4. On a circle, 60 red points and one blue point are marked. All possible polygons with vertices at the marked points are considered. Which type of polygons is more numerous, and by how many: those with a blue vertex, or those without it?
Answer: The number of polygons with a blue vertex is 1770 more than the number of polygons without a blue vertex.

Solution. Let's call polygons with a blue vertex blue, and those without a blue vertex red. Take an arbitrary red polygon. Adding a blue vertex to it gives exactly one blue polygon. Thus, any blue polygon with at least four vertices can be obtained this way. Therefore, the difference between the number of blue polygons and the number of red polygons is equal to the number of blue triangles. The number of blue triangles is the same as the number of ways to choose two red points (two red vertices of a blue triangle), i.e., $\frac{60 \cdot 59}{2}=1770$.

Evaluation. 14 points for a correct solution.

## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences

## Final Stage 2017-2018 academic year

## Problems, Answers, and Evaluation Criteria

## 7th Grade Variant 2

 physics5. A person is walking parallel to a railway track at a constant speed. A train also passes by at a constant speed. The person noticed that depending on the direction of the train's movement, it passes by in either $t_{1}=2 \text{ min}$ or $t_{2}=4$ min. Determine how long it would take the person to walk from one end of the train to the other.

## (15 points)

Answer: 8 min

Solution. When the train and the person are moving towards each other, $l=\left(v_{n}+v_{4}\right) \cdot t_{1} \quad$ (3 points), where $l$ is the length of the train, $v_{n}$ is its speed, and $v_{4}$ is the person's speed. If the directions of the train and the person are the same, then $l=\left(v_{n}-v_{4}\right) \cdot t_{2} \quad$ (3 points). In the situation where the person is walking along the train: $l=v_{u} \cdot t_{3}$ (3 points). Solving this system of equations, we get the final answer: $t_{3}=\frac{2 t_{1} t_{2}}{t_{2}-t_{1}}=\frac{2 \cdot 2 \cdot 4}{4-2}=8$ minutes (6 points).