8.4. Twelve chairs are arranged in a row. Sometimes a person sits on one of the free chairs. In this case, exactly one of his neighbors (if they were there) stands up and leaves. What is the maximum number of people that can be sitting simultaneously, if initially all the chairs were empty?
Answer: 11.

Solution. Evaluation. Note that it is impossible to occupy all chairs simultaneously, as at the moment when a person sits on the last unoccupied chair, one of his neighbors will stand up. Therefore, the number of people sitting simultaneously cannot exceed 11.

Example. We will show how to seat 11 people. Number the chairs from 1 to 12. The first chair is easy to occupy. The second chair can be occupied in two stages. In the first stage, a person sits on the third chair, and in the second stage, we seat a person on the second chair, and the person on the third chair stands up. We proceed similarly: if chairs numbered from 1 to \( k \) are occupied, we first seat a person on the chair numbered \( k+2 \), and then seat a person on the chair numbered \( k+1 \), thereby freeing the chair numbered \( k+2 \). After this operation is performed for all \( k \) from 1 to 10, the chairs numbered from 1 to 11 will be occupied, and the twelfth chair will be free.

## Evaluation Criteria:

+ A complete and justified solution is provided

± The correct answer is provided and it is explained how to seat 11 (a correct example is given), but it is not explained why there cannot be 12 sitting (no evaluation)
干 The correct answer is provided and it is explained why there cannot be 12 sitting, but the seating example for 11 is not provided or is incorrect

- Only the answer is provided
- The problem is not solved or is solved incorrectly

8.5. Inside an equilateral triangle \( ABC \), a point \( M \) is marked. Prove that it is possible to choose a point \( C_{1} \) on side \( AB \), a point \( A_{1} \) on side \( BC \), and a point \( B_{1} \) on side \( AC \) such that the lengths of the sides of triangle \( A_{1} B_{1} C_{1} \) are equal to the segments \( MA, MB \), and \( MC \).

Solution. Mark points \( C_{1} \) on side \( AB \), \( A_{1} \) on side \( BC \), and \( B_{1} \) on side \( AC \) such that \( M C_{1} \parallel BC, M A_{1} \parallel AC, M B_{1} \parallel AB \) (see Fig. 8.5). Then the segments \( M A_{1}, M B_{1} \), and \( M C_{1} \) will divide the given triangle into three trapezoids. From the parallelism, it follows that each angle at the larger base of these trapezoids is \( 60^{\circ} \), so these trapezoids are isosceles. Consequently, in each trapezoid, the diagonals are equal: \( B_{1} C_{1} = MA \), \( A_{1} C_{1} = MB \), \( A_{1} B_{1} = MC \), which is what we needed to prove.

Evaluation Criteria:

![](https://cdn.mathpix.com/cropped/2024_05_06_83dbb8f64176429cc0ceg-3.jpg?height=400&width=426&top_left_y=417&top_left_x=1523)

Fig. 8.5

+ A complete and justified solution is provided

± A correct construction is provided, but it is not explained (or explained incorrectly) why the sides of the resulting triangle are the desired ones

- The problem is not solved or is solved incorrectly

The presence of any reasoning about the number of ways to mark the desired points in the solution does not affect the evaluation result.