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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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    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