, 2009

Solution:
From the conditions we have
$$36 = (x^{2} + y^{2})(z^{2} + t^{2}) = (x t + y z)^{2} + (x z - y t)^{2} \geq 36 + (x z - y t)^{2}$$
and this implies $x z - y t = 0$.
Now it is clear that
$$x^{2} + z^{2} + y^{2} + t^{2} = (x + z)^{2} + (y - t)^{2} = 13$$
and the maximum value of $z + x$ is $\sqrt{13}$. It is achieved for $x = \frac{4}{\sqrt{13}}, y = t = \frac{6}{\sqrt{13}}$ and $z = \frac{9}{\sqrt{13}}$.

Solution:
From inequality $x t + y z \geq 6$ and problem conditions we have:
$$\begin{gathered} (x t + y z)^{2} - 36 \geq 0 \Leftrightarrow \\ (x t + y z)^{2} - (x^{2} + y^{2})(z^{2} + t^{2}) \geq 0 \Leftrightarrow \\ 2 x y z t - x^{2} y^{2} - y^{2} t^{2} \geq 0 \Leftrightarrow \\ -(x z - y t)^{2} \geq 0 \end{gathered}$$
From here we have $x z = y t$.
Furthermore,
$$x^{2} + y^{2} + z^{2} + t^{2} = (x + z)^{2} + (y - t)^{2} = 13$$
and it follows that
$$(x + z)^{2} \leq 13$$
Thus,
$$x + z \leq \sqrt{13}$$
Equality $x + z = \sqrt{13}$ holds if we have $y = t$ and $z^{2} - x^{2} = 5$, which leads to $z - x = \frac{5}{\sqrt{13}}$. Therefore, $x = \frac{4}{\sqrt{13}}, y = t = \frac{6}{\sqrt{13}}, z = \frac{9}{\sqrt{13}}$.