(x)+3\cos(x)$? that's not even true at $x=0$. – lulu Sep 11 at 10:31
• I agree with @lulu: it should be $4\cos^3x\color{blue}{-}3\cos x$. – J.G. Sep 11 at 10:31
• 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. – Toby Mak Sep 11 at 10:32
• Lagrange method yields the restrictions $$3\sin (3x_j) = \lambda \sin x_j ,\quad 1\leq j\leq 10$$ – Alvin Lepik Sep 11 at 10:33
• See also here: artofproblemsolving.com/community/c7h1747710p11382944 – Michael Rozenberg Sep 11 at 11:28

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.

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