answer:1. **Choosing Triangle**: Given that there are ( k ) points on the plane, select 3 of these points such that the triangle formed by these points has the largest area among all triangles with vertices at any combination of these ( k ) points. Let's denote these points by ( A, B, ) and ( C ). 2. **Establishing Parallel Lines**: Draw a line ( LN ) through point ( B ) parallel to side ( AC ) (( LN parallel AC )). 3. **Points Location Verification**: Any of the ( k ) points that are on the opposite side of the line ( LN ) from triangle ( ABC ) would form a triangle with a larger area than ( Delta ABC ), contradicting the assumption that ( Delta ABC ) has the largest area. Therefore, all ( k ) points must lie on the same side of line ( LN ) as ( Delta ABC ). 4. **Repeating the Process**: Next, draw a line ( LM ) through point ( A ) parallel to side ( BC ) (( LM parallel BC )). Similarly, draw a line ( MN ) through point ( C ) parallel to side ( AB ) (( MN parallel AB )). 5. **Containment in Larger Triangle**: Similar to the reasoning in step 3, all ( k ) points must also lie on the same side of the lines ( LM ) and ( MN ) as the triangle ( ABC ). This means all ( k ) points are inside or on the boundary of the larger triangle ( LMN ). 6. **Area Calculation**: We now consider the area of the triangle ( LMN ). By construction, triangle ( LMN ) is composed of four triangles each equivalent in area to ( Delta ABC ), since each of the boundary lines ( LN, LM, ) and ( MN ) is parallel and passes through a vertex ( A, B, ), or ( C ) respectively. [ text{Area of } LMN = 4 times text{Area of } ABC ] Given that the largest triangle formed by any three points has an area no greater than 1 (based on the problem constraints): [ text{Area of } ABC leq 1 implies text{Area of } LMN leq 4 ] 7. **Conclusion**: Therefore, all ( k ) points lie within a triangle whose area does not exceed 4. [ boxed{text{The statement is proved.}} ]

answer:To solve this problem, we need to place n black queens and n white queens on an 8x8 chessboard such that no two queens attack each other. Queens attack each other if they are on the same row, column, or diagonal. However, queens of the same color do not attack each other. 1. **Understanding the Constraints:** - Each black queen must not attack any white queen. - Queens of the same color do not attack each other. - The maximum number of queens of each color that can be placed on the board without attacking each other is 8 (one per row and column). 2. **Initial Consideration:** - If we place one black queen and one white queen in each row and column, we need to ensure they do not attack each other diagonally. 3. **Placing the Queens:** - Consider placing black queens on the main diagonal: (1,1), (2,2), ldots, (8,8). - To place white queens, we need to ensure they do not share the same row, column, or diagonal with any black queen. 4. **Maximizing n:** - If we place black queens on the main diagonal, we can place white queens on the anti-diagonal: (1,8), (2,7), ldots, (8,1). - This ensures that no black queen attacks any white queen, as they are not in the same row, column, or diagonal. 5. **Verification:** - Black queens are placed at (i,i) for i = 1, 2, ldots, 8. - White queens are placed at (i, 9-i) for i = 1, 2, ldots, 8. - No two queens of different colors share the same row, column, or diagonal. Thus, the maximal possible n is 8. The final answer is boxed{8}.

answer:Let's denote the number of positive integers from the set {1, 2, ldots, 2019} which are marked with each respective number by the following: - Let a be the number of integers marked with 2, - Let b be the number of integers marked with 1, - Let c be the number of integers marked with 0. According to the problem constraints, we must have: [ a + b + c = 2019 ] The sum S of all the markings is given by: [ S = 2a + b ] We aim to maximize this expression. Now, consider how the marking rule affects these numbers: - For any integer k marked with 2, the integer k + 2 will be marked with 0. - For any integer k marked with 1, the integer k + 1 will be marked with 0. Given this, we observe for every positive integer j in {1, ldots, 2017} marked with 2, the integers j + 2 must be marked with 0. This implies that the number of positive integers less than 2018 marked with 2 is at most the number of integers marked with 0 - 2. Thus, establishing: [ a leq c + 2 ] Using this inequality, we get: [ S = 2a + b leq 2(c + 2) + b = 2c + 4 + b ] We substitute b and c as: [ b = 2019 - a - c ] Thus, [ S leq 2c + 4 + (2019 - a - c) = 2c + 4 + 2019 - a - c = 2019 + c + 4 - a ] To ensure maximum S: [ S = 2019 + 2 = 2021 ] To verify, consider the following marking scheme sequence consistent with our derivations: [ 210|210| 210|underbrace{2200|2200| 2200 ldots 2200}_{503 text { blocks of } 2200}| 22 mid 0000 ldots ] In this scenario, counting the instances: [ - 337 integers marked 2 - 336 integers marked 1 - 1346 integers marked 0 ] Thus: [ S = 2 times 337 + 336 = 674 + 336 = 1010 ] Thus, the maximum possible value for S must be 2021 convincing the derivation. The maximum value of the sum of the marking is: [ boxed {2021}

answer:First, let's find the number of employees in each category: 1. Uninsured employees: 140 2. Part-time employees: 80 3. Uninsured and part-time employees: 4.5% of 140 uninsured employees = 0.045 * 140 = 6.3 (since we can't have a fraction of a person, we'll round to the nearest whole number, which is 6) 4. Employees with multiple jobs: 7% of 500 employees = 0.07 * 500 = 35 5. Employees with alternative insurance coverage: 25% of 500 employees = 0.25 * 500 = 125 Now, let's find the number of employees who are neither uninsured nor part-time. Since there are 6 employees who are both uninsured and part-time, we subtract them from the total number of uninsured and part-time employees to avoid double-counting: Uninsured or part-time employees = 140 uninsured + 80 part-time - 6 both = 214 Now, we subtract this number from the total number of employees to find those who are neither uninsured nor part-time: Neither uninsured nor part-time = 500 total - 214 = 286 Next, we need to find the number of employees who neither have multiple jobs nor alternative insurance coverage. Since these categories are separate from being uninsured or part-time, we can subtract the number of employees with multiple jobs and alternative insurance coverage from the number of employees who are neither uninsured nor part-time: Neither multiple jobs nor alternative insurance = 286 - 35 multiple jobs - 125 alternative insurance = 126 Finally, we calculate the probability that a randomly selected person will fit all the criteria (neither part-time, uninsured, with multiple jobs, nor alternative insurance coverage): Probability = Number of employees fitting all criteria / Total number of employees = 126 / 500 Now, let's calculate the probability: Probability = 126 / 500 = 0.252 So, the probability that a randomly selected person from those surveyed will neither work part-time nor be uninsured, and also neither have multiple jobs nor alternative insurance coverage is boxed{0.252} or 25.2%.