Discount Putnam

Geometry Level 4

if ∑ i = 1 9 cos ⁡ ( x i ) = 0 \displaystyle \sum_{i = 1}^{9}\cos(x_i) = 0 for real numbers x 1 , x 2 , . . . , x 9 x_1, x_2, ..., x_9 , find the maximum possible value of ∑ i = 1 9 cos ⁡ ( 3 x i ) \displaystyle \sum_{i = 1}^{9}\cos(3x_i)


The answer is 9.

Pedro Cardoso by Pedro Cardoso , 20, Brazil

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3 solutions

Pedro Cardoso
Jul 31, 2019

The maximum possible value for a sum like ∑ i = 1 9 cos ⁡ ( y i ) \displaystyle \sum_{i=1}^{9}\cos(y_i) for a sequence y 1 , y 2 , . . . , y 9 y_1, y_2, ..., y_9 is, of course, 9 9 as the maximum value of cos ⁡ ( x ) \cos(x) is 1 1 at x ≡ 0 x \equiv 0 .

Now, note that cos ⁡ ( 0 ) + cos ⁡ ( 2 π 3 ) + cos ⁡ ( 4 π 3 ) = 0 \cos(0) + \cos(\frac{2\pi}{3}) + \cos(\frac{4\pi}{3}) = 0 and 3 ⋅ 0 ≡ 3 ⋅ 2 π 3 ≡ 3 ⋅ 4 π 3 ≡ 0 3 \cdot 0 \equiv 3 \cdot \frac{2\pi}{3} \equiv 3 \cdot \frac{4\pi}{3} \equiv 0 , so, with the sequence ( 0 , 0 , 0 , 2 π 3 , 2 π 3 , 2 π 3 , 4 π 3 , 4 π 3 , 4 π 3 ) (0, 0, 0, \frac{2\pi}{3}, \frac{2\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{4\pi}{3}, \frac{4\pi}{3}) , we satisfy the first sum, and the second sum is 9 \boxed 9

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Here are 9 terms. Question - what is the answer for 10 terms? I suppose 480/49.

Yuriy Kazakov - 1 year, 10 months ago

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Yes. I think recently I have seen a problem posted about the original problem, but I can't find it anymore, maybe it has been deleted. Anyway, it's quite interesting that when the number of terms isn't a multiple of 3 3 the problem becomes much harder. A solution to the original problem can be found here , but I found it a little hard to follow.

My solution envolved:

  • Using the cos ⁡ ( 3 x ) \cos(3x) identity to show ∑ i = 0 10 cos ⁡ ( 3 x i ) = 4 ∑ i = 0 10 cos ⁡ 3 ( x i ) \displaystyle \sum_{i=0}^{10}\cos(3x_i) = 4 \displaystyle \sum_{i=0}^{10}\cos^3(x_i) .
  • Letting cos ⁡ ( x i ) = y i \cos(x_i) = y_i
  • Saying that, for some sequence that satisfies the first sum, there are a a positive numbers and 10 − a 10 - a negative numbers

Now you only need to prove a couple of things:

  • a < 5 a < 5 (i.e There are more negatives than positives)
  • By replacing all the negative numbers by their average, the first sum is still 0 0 , but the second sum is bigger
  • By replacing all the positive numbers by their average, then scaling all the numbers up so that all the positive numbers become 1 1 , the first sum is still 0 0 , but the second sum is bigger.

This proves there can only be two values in the sequence: 1 1 , repeated a a times, and − v -v , repeated 10 − a 10-a times. Now you can just test what is the maximum sum by checking all the values of a a from 1 1 to 10 10 .

It turns out a = 3 a = 3 , so the first sum is 3 − 7 v = 0 3 - 7v = 0 and v = 3 7 v = \frac{3}{7} , so the second sum is 4 ( 3 − 7 ( 3 7 ) 3 ) = 480 49 4 \left(3 - 7 \left(\frac{3}{7}\right)^3 \right) = \boxed{ \frac{480}{49} }

Pedro Cardoso - 1 year, 10 months ago

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Fine. Solution by Lagrange method from Mark Hennings was here - but it is delete.

Yuriy Kazakov - 1 year, 10 months ago

We want cos ⁡ ( 3 θ ) = 1 \cos \left( 3\theta \right)= 1 that's true if θ = 2 π 3 \theta = \frac{2\pi}{3} and recall that cos ⁡ ( θ ) = − cos ⁡ ( π − θ ) \cos \left( \theta \right) = - \cos \left( \pi - \theta \right) and so cos ⁡ ( 2 π 3 ) = − cos ⁡ ( π 3 ) \cos \left( \frac{2\pi}{3} \right) = - \cos \left( \frac{\pi}{3} \right) And cos ⁡ ( π 3 ) \cos \left( \frac{\pi}{3} \right) is precisely half the base of an equilateral triangle, so cos ⁡ ( π 3 ) = 1 2 \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} thus we have cos ⁡ ( 2 π 3 ) = − 1 2 ⇔ 2 ⋅ cos ⁡ ( 2 π 3 ) = − 1 \cos \left( \frac{2\pi}{3} \right) = - \frac{1}{2} \Leftrightarrow 2 \cdot \cos \left( \frac{2\pi}{3} \right) = -1 We note that cos ⁡ ( 0 ) = 1 \cos \left( 0 \right) = 1 and so then cos ⁡ ( 2 π 3 ) + cos ⁡ ( 2 π 3 ) + cos ⁡ ( 0 ) = 0 ⇔ 3 ⋅ ( cos ⁡ ( 2 π 3 ) + cos ⁡ ( 2 π 3 ) + cos ⁡ ( 0 ) ) = 0 \cos \left( \frac{2\pi}{3} \right) + \cos \left( \frac{2\pi}{3} \right) + \cos \left( 0 \right) = 0 \Leftrightarrow 3 \cdot \left(\cos \left( \frac{2\pi}{3} \right) + \cos \left( \frac{2\pi}{3} \right) + \cos \left( 0 \right)\right) = 0 Now when we triple each angle, every cosine becomes one and we get 3 ( cos ⁡ ( 2 π ) + cos ⁡ ( 2 π ) + cos ⁡ ( 0 ) ) = 9 3\left( \cos \left( 2\pi \right) + \cos \left( 2\pi \right) + \cos \left( 0 \right) \right) = 9

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Wang Xingyu
Sep 6, 2019

1.Use the formula of cos(3x)

∑ i = 1 9 c o s ( 3 x i ) = ∑ i = 1 9 ( 4 c o s 3 x i − 3 c o s ( x i ) ) = ∑ i = 1 9 4 c o s 3 x i \sum_{i=1}^{9} cos(3x_{i}) = \sum_{i=1}^{9} (4cos^{3}x_{i} - 3cos(x_{i})) = \sum_{i=1}^{9} 4cos^{3}x_{i}

2.Considering there're 9 x i x_{i} , to find the maximum, must in the following circumstance.

x i x_{i} =1, 1, 1, 1, − k 9 − k \frac{-k}{9-k} , ..... − k 9 − k \frac{-k}{9-k}

k is the number of 1

so ∑ i = 1 9 4 c o s 3 x i = k + ( − k ) 3 ( 9 − k ) 2 \sum_{i=1}^{9} 4cos^{3}x_{i} = k + \frac{(-k)^{3}}{(9-k)^{2}}

obviously, choose k=3, the sum is 9

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