On the level terrain, the vertices of the square $ABCD$ are marked with trees. The tree at $A$ is $7 \mathrm{~m}$ tall, the one at $B$ is $13 \mathrm{~m}$ tall, and the one at $C$ is $17 \mathrm{~m}$ tall.

What is the minimum length of the side of the square, given that there is a point inside it that is equidistant from the tops of all 4 trees? How tall is the tree at vertex $D$?
The geometric place of points equidistant from the tops of trees standing at points $A$ and $B$ is the perpendicular bisector plane of the segment defined by the tops of these trees. This plane intersects a segment $EF$ perpendicular to the side $AB$ of the square $ABCD$. Let $BE = x$ and the side of the square be $a$.

![](https://cdn.mathpix.com/cropped/2024_05_02_2a8c7ca13dcfa5fa83d0g-1.jpg?height=727&width=1003&top_left_y=314&top_left_x=542)

Then,

$$
13^{2} + x^{2} = 7^{2} + (a - x)^{2}
$$

from which

$$
x = \frac{a^{2} - 120}{2a}
$$

The points in the square $ABCD$ that are equidistant from the tops of the trees at points $B$ and $C$ similarly lie on a segment $GH$ perpendicular to $BC$. Let $CG = y$, then

$$
17^{2} + y^{2} = 13^{2} + (a - y)^{2}
$$

from which

$$
y = \frac{a^{2} - 120}{2a}, \quad \text{so} \quad x = y
$$

This means that the point $O$ equidistant from the tops of the three trees lies on the diagonal $AC$ of the square. Thus, the distance from the top of the tree at $D$ is equal to the other three only if a 13-meter tall tree also stands at $D$, as it does at $B$.

The point equidistant from the tops of the four trees lies inside the square if $x = y > 0$, i.e., if $a > \sqrt{120} \, \text{m} = 10.95 \, \text{m}$.

Pászthory Róbert (Budapest, I. László Gymn., III. o. t.) Hunyady László (Szombathely, Nagy Lajos Gymn., III. o. t.)