Balti Tee 1990 · Ülesanne 7

Solution:

Let $A$, $B$, $C$, $D$ and $E$ be the vertices of the pentagon (in order), and take any point $O$ as origin. Let $M$ be the intersection point of the medians of the triangle $CDE$, and let $N$ be the midpoint of the segment $AB$. We have

$$\overline{OM} = \frac{1}{3}(\overline{OC} + \overline{OD} + \overline{OE})$$

and

$$\overline{ON} = \frac{1}{2}(\overline{OA} + \overline{OB})$$

The segment $NM$ may be written as

$$\overline{ON} + t(\overline{OM} - \overline{ON}), \quad 0 \leq t \leq 1$$

Taking $t = \frac{3}{5}$ we get the point

$$P = \frac{1}{5}(\overline{OA} + \overline{OB} + \overline{OC} + \overline{OD} + \overline{OE}),$$

the centre of gravity of the pentagon. Choosing a different side of the pentagon, we clearly get the same point $P$, which thus lies on all such line segments.