Gizi and Tibi are playing the following game. Gizi selects a point $X$ on the side $AB$ of triangle $ABC$, then Tibi chooses a point $Y$ on the side $BC$. Finally, Gizi finds a point $Z$ on $AC$. Gizi aims to make the area of triangle $XYZ$ as large as possible, while Tibi aims to make it as small as possible. What will be the area of triangle $XYZ$ if both play optimally? (H)
We will show that if both play well, then $T_{X Y Z}=\frac{1}{4} T_{A B C}$. For this, it is sufficient to prove the following two statements:

(i) Tibi can play in such a way that $T_{X Y Z} \leq \frac{1}{4} T_{A B C}$;

(ii) Gizi can play in such a way that $T_{X Y Z} \geq \frac{1}{4} T_{A B C}$.

Tibi's strategy is as follows: if Gizi selects point $X$ on side $A B$, then Tibi will place $Y$ on side $B C$ such that $X Y$ is parallel to $A C$ (this can always be done). If the height of triangle $A B C$ from $B$ is denoted by $m_{b}$, the height of triangle $X B Y$ from $B$ by $m_{b}^{\prime}$, and the ratio $\frac{X B}{A B}$ by $t$, then no matter where Gizi chooses point $Z$, the height of triangle $X Y Z$ from $Z$ will always be $m_{z}=m_{b}-m_{b}^{\prime}$ (Figure 1). However, due to the similarity of triangles $A B C$ and $X B Y$, we have $m_{b}^{\prime}=t \cdot m_{b}$ and $X Y=t \cdot A C$. Therefore,

\[
T_{X Y Z}=\frac{1}{2} X Y \cdot m_{z}=\frac{1}{2}(t \cdot A C)\left(m_{b}-t \cdot m_{b}\right)=t(1-t) \cdot \frac{1}{2} A C \cdot m_{b}=t(1-t) \cdot T_{A B C}
\]

However, $t(1-t)=\frac{1}{4}-\left(t-\frac{1}{2}\right)^{2} \leq \frac{1}{4}$, which proves statement (i).

Gizi's strategy is as follows: she selects $X$ on side $A B$ and $Z$ as the midpoint of side $A C$. In this case, no matter where Tibi chooses point $Y$ (Figure 2), the height of triangle $X Y Z$ from $Y$ will be exactly half the height of triangle $A B C$ from $A$, because $X Z$ is the midline of triangle $A B C$; and for the same reason, $X Z=\frac{1}{2} B C$ also holds. Thus, in this case, $T_{X Y Z}=\frac{1}{4} T_{A B C}$. This proves statement (ii).