Czech Republic competition problems

Consider an arbitrary line segment of the polyline and denote by $d$ its length and by $x$ and $y$ the lengths of its perpendicular projections on the sides of lengths 2 and 3, respectively. Cauchy-Schwarz inequality gives us
$$(2x + 3y)^2 \leq (2^2 + 3^2)(x^2 + y^2) = 13d^2,$$
which means $2x + 3y \leq d \cdot \sqrt{13}$. Denote by $X$ and $Y$ the total length of all the perpendicular projections of all the line segments on the sides of lengths 2 and 3, respectively. Summing up our estimations for each line segment gives us $2X + 3Y \le 36 \cdot \sqrt{13} < 130$. But then either $2X < 40$, or $3Y < 90$. In the first case we would have $X < 20$, so on the side of length 2 there is a point that is contained in fewer than 10 projections. A line perpendicular to this side at this point intersects the polyline at most 9 times. The other case is analogous.