
6. In the real numbers, solve the following system of equations:

$$
\begin{aligned}
& \sin ^{2} x+\cos ^{2} y=\tan ^{2} z, \\
& \sin ^{2} y+\cos ^{2} z=\tan ^{2} x, \\
& \sin ^{2} z+\cos ^{2} x=\tan ^{2} y .
\end{aligned}
$$

(Pavel Calábek)


Solution. Substituting $\cos ^{2} x=a, \cos ^{2} y=b, \cos ^{2} z=c$ leads to the system

$$
\begin{aligned}
& 1-a+b=\frac{1}{c}-1, \\
& 1-b+c=\frac{1}{a}-1, \\
& 1-c+a=\frac{1}{b}-1,
\end{aligned}
$$

where $a, b, c \in(0,1\rangle$.

Adding these equations together yields

$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6
$$

so the harmonic mean of the numbers $a, b, c$ is $\frac{1}{2}$.

Multiplying the equations by the numbers $c$, $a$, and $b$, respectively, we get

$$
\begin{aligned}
c-a c+b c & =1-c \\
a-a b+a c & =1-a \\
b-b c+a b & =1-b
\end{aligned}
$$

which sums to $2(a+b+c)=3$. Therefore, the arithmetic mean of the numbers $a, b$, $c$ is $\frac{1}{2}$, too. Since the arithmetic and harmonic means are equal, it must be that $a=$ $b=c=\frac{1}{2}$. We can easily verify that this triple satisfies the system (1). The solutions of the original system are thus exactly the triples $\left(\frac{1}{4} \pi+\frac{1}{2} k \pi, \frac{1}{4} \pi+\frac{1}{2} l \pi, \frac{1}{4} \pi+\frac{1}{2} m \pi\right)$, where $k, l, m$ are integers.

Jiné řešení. We use the substitution from the above solution. The system (1) is cyclic; if a triple $(p, q, r)$ satisfies it, then so do the triples $(q, r, p)$ and $(r, p, q)$. Therefore, it suffices to find the solutions for which $a \geqslant b, a \geqslant c$, and all the other solutions can then be obtained by cyclic exchange.

Let $a \geqslant b, a \geqslant c$. It follows from the first equation that $1 / c=2-a+b \leqslant 2$, so $c \geqslant \frac{1}{2}$. Similarly, it follows from the third equation that $1 / b=2-c+a \geqslant 2$, so $b \leqslant \frac{1}{2}$, and thus $b \leqslant c$. ¿From the second equation, we have $1 / a=2-b+c \geqslant 2$, so $a \leqslant \frac{1}{2}$. Altogether,

$$
\frac{1}{2} \geqslant a \geqslant c \geqslant \frac{1}{2}
$$

so $a=c=\frac{1}{2}$. Now, any of the equations of the system (1) yields $b=\frac{1}{2}$. Just as in the previous case, we verify that the found triple satisfies the system (1); so the solutions of the system are the triples $\left(\frac{1}{4} \pi+\frac{1}{2} k \pi, \frac{1}{4} \pi+\frac{1}{2} l \pi, \frac{1}{4} \pi+\frac{1}{2} m \pi\right)$, where $k, l$, $m$ are integers.

## First Round of the 62nd Czech and Slovak Mathematical Olympiad (December 6th, 2012)

![](https://cdn.mathpix.com/cropped/2024_04_17_5816354d2b4b35f9c465g-13.jpg?height=226&width=537&top_left_y=475&top_left_x=742)
