answer:# Solution: Part (1): Finding the value of a Given the function f(x) = |x-4| + |x-a|, we know that the minimum value of f(x) is a. By the properties of absolute values, we can write: f(x) = |x-4| + |x-a| geq |(x-4) - (x-a)| = |4-a| Since the minimum value of f(x) is given to be a, we have: |4-a| = a This equation holds true when a leq 4. Solving for a, we consider two cases due to the absolute value: 1. When 4-a geq 0 (i.e., a leq 4), we have 4-a = a, leading to 2a = 4, thus a = 2. 2. When 4-a < 0 (which contradicts a leq 4), we would have a-4 = a, which is not possible. Therefore, the only solution is a = 2. boxed{a = 2} Part (2): Solving the inequality f(x) leq 5 Given a = 2, we rewrite f(x) as: f(x) = |x-4| + |x-2| We consider the behavior of f(x) in different intervals based on the critical points x = 2 and x = 4: 1. **For x leq 2:** f(x) = -(x-4) - (x-2) = -2x + 6 Solving -2x + 6 leq 5 gives: -2x leq -1 implies x geq frac{1}{2} 2. **For 2 < x < 4:** f(x) = (x-4) - (x-2) = -2 Since -2 leq 5 always holds, this interval satisfies the inequality. 3. **For x geq 4:** f(x) = (x-4) + (x-2) = 2x - 6 Solving 2x - 6 leq 5 gives: 2x leq 11 implies x leq frac{11}{2} Combining these intervals, the solution set for f(x) leq 5 is: boxed{left{x mid frac{1}{2} leq x leq frac{11}{2} right}}

answer:1. **Length of segment MN:** - MN is still parallel to AB and half its length, and AB does not change. Moving P perpendicularly only changes the orientation of MN relative to P. - **Conclusion:** The length of MN does not change. 2. **Perimeter of triangle PAQ:** - The perimeter of triangle PAQ is PA + AQ + PQ. - As P moves perpendicular to AB, PA changes in length but PQ remains equal to PB (kept fixed by the problem setup). - **Conclusion:** The perimeter of triangle PAQ changes. 3. **Area of triangle PAB:** - The base AB remains constant, but as P moves perpendicular to AB, the height from P to line AB changes (increases or decreases). - **Conclusion:** The area of triangle PAB changes. 4. **Area of trapezoid ABNM:** - The bases AB and MN don't change, but the height from AB to MN changes as it is dependent on P's distance from AB. - **Conclusion:** The area of trapezoid ABNM changes. Only the length of segment MN remains unchanged. Therefore, the correct answer is textbf{(D) 3}. The final answer is boxed{textbf{(D)} 3}

answer:1. Given the conditions: [ a^2 + b = b^2 + c = c^2 + a ] we can derive the following expressions: [ a^2 - b^2 = c - b, quad b^2 - c^2 = a - c, quad c^2 - a^2 = b - a ] 2. We need to find the value of the expression: [ a(a^2 - b^2) + b(b^2 - c^2) + c(c^2 - a^2) ] 3. Substituting the expressions derived in step 1, we get: [ a(a^2 - b^2) = a(c - b) ] [ b(b^2 - c^2) = b(a - c) ] [ c(c^2 - a^2) = c(b - a) ] 4. Therefore, the original expression can be rewritten as: [ a(a^2 - b^2) + b(b^2 - c^2) + c(c^2 - a^2) = a(c - b) + b(a - c) + c(b - a) ] 5. Notice that (a(c - b) + b(a - c) + c(b - a)) is a cyclic sum. We can simplify it as follows: [ a(c - b) + b(a - c) + c(b - a) ] 6. Rearranging the terms within the sum, we get: [ a(c - b) + b(a - c) + c(b - a) = (ac - ab) + (ba - bc) + (cb - ca) ] 7. Considering the above, the terms pair up and cancel each other out: [ ac - ab + ba - bc + cb - ca = 0 ] 8. Hence, we conclude that: [ a(a^2 - b^2) + b(b^2 - c^2) + c(c^2 - a^2) = 0 ] # Conclusion: [ boxed{0} ]

answer:1. **Initial Condition**: - We start with three piles of 40 stones each. Thus, initially, there are 3 piles. 2. **Operation Description**: - Each move consists of combining two piles and then dividing the combined pile into four new piles. Therefore, each move replaces 2 piles with 4 piles, effectively increasing the total number of piles by 2. 3. **Parity Analysis**: - Since the initial number of piles is 3 (which is odd), and each move increases the number of piles by 2, the number of piles will always remain odd after any number of moves. This is inferred from the fact that adding an even number (2) to an odd number (3) results in another odd number. Mathematically, if ( n ) is the number of piles, then: [ n_{text{initial}} = 3 ] After ( k ) steps: [ n_{k} = 3 + 2k ] 4. **End Condition**: - The game ends when no more moves can be made. This occurs when the two largest piles combined have 3 or fewer stones in total. In that case, the smallest possible combined pile sum is 3 stones, requiring the remaining stones to be divided such that all new piles have 1 stone each (since 3 stones must be distributed). 5. **Calculation of Final Moves and Determination of Winner**: - To reach this end condition, consider how many times we perform the operation until we cannot continue. - The final condition requires there to be piles each containing roughly single stones, with a maximum sum of stone counts in any two piles being 3. - This necessitates having a large number of single-stone piles. Indeed, we need at least 119 piles of one stone each to ensure that combining any two piles results in no more than 3 stones. To reach this stage, starting from 3 piles, the number of additional piles needed is: [ 119 - 3 = 116 ] - Since each move increases the number of piles by 2: [ frac{116}{2} = 58 ] moves are required. - This means ( 58 ) moves must be made to achieve the final configuration. Since ( 58 ) is an even number, the 58th move will be made by Vasya (since he goes after Petya's first move). # Conclusion boxed{text{Vasya}} will win the game regardless of what strategy Petya employs, because the game finishes after an even number of moves which means the last move will always be made by Vasya.