# Finding equal sums of constrained trig functions

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 $\pi/4$. – Tom Stephens Aug 18 '10 at 22:18
Also, this question might be a good example of needing a policy regarding questions which might be homework. – Aryabhata Aug 18 '10 at 22:19
It isn't homework, it's a problem from a math olympiad. – Weltschmerz Aug 18 '10 at 22:23
@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. – Aryabhata Aug 18 '10 at 22:24
It might be easier if you just used $\sin a + \cos a = \sqrt{2} \sin(a + \frac{\pi}{4})$ – Aryabhata Aug 19 '10 at 0:44