Problem 1669

Given a regular triangle $OAB$ with side length $1$. Let $M$ and $N$ be the points that divide $AB$ into three equal parts, with $M$ being the closest to $A$. We need to find the real numbers $x$ and $y$ such that $\overrightarrow{m} = x \overrightarrow{a} + y \overrightarrow{b}$, and then find the dot product of $\overrightarrow{m}$ and $\overrightarrow{n}$.

1. **Finding $\overrightarrow{m}$ in terms of $\overrightarrow{a}$ and $\overrightarrow{b}$:**

Since $M$ divides $AB$ into three equal parts, we can use the section formula. The section formula states that if a point $P$ divides the line segment joining points $A$ and $B$ in the ratio $m:n$, then the position vector of $P$ is given by:
$$\overrightarrow{P} = \frac{n \overrightarrow{A} + m \overrightarrow{B}}{m+n}$$

Here, $M$ divides $AB$ in the ratio $1:2$. Therefore,
$$\overrightarrow{M} = \frac{2 \overrightarrow{A} + 1 \overrightarrow{B}}{1+2} = \frac{2 \overrightarrow{A} + \overrightarrow{B}}{3}$$

Thus, $x = \frac{2}{3}$ and $y = \frac{1}{3}$.

2. **Finding $\overrightarrow{n}$ in terms of $\overrightarrow{a}$ and $\overrightarrow{b}$:**

Similarly, $N$ divides $AB$ in the ratio $2:1$. Therefore,
$$\overrightarrow{N} = \frac{1 \overrightarrow{A} + 2 \overrightarrow{B}}{1+2} = \frac{\overrightarrow{A} + 2 \overrightarrow{B}}{3}$$

3. **Finding the dot product $(\overrightarrow{m}, \overrightarrow{n})$:**

We have:
$$\overrightarrow{m} = \frac{2 \overrightarrow{A} + \overrightarrow{B}}{3}, \quad \overrightarrow{n} = \frac{\overrightarrow{A} + 2 \overrightarrow{B}}{3}$$

The dot product is given by:
$$(\overrightarrow{m}, \overrightarrow{n}) = \left( \frac{2 \overrightarrow{A} + \overrightarrow{B}}{3} \right) \cdot \left( \frac{\overrightarrow{A} + 2 \overrightarrow{B}}{3} \right)$$

Simplifying, we get:
$$(\overrightarrow{m}, \overrightarrow{n}) = \frac{1}{9} \left( (2 \overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} + 2 \overrightarrow{B}) \right)$$

Expanding the dot product:
$$(2 \overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} + 2 \overrightarrow{B}) = 2 \overrightarrow{A} \cdot \overrightarrow{A} + 4 \overrightarrow{A} \cdot \overrightarrow{B} + \overrightarrow{B} \cdot \overrightarrow{A} + 2 \overrightarrow{B} \cdot \overrightarrow{B}$$

Since $\overrightarrow{A} \cdot \overrightarrow{A} = 1$ (magnitude of $\overrightarrow{A}$ is 1), $\overrightarrow{B} \cdot \overrightarrow{B} = 1$ (magnitude of $\overrightarrow{B}$ is 1), and $\overrightarrow{A} \cdot \overrightarrow{B} = \overrightarrow{B} \cdot \overrightarrow{A} = \cos(120^\circ) = -\frac{1}{2}$ (since $OAB$ is an equilateral triangle with $120^\circ$ between $\overrightarrow{A}$ and $\overrightarrow{B}$):

$$(2 \overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} + 2 \overrightarrow{B}) = 2 \cdot 1 + 4 \cdot \left( -\frac{1}{2} \right) + \left( -\frac{1}{2} \right) + 2 \cdot 1$$
$$= 2 - 2 - \frac{1}{2} + 2 = 1.5$$

Therefore,
$$(\overrightarrow{m}, \overrightarrow{n}) = \frac{1.5}{9} = \frac{1}{6}$$

The final answer is $\boxed{ \frac{1}{6} }$