11.4. For what values of $x$ and $y$ is the equality

$$
x^{2}+(1-y)^{2}+(x-y)^{2}=\frac{1}{3} ?
$$
Answer: when $x=\frac{1}{3}, y=\frac{2}{3}$.

Solution. First method. After expanding the brackets and combining like terms, we get:

$$
\begin{aligned}
& x^{2}+(1-y)^{2}+(x-y)^{2}=\frac{1}{3} \quad \Leftrightarrow \quad 2 x^{2}+2 y^{2}-2 x y-2 y+\frac{2}{3}=0 \quad \Leftrightarrow \\
& \Leftrightarrow \quad \frac{1}{2}\left(4 x^{2}-4 x y+y^{2}\right)+\frac{3}{2}\left(y^{2}-\frac{4}{3} y+\frac{4}{9}\right)=0 \quad \Leftrightarrow \quad \frac{1}{2}(2 x-y)^{2}+\frac{3}{2}\left(y-\frac{2}{3}\right)^{2}=0 \quad \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 2 x - y = 0 , } \\
{ y - \frac { 2 } { 3 } = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{1}{3} \\
y=\frac{2}{3}
\end{array}\right.\right.
\end{aligned}
$$

Second method. Rewrite the equation as: $x^{2}-y x+\left(y^{2}-y+\frac{1}{3}\right)=0$ and consider it as a quadratic equation in $x$. Then $D=y^{2}-4 y^{2}+4 y-\frac{4}{3}=-3 y^{2}+4 y-\frac{4}{3}=-3\left(y^{2}-\frac{4}{3} y+\frac{4}{9}\right)=$ $=-3\left(y-\frac{2}{3}\right)^{2}$. The quadratic equation has roots if $D=-3\left(y-\frac{2}{3}\right)^{2} \geqslant 0$, that is, if $y=\frac{2}{3}$. Substituting this value into the original equation, we get $x=\frac{1}{3}$.

Third method. Using the inequality between the quadratic mean and the arithmetic mean (for non-negative numbers) and the fact that $|a| \geqslant a$, we get:

$$
\sqrt{\frac{x^{2}+(1-y)^{2}+(x-y)^{2}}{3}} \geqslant \frac{|x|+|1-y|+|x-y|}{3} \geqslant \frac{x+(1-y)+(y-x)}{3}=\frac{1}{3}
$$

Therefore, $x^{2}+(1-y)^{2}+(x-y)^{2} \geqslant \frac{1}{3}$.

Equality is achieved if and only if $x=1-y=y-x$, that is, when $x=\frac{1}{3}, y=\frac{2}{3}$.

Fourth method. Let $a=x, b=1-y, c=y-x$, then $a+b+c=1$ and $a^{2}+b^{2}+c^{2}=\frac{1}{3}$.

Consider the vectors: $\vec{m}(a ; b ; c), \vec{n}(1 ; 1 ; 1)$. Their magnitudes: $|\vec{m}|=\sqrt{a^{2}+b^{2}+c^{2}}=\frac{1}{\sqrt{3}}, \vec{n}=\sqrt{3}$.

Thus, $\vec{m} \cdot \vec{n}=|\vec{m}| \cdot|\vec{n}|$, hence the vectors are collinear. Then their coordinates are proportional: $\frac{a}{1}=\frac{b}{1}=\frac{c}{1}$, that is, $a=b=c=\frac{1}{3}$. Therefore, $x=1-y=y-x$, hence $x=\frac{1}{3}, y=\frac{2}{3}$.

## Evaluation Criteria

+ Correct answer and a fully justified solution (by any method)

$\pm$ Correct answer and a generally correct solution, but with minor gaps in justification

干 Correct solution approach, but errors that led to an incorrect answer

- Only the answer is provided
- The problem is not solved or is solved incorrectly