In a unit-square grid, points $A, B, C$, and $D$ are the vertices of a $10 \times 10$ square. At points $A, B$, and $C$, there is a doll each. We moved these dolls to $B$, $A$, and $D$ respectively through some sequence of steps; in each step, we moved one arbitrarily chosen doll to an adjacent grid point, without stepping out of the square. In every position, the area of the triangle formed by the dolls was at least $t$. What is the largest possible value of $t$?

Pap László, Szeged
Some of the solvers found that the first sentence in the problem statement was a short, conventional formulation, which they did not consider precise. Therefore, now in the solution, we clarify that the points $A, B, C, D$ - in the described order - are (different) vertices of a $10 \times 10$ grid square, and the sides of this square are parallel to the main lines (the lines forming the grid).

![](https://cdn.mathpix.com/cropped/2024_05_02_6188ea1c02169f861be4g-1.jpg?height=702&width=509&top_left_y=316&top_left_x=797)

Figure 1

From the fact that which doll needs to end up in which vertex, it can be determined that the direction of the traversal of the triangle formed by the dolls has changed by the end of the sequence of steps. This means that one of the dolls has crossed the line connecting the other two dolls. Let these two dolls be $E$ and $F$, and the one that crossed the line $EF$ be $G$. Let the position of $G$ before the crossing be $G_{1}$, and after the crossing be $G_{2}$. Since the distance $G_{1} G_{2}$ is 1 unit, the smallest distance from point $G$ to the line $EF$, let this be $m$, satisfies $m \leq 0.5 \cdot \sin \alpha$, where $\alpha$ is the angle between the lines $EF$ and $G_{1} G_{2}$. Using the notation from Figure 1, $\sin \alpha = \frac{y}{EF}$, and thus the area $t$ of triangle $EFG$ is:

$$
t = \frac{EF \cdot m}{2} \leq \frac{EF \cdot 0.5 \cdot \frac{y}{EF}}{2} = \frac{y}{4}
$$

Since the points $E, F, G$ are on a $10 \times 10$ grid square, $y \leq 10$, and thus $t \leq 2.5$.

We will show that $t = 2.5$ is possible. In Figure 2, the doll standing on point $A$ was moved to $B$ via points $M$ and $N$, the doll standing on $B$ was moved to $A$ via point $P$, and the doll standing on $C$ was moved to $D$ via the shortest path. During this process, the areas of triangles $PG_{1}D$ and $PG_{2}D$ were 2.5, and the areas of the other triangles formed by the dolls were more than 2.5.

![](https://cdn.mathpix.com/cropped/2024_05_02_6188ea1c02169f861be4g-1.jpg?height=700&width=708&top_left_y=1645&top_left_x=253)

Figure 2

![](https://cdn.mathpix.com/cropped/2024_05_02_6188ea1c02169f861be4g-2.jpg?height=477&width=466&top_left_y=127&top_left_x=1250)

Figure 3

Gergely Levente (Debrecen, Fazekas M. Gymnasium, 4th grade)

Note. The original problem statement allowed for a $10 \times 10$ grid square as shown in Figure 3. Similarly to the given solution, we find that in this case, the maximum value of $t$ is 3.

Burcsi Péter (Pápa, Türr István Gymnasium, 2nd grade) Koblinger Egmont (Fazekas M. Főv. Gyak. Gymnasium, 3rd grade)