Mathematical Olympiad in Ukraine, Third Round (January 23, ), 2010

Only $x = 0$ and $x = 1$.

Obviously $x \in [0, 1]$ and $0 \le (\sqrt{x})^{2009} + (\sqrt{1-x})^{2010} < 1$. On the other hand we have $x \le 1$ and $1-x \le 1$ which implies $(\sqrt{x})^{2009} + (\sqrt{1-x})^{2010} \le x < (1-x) = 1$. The case $(\sqrt{x})^{2009} + (\sqrt{1-x})^{2010} = 0$ is impossible, so we have to consider only the case $(\sqrt{x})^{2009} + (\sqrt{1-x})^{2010} = 1$. Equality in the previous inequality obtains iff $(\sqrt{x})^{2009} = x$ and $(\sqrt{1-x})^{2010} = 1-x$. This is possible only when $x = 0$ or $x = 1$. It's easy to see that both values satisfy the statement.