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let $w$ is postive integer,if there exist $a,b,c(\pi\le a<b<c\le 2\pi)$such $$\sin{(aw)}+\sin{(bw)}+\sin{(cw)}=3$$ Find the $w$ range.

My attempt： since $$\sin{(wa)}\le 1,\sin{(wb)}\le 1,\sin{(wc)}\le 1$$ then $$\sin{(wa)}=\sin{(wb)}=\sin{(wc)}=1$$

$$aw=\dfrac{\pi}{2}+2k_{1}\pi,k_{1}\in Z$$ $$bw=\dfrac{\pi}{2}+2k_{2}\pi,k_{2}\in Z$$ $$cw=\dfrac{\pi}{2}+2k_{3}\pi,k_{3}\in Z$$

what approaches do you think, I could take to solving the next step?

  • Doubting that sums of 3 sines would exceed 3 in a broad range, I selected 2 angles symmetrical about $ 3π/2 $ as $(pπ,3π/2,(3−p)π)$ and got approximate numerical solution for single points (p= 1.14694, 0.928) but without any range . That is the guess is that there could be only maxima points to satisfy the given equation. – Narasimham Oct 12 '15 at 15:12

3 Answers3

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We have $$\begin{align}&\sin(aw)+\sin(bw)+\sin(cw)=3\\\\&\iff \sin(aw)=\sin(bw)=\sin(cw)=1\\\\&\small\iff\text{There exist three integers $k,l,m$ such that $aw=\frac{\pi}{2}+2k\pi,bw=\frac{\pi}{2}+2l\pi,cw=\frac{\pi}{2}+2m\pi$}\\\\&\small\iff\text{There exist three integers $k,l,m$ such that $\pi\le\frac{\pi+4k\pi}{2w}\lt\frac{\pi+4l\pi}{2w}\lt\frac{\pi+4m\pi}{2w}\le 2\pi$}\\\\&\small\iff \text{There exist three integers $k,l,m$ such that $2\le\frac{1+4k}{w}\lt\frac{1+4l}{w}\lt\frac{1+4m}{w}\le 4$}\end{align}$$

So, let us find a condition for $w$ such that there exist at least three distinct positive integers $n$ satisfying $2\le\frac{1+4n}{w}\le 4$, i.e. $\frac{2w-1}{4}\le n\le \frac{4w-1}{4}$.

For $w=2N$ where $N$ is a positive integer, the condition is $$\left\lfloor\frac{4w-1}{4}\right\rfloor-\left\lceil\frac{2w-1}{4}\right\rceil+1\ge 3\iff (2N-1)-N+1\ge 3\iff N\ge 3.$$

For $w=2N-1$ where $N$ is a positive integer, the condition is $$\left\lfloor\frac{4w-1}{4}\right\rfloor-\left\lceil\frac{2w-1}{4}\right\rceil+1\ge 3\iff (2N-2)-N+1\ge 3\iff N\ge 4.$$

Hence, the answer is $\color{red}{w\ge 6}$.

mathlove
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2

Hint:

You need to solve the system

$$\frac{4k_1+1}a=\frac{4k_2+1}b=\frac{4k_3+1}c.$$

Then the numbers $\frac1a,\frac1b,\frac1c$ must be proportional to three integers (otherwise there is no solution) and you end-up with a classical system of linear Diophantine equations

$$\alpha k_1-\beta k_2=d,\\\alpha k_1-\gamma k_3=d'.$$

You will need to find the solutions in $k_1$ that coincide.

1

we also solve if $w$ be postive real numbers,I have done This result is $$w\in [\dfrac{17}{4},\dfrac{9}{2}]\bigcup [\dfrac{21}{4},+\infty) $$ This problem is equivl this:there $p,q,r\in N^{+}$,such $$w\pi\le 2p\pi+\dfrac{\pi}{2}<2q\pi+\dfrac{\pi}{2}<2r\pi+\dfrac{\pi}{2}\le 2w\pi$$ case1 if $w\ge 6$ it is clear,because $(w\pi,2w\pi)\subset (6\pi,12\pi)$,and $\dfrac{12\pi-6\pi}{3}=2\pi$,

case 2: if $0<w<6$,then $(w\pi,2w\pi)\subset (0,12\pi)$,

then

2.1 if $$w\pi\le\dfrac{\pi}{2}<\dfrac{5\pi}{2}<\dfrac{9\pi}{2}\le 2w\pi$$ then $$\Longrightarrow w\le\dfrac{1}{2},\rm{and} w\ge\dfrac{9}{4}$$ that's impossible.

2.2 if $$w\pi\le\dfrac{5\pi}{2}<\dfrac{9\pi}{2}\le\dfrac{13\pi}{2}\le 2w\pi$$ then $$\Longrightarrow w\ge \dfrac{13}{4},\rm{and} w\le \dfrac{5}{2}$$ also impossible

$2.3$ if $$w\pi\le\dfrac{9\pi}{2}<\dfrac{13\pi}{2}<\dfrac{17}{2}\pi\le 2w\pi$$ then $$\Longrightarrow \dfrac{17}{4}\le w\le\dfrac{9}{2}$$

$2.4$ if $$w\pi\le\dfrac{13\pi}{2}<\dfrac{17\pi}{2}<\dfrac{21\pi}{2}\le 2w\pi$$ then $$\dfrac{21}{4}\le w\le\dfrac{13}{2}$$

so $$w\in [\dfrac{17}{4},\dfrac{9}{2}]\bigcup [\dfrac{21}{4},+\infty) $$

Note: if $w\in Z^{+}$,then $w\ge 6,w\in Z^{+}$$

math110
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