Given a triangle $ABC$. The first player selects a point $P$ on side $AB$, the second player selects a point $Q$ on side $BC$, then the first player selects a point $R$ on side $AC$.

The goal of the first player is to make the area of triangle $PQR$ as large as possible, while the goal of the second player is to make it as small as possible. What is the largest area that the first player can definitely achieve?
I. First, we show that the first player can achieve a quarter of the area $t$ of the triangle $ABC$. Choose $P$ on $AB$ and $R$ at the midpoint of $AC$, then $PR = \frac{1}{2} BC$, and regardless of the choice of $Q$, $m_{Q} = \frac{1}{2} m_{a}$, which means $t_{PQR} = \frac{1}{4} t_{ABC}$.

II. We show that the second player can ensure that the area of the triangle $PQR$ is not greater than $t / 4$. Play in such a way that $PQ \parallel AC$ (if $P \equiv B$, then $Q \equiv B$), then $\triangle PQB \cong \triangle ACB$, and by introducing $PQ = \lambda \cdot AC$,

$$
t_{PQR} = \frac{\lambda \cdot AC \cdot m_{b}(1-\lambda)}{2}
$$

From this, using the inequality between the arithmetic and geometric means, we get:

$$
t_{PQR} = \frac{AC \cdot m_{b}}{2} \cdot \lambda \cdot (1-\lambda) \leq t_{ABC} \cdot \frac{1}{4}
$$

Thus, the largest achievable area is one quarter of the area of triangle $ABC$.