Variant 132. 9.6 meters. Variant 133. 2.0 seconds. Variant 134. 10.0 meters.

Evaluation criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; $\mathbf{1 5}$ points - mostly correct solution, but with defects (for example, the choice of value $\alpha$ is not justified); 10 points - solution is mostly correct, but contains computational errors; 5 points - correct formulas for range and flight time are written, but there is no significant progress.

## Problem 3 (Variant 131)

Three pulleys with parallel axes and identical radii $r=2$ cm are to be connected by a flat belt drive. The distance between the axes of rotation of pulleys $O_{1}$ and $O_{2}$ is 12 cm, and the distance between the axes of rotation of pulleys $O_{1}$ and $O_{3}$ is $10 \mathrm{~cm}$. The distance from the axis $O_{3}$ to the plane in which the axes $O_{1}$ and $O_{2}$ are located is 8 cm. Determine the length of the belt for the drive, which is made by sewing the ends of an inextensible rubberized cord (assume that the length of the belt is equal to the length of this cord). Will a cord 54 cm long always be sufficient for its manufacture?
Solution. 1) First, we solve the geometric problem of finding the side $O_{2}O_{3}$. According to the condition, $\sin \angle O_{1}=\frac{8}{10}=\frac{4}{5}$; therefore, $\cos \angle O_{1}= \pm \frac{3}{5}$. The third side is found using the cosine theorem and equals either 10 cm or $2 \sqrt{97}$ cm.

2) The perimeter of the figure will be the sum of three straight segments and the sum of three arcs. The sum of the straight segments equals the perimeter of the triangle $O_{1} O_{2} O_{3}$.
3) The sum of the lengths of the three arcs equals $r \cdot(\pi-\alpha)+r \cdot(\pi-\beta)+r \cdot(\pi-\gamma)$ (where $\alpha, \beta, \gamma$ are the angles of the triangle $O_{1} O_{2} O_{3}$), which is equal to $r \cdot(3 \pi-\alpha-\beta-\gamma)=r \cdot(3 \pi-\pi)=2 \pi r$.
4) In the case of an acute triangle, the required length of the belt is $12+10+10+2 \pi \cdot 2=32+4 \pi$ cm (for reference: $32+4 \pi \approx 44.57$). This is obviously less than 54 (since $32+4 \pi < 54 \Leftrightarrow \sqrt{97}+2 \pi > 16 \Leftrightarrow \sqrt{97}+2 \pi > 9.8+6.2=16$).

Answer: a) $32+4 \pi$ cm or $22+2 \sqrt{97}+4 \pi$ cm; b) not always.

## Answers of other variants:

Variant 132. a) $36+6 \pi$ cm or $23+3 \sqrt{41}+6 \pi$ cm; b) not always.

Variant 133. a) $32+6 \pi$ cm or $22+2 \sqrt{97}+6 \pi$ cm; b) not always.