Problem 10. Find all $k>0$ such that there is a function $f:[0,1] \times[0,1] \rightarrow$ $[0,1]$ satisfying the following conditions:
a) $f(f(x, y), z)=f(x, f(y, z))$;
b) $f(x, y)=f(y, x)$;
c) $f(x, 1)=x$;
d) $f(z x, z y)=z^{k} f(x, y)$,

for any $x, y, z \in[0,1]$.
10. Using (d) for $y=1$ and (c) we see that $f(z x, z)=z^{k} x$. Set $z x=y$. Then $f(y, z)=y z^{k-1}$ for $y \leq z$. Conversely, if $y \leq z$, then we set $y=z x$, $x \in[0,1]$, and $f(y, z)=f(z x, z)=z^{k} x=y z^{k-1}$.

Now let $x \leq y \leq z, x, y, z \in(0,1)$. Then using (a), we get that $f\left(x y^{k-1}, z\right)=$ $f\left(x, y z^{k-1}\right)$. Hence it follows from the above that

$$
\left\{x y^{k-1} z^{k-1}, x^{k-1} y^{(k-1)^{2}} z\right\} \cap\left\{x y^{k-1} z^{(k-1)^{2}}, x^{k-1} y z^{k-1}\right\} \neq \phi
$$

This easily implies that either $k=1$ and $f(x, y)=\min \{x, y\}$ or $k=2$ and $f(x, y)=x y$. A direct verification shows that both functions are solutions for the respective $k$.