A point $B$ and an equilateral triangle with side lengths of $1 \mathrm{~cm}$ are given. The distances from point $B$ to two vertices of this triangle are $2 \mathrm{~cm}$ and $3 \mathrm{~cm}$.

Calculate the distance from point $B$ to the third vertex of the triangle.

(J. Tkadlec)
The vertices of the triangle are denoted as $C, D, E$, and the distances from the problem statement are assigned to the lengths of segments $B C$ and $B D$. Since $1+2=3$, point $B$ lies on the line $C D$. The distance from point $B$ to the third vertex $E$ is the hypotenuse of the marked right triangle $B F E$:

![](https://cdn.mathpix.com/cropped/2024_04_17_5ebc08916757a9a5715ag-3.jpg?height=357&width=916&top_left_y=718&top_left_x=567)

Segment $E F$ is the height of the equilateral triangle $C D E$, thus

$$
|E F|=\frac{\sqrt{3}}{2} \text{ cm}
$$

This is a known relationship that can be found in tables or calculated using the Pythagorean theorem in triangle $C F E$. Point $F$ is the midpoint of segment $C D$, so

$$
|B F|=|B C|+|C F|=\frac{5}{2} \text{ cm}
$$

The length of the hypotenuse of triangle $B F E$ is, according to the Pythagorean theorem,

$$
|B E|=\sqrt{\left(\frac{5}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\sqrt{\frac{28}{4}}=\sqrt{7}(\text{ cm}).
$$

The distance from point $B$ to the third vertex of the triangle is $\sqrt{7} \text{ cm}$.

Evaluation. 1 point for determining that $B$ lies on the line $C D$; 3 points for calculating $|B E|$; 2 points for the quality of the commentary.