206. Given a triangle $ABC$ with an area of 1. The first player chooses a point $X$ on side $AB$, the second player chooses a point $Y$ on side $BC$, and then the first player chooses a point $Z$ on side $AC$. The goal of the first player is to obtain a triangle $XYZ$ with the largest possible area, while the goal of the second player is to obtain the smallest possible area. What is the largest area that the first player can ensure for himself?
206. Answer: 1/4. First of all, note that the second player can ensure that \( S_{X Y Z} \leqslant 1 / 4 \) regardless of the first player's move. For this, it is sufficient for the second player to choose \( Y \) such that \( X Y \| A C \) (Fig. 84). Then, for any point \( Z \) on the base \( A C \), the inequality

\[
\frac{S_{X Y Z}}{S_{A B C}} = \frac{X Y}{A C} \cdot \frac{H-h}{H} = \frac{h(H-h)}{H^{2}} \leqslant \frac{1}{4}
\]

will be satisfied.

On the other hand, the first player, by choosing \( X \) and \( Z \) as the midpoints of sides \( A B \) and \( A C \), ensures that \( S_{X Y Z} = 1 / 4 \) regardless of the second player's choice.

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-192.jpg?height=308&width=450&top_left_y=484&top_left_x=113)

Fig. 84

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-192.jpg?height=348&width=349&top_left_y=426&top_left_x=636)

Fig. 85

\(\nabla\) A more difficult analogous problem is where the discussion is about the perimeters of triangles (see "Kvant", 1976, No. 4, p. 32).