Nóra and Kristóf play the following game: on a given triangle $A B C$, Nóra selects a point $N_{1}$ on the side $A B$. Then Kristóf chooses a point $K$ on the side $B C$. Finally, Nóra marks a point $N_{2}$ on the side $C A$. Nóra aims to maximize the area of the triangle $N_{1} K N_{2}$, while Kristóf aims to minimize it. What will be the area of the triangle $N_{1} K N_{2}$ if both play optimally?
Solution. If Nóra chose point $N_{1}=A$ or $N_{1}=B$, then Kristóf could choose point $K$ sufficiently close to point $C$ or $B$, respectively, to make the area of triangle $N_{1} K N_{2}$ arbitrarily close to 0, regardless of where Nóra marks point $N_{2}$ on side $A C$.

Therefore, Nóra must choose point $N_{1}$ inside the segment $A B$.

Suppose that in Nóra's choice, $A N_{1}=\lambda A B (0 < \lambda < 1)$. If Kristóf chooses $K$ such that $B K > \lambda B C$, then for example, if Nóra chooses $N_{2}^{\prime \prime}=A$,

$$
T\left(N_{1} K N_{2} \triangle\right)=T\left(N_{1} K A \triangle\right) < \lambda T_{A B C \Delta}
$$

Similarly, if Kristóf chooses $K$ such that $B K < \lambda B C$, then for example, if Nóra chooses $N_{2}^{\prime \prime}=C$,

$$
T\left(N_{1} K N_{2} \triangle\right)=T\left(N_{1} K C \triangle\right) < (1-\lambda) T_{A B C \Delta}
$$

Thus, we have shown that Kristóf must choose point $K$ on side $B C$ such that $N_{1} K \parallel A C$, and in this case, the area of triangle $N_{1} K N_{2}$ is independent of the point $N_{2}$ that Nóra marks on side $C A$.

Therefore, in the correct game strategy, the maximum area of triangle $N_{1} K N_{2}$ depends on the correct choice of $\lambda$.

Since $N_{1} K \parallel A C$, triangles $N_{1} B K$ and $A B C$ are similar because $\angle K N_{1} B = \angle C A B$ and $\angle B K N_{1} = \angle B C A$ (see figure 3).

Since $N_{1} B = (1-\lambda) A B$, by the similarity established earlier, $N_{1} K = (1-\lambda) A C = (1-\lambda) b$, and the height from vertex $B$ in triangle $N_{1} B K$ is $(1-\lambda) \cdot m_{b}$.

Thus, the distance between the parallel sides $N_{1} K$ and $A C$ is $m_{b} - (1-\lambda) m_{b} = \lambda m_{b}$, so the area of triangle $N_{1} K N_{2}$ is:

$$
\frac{(1-\lambda) b \cdot \lambda m_{b}}{2} = \lambda(1-\lambda) T_{A B C \Delta}
$$

By the arithmetic mean-geometric mean inequality:

$$
\lambda(1-\lambda) \leq \left[\frac{\lambda + (1-\lambda)}{2}\right]^{2} = \frac{1}{4}
$$

Equality holds when $\lambda = 1 - \lambda$, i.e., $\lambda = \frac{1}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_7d7293666aa96bf58121g-2.jpg?height=518&width=596&top_left_y=126&top_left_x=754)

Therefore, in the correct game strategy:

- Nóra chooses the midpoint of side $A B$ as $N_{1}$;
- Kristóf chooses the midpoint of side $B C$ as $K$ (so that $N_{1} K$ is the midline parallel to $A C$);
- Finally, Nóra marks any point $N_{2}$ on side $A C$.

Thus, the area of triangle $N_{1} K N_{2}$ is $\frac{1}{4}$ of the area of triangle $A B C$.