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Problem 63 of the 2001 St. Petersburg Mathematical Olympiad, Second Round, 11th grade:

Are there three different numbers $x, y, z$ in $[0,\pi/2]$ such that the numbers $\sin x$, $\sin y$, $\sin z$, $\cos x$, $\cos y$, $\cos z$ can be divided into three pairs with equal sums?

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Without loss of generality each of $x$, $y$, $z$ are $\le\pi/4$, since if for example $x\gt\pi/4$, then we can replace $x$ with $\pi/2-x$ which will not change the set $\{\cos(x),\sin(x)\}$.

So now $0\lt x \lt y \lt z \le \pi/4$, and we have the following ordering:

$\sin(x)\lt\sin(y)\lt\sin(z)\lt\cos(z)\lt\cos(y)\lt\cos(x)$

Hence the only possible pairings must be:

$A=\sin(x)+\cos(x)$

$B=\sin(y)+\cos(y)$

$C=\sin(z)+\cos(z)$

With $A=B=C$ (any other pairing will make one side heavier, term by term).

Now note $f(t)=\sin(t)+\cos(t)$ is a concave function since $\frac{d^2f(t)}{dt^2}=-\sin(t)-\cos(t)<0$ for $t \in (0,\pi/4)$.

So it follows that $B\gt \frac{A+C}{2}$.

So they can never be equal.

Edit: Fixed glitch spotted by Tom

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    erm, maybe I'm out of coffee, but I think your string of inequalities at the top is only true up to $\pi/4$.2010-08-18
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    Also, this question might be a good example of needing a policy regarding questions which might be homework.2010-08-18
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    It isn't homework, it's a problem from a math olympiad.2010-08-18
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    @Welt: Why don't you mention the source then? People are reluctant to answer questions which look like homework. If you upfront mention the source, people won't have these doubts, and they will be able to restrict the answers to the appropriate level.2010-08-18
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    No problem! St. Petersburg Mathematics Olympiad for school students, 11th class, somewhere at http://www.pdmi.ras.ru/~olymp/index.html If you want the exact year, I can look it up.2010-08-18
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    @Tom you are right about $\pi/4$ - but I hope I am not wrong in saying essentially the proof will remain the same. (I guess I will refrain from editing till the homework policy issue is settled... and I need to sleep.)2010-08-18
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    @Stephens: I too think those inequalities are valid only in [0,pi4].2010-08-18
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    Fixed the proof. Thanks Tom.2010-08-19
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    Btw, I'm curious, is homework forbidden by policy? The FAQ does not seem to list it.2010-08-19
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    It might be easier if you just used $\sin a + \cos a = \sqrt{2} \sin(a + \frac{\pi}{4})$2010-08-19
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    @KalEl: [This meta question](http://meta.math.stackexchange.com/questions/106/what-is-the-proper-way-to-handle-homework-questions) discusses homework.2010-08-19
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    Thanks Kaestur. I will outline my view on HW here. As not answering HW is not part of policy, my viewpoint is that I will choose to answer any questions that I find interesting. If it turns out to be a HW question it will most likely help than harm the person asking it. If he is foolhardy enough to copy it without understanding anything, that's his business. I have come here for my own interest in math.2010-08-20