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Topic: Trigonometry Problem (Read 133 times) 

navdeep1771
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Trigonometry Problem
« on: Jan 10^{th}, 2018, 7:31am » 
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If A1 + A2 + A3 + A4 + A5 + A6 + A7 + A8 + A9 + A10 = 300° Find the maximum value of: (SinA1 + SinA2 +.....+ SinA10) [Note: It's a very challenging math problem. My math teacher created this problem when he was alone in his cabin and next day gave this in class. Even my friend who was qualified for IMO, couldn't solve this problem! Let's see if anyone of you can solve it to exact answer with a logical solution].

« Last Edit: Jan 11^{th}, 2018, 7:41am by navdeep1771 » 
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towr
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Re: Trigonometry Problem
« Reply #1 on: Jan 10^{th}, 2018, 10:29am » 
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It's fairly easy to get at least 9 270 x 4 + 90 x 4 + 510 x 2 Of course even if you could easily get to the optimal solution from there, proving it is yet another matter. It's a bit too long ago for me to pull out Lagrangian multipliers, or even be certain that's the (or a) way to go.

« Last Edit: Jan 10^{th}, 2018, 10:55am by towr » 
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navdeep1771
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Re: Trigonometry Problem
« Reply #2 on: Jan 11^{th}, 2018, 5:17am » 
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Yeah, you are going right. It's just the first step in the optimal solution. Well, obviously answer is more than 9 and even more than 9.5! And let me notify you that "you can approach to the exact answer by high school mathematics only" no need think about any complex theorems etc.


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Grimbal
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Re: Trigonometry Problem
« Reply #3 on: Jan 11^{th}, 2018, 8:16am » 
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If you take 8x90 + 2x270 the sum is 180 This is the first approximation To reach 300 you need to add 120. Assuming bluntly that the optimum is to add the same amount to each angle, you add 12 to each of them You get 8x102 + 2x258. The sines sum up to 9.781476007

« Last Edit: Jan 11^{th}, 2018, 8:16am by Grimbal » 
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navdeep1771
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Re: Trigonometry Problem
« Reply #4 on: Jan 11^{th}, 2018, 8:35am » 
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Although your answer is correct but your way does not ensures that this value will be guaranteed maximum. Basically you did some statistics and used your clever mind to get to the answer. Perhaps motive of the problem was a complete mathematical solution which guarantees that the maximum is 0.9781476007

« Last Edit: Jan 11^{th}, 2018, 8:48am by navdeep1771 » 
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dudiobugtron
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Re: Trigonometry Problem
« Reply #5 on: Jan 11^{th}, 2018, 4:45pm » 
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As others have observed, the angles can be negative, effectively meaning that the angles really only need to sum to 300 mod 360. For the variables we are interested in, the value of sin(x) gets increasingly lower* as you get further away from x = 90. So therefore for any two of the variables pairwise (say, Ai and Aj), if the two variables are different (say Ai < Aj), you can get a higher sum of the sines for the same sum of angles by increasing Ai and decreasing Aj by the same amount until they are both equal. *As others have shown, the result is greater than 9, meaning none of the sin values can be negative. And the second derivative sin'' is negative when sin is positive. For this reason, the maximum will occur (no surprises here for anyone who has done maximisation with interchangeable variables before) when all of the variables are equal to each other (mod 360). Otherwise you could easily get a higher sum of sines by choosing two nonequal variables and making them equal as above, which would be a contradiction. The maximum value of sin(x) comes when x = 90 mod 360. 90 x 10 = 900 = 180 mod 360. To get it up to 300 mod 360 we need to add 120 to the total. Since all the variables will be the same, and as close to 90 as possible, this means adding the same amount (12) to each variable. So then you get Grimbal's answer. Another way to see this is to look at all of the values which would give 300 mod 360 (namely 30 + 36n; ie: 30, 66, 102, 138 etc...), and see which one gets closest to 90  ie: 102.  Slightly more robust proof of my 'pairwise' assertion: We know that the maximum value will be at least 9. The maximum each variable can contribute to the sum is 1, so even if 9 of them contribute 1 each, the 10th must contribute at least 0. This means each of the variables must be in between 0 and 180 (mod 360). Take any two of the variables (Ai and Aj), and call their sum Ai + Aj = S. If we keep S constant, then the maximum value for V = sin(Ai) + sin(Aj) = sin(Ai) + sin(SAi) occurs when dV/d(Ai) = 0, ie when: cos(Ai) + cos(SAi) = 0 cos(Ai) = cos(SAi) since Ai is between 0 and 180 (mod 360), where cos is 11, this means that Ai = SAi = Aj (mod 360).

« Last Edit: Jan 11^{th}, 2018, 5:36pm by dudiobugtron » 
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Grimbal
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Re: Trigonometry Problem
« Reply #6 on: Jan 12^{th}, 2018, 6:29am » 
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My blunt assumption was based on the fact that the sine function is convex between 0 and 180°.


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navdeep1771
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Re: Trigonometry Problem
« Reply #7 on: Jan 26^{th}, 2018, 7:32pm » 
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You guys (Grimbal and towr) are very clever. The solution I have (which my teacher explained) is a way long than of Grimbal's solution. That's because my teacher used pure trigonometry to solve it. Well, I am coming with some more challening math problems, keep connected to this forum.

« Last Edit: Jan 26^{th}, 2018, 7:32pm by navdeep1771 » 
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