4. For what values of $x$ and $y$ is the equality $x^{2}+(1-y)^{2}+(x-y)^{2}=1 / 3$ true?
Answer: $\mathrm{x}=1 / 3, \mathrm{y}=2 / 3$.

Solution. After expanding the brackets and combining like terms, we get $0=2 x^{2}+2(1-y)^{2}+2(x-y)^{2}-2 / 3=4 x^{2}+4 y^{2}-4 x y-4 y+4 / 3=(2 x-y)^{2}+3(y-2 / 3)^{2}$. From here $2 x$ $=y=2 / 3$.