Trigonometry Trigonometry Math Forum

 July 20th, 2014, 04:40 PM #21 Global Moderator   Joined: Dec 2006 Posts: 20,476 Thanks: 2039 The correct answer is 25, given a reasonable interpretation of the question! Now you have to (a) identify the errors already made, (b) identify the intended interpretation, and (c) show that the answer, based on that interpretation, is indeed 25. I'll let this one run for a bit to see whether anyone cracks it.
July 22nd, 2014, 03:12 AM   #22
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 Originally Posted by Math Message Board tutor Just refer to someone by the exact spelling of their username. Then no one on here should object to that.
Thanks for the advice Math Message Bo!

 July 28th, 2014, 03:11 PM #23 Global Moderator   Joined: Dec 2006 Posts: 20,476 Thanks: 2039 Here's a gentle hint: it's not difficult to get the right answer (25), but it's very easy to make a slip and hence get the wrong answer (100).
July 28th, 2014, 06:01 PM   #24
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 Originally Posted by tahirimanov OK. $\displaystyle \sin ( \sqrt{ax - x^2} ) = 0$ and $\displaystyle x_1 + x_2 + .... x_n = 100$. $\displaystyle x_i$ is the root (solution). i=1,2,3,....n. $\displaystyle \sqrt{ax - x^2} = k \pi . \, \, k \in \mathbb{Z}$ $\displaystyle ax-x^2 = k^2 \pi^2 \Rightarrow x^2 - ax + k^2 \pi^2 = 0$ $\displaystyle x=\frac{a \pm \sqrt{a^2 - 4 k^2 \pi^2}}{2}$
$\displaystyle \sqrt{ax - x^2} = k \pi . \, \, k \in \mathbb{Z}$

But the primary solutions are for $k = 0$ and $k = 1$.

Thus, solutions are:
$$\text{For k=0:} \qquad x=\frac{a \pm \sqrt{a^2 - 4 k^2 \pi^2}}{2} \\ \text{For k=1:} \qquad x=\frac{a \pm \sqrt{a^2 - 4 k^2 \pi^2}}{2} \\$$

When we sum the four solutions we get $2a = 100$.

So I'm now missing four solutions.

 August 3rd, 2014, 07:54 PM #25 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications You get a as the sum of the 2 solutions for each k. So you have to find a such that as k gets too large, the solutions become complex, and the real solutions sum to 100 (don't forget that we are taking the sine, so I assume that they are not looking for complex numbers in this case). With a=25, for k>3, the solutions are no longer real. So k=0,1,2,3. So 4a=100, a=25. With a=100, k can go up to 15 so things don't add up because you would have 15a=100. Similar for a=50, a=100/3, and too few k for a=20. Thanks from Pilosa
August 4th, 2014, 07:13 AM   #26
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Correction:

Quote:
 With a=100, k can go up to 15 so things don't add up because you would have 15a=100
Actually, it would be 16a=100 since k=0,1,2...14,15.

Also, to be more clear, since each solution of k gives a sum of a:

Letting k=0 gives a=100, but valid k are 0,1,2...14,15; giving 16a=100. Contradiction.

Letting k=0,1 gives a=100/2, but valid k are 0,1,2...7; giving 8a=100. Contradiction.

Letting k=0,1,2 gives a=100/3, but valid k are 0,1,2...5; giving 6a=100. Contradiction.

Letting k=0,1,2,3 gives a=100/4, and valid k are 0,1,2...3; giving 4a=100. Works., so 25 is a solution.

Letting k=0,1,2,3,4 gives a=100/5, but valid k are 0,1,2,3; giving 4a=100. Contradiction.

Etc.

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