🧠:[<think>Okay, let's try to figure out this problem. So, we have a 100x100 grid paper, and we need to connect the two opposite corners with broken lines that follow the grid lines. The goal is to find the smallest number of such broken lines needed so that all the vertices (the corners of the cells) are covered by the union of these lines. First, let me understand the problem better. A broken line that connects the opposite corners must be a shortest path, right? So, in a grid, the shortest path from one corner to the opposite would be moving right and down (or left and up, depending on the direction) without any backtracking. For a 100x100 grid, such a path would consist of 100 moves right and 100 moves down, totaling 200 steps. Each step is along an edge of a cell. So, each broken line has 200 edges and passes through 201 vertices (since each edge connects two vertices). But the question is, how many such broken lines do we need so that every vertex in the grid is covered by at least one of these lines? The union of their paths must include all the vertices. So, we need to cover all (100+1)x(100+1) = 101x101 = 10201 vertices. Each broken line covers 201 vertices. If we naively divide 10201 by 201, we get approximately 50.75, so at least 51 lines. But this is probably a lower bound and the actual answer might be higher because the broken lines overlap at many vertices, especially along the main diagonal.Wait, but maybe the overlapping can be minimized by cleverly arranging the paths. However, since each path must go from one corner to the opposite, they all have to traverse the grid from, say, (0,0) to (100,100). The key is to find paths that cover different parts of the grid so that their union covers all vertices.Let me think. In a grid, each vertex can be represented by coordinates (i,j) where i and j range from 0 to 100. A shortest path from (0,0) to (100,100) will consist of moving right (R) and down (D) steps in some order. The number of such paths is C(200,100), which is a huge number, so we need to select a subset of these paths that cover all the vertices.Each vertex (i,j) lies on some shortest path. For example, the vertex (i,j) is on any path that has exactly i right moves and j down moves up to that point, and then the remaining (100 - i) right and (100 - j) down moves. So, to cover all vertices, we need to ensure that for every (i,j), there is at least one path in our set that passes through (i,j).Alternatively, maybe we can model this as covering all the vertices with paths, where each path is a shortest path from (0,0) to (100,100). The question then becomes, what's the minimum number of such paths needed to cover all vertices.This seems similar to a covering problem in graph theory. Each path is a set of vertices, and we want the minimum number of such sets whose union is the entire vertex set.Alternatively, perhaps there's a way to use the structure of the grid to determine how the paths can cover the vertices. Let's consider smaller grids first to see if we can find a pattern.Take a 1x1 grid (which is 2x2 vertices). The two opposite corners can be connected by two paths: one going right then down, and the other going down then right. These two paths cover all four vertices. So, the answer here is 2.For a 2x2 grid (3x3 vertices). The shortest paths from (0,0) to (2,2) have lengths of 4 steps. Each path will cover 5 vertices. The total number of vertices is 9. If we take two paths, they can cover at most 5 + 5 - overlap vertices. The overlap must be at least 1 (since they both start and end at the same points). So, 5 + 5 - 1 = 9. So, two paths might cover all vertices. Let's check. For example, the path R, R, D, D and the path D, D, R, R. The first path goes right twice then down twice, covering vertices (0,0), (1,0), (2,0), (2,1), (2,2). The second path goes down twice then right twice, covering (0,0), (0,1), (0,2), (1,2), (2,2). Together, these cover all vertices except (1,1). So, they miss the center. Therefore, two paths are not enough. How about three paths? Let's take another path that goes R, D, R, D. This covers (0,0), (1,0), (1,1), (2,1), (2,2). Similarly, another path D, R, D, R covers (0,0), (0,1), (1,1), (1,2), (2,2). So, adding these two paths to the first two would cover all vertices. But actually, perhaps three paths suffice. Let me see:First path: R, R, D, D covers (0,0), (1,0), (2,0), (2,1), (2,2).Second path: D, D, R, R covers (0,0), (0,1), (0,2), (1,2), (2,2).Third path: R, D, R, D covers (0,0), (1,0), (1,1), (2,1), (2,2).Fourth path: D, R, D, R covers (0,0), (0,1), (1,1), (1,2), (2,2).But actually, even with three paths, maybe we can cover everything. If we take the first, second, and third paths, they cover all except (1,2) and (0,1). Wait, the second path covers (0,1) and (1,2) is covered by the fourth path. Hmm, maybe three paths aren't enough. Let me list all the vertices:(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2).First path: covers (0,0), (1,0), (2,0), (2,1), (2,2).Second path: (0,0), (0,1), (0,2), (1,2), (2,2).Third path: (0,0), (1,0), (1,1), (2,1), (2,2).So, combining these three, we cover:From first: (0,0), (1,0), (2,0), (2,1), (2,2).Second: (0,0), (0,1), (0,2), (1,2), (2,2).Third: (0,0), (1,0), (1,1), (2,1), (2,2).Missing vertices: (1,1) is covered by the third path? Wait, (1,1) is in the third path. Wait, no: third path is (0,0), (1,0), (1,1), (2,1), (2,2). So (1,1) is covered. Then, what's missing? Let's see:First path covers: 0,0;1,0;2,0;2,1;2,2.Second path: 0,0;0,1;0,2;1,2;2,2.Third path: 0,0;1,0;1,1;2,1;2,2.So combined, they cover:0,0;1,0;2,0;2,1;2,2 (from first)0,1;0,2;1,2 (from second)1,1 (from third)So all vertices are covered. So three paths suffice for a 2x2 grid.But initially, with two paths, we were missing some. So, for 2x2 grid, the answer is 3?Wait, but maybe there's a smarter way. Alternatively, maybe the minimal number is 2. Hmm. Wait, if we take the two diagonal paths: (0,0) -> (1,1) -> (2,2) and (0,0) -> (1,0) -> (1,1) -> (1,2) -> (2,2). Wait, but these are not shortest paths. The first path would require moving right and down alternately, but in a 2x2 grid, moving from (0,0) to (2,2) requires two rights and two downs, so the path through (1,1) is a valid shortest path. Similarly, another path going through (1,0) and (1,1) and (1,2) is also a shortest path? Wait, but moving from (0,0) to (1,0) is a right, then from (1,0) to (1,1) is a down, then (1,1) to (1,2) is a down, then (1,2) to (2,2) is a right. That's two rights and two downs, so that's a valid shortest path.So this path would cover (0,0), (1,0), (1,1), (1,2), (2,2). If we combine this with the diagonal path (0,0), (1,1), (2,2), does that cover all vertices? The diagonal path would go (0,0), (1,1), (2,2). So, the vertices covered are (0,0), (1,1), (2,2). The other path covers (0,0), (1,0), (1,1), (1,2), (2,2). So together, they cover (0,0), (1,0), (1,1), (1,2), (2,2), plus (0,0), (1,1), (2,2). So missing are (0,1), (0,2), (2,0), (2,1). Wait, no. Wait, the second path is (0,0) -> (1,0) -> (1,1) -> (1,2) -> (2,2). So it doesn't cover (0,1) or (0,2) or (2,0). So two paths aren't enough.Alternatively, perhaps if we take another path. Let's see, maybe (0,0) -> (0,1) -> (1,1) -> (2,1) -> (2,2). That's a path: right, down, right, down. Wait, no. From (0,0) to (0,1) is up? Wait, in grid terms, usually (0,0) is bottom-left, so moving right increases the x-coordinate, and moving up increases the y-coordinate. Wait, but maybe in the problem, it's a square grid, so moving right and down. Wait, actually, the problem says "connect the two opposite corners of a square sheet of grid paper". So, maybe the grid is considered with coordinates such that moving right is in one direction and moving up is in another. But the problem says "follow the sides of the cells", so regardless, a broken line can go right or up, but to connect opposite corners, you need a combination of right and up moves. Wait, perhaps I need to clarify.Wait, in a grid, to go from (0,0) to (100,100), you need 100 right moves and 100 up moves (if we consider the grid as a matrix). Alternatively, if it's a grid on a square sheet, the opposite corners could be, say, bottom-left to top-right, which would require moving right and up. But the problem mentions "broken lines that follow the sides of the cells and connect the two opposite corners... by the shortest path". So, the shortest path would consist of moving right and up, totaling 200 steps: 100 right and 100 up. But in that case, each path is a staircase going from bottom-left to top-right.But in the initial problem statement, the user mentioned a 100x100 grid, so perhaps the grid has 100 cells along each side, meaning 101 vertices along each side. So, the coordinates go from (0,0) to (100,100). Each step is either right or up.So, in this case, each shortest path is a sequence of 100 rights and 100 ups, arranged in some order. Each such path will pass through 201 vertices (including start and end). The problem is to cover all 101x101 = 10201 vertices with the union of such paths. What's the minimal number of paths required?For the 1x1 grid (2x2 vertices), the answer was 2. For a 2x2 grid (3x3 vertices), the answer seems to be 3. Maybe there's a pattern here. For an n x n grid (from (0,0) to (n,n)), maybe the minimal number of paths required is n+1? For n=1, 2; for n=2, 3. If this holds, then for n=100, it would be 101. But let's check for n=3.Take a 3x3 grid (4x4 vertices). If the pattern holds, we need 4 paths. Let's see. Each path has 6 steps (3 right and 3 up), passing through 7 vertices. The total number of vertices is 16. If we have 4 paths, each covering 7 vertices, with overlaps, we need to cover 16 vertices. The maximum coverage without overlap would be 4*7 - overlaps. But it's complicated. Let's try to construct 4 paths that cover all vertices.First path: all rights first, then all ups: (0,0) -> (3,0) -> (3,3). This covers (0,0), (1,0), (2,0), (3,0), (3,1), (3,2), (3,3).Second path: all ups first, then rights: (0,0) -> (0,3) -> (3,3). Covers (0,0), (0,1), (0,2), (0,3), (1,3), (2,3), (3,3).Third path: a diagonal path: (0,0) -> (1,1) -> (2,2) -> (3,3). This covers (0,0), (1,1), (2,2), (3,3). But wait, a diagonal path in a grid where you move right and up can't go strictly diagonally; each step is either right or up. So, actually, the diagonal path would need to alternate right and up moves. For a 3x3 grid, a path like R, U, R, U, R, U. That would go (0,0) -> (1,0) -> (1,1) -> (2,1) -> (2,2) -> (3,2) -> (3,3). So this covers (0,0), (1,0), (1,1), (2,1), (2,2), (3,2), (3,3). Similarly, another diagonal path could be U, R, U, R, U, R: (0,0) -> (0,1) -> (1,1) -> (1,2) -> (2,2) -> (2,3) -> (3,3). So, combining the first two paths (all rights then ups and all ups then rights), we cover the bottom edge, top edge, left edge, right edge, and the main diagonal. Then, adding these two diagonal paths, which cover some inner vertices.But even with these four paths, are all vertices covered? Let's check:First path: (0,0), (1,0), (2,0), (3,0), (3,1), (3,2), (3,3).Second path: (0,0), (0,1), (0,2), (0,3), (1,3), (2,3), (3,3).Third path: (0,0), (1,0), (1,1), (2,1), (2,2), (3,2), (3,3).Fourth path: (0,0), (0,1), (1,1), (1,2), (2,2), (2,3), (3,3).Combined, these cover:From first: bottom row and rightmost column.From second: leftmost column and top row.From third: (1,0), (1,1), (2,1), (2,2), (3,2).From fourth: (0,1), (1,1), (1,2), (2,2), (2,3).So, missing vertices:(1,3): Covered by second path? Second path covers (1,3) as part of (1,3), (2,3), (3,3).Wait, second path is (0,0), (0,1), (0,2), (0,3), (1,3), (2,3), (3,3). Yes, (1,3) is there.What about (2,0)? Covered by first path: (2,0) is in first path.(2,1): Covered by third path.(1,2): Covered by fourth path.(0,1), (0,2), (0,3): Covered by second path.(3,1), (3,2): Covered by first path.(1,1): Covered by third and fourth paths.(2,2): Covered by third and fourth paths.So, is there any vertex missing?Wait, (1,0) is covered by first and third paths.(0,0) and (3,3) are covered by all.Wait, seems like all vertices are covered. So with four paths, we cover all 16 vertices in a 3x3 grid. So, the pattern seems n+1, where n is the grid size. For 1x1 grid (2x2 vertices), 2 paths; 2x2 grid (3x3 vertices), 3 paths; 3x3 grid (4x4 vertices), 4 paths. Hence, for an n x n grid (from (0,0) to (n,n)), the minimal number of paths needed is n+1.Therefore, for a 100x100 grid (101x101 vertices), the minimal number of paths required would be 100 + 1 = 101.But let me verify this pattern with another example. Let's take n=1, which is a 1x1 grid (2x2 vertices). The minimal number is 2. 1+1=2, which matches. For n=2, 3x3 vertices, minimal number is 3. 2+1=3, which matches. For n=3, 4x4 vertices, minimal number is 4. 3+1=4, which matches. So, the pattern holds. Therefore, by induction or generalization, for an n x n grid, the minimal number of paths needed is n+1. Hence, for a 100x100 grid, the answer is 101.But why is this the case? Let me think about why n+1 paths are sufficient. If we take all the paths that correspond to moving right some number of times, then up, then right, then up, etc., but arranged in such a way that each path covers a different "diagonal" or band of the grid. Alternatively, if we consider the grid as a matrix, and each path corresponds to a permutation of rights and ups, such that each path covers different rows and columns.Wait, another way to think about it: Each vertex (i,j) in the grid lies on some shortest path from (0,0) to (n,n). To cover all vertices, we can parameterize the paths by a parameter k, where each path takes k steps to the right first, then n steps up, then n - k steps to the right. Wait, no, that might not work. Alternatively, perhaps using different "slopes".But in grid paths, all slopes are effectively combinations of right and up moves. Another approach: For each diagonal, where a diagonal is defined by i + j = constant, we can cover all the vertices on that diagonal with a single path. However, a single path can cover only one vertex per diagonal if it's a strictly alternating right and up path. But in reality, a path can cover multiple vertices on different diagonals.Wait, maybe the key idea is that to cover all the vertices, you need to have paths that go through every possible "lane" in the grid. Since each path can only go right or up, to cover all vertical and horizontal lanes, you need enough paths such that every column and row is traversed in different ways. But I'm not sure.Alternatively, think of the grid as a bipartite graph. Each vertex can be colored black or white in a checkerboard pattern. Each shortest path from (0,0) to (n,n) will alternate between black and white vertices, starting and ending on the same color (since (0,0) and (n,n) are the same color if n is even). Wait, in a 100x100 grid, n=100, which is even, so (0,0) and (100,100) are the same color. Therefore, each path will go through 101 vertices of one color and 100 of the other. Wait, but the total number of vertices is 101x101. If n is even, then (0,0) and (n,n) are the same color, so the number of black and white vertices will differ by 1. Specifically, in a 100x100 grid (101x101 vertices), there are (101*101 + 1)/2 = 5101 black vertices and 5100 white vertices, or vice versa. Each path, starting and ending on black, will have 101 black vertices and 100 white vertices. Therefore, to cover all black vertices, which are 5101, each path can cover 101 black vertices. So, the minimal number of paths needed is at least ceiling(5101 / 101) = ceiling(50.5049...) = 51. Similarly, for white vertices, ceiling(5100 / 100) = 51. Therefore, the minimal number of paths needed is at least 51. But this contradicts our earlier pattern of n+1=101. Wait, this suggests a lower bound of 51, but our previous examples suggested n+1. So, there's a discrepancy here.Wait, this is an important point. Let me recast the problem in terms of graph coloring. In a grid where each vertex is colored like a chessboard, black and white alternately, then any path from (0,0) to (100,100) must alternate colors. Since (0,0) is black, (100,100) is also black (as 100 is even), so each path will visit black and white vertices alternately, starting and ending with black. Thus, each path covers 101 black vertices and 100 white vertices. The total number of black vertices is (101x101 +1)/2 = (10201 +1)/2 = 10202/2 = 5101. The number of white vertices is 5100. So, to cover all black vertices, each path can cover 101 black vertices. Thus, the minimal number of paths required is at least ceiling(5101 / 101) = 51. Similarly, for white vertices, ceiling(5100 / 100) = 51. So, the lower bound is 51. However, our previous examples suggested a pattern of n+1, which for n=100 would be 101. But here, the lower bound is 51, which is much lower. Therefore, there's a conflict. Which one is correct?Wait, maybe my initial reasoning with smaller grids was wrong. Let's check the 2x2 grid (3x3 vertices). Total vertices: 9. Number of black vertices: 5, white vertices: 4. Each path covers 3 black and 2 white vertices (since a path from (0,0) to (2,2) has 3 black vertices and 2 white vertices). So, to cover all 5 black vertices, we need at least ceiling(5/3) = 2 paths, and to cover white vertices, ceiling(4/2) = 2. So, the lower bound is 2. But as we saw earlier, 2 paths are insufficient because they only cover 3 + 3 - 1 = 5 black vertices (but actually, each path covers 3 black, so 2 paths would cover 6 black vertices, but there's overlap). Wait, no. Wait, each path covers 3 black vertices, but since they start and end at the same black vertex, (0,0) and (2,2), so overlapping at least two black vertices. Wait, maybe my earlier analysis was off.Wait, in the 2x2 grid, each path has 3 black vertices and 2 white vertices. The total number of black vertices is 5. If we have two paths, they can cover at most 3 + 3 - overlap. The overlap must be at least 1 (the start and end). So, 3 + 3 - 1 = 5. So, two paths can exactly cover all 5 black vertices. However, each path also covers 2 white vertices. Two paths would cover 2 + 2 - overlap. The overlap on white vertices could be 1 (the middle vertex, maybe). So, 2 + 2 - 1 = 3. But there are 4 white vertices, so two paths would leave one white vertex uncovered. Hence, the lower bound considering white vertices is ceiling(4 / 2) = 2, but due to overlaps, we actually need more. Therefore, the lower bound isn't tight. Therefore, the actual minimal number is higher. For the 2x2 grid, the minimal number is 3, which is higher than the lower bound of 2.Therefore, the lower bound from the bipartite analysis isn't tight because overlaps can't be avoided. Hence, the actual minimal number is higher. So, returning to the original problem, the lower bound of 51 is perhaps not tight, and the actual answer could be 101, following the n+1 pattern. But how to confirm this?Alternatively, think of each path as a permutation. Wait, another approach: For each row, we need to cover all the vertices in that row. Each path can cover at most one vertex per column in a given row. Wait, not necessarily. A path moving right along a row would cover multiple vertices in that row. Wait, no. If a path is moving from (0,0) to (100,100), it can only go right or up. So, once it moves up from a row, it can't come back down. Therefore, in each row, a path can cover at most one vertex (the one where it enters the row from the right and then moves up). Wait, no. Let's clarify.Consider a single row, say row j. A path moving from left to right and up would enter the row from the left, move right some number of times, then move up. So, in row j, the path covers all the vertices from (i,j) where i ranges from some starting point to the point where it moves up. Wait, no. Once the path is in row j, it can move right multiple times, covering multiple vertices in that row before moving up. For example, a path that moves right 10 times, then up once, would cover 11 vertices in row 0 (from (0,0) to (10,0)), then move up to (10,1). So, in row 0, it covers 11 vertices.But if we need to cover all 101 vertices in each row, how many paths do we need? If each path can cover up to 101 vertices in a row (if it goes all the way right before moving up), then one path could cover an entire row. But since the paths have to go from (0,0) to (100,100), they can't stay in a single row forever. The maximum number of vertices a path can cover in a single row is 101 (if it goes all the way right in row 0), but then it has to move up. Similarly, in other rows, a path can cover at most 100 - j vertices if it moves right from some column i to column 100 in row j. But this is getting complicated.Alternatively, consider that each path has to move from left to right and bottom to top. To cover all the vertices in each row, you need multiple paths such that their right moves in each row cover all the columns. For example, in row 0, to cover all columns from 0 to 100, a single path that goes all the way right in row 0 would suffice. Similarly, in row 1, to cover all columns, you need some path that moves right through row 1. But a path that goes all the way right in row 1 would have to reach row 1 first, which requires moving up from row 0. So, to cover all columns in row 1, a path could move up to row 1 at some column and then move right. If we have different paths that move up to row 1 at different columns, then their right moves in row 1 would cover different columns.Therefore, if we have a set of paths where each path moves up to a certain column in each row, thereby covering the remaining columns to the right in that row, then by varying the column where they move up, we can cover all columns in all rows.Similarly, the same logic applies to columns. So, to cover all the vertices, we need that for each row j and column i, there exists a path that goes through (i,j). To achieve this, perhaps we need as many paths as the number of rows or columns. Since there are 101 rows and 101 columns, but since the paths are from (0,0) to (100,100), the number of required paths might be 101. Each path could be responsible for covering a specific diagonal or a specific set of columns in each row.Alternatively, arrange the paths such that each path corresponds to a different "diagonal" offset. For instance, the first path goes straight right then up, the second path goes up one, right one, up one, etc., the third path goes up two, right two, and so on. However, in a grid, moving in such diagonals isn't straightforward because each step is either right or up.Wait, here's a construction. Let’s consider 101 paths where each path is shifted by one column in each row. For example, the first path goes all the way right in row 0, then all the way up. The second path goes right one column, up one row, right one column, up one row, etc. The third path goes right two columns, up one row, right two columns, up one row, etc. But this might not work because you can't move right more than 100 columns.Alternatively, use cyclic permutations. If we have 101 paths, each starting by moving up a different number of times before moving right. For example, path k (0 ≤ k ≤ 100) moves up k times, then right 100 times, then up (100 - k) times. Wait, no. The total moves should be 100 rights and 100 ups. If a path moves up k times, then right 100 times, then up (100 - k) times, then total up moves would be k + (100 - k) = 100, and right moves 100. So, that works. But such a path would go from (0,0) up k times to (0,k), then right 100 times to (100,k), then up (100 - k) times to (100,100). This path would cover the entire right edge from (100,k) to (100,100), and the entire horizontal line from (0,k) to (100,k). However, this path only covers the vertical line from (0,0) to (0,k). Therefore, in row 0, it only covers (0,0) to (0,k). In other rows, it only covers the horizontal line at y = k. Therefore, if we take k from 0 to 100, then we have 101 paths, each covering a horizontal line at y = k, and parts of the leftmost column. However, this would leave many vertices uncovered. For example, the vertices not on any of the horizontal lines y = k or the rightmost column. Therefore, this approach is insufficient.Alternatively, consider a different construction. If we take all paths that at each step, either move right or up, but arrange their sequences such that every possible combination of right and up moves is covered. But since the number of paths is factorial, this is not feasible.Wait, let me think again about the small example. For a 2x2 grid, we needed 3 paths. If we model this, each path corresponds to a different way of cutting through the grid. If we have n+1 paths for an n x n grid, maybe each path is responsible for covering a specific set of diagonals. For example, in a 100x100 grid, each of the 101 paths could correspond to a diagonal offset from the main diagonal. By offsetting each path by one, they can cover all the vertices. Alternatively, imagine the grid as a matrix where each cell is identified by its row and column. To cover all cells (i,j), you need paths that pass through each (i,j). Each path can be associated with a permutation of the steps, but in such a way that every (i,j) is included in at least one path. Wait, here's an idea inspired by the concept of covering codes or covering sets. For each vertex (i,j), we need a path that includes (i,j). The number of paths required would be the minimal number such that every vertex is included in at least one path. If we can design the paths such that each path "covers" a distinct set of vertices with minimal overlap, then we can minimize the number of paths.But how to formalize this? Let's consider that each path is defined by a sequence of right and up moves. For a vertex (i,j), it is on a path if the path has exactly i right moves and j up moves before reaching (i,j), and then the remaining moves. So, for each (i,j), there exists a path in our set that has i right moves and j up moves in some order up to (i,j), and then continues to (100,100). To cover all (i,j), we need that for every (i,j), there's a path that goes through (i,j). One way to ensure this is to have a path for each possible (i,j), but that's too many. Instead, we need a set of paths such that every (i,j) is on at least one of them. Perhaps if we fix certain points where the paths turn from right to up. For example, if a path turns from right to up at column k, then it will cover all vertices (k, j) for j from 0 to 100. Wait, no. If a path goes right k times, then up 100 times, then right (100 - k) times. Wait, no, in a path from (0,0) to (100,100), the total rights and ups must each be 100. So, if a path goes right k times, then up 100 times, that would overshoot the target. It needs to go right k, then up 100, then right (100 - k), but that totals 100 rights and 100 ups, arriving at (100,100). Such a path would pass through (k, 0) to (k, 100) vertically, and then from (k,100) to (100,100) horizontally. Wait, but actually, moving right k times gets to (k, 0), then moving up 100 times gets to (k, 100), then moving right (100 - k) times gets to (100, 100). Therefore, this path covers the vertical line x = k from y=0 to y=100, and the horizontal line y=100 from x=k to x=100. Similarly, if we have paths for each k from 0 to 100, turning from right to up at column k, then the union of all these paths would cover all vertical lines x = 0 to x=100 (since each x=k is covered by the k-th path) and the horizontal line y=100. But this still leaves all the other horizontal lines (y < 100) only covered at x=0 to x=k for each path. Wait, no. Each path covers the vertical line x=k, but for y from 0 to 100, and the horizontal line y=100 from x=k to x=100. Therefore, combining all these paths, we cover all vertical lines x=0 to x=100 completely, and the horizontal line y=100 completely. However, the other horizontal lines (y < 100) are only partially covered. For example, for y=50, the only covered points are x=0,1,...,100 (since each vertical line x=k covers (k,50)), but wait, no. Each path with k=0 to 100 covers x=k from y=0 to y=100, so every horizontal line y=50 is covered by all x=0 to 100. Wait, this is true! If we have a path for each k from 0 to 100, each path covers the vertical line x=k entirely. Therefore, the union of all these paths would cover every vertical line x=k, which in turn covers every horizontal line y=j entirely, because for each y=j, every x=k is covered by the k-th path. Similarly, the horizontal line y=100 is covered by all paths from x=k to x=100. Therefore, in this construction, using 101 paths (k=0 to 100), we cover all vertices. But wait, if we have 101 paths, each turning at a different column k, then each path covers x=k vertical line and part of y=100. However, since every x and y is covered by the vertical lines, which are entirely covered by each respective path, this means all vertices are indeed covered. Therefore, this construction works and requires 101 paths.Is this the minimal number? Since our lower bound from the bipartite analysis gave 51, but this construction gives 101, which is much higher. However, in the small cases, like n=1 and n=2, this construction would yield n+1 paths, which matches the minimal number we found earlier. For example, in n=1 (2x2 grid), two paths: one going right then up, covering x=0 vertical line and y=1 horizontal line; the other going up then right, covering x=1 vertical line and y=0 horizontal line. But wait, no. For n=1, the two paths would be:Path k=0: right 0, up 1, right 1. Wait, no. For n=1, a 1x1 grid, the paths are:First path: right 1, up 1. So, (0,0)->(1,0)->(1,1).Second path: up 1, right 1. So, (0,0)->(0,1)->(1,1).These two paths cover all vertical lines x=0 and x=1, and horizontal lines y=0 and y=1. Hence, all four vertices. But in this case, two paths suffice, which matches n+1=2.Similarly, for n=2, three paths:Path k=0: right 0, up 2, right 2. But in reality, the path would be (0,0) -> (0,2) -> (2,2).Path k=1: right 1, up 2, right 1. (0,0) -> (1,0) -> (1,2) -> (2,2).Path k=2: right 2, up 2. (0,0) -> (2,0) -> (2,2).These three paths would cover all vertical lines x=0, x=1, x=2 and the horizontal line y=2. But for other horizontal lines y=0 and y=1:- y=0: covered by x=0 (path k=0), x=1 (path k=1), x=2 (path k=2).- y=1: covered by x=0 (path k=0 up to y=2), but actually, path k=0 only covers x=0 from y=0 to y=2. Similarly, path k=1 covers x=1 from y=0 to y=2, and path k=2 covers x=2 from y=0 to y=2. Hence, all horizontal lines are covered. Therefore, three paths cover all vertices for n=2. This matches n+1=3.Therefore, this construction works for smaller grids and provides the minimal number of paths as n+1. Hence, for the 100x100 grid, the answer should be 101. Therefore, the minimal number of paths required is 101.But wait, the user mentioned "the union contains all the vertices of the cells". In the construction above, with 101 paths each covering a vertical line x=k, this indeed covers all vertices, since every vertex (i,j) lies on the vertical line x=i, which is covered by the i-th path. Hence, the union of these 101 paths includes all vertices. Therefore, the answer should be 101.But why was the lower bound from the bipartite analysis suggesting 51? Because it considered covering the black and white vertices separately, but overlaps are inevitable, and the actual number needed is higher. Therefore, the correct answer is 101.Final Answerboxed{101}</think>To solve the problem of finding the smallest number of broken lines needed to cover all vertices of a 100x100 grid paper when connecting the opposite corners with the shortest path, we need to consider the structure of the grid and the coverage of each path.1. Understanding the Grid and Paths: - A 100x100 grid has 101x101 vertices. - Each shortest path from one corner to the opposite corner consists of 100 right moves and 100 down (or up) moves, covering 201 vertices.2. Key Observations: - Each vertex (i, j) must be covered by at least one path. - A path can be parameterized by the column where it transitions from right to up moves. For each column k, a path can be constructed that moves right k times, then up 100 times, and then right (100 - k) times.3. Constructing Paths: - For each column k (from 0 to 100), construct a path that moves right k times, then up 100 times, and finally right (100 - k) times. This path covers the vertical line x = k from y = 0 to y = 100, and the horizontal line y = 100 from x = k to x = 100. - This construction ensures that every vertex (i, j) is covered by the vertical line x = i, which is part of the i-th path.4. Verification: - Each vertical line x = k is covered by the k-th path, ensuring all vertices in that column are covered. - Horizontal lines are covered by the vertical lines intersecting them, as each vertical line x = k covers all vertices in the column from y = 0 to y = 100.5. Conclusion: - The minimal number of paths required to cover all vertices in a 100x100 grid is 101, as each path covers a unique vertical line and the union of all these paths covers all vertices.Thus, the smallest number of broken lines needed is (boxed{101}).,<think>Okay, so I need to figure out the smallest number of broken lines that cover all the vertices in a 100x100 grid, connecting the opposite corners with the shortest path. Hmm, let's start by understanding the problem better.First, a broken line that follows the grid lines and connects two opposite corners (let's say from (0,0) to (100,100)) using the shortest path. The shortest path in such a grid would consist of moving right and up only, right? So each broken line would have 100 right moves and 100 up moves, totaling 200 steps. But since it's a broken line, it's just a polygonal path connecting the vertices, so it alternates between horizontal and vertical segments. Wait, but actually, in a grid, each step is either horizontal or vertical, so the broken line would be a sequence of horizontal and vertical line segments moving from the start to the end.Now, the problem is asking for the minimal number of such broken lines needed so that every vertex in the 100x100 grid is covered by at least one of these lines. So, every intersection point (i,j) where i and j are integers from 0 to 100 must be on at least one of these broken lines.First, let's think about a smaller grid, maybe 2x2, to get some intuition.In a 2x2 grid, the opposite corners are (0,0) and (2,2). The shortest path broken lines would be the two possible paths: right, right, up, up or up, up, right, right, but in terms of lines, they would go either (0,0)->(2,0)->(2,2) or (0,0)->(0,2)->(2,2). But there are also diagonal paths? Wait, no, the broken lines have to follow the grid lines. So they can't go diagonally. So for 2x2 grid, the two main diagonals (but along the grid lines) would cover all vertices. Wait, but each broken line can only move right or up. So starting from (0,0), you can only go right or up. Similarly, starting from (2,2), you can go left or down. But the problem says connecting the two opposite corners by the shortest path, so we have to go from (0,0) to (2,2) with the minimal steps, which is 4 steps (2 right, 2 up). So each broken line is a staircase-like path.In the 2x2 grid, how many such paths do we need to cover all vertices?Each vertex is at (i,j) where i and j are 0,1,2. Let's list them:(0,0), (0,1), (0,2)(1,0), (1,1), (1,2)(2,0), (2,1), (2,2)A single shortest path would go from (0,0) to (2,2) moving right and up. Let's say the path: (0,0) → (1,0) → (2,0) → (2,1) → (2,2). This path covers (0,0), (1,0), (2,0), (2,1), (2,2). Similarly, another path could be (0,0) → (0,1) → (0,2) → (1,2) → (2,2). This covers (0,0), (0,1), (0,2), (1,2), (2,2). But these two paths together still don't cover (1,1), (1,0), (1,2) is covered, (2,1) is covered, but (1,1) is not. So maybe we need a third path that goes through (1,1). For example, (0,0) → (1,0) → (1,1) → (1,2) → (2,2). That path covers (0,0), (1,0), (1,1), (1,2), (2,2). Now, with these three paths, do we cover all vertices?First path: covers (0,0), (1,0), (2,0), (2,1), (2,2)Second path: (0,0), (0,1), (0,2), (1,2), (2,2)Third path: (0,0), (1,0), (1,1), (1,2), (2,2)So missing vertices: (0,1) is covered by the second path. (1,1) is covered by the third path. (2,1) is covered by the first path. (1,2) is covered by second and third. So all vertices are covered.But is there a way to do it with two paths? Let's see.Suppose one path is the first one: (0,0) to (2,0) to (2,2). Second path: (0,0) to (0,2) to (2,2). These two cover all except (1,1), (1,0), (0,1), (1,2), (2,1). Wait, actually, no. (1,0) is covered by the first path? Wait, the first path is (0,0) → (1,0) → (2,0) → (2,1) → (2,2). So (1,0) is covered. Similarly, (0,1) is covered by the second path: (0,0) → (0,1) → (0,2) → (1,2) → (2,2). So (0,1), (0,2), (1,2). Then (1,0) and (2,1) are covered by the first path, (1,1) is left. So (1,1) is not covered. So we need a third path that goes through (1,1). So in a 2x2 grid, the minimal number is 3.But wait, maybe another path? What if we take a different path: (0,0) → (1,0) → (1,1) → (2,1) → (2,2). This covers (0,0), (1,0), (1,1), (2,1), (2,2). If we have this path along with the previous two, maybe? Wait, but if we have two paths:First path: (0,0) → (1,0) → (1,1) → (2,1) → (2,2)Second path: (0,0) → (0,1) → (1,1) → (1,2) → (2,2)These two paths together cover:First path: (0,0), (1,0), (1,1), (2,1), (2,2)Second path: (0,0), (0,1), (1,1), (1,2), (2,2)So missing vertices: (0,2), (2,0), (1,0) is covered, (0,1) is covered. Wait, (0,2) is missing, (2,0) is missing, (1,2) is covered by the second path. So to cover (0,2) and (2,0), we need another path. So maybe a third path going along the edges: (0,0) → (0,1) → (0,2) → (1,2) → (2,2). But then that would be the same as the second path. Hmm. So maybe even with different paths, you still need three.Alternatively, what if you use a path that zigzags? But in a 2x2 grid, there's not much room. So maybe three is the minimum.So in the 2x2 grid, it's three. Then in a 1x1 grid, which is just a square with four vertices, the two corners connected by a diagonal. But since we can't use diagonals, the shortest path would be two steps: right then up or up then right. So two paths would cover all four vertices. Wait, in a 1x1 grid (from (0,0) to (1,1)), the two paths are (0,0) → (1,0) → (1,1) and (0,0) → (0,1) → (1,1). These two cover all four vertices. So for 1x1 grid, it's 2.Similarly, for a 3x3 grid, maybe the pattern continues? But how?Wait, perhaps the minimal number of paths is equal to the grid size? For a 1x1 grid (2x2 vertices), it's 2. Wait, 2x2 grid (3x3 vertices) needs 3. So for n x n grid (which has (n+1)x(n+1) vertices), the minimal number of paths is n+1. But in the 1x1 case, n=1, so 2. 2x2 grid, n=2, so 3. Hmm, maybe. So for a 100x100 grid, would it be 101? But wait, let's check with 3x3.Take a 3x3 grid (4x4 vertices). If the minimal number is 4, how would that work?Each path is a shortest path from (0,0) to (3,3), consisting of 3 rights and 3 ups. To cover all 16 vertices. Let's see.If we use 4 paths, each going through different diagonals? Wait, but how? For example, the first path goes all right then all up. The second path goes up then right. Then two more paths that snake through the middle? Maybe.Alternatively, perhaps by shifting the paths.Wait, maybe if we parameterize the paths by their "slope". For each path, if we vary the number of right steps before switching to up steps, we can cover different vertices.But in a grid, each path is determined by the order of rights and ups. If we can arrange multiple paths such that every vertex is on at least one path.Alternatively, think of the grid as a graph where each vertex is a point (i,j), and edges are right or up moves. Then the problem reduces to covering all vertices with the minimal number of paths from (0,0) to (n,n).This is similar to a covering problem. In graph terms, what's the minimum number of paths needed to cover all vertices of a grid graph, where each path is a shortest path from (0,0) to (n,n).I recall that in such grid graphs, the minimum number of paths required to cover all vertices is equal to the maximum size of an antichain. By Dilworth's theorem, in a poset, the minimum number of chains needed to cover the poset equals the size of the largest antichain.But is this applicable here? Let's see.In the grid, we can define a partial order where (i,j) ≤ (k,l) if i ≤ k and j ≤ l. Then, a chain in this poset would correspond to a sequence of points where each is to the right or above the previous, which is exactly a path from (0,0) to (n,n). An antichain would be a set of points where no two are comparable, meaning no two points lie on the same chain.In this poset, the largest antichain is given by Sperner's theorem? Wait, Sperner's theorem is about subsets, but in the grid poset, the largest antichain is the set of points (i,j) where i + j = n, which has size n+1. Wait, for an n x n grid, the number of points where i + j = n is n+1. But actually, for a grid from (0,0) to (n,n), the middle layer where i + j = n would have n+1 points. Hmm, is that the largest antichain?Wait, actually, in the grid poset (0,0) to (n,n), the size of the largest antichain is indeed the binomial coefficient C(2n, n), but that's for subsets. Wait, maybe not. Let me think again.Actually, in the grid poset, the poset is a two-dimensional grid with component-wise ordering. The size of the largest antichain in such a grid is given by the de Bruijn-Tengbergen-Kruyswijk theorem, which states that the largest antichain in the product of two chains of lengths m and n is the set of elements with coordinates (i,j) where i + j is equal to the floor or ceiling of (m + n)/2. For a square grid n x n, the largest antichain is the middle layer, which has size n + 1 when n is even. Wait, for example, in a 2x2 grid (n=2), the middle layer is i + j = 2, which has 3 points: (0,2), (1,1), (2,0). But in our grid, we have (n+1)x(n+1) points. Wait, maybe for a grid from (0,0) to (n,n), the largest antichain is indeed the set of points where i + j = n, which has n + 1 points. So if we use Dilworth's theorem, the minimal number of chains (paths) needed to cover the poset is equal to the size of the largest antichain. Therefore, the minimal number is n + 1.So for a 100x100 grid, which is a grid from (0,0) to (100,100), the largest antichain has 101 points (since i + j = 100 gives 101 points: (0,100), (1,99), ..., (100,0)), so according to Dilworth's theorem, the minimal number of chains (shortest paths) needed is 101.Therefore, the answer should be 101. But let me verify with the small cases.For the 1x1 grid (n=1), the largest antichain would be 2 points (i + j = 1: (0,1) and (1,0)), so minimal number of chains is 2, which matches the earlier result.For the 2x2 grid (n=2), largest antichain is 3 points, so minimal chains needed is 3, which matches the 2x2 case where we needed 3 paths.Therefore, by Dilworth's theorem, the minimal number of paths is equal to the size of the largest antichain, which is n + 1 for an n x n grid. Therefore, for a 100x100 grid, the minimal number is 101.But let me think again: in the grid, each path from (0,0) to (n,n) corresponds to a chain in the poset. Since the poset can be covered by 101 chains, each chain corresponds to a path, and each antichain of size 101 requires at least 101 chains. Hence, the minimal number is 101.Therefore, the answer is boxed{101}.Final Answerboxed{101}</think>To determine the smallest number of broken lines needed to cover all vertices of a 100x100 grid, connecting the opposite corners by the shortest path, we can use Dilworth's theorem from poset theory. A broken line following the grid lines from (0,0) to (100,100) corresponds to a chain in the poset where each point (i,j) is ordered by component-wise comparison. The largest antichain in this poset is the set of points where the sum of the coordinates is constant, specifically i + j = 100. This antichain contains 101 points. By Dilworth's theorem, the minimum number of chains (or paths) needed to cover the poset is equal to the size of the largest antichain. Therefore, the minimal number of broken lines required to cover all vertices in the 100x100 grid is 101.[boxed{101}]]

🔑:77.16. Answer. 101 broken lines. The fact that there are no fewer than 101 broken lines follows from the fact that if we consider all the vertices lying on the diagonal connecting two other opposite vertices of the square (there are 101 of them), then no broken line of the specified type can contain two vertices from this set. An example for 101 broken lines is obvious.

🧠:[<think>Okay, let's tackle this problem step by step. The function given is ( f(x) = sin x + cos x ), and there are three parts to solve: finding ( fleft(frac{pi}{2}right) ), determining the smallest positive period of ( f(x) ), and calculating the minimum value of another function ( g(x) ). Let's start with part I.Part I: Find ( fleft(frac{pi}{2}right) ).This seems straightforward. We just need to substitute ( x = frac{pi}{2} ) into the function ( f(x) ).So, ( fleft(frac{pi}{2}right) = sinleft(frac{pi}{2}right) + cosleft(frac{pi}{2}right) ).From trigonometry, I remember that ( sinleft(frac{pi}{2}right) = 1 ) and ( cosleft(frac{pi}{2}right) = 0 ). Therefore, substituting these values in:( fleft(frac{pi}{2}right) = 1 + 0 = 1 ).That was simple enough. Let me double-check to make sure I didn't mix up sine and cosine. Nope, sine of ( pi/2 ) is definitely 1, cosine is 0. So part I should be 1.Part II: Determine the smallest positive period of ( f(x) ).Hmm, the function is ( f(x) = sin x + cos x ). Both sine and cosine functions individually have a period of ( 2pi ). But when you add two periodic functions, the period of the resulting function can sometimes be the least common multiple (LCM) of their individual periods. However, since both sine and cosine here have the same period ( 2pi ), their sum should also have the same period, right? But wait, maybe there's a smaller period. Let me think.Alternatively, I recall that ( sin x + cos x ) can be rewritten as a single sine or cosine function using the amplitude-phase form. Maybe that would help in determining the period.Let me try that. The general identity is ( a sin x + b cos x = R sin(x + phi) ), where ( R = sqrt{a^2 + b^2} ). In our case, ( a = 1 ) and ( b = 1 ), so ( R = sqrt{1 + 1} = sqrt{2} ). Then ( phi ) is the phase shift such that ( cos phi = frac{a}{R} = frac{1}{sqrt{2}} ) and ( sin phi = frac{b}{R} = frac{1}{sqrt{2}} ). Therefore, ( phi = frac{pi}{4} ).So, ( f(x) = sqrt{2} sinleft(x + frac{pi}{4}right) ). The sine function has a period of ( 2pi ), so shifting it by ( frac{pi}{4} ) doesn't change its period. Therefore, the period of ( f(x) ) is the same as the period of ( sin x ), which is ( 2pi ).Wait, but maybe there's a smaller period? Let me check. Suppose there exists a positive number ( T ) such that ( f(x + T) = f(x) ) for all ( x ). Then ( sqrt{2} sinleft(x + T + frac{pi}{4}right) = sqrt{2} sinleft(x + frac{pi}{4}right) ). Dividing both sides by ( sqrt{2} ), we get ( sinleft(x + T + frac{pi}{4}right) = sinleft(x + frac{pi}{4}right) ). For this equality to hold for all ( x ), the arguments must differ by an integer multiple of ( 2pi ), or differ by an odd multiple of ( pi ) but with sine being equal. However, since sine is periodic with period ( 2pi ), the smallest period is ( 2pi ). But let's test with ( T = pi ). Let's see if ( f(x + pi) = f(x) ):( f(x + pi) = sin(x + pi) + cos(x + pi) = -sin x - cos x = -f(x) ). So, ( f(x + pi) = -f(x) ), which is not equal to ( f(x) ). Therefore, ( pi ) is not a period. What about ( T = pi/2 )? Let's compute ( f(x + pi/2) ):( f(x + pi/2) = sin(x + pi/2) + cos(x + pi/2) = cos x - sin x ), which is not equal to ( f(x) = sin x + cos x ). So, not equal. So, the period must be ( 2pi ).But wait, when I rewrote ( f(x) ) as ( sqrt{2} sinleft(x + frac{pi}{4}right) ), which is a sine function with the same period ( 2pi ). So that confirms it. Therefore, the smallest positive period is ( 2pi ).But I feel like sometimes when you combine functions, you might get a smaller period, but in this case, since both sine and cosine have the same period, and they are combined linearly, the period remains the same. So, part II's answer is ( 2pi ).Part III: Calculate the minimum value of the function ( g(x) = fleft(x + frac{pi}{4}right) + fleft(x + frac{3pi}{4}right) ).This seems a bit more involved. Let's first write down what ( g(x) ) is. Substitute the definition of ( f ):( g(x) = left[ sinleft(x + frac{pi}{4}right) + cosleft(x + frac{pi}{4}right) right] + left[ sinleft(x + frac{3pi}{4}right) + cosleft(x + frac{3pi}{4}right) right] ).Alternatively, since ( f(x) = sqrt{2} sinleft(x + frac{pi}{4}right) ), then substituting:( g(x) = sqrt{2} sinleft( x + frac{pi}{4} + frac{pi}{4} right) + sqrt{2} sinleft( x + frac{3pi}{4} + frac{pi}{4} right) ).Wait, hold on. If ( f(x) = sqrt{2} sinleft(x + frac{pi}{4}right) ), then ( fleft(x + aright) = sqrt{2} sinleft( x + a + frac{pi}{4} right) ). So, substituting ( a = frac{pi}{4} ), we get:( fleft(x + frac{pi}{4}right) = sqrt{2} sinleft( x + frac{pi}{4} + frac{pi}{4} right) = sqrt{2} sinleft( x + frac{pi}{2} right) ).Similarly, ( fleft(x + frac{3pi}{4}right) = sqrt{2} sinleft( x + frac{3pi}{4} + frac{pi}{4} right) = sqrt{2} sinleft( x + pi right) ).Therefore, ( g(x) = sqrt{2} sinleft( x + frac{pi}{2} right) + sqrt{2} sinleft( x + pi right) ).Simplify each term:( sinleft( x + frac{pi}{2} right) = cos x ), since shifting sine by ( pi/2 ) gives cosine.( sinleft( x + pi right) = -sin x ), since shifting sine by ( pi ) inverts it.Therefore, substituting back:( g(x) = sqrt{2} cos x + sqrt{2} (-sin x) = sqrt{2} (cos x - sin x) ).Alternatively, ( g(x) = sqrt{2} (cos x - sin x) ).Now, to find the minimum value of ( g(x) ). Let's analyze this expression.We can write ( cos x - sin x ) in the amplitude-phase form again. Let's do that.( cos x - sin x = R cos(x + phi) ), where ( R = sqrt{1^2 + (-1)^2} = sqrt{2} ). Let's verify:( R cos(x + phi) = sqrt{2} cos(x + phi) ).Expanding ( sqrt{2} cos(x + phi) = sqrt{2} (cos x cos phi - sin x sin phi) ).Comparing coefficients with ( cos x - sin x ):Coefficient of ( cos x ): ( sqrt{2} cos phi = 1 )Coefficient of ( -sin x ): ( -sqrt{2} sin phi = -1 )Thus,( sqrt{2} cos phi = 1 Rightarrow cos phi = frac{1}{sqrt{2}} )( sqrt{2} sin phi = 1 Rightarrow sin phi = frac{1}{sqrt{2}} )Therefore, ( phi = frac{pi}{4} ). Hence,( cos x - sin x = sqrt{2} cosleft(x + frac{pi}{4}right) ).Therefore, ( g(x) = sqrt{2} times sqrt{2} cosleft(x + frac{pi}{4}right) = 2 cosleft(x + frac{pi}{4}right) ).Now, the minimum value of ( cos theta ) is -1, so the minimum value of ( 2 costheta ) is -2. Therefore, the minimum value of ( g(x) ) is -2.Wait, let me make sure I didn't skip any steps. Let's verify:Starting from ( g(x) = 2 cosleft(x + frac{pi}{4}right) ). Since the cosine function oscillates between -1 and 1, multiplying by 2 scales it to oscillate between -2 and 2. Therefore, the minimum value is indeed -2. So the minimum of ( g(x) ) is -2.Alternatively, if I didn't use the amplitude-phase form, I could take the derivative of ( g(x) ) to find critical points, but since we already converted ( g(x) ) into a single cosine function with amplitude 2, it's straightforward. The maximum is 2, the minimum is -2. Therefore, the answer is -2.But let me check again the steps in case I made an error.Original ( g(x) = f(x + pi/4) + f(x + 3pi/4) ). Then, expanding each term:First term: ( f(x + pi/4) = sin(x + pi/4) + cos(x + pi/4) ).Second term: ( f(x + 3pi/4) = sin(x + 3pi/4) + cos(x + 3pi/4) ).Let me compute each sine and cosine term separately.First term:( sin(x + pi/4) = sin x cos pi/4 + cos x sin pi/4 = frac{sqrt{2}}{2} (sin x + cos x) )( cos(x + pi/4) = cos x cos pi/4 - sin x sin pi/4 = frac{sqrt{2}}{2} (cos x - sin x) )Therefore, adding them:( sin(x + pi/4) + cos(x + pi/4) = frac{sqrt{2}}{2} (sin x + cos x + cos x - sin x) = frac{sqrt{2}}{2} (2 cos x) = sqrt{2} cos x ).Similarly, second term:( sin(x + 3pi/4) = sin x cos 3pi/4 + cos x sin 3pi/4 = sin x (- frac{sqrt{2}}{2}) + cos x (frac{sqrt{2}}{2}) )( cos(x + 3pi/4) = cos x cos 3pi/4 - sin x sin 3pi/4 = cos x (-frac{sqrt{2}}{2}) - sin x (frac{sqrt{2}}{2}) )Adding these:( sin(x + 3pi/4) + cos(x + 3pi/4) = -frac{sqrt{2}}{2} sin x + frac{sqrt{2}}{2} cos x - frac{sqrt{2}}{2} cos x - frac{sqrt{2}}{2} sin x )Simplify:Combine like terms:For ( sin x ): ( -frac{sqrt{2}}{2} - frac{sqrt{2}}{2} = -sqrt{2} sin x )For ( cos x ): ( frac{sqrt{2}}{2} - frac{sqrt{2}}{2} = 0 )Therefore, the second term simplifies to ( -sqrt{2} sin x ).Therefore, ( g(x) = sqrt{2} cos x - sqrt{2} sin x = sqrt{2} (cos x - sin x) ), which matches what we had earlier. Then converting that into a single cosine function, we get ( 2 cos(x + pi/4) ), whose minimum is -2. So, that's correct.Alternatively, another way to check is to note that ( sqrt{2} (cos x - sin x) ) is of the form ( A cos x + B sin x ), which can be written as ( C cos(x + phi) ), with amplitude ( C = sqrt{A^2 + B^2} ). Here, ( A = sqrt{2} ), ( B = -sqrt{2} ), so ( C = sqrt{ (sqrt{2})^2 + (-sqrt{2})^2 } = sqrt{2 + 2} = sqrt{4} = 2 ). Therefore, the amplitude is 2, so the range is from -2 to 2, so minimum is -2.Hence, part III's answer is -2.Double-checking All Parts:I. Substituted ( pi/2 ) into sin and cos, got 1 and 0, so 1. Correct.II. Rewrote the function as a single sine wave with the same period, confirmed that shifting by pi or pi/2 didn't work, so period is 2pi. Correct.III. Expanded g(x), converted into a single cosine function with amplitude 2, so min is -2. Correct.All steps seem solid, no mistakes detected.Final AnswerI. (boxed{1})II. (boxed{2pi})III. (boxed{-2})</think>Given the function ( f(x) = sin x + cos x ):I. Find the value of ( fleft(frac{pi}{2}right) ).Substitute ( x = frac{pi}{2} ) into the function:[fleft(frac{pi}{2}right) = sinleft(frac{pi}{2}right) + cosleft(frac{pi}{2}right) = 1 + 0 = 1]II. Determine the smallest positive period of the function ( f(x) ).Rewrite ( f(x) ) using the amplitude-phase form:[f(x) = sqrt{2} sinleft(x + frac{pi}{4}right)]The period of ( sin(x + frac{pi}{4}) ) is ( 2pi ), so the smallest positive period of ( f(x) ) is ( 2pi ).III. Calculate the minimum value of the function ( g(x) = fleft(x + frac{pi}{4}right) + fleft(x + frac{3pi}{4}right) ).First, express ( g(x) ) in terms of sine and cosine:[g(x) = left[ sinleft(x + frac{pi}{4}right) + cosleft(x + frac{pi}{4}right) right] + left[ sinleft(x + frac{3pi}{4}right) + cosleft(x + frac{3pi}{4}right) right]]Simplify each term:[sinleft(x + frac{pi}{4}right) + cosleft(x + frac{pi}{4}right) = sqrt{2} cos x][sinleft(x + frac{3pi}{4}right) + cosleft(x + frac{3pi}{4}right) = -sqrt{2} sin x]Combine the terms:[g(x) = sqrt{2} cos x - sqrt{2} sin x = sqrt{2} (cos x - sin x)]Rewrite using the amplitude-phase form:[cos x - sin x = sqrt{2} cosleft(x + frac{pi}{4}right)]Thus,[g(x) = 2 cosleft(x + frac{pi}{4}right)]The minimum value of ( 2 cosleft(x + frac{pi}{4}right) ) is (-2). Final AnswerI. (boxed{1})II. (boxed{2pi})III. (boxed{-2}),<think>Alright, let's tackle this problem step by step. The function given is f(x) = sin x + cos x. There are three parts to the problem: finding f(π/2), determining the smallest positive period of f(x), and calculating the minimum value of another function g(x) which is a combination of shifted f(x) terms. Let me start with part (I).Part I: Find the value of f(π/2).Okay, so f(x) is sin x + cos x. To find f(π/2), I just need to substitute x with π/2 in the function. Let me compute that.sin(π/2) is 1, right? Because sine of π/2 is 1. Then cos(π/2) is 0, since cosine of π/2 is 0. So adding those together: 1 + 0 = 1. Therefore, f(π/2) = 1. That seems straightforward.Wait, let me double-check. Maybe I mixed up sine and cosine? No, no. Cosine of π/2 is indeed 0, and sine is 1. Yes, that's correct. So part (I) answer is 1.Part II: Determine the smallest positive period of the function f(x).Hmm, the function is sin x + cos x. Both sine and cosine have a period of 2π, but since they are being added together, the period of the sum should also be 2π, right? But wait, sometimes when functions are combined, if there's a common period smaller than 2π, that could be the period. However, both sin x and cos x have the same period of 2π, so their sum should also have period 2π. But let me verify.Alternatively, maybe we can rewrite f(x) as a single sinusoidal function. Let me recall that A sin x + B cos x can be written as C sin(x + φ) or C cos(x + θ), which might make the period more evident. Let's try that.Given f(x) = sin x + cos x. Let's express this in the form R sin(x + α), where R is the amplitude and α is the phase shift.The formula for combining sine and cosine terms: R = √(A² + B²), where A and B are coefficients. Here, A = 1 (coefficient of sin x) and B = 1 (coefficient of cos x). So R = √(1² + 1²) = √2.Then, α is given by tan α = B/A. Wait, actually, if we write it as R sin(x + α), the formula is:sin x + cos x = R sin(x + α) = R sin x cos α + R cos x sin α.Comparing coefficients:1 = R cos α (coefficient of sin x)1 = R sin α (coefficient of cos x)So, we have:R cos α = 1R sin α = 1Since R = √2, then cos α = 1/√2 and sin α = 1/√2. Therefore, α is π/4 (45 degrees) because both sine and cosine are positive and equal. Therefore, f(x) can be written as √2 sin(x + π/4). The standard sine function has a period of 2π, so if it's sin(x + π/4), the period remains 2π. Therefore, the smallest positive period of f(x) is 2π. But wait, hold on. Sometimes when functions are combined with different periods, the period can be the least common multiple (LCM) of their individual periods. However, in this case, both sine and cosine have the same period, so their sum should also have the same period. Therefore, the period is 2π. But wait, another thought. If the function is transformed horizontally, like sin(x + π/4), does that affect the period? No, horizontal shifts don't affect the period. So even after the phase shift, the period remains 2π. Therefore, the answer should be 2π.But let me check by testing the function at some points. Let's compute f(0) and f(2π). f(0) = sin 0 + cos 0 = 0 + 1 = 1. f(2π) = sin(2π) + cos(2π) = 0 + 1 = 1. What about f(π/2)? That's 1 + 0 = 1, which we already calculated. But to check periodicity, we need to see if f(x + T) = f(x) for the smallest T.But if we take T = π, does f(x + π) equal f(x)? Let's compute f(x + π) = sin(x + π) + cos(x + π) = -sin x - cos x = -f(x). So that's not equal to f(x), so T = π is not a period. What about T = π/2? f(x + π/2) = sin(x + π/2) + cos(x + π/2) = cos x - sin x. Which is not equal to f(x). So that's different. Therefore, the period is indeed 2π. So part (II) answer is 2π.Wait, but there's another approach. If we write f(x) as √2 sin(x + π/4), then its period is 2π. So that confirms it. So yes, the period is 2π.Part III: Calculate the minimum value of the function g(x) = f(x + π/4) + f(x + 3π/4).Hmm, okay. Let's first write out what g(x) is. Since f(x) = sin x + cos x, then:g(x) = f(x + π/4) + f(x + 3π/4) = [sin(x + π/4) + cos(x + π/4)] + [sin(x + 3π/4) + cos(x + 3π/4)].So let's compute each term separately.First term: sin(x + π/4) + cos(x + π/4).Second term: sin(x + 3π/4) + cos(x + 3π/4).Alternatively, maybe we can use the expression we found earlier for f(x) as √2 sin(x + π/4). Let's see.Since f(x) = √2 sin(x + π/4), then:g(x) = √2 sin(x + π/4 + π/4) + √2 sin(x + 3π/4 + π/4) ?Wait, wait, hold on. Wait, f(x + a) = √2 sin((x + a) + π/4). So f(x + π/4) = √2 sin(x + π/4 + π/4) = √2 sin(x + π/2). Similarly, f(x + 3π/4) = √2 sin(x + 3π/4 + π/4) = √2 sin(x + π). So then:g(x) = √2 sin(x + π/2) + √2 sin(x + π).Simplify each term:sin(x + π/2) is equal to cos x. Because sin(x + π/2) = cos x.sin(x + π) is equal to -sin x. Because sin(x + π) = -sin x.Therefore, substituting back:g(x) = √2 cos x + √2 (-sin x) = √2 (cos x - sin x).So g(x) = √2 (cos x - sin x). Now, we need to find the minimum value of this function.Alternatively, we can express cos x - sin x as another sinusoidal function. Let me do that.Let’s write cos x - sin x as R cos(x + α). Let's find R and α.Express R cos(x + α) = R cos x cos α - R sin x sin α. Comparing to cos x - sin x, we have:R cos α = 1 (coefficient of cos x)-R sin α = -1 (coefficient of sin x) => R sin α = 1So, R cos α = 1 and R sin α = 1. Then, R² = 1² + 1² = 2 => R = √2.Then, tan α = (R sin α)/(R cos α) = 1/1 = 1 => α = π/4. Therefore:cos x - sin x = √2 cos(x + π/4).Therefore, g(x) = √2 * √2 cos(x + π/4) = 2 cos(x + π/4).Wait, wait. Wait, if cos x - sin x = √2 cos(x + π/4), then multiplying by √2 gives √2*(cos x - sin x) = 2 cos(x + π/4). Wait, no:Wait, actually, let's step back. I have g(x) = √2 (cos x - sin x). Then, if cos x - sin x = √2 cos(x + π/4), then:g(x) = √2 * √2 cos(x + π/4) = 2 cos(x + π/4). Is that correct?Wait, let's check:√2 cos(x + π/4) = √2 [cos x cos π/4 - sin x sin π/4] = √2 [ (cos x)(√2/2) - (sin x)(√2/2) ] = √2*(√2/2)(cos x - sin x) = (2/2)(cos x - sin x) = cos x - sin x. So yes, cos x - sin x = √2 cos(x + π/4). Therefore, g(x) = √2*(cos x - sin x) = √2*(√2 cos(x + π/4)) = 2 cos(x + π/4). Therefore, g(x) simplifies to 2 cos(x + π/4). Now, the minimum value of 2 cos(x + π/4) is simply -2, since the cosine function oscillates between -1 and 1, so multiplying by 2 scales it to between -2 and 2. Therefore, the minimum value is -2. But let's confirm this. Alternatively, since g(x) is expressed as 2 cos(x + π/4), its amplitude is 2, so indeed, the minimum is -2. However, let me check if there might be any error in the transformation steps.Starting from g(x):g(x) = f(x + π/4) + f(x + 3π/4)f(x + π/4) = sin(x + π/4) + cos(x + π/4)f(x + 3π/4) = sin(x + 3π/4) + cos(x + 3π/4)Compute each term:First, compute sin(x + π/4) + cos(x + π/4):Using angle addition formulas:sin(x + π/4) = sin x cos π/4 + cos x sin π/4 = (√2/2)(sin x + cos x)cos(x + π/4) = cos x cos π/4 - sin x sin π/4 = (√2/2)(cos x - sin x)Therefore, f(x + π/4) = (√2/2)(sin x + cos x) + (√2/2)(cos x - sin x) = (√2/2)(sin x + cos x + cos x - sin x) = (√2/2)(2 cos x) = √2 cos x.Similarly, compute f(x + 3π/4):sin(x + 3π/4) = sin x cos 3π/4 + cos x sin 3π/4 = sin x*(-√2/2) + cos x*(√2/2) = (√2/2)(-sin x + cos x)cos(x + 3π/4) = cos x cos 3π/4 - sin x sin 3π/4 = cos x*(-√2/2) - sin x*(√2/2) = (-√2/2)(cos x + sin x)Therefore, f(x + 3π/4) = (√2/2)(-sin x + cos x) + (-√2/2)(cos x + sin x) = √2/2 (-sin x + cos x - cos x - sin x) = √2/2 (-2 sin x) = -√2 sin x.Thus, adding f(x + π/4) + f(x + 3π/4) = √2 cos x - √2 sin x = √2 (cos x - sin x) which is the same as before. Then, as before, we write this as 2 cos(x + π/4). Hence, the minimum value is -2. Wait, but let's confirm this conclusion once more. The function 2 cos(x + π/4) has a range of [-2, 2], so the minimum is indeed -2. Therefore, the answer for part (III) is -2.But hold on, let's check if there might be any mistake in the transformation. Let's compute g(x) numerically at a specific point to verify.Take x = -π/4. Then:g(-π/4) = f(-π/4 + π/4) + f(-π/4 + 3π/4) = f(0) + f(π/2) = (sin 0 + cos 0) + (sin π/2 + cos π/2) = (0 + 1) + (1 + 0) = 1 + 1 = 2. According to our expression, 2 cos(-π/4 + π/4) = 2 cos(0) = 2*1 = 2. Correct.Another point, x = 3π/4 - π/4 = π/2. Wait, let's take x = π/4. Then:g(π/4) = f(π/4 + π/4) + f(π/4 + 3π/4) = f(π/2) + f(π) = (1 + 0) + (0 + (-1)) = 1 - 1 = 0.According to 2 cos(π/4 + π/4) = 2 cos(π/2) = 0. Correct.Another test: x = 0. Then:g(0) = f(π/4) + f(3π/4). Compute f(π/4) = sin(π/4) + cos(π/4) = √2/2 + √2/2 = √2. f(3π/4) = sin(3π/4) + cos(3π/4) = √2/2 - √2/2 = 0. Therefore, g(0) = √2 + 0 = √2 ≈ 1.414. According to 2 cos(0 + π/4) = 2*(√2/2) = √2. Correct.Now, to check the minimum value. Let's compute when cos(x + π/4) = -1. That happens when x + π/4 = π + 2πk, i.e., x = 3π/4 + 2πk. Then, 2 cos(x + π/4) = 2*(-1) = -2. Let's check x = 3π/4.g(3π/4) = f(3π/4 + π/4) + f(3π/4 + 3π/4) = f(π) + f(3π/2). f(π) = 0 + (-1) = -1. f(3π/2) = -1 + 0 = -1. Therefore, g(3π/4) = -1 + (-1) = -2. Which matches our conclusion. Therefore, the minimum value is indeed -2.Summary:- (I) f(π/2) = 1- (II) The smallest positive period is 2π- (III) The minimum value of g(x) is -2Final AnswerI. boxed{1}II. boxed{2pi}III. boxed{-2}</think>Given the function ( f(x) = sin x + cos x ):(I) Find the value of ( fleft(frac{pi}{2}right) ):Substitute ( x = frac{pi}{2} ) into the function:[fleft(frac{pi}{2}right) = sinleft(frac{pi}{2}right) + cosleft(frac{pi}{2}right) = 1 + 0 = 1]Thus, the value is (boxed{1}).(II) Determine the smallest positive period of the function ( f(x) ):The function ( f(x) = sin x + cos x ) can be rewritten using a phase shift:[f(x) = sqrt{2} sinleft(x + frac{pi}{4}right)]The period of ( sin(x + frac{pi}{4}) ) is ( 2pi ), so the smallest positive period of ( f(x) ) is ( 2pi ). Therefore, the smallest positive period is (boxed{2pi}).(III) Calculate the minimum value of the function ( g(x) = fleft(x + frac{pi}{4}right) + fleft(x + frac{3pi}{4}right) ):First, express ( g(x) ) using the given function ( f ):[g(x) = fleft(x + frac{pi}{4}right) + fleft(x + frac{3pi}{4}right)]Substitute ( f(x) = sin x + cos x ):[g(x) = left[sinleft(x + frac{pi}{4}right) + cosleft(x + frac{pi}{4}right)right] + left[sinleft(x + frac{3pi}{4}right) + cosleft(x + frac{3pi}{4}right)right]]Simplify each term using trigonometric identities:[sinleft(x + frac{pi}{4}right) + cosleft(x + frac{pi}{4}right) = sqrt{2} cos x][sinleft(x + frac{3pi}{4}right) + cosleft(x + frac{3pi}{4}right) = -sqrt{2} sin x]Combine the terms:[g(x) = sqrt{2} cos x - sqrt{2} sin x = sqrt{2} (cos x - sin x)]Rewrite using a single sinusoidal function:[cos x - sin x = sqrt{2} cosleft(x + frac{pi}{4}right)]Thus,[g(x) = sqrt{2} cdot sqrt{2} cosleft(x + frac{pi}{4}right) = 2 cosleft(x + frac{pi}{4}right)]The minimum value of ( 2 cosleft(x + frac{pi}{4}right) ) is (-2). Therefore, the minimum value is (boxed{-2}).]

🔑:- (I) Since f(x) = sin x + cos x, we have fleft(frac{pi}{2}right) = sin frac{pi}{2} + cos frac{pi}{2} = 1 + 0 = 1. Therefore, fleft(frac{pi}{2}right) = boxed{1}.- (II) The function f(x) = sin x + cos x can be rewritten using the sum to product identity as f(x) = sqrt{2} sin left(x + frac{pi}{4}right). Since sine has a fundamental period of 2pi, the smallest positive period of f(x) is also 2pi. Hence, the smallest positive period of f(x) is boxed{2pi}.- (III) To find the minimum value of g(x), we write g(x) in terms of sine and cosine functions:g(x) = fleft(x + frac{pi}{4}right) + fleft(x + frac{3pi}{4}right)= sqrt{2} sin left(x + frac{pi}{4} + frac{pi}{4}right) + sqrt{2} sin left(x + frac{3pi}{4} + frac{pi}{4}right)= sqrt{2} cos x + sqrt{2} sin left(x + piright)= sqrt{2} cos x - sqrt{2} sin x text{ (since } sin(x + pi) = -sin x text{)}= sqrt{2}(cos x - sin x)To express this in terms of a single trigonometric function, we use the angle addition formula for cosine:cos(x - y) = cos x cos y + sin x sin ySetting y = frac{pi}{4}, we have cos y = sin y = frac{1}{sqrt{2}}, thusg(x) = sqrt{2} cdot frac{1}{sqrt{2}}(cos x - sin x) = 2cos left(x + frac{pi}{4}right)The minimum value of the cosine function is -1, so the minimum of 2cos left(x + frac{pi}{4}right) is -2, occurring when x + frac{pi}{4} = (2k+1)pi for any integer k, which simplifies to x = (2k+1)pi - frac{pi}{4}.Therefore, the minimum value of g(x) is boxed{-2}.

🧠:[<think>Okay, so I have this problem about Da Mao, Er Mao, and San Mao's weights. Let me try to figure it out step by step. The problem states their combined weights in different pairs and how much more they weigh than the third person. I need to find their total weight together. Hmm, let's break it down.First, let me assign variables to each person's weight to make it easier. Let's say:- D = Da Mao's weight- E = Er Mao's weight- S = San Mao's weightThe problem gives three statements:1. Da Mao and Er Mao together weigh 23 kilograms more than San Mao.2. Er Mao and San Mao together weigh 27 kilograms more than Da Mao.3. San Mao and Da Mao together weigh 37 kilograms more than Er Mao.And the question is asking for the total weight of all three together, which would be D + E + S. So, I need to find D + E + S.Let me translate each statement into equations.First statement: D + E = S + 23. Because Da Mao and Er Mao together (D + E) weigh 23 kg more than San Mao (S + 23).Second statement: E + S = D + 27. Similarly, Er Mao and San Mao together (E + S) weigh 27 kg more than Da Mao (D + 27).Third statement: S + D = E + 37. San Mao and Da Mao together (S + D) weigh 37 kg more than Er Mao (E + 37).So now I have three equations:1. D + E = S + 23 ...(1)2. E + S = D + 27 ...(2)3. S + D = E + 37 ...(3)And I need to find D + E + S.Hmm, so maybe if I can solve these equations simultaneously, I can find each person's weight or find the total directly.Let me see. Let's write down all three equations:Equation (1): D + E - S = 23Equation (2): -D + E + S = 27Equation (3): D - E + S = 37So, writing them in standard form with variables on the left and constants on the right. Now, this looks like a system of three equations with three variables.Maybe I can use elimination to solve for D, E, S. Let's try that.First, let's list the equations again:1. D + E - S = 23 ...(1)2. -D + E + S = 27 ...(2)3. D - E + S = 37 ...(3)Let me try adding equations (1) and (2) together to eliminate D.Equation (1) + Equation (2):(D + E - S) + (-D + E + S) = 23 + 27Simplify:D - D + E + E + (-S + S) = 50So, 0D + 2E + 0S = 50Which simplifies to 2E = 50 => E = 25.Oh, that's great! So Er Mao weighs 25 kg.Now that I know E = 25, I can substitute this value back into the equations to find D and S.Let's substitute E = 25 into equation (1):Equation (1): D + 25 - S = 23So, D - S = 23 - 25 = -2Therefore, D - S = -2 => D = S - 2 ...(1a)Similarly, substitute E = 25 into equation (2):Equation (2): -D + 25 + S = 27So, -D + S = 27 - 25 = 2Which gives: -D + S = 2 => S = D + 2 ...(2a)Wait, from equation (1a) we have D = S - 2, and from equation (2a) we have S = D + 2. These two equations are the same. So substituting D from (1a) into (2a):S = (S - 2) + 2 => S = S. Hmm, that's an identity. So it doesn't help us further. We need another equation to find D and S.Let's use equation (3) with E = 25.Equation (3): D - 25 + S = 37So, D + S = 37 + 25 = 62Therefore, D + S = 62 ...(3a)Now, from equation (1a): D = S - 2Substitute D = S - 2 into equation (3a):(S - 2) + S = 62So, 2S - 2 = 62 => 2S = 64 => S = 32Now that S = 32, we can find D from equation (1a):D = 32 - 2 = 30So, Da Mao weighs 30 kg, Er Mao 25 kg, and San Mao 32 kg.Let me check if these values satisfy all three original equations.First equation: D + E = 30 + 25 = 55. S + 23 = 32 + 23 = 55. Correct.Second equation: E + S = 25 + 32 = 57. D + 27 = 30 + 27 = 57. Correct.Third equation: S + D = 32 + 30 = 62. E + 37 = 25 + 37 = 62. Correct.All equations are satisfied. So the individual weights are:Da Mao: 30 kgEr Mao: 25 kgSan Mao: 32 kgTherefore, the total weight together is 30 + 25 + 32 = 87 kg.Wait, but let me make sure. The problem is asking for the total weight of all three together. So yes, D + E + S = 30 + 25 + 32 = 87. So 87 kilograms.But let me cross-verify this total with another approach. Sometimes, when solving systems of equations, adding all three equations can lead directly to the total.Original equations:1. D + E = S + 232. E + S = D + 273. S + D = E + 37If we add all three equations together:(D + E) + (E + S) + (S + D) = (S + 23) + (D + 27) + (E + 37)Left side: 2D + 2E + 2SRight side: D + E + S + 23 + 27 + 37 = D + E + S + 87So:2D + 2E + 2S = D + E + S + 87Subtract (D + E + S) from both sides:D + E + S = 87Which is the total weight. So, indeed, the total weight of all three together is 87 kg. That's the answer.So even without solving for each individual's weight, adding all three equations gives the total directly. That's a nice check. So regardless of the individual weights, the total is 87.Therefore, the answer is 87 kilograms.Final AnswerThe total weight of Da Mao, Er Mao, and San Mao together is boxed{87} kilograms.</think>Let ( D ) be Da Mao's weight, ( E ) be Er Mao's weight, and ( S ) be San Mao's weight. We are given the following equations based on the problem statements:1. ( D + E = S + 23 )2. ( E + S = D + 27 )3. ( S + D = E + 37 )We need to find ( D + E + S ).First, we rewrite the equations in standard form:1. ( D + E - S = 23 )2. ( -D + E + S = 27 )3. ( D - E + S = 37 )Adding equations (1) and (2) to eliminate ( D ):[(D + E - S) + (-D + E + S) = 23 + 27 2E = 50 E = 25]Substituting ( E = 25 ) into the equations:1. ( D + 25 - S = 23 implies D - S = -2 implies D = S - 2 )2. ( -D + 25 + S = 27 implies -D + S = 2 implies S = D + 2 )Using the third equation with ( E = 25 ):[D - 25 + S = 37 implies D + S = 62]Substituting ( D = S - 2 ) into ( D + S = 62 ):[(S - 2) + S = 62 2S - 2 = 62 2S = 64 S = 32]Then, ( D = 32 - 2 = 30 ).Verifying the weights:- ( D = 30 )- ( E = 25 )- ( S = 32 )Adding all three original equations:[(D + E) + (E + S) + (S + D) = (S + 23) + (D + 27) + (E + 37) 2D + 2E + 2S = D + E + S + 87 D + E + S = 87]Thus, the total weight of Da Mao, Er Mao, and San Mao together is (boxed{87}) kilograms.,<think>Okay, so I need to solve this problem about Da Mao, Er Mao, and San Mao's weights. Let me read it again carefully:"Da Mao and Er Mao together weigh 23 kilograms more than San Mao. Er Mao and San Mao together weigh 27 kilograms more than Da Mao. San Mao and Da Mao together weigh 37 kilograms more than Er Mao. Da Mao, Er Mao, and San Mao together weigh ______ kilograms."Hmm, okay. So we have three people: Da Mao (let's call him D), Er Mao (E), and San Mao (S). We need to find their combined weight, which is D + E + S.The problem gives three statements:1. D + E = S + 232. E + S = D + 273. S + D = E + 37And we need to find D + E + S.So, it's a system of three equations with three variables. Let me write them down:1. D + E - S = 23 (from the first statement)2. -D + E + S = 27 (from the second statement)3. D - E + S = 37 (from the third statement)Now, we need to solve this system. Let me label the equations for clarity:Equation (1): D + E - S = 23 Equation (2): -D + E + S = 27 Equation (3): D - E + S = 37 One way to solve this is to add all three equations together. Let's see what happens when we do that.Adding equation (1), (2), and (3):(D + E - S) + (-D + E + S) + (D - E + S) = 23 + 27 + 37Let me compute the left side step by step:First, combine D terms: D - D + D = D Then E terms: E + E - E = E Then S terms: -S + S + S = SSo, left side becomes D + E + S.Right side: 23 + 27 is 50, 50 + 37 is 87.Therefore, D + E + S = 87.Wait, is that it? So the total weight of all three is 87 kilograms? Hmm, that seems straightforward. Let me check if this is correct.But just to be thorough, maybe I should solve for each individual's weight as well to verify.Let's see. If D + E + S = 87, then perhaps we can find each variable.From equation (1): D + E = S + 23 But D + E = 87 - S (since total is 87). So:87 - S = S + 23 87 - 23 = S + S 64 = 2S S = 32So San Mao weighs 32 kg.Now from equation (2): E + S = D + 27 E + 32 = D + 27 So, E = D + 27 - 32 E = D - 5From equation (3): S + D = E + 37 32 + D = E + 37 But since E = D -5, substitute:32 + D = (D -5) + 37 32 + D = D -5 +37 32 + D = D +32 Subtract D from both sides: 32 =32 Hmm, that's an identity. So it doesn't give new information. That means that once we have E = D -5 and S =32, the third equation is automatically satisfied.So, we need another equation to find D and E. Wait, but maybe from the total.Total D + E + S =87 S=32, so D + E = 87 -32=55 But from equation (1): D + E = 55, which is equal to S +23=32 +23=55. So that checks out.But we also have E = D -5. So D + (D -5) =55 2D -5=55 2D=60 D=30Therefore, Da Mao is 30 kg. Then Er Mao is D -5=25 kg.So, checking:D=30, E=25, S=32.Check equation (1): 30 +25=55, which should be S +23=32 +23=55. Correct.Equation (2): E + S=25 +32=57, which should be D +27=30 +27=57. Correct.Equation (3): S + D=32 +30=62, which should be E +37=25 +37=62. Correct.Everything checks out. Therefore, the total weight is 87 kg. So the answer is 87. But let me just make sure I didn't make any mistakes in the algebra.Wait, when I added all three equations, I got D + E + S =87. Then solved for S first, using equation (1). Then found E in terms of D using equation (2), then substituted into equation (3) which gave an identity, meaning the system is consistent. Then found D from total minus S, then E. So seems correct.Alternatively, another way to check is to plug in D=30, E=25, S=32 into all original equations:1. D + E - S =30+25-32=23. Correct. 2. -D + E + S= -30 +25 +32=27. Correct. 3. D - E + S=30 -25 +32=37. Correct. All equations are satisfied. So the answer is 87. Therefore, the combined weight is 87 kilograms.Final AnswerThe combined weight of Da Mao, Er Mao, and San Mao is boxed{87} kilograms.</think>Given the problem of finding the combined weight of Da Mao (D), Er Mao (E), and San Mao (S), we start with the following equations based on the problem statements:1. ( D + E - S = 23 )2. ( -D + E + S = 27 )3. ( D - E + S = 37 )To find the combined weight ( D + E + S ), we add all three equations together:[(D + E - S) + (-D + E + S) + (D - E + S) = 23 + 27 + 37]Simplifying the left side:[(D - D + D) + (E + E - E) + (-S + S + S) = D + E + S]The right side simplifies to:[23 + 27 + 37 = 87]Thus, we have:[D + E + S = 87]To verify, we solve for each individual weight. From equation 1:[D + E = S + 23]Since ( D + E + S = 87 ), substituting ( D + E ) gives:[87 - S = S + 23 implies 64 = 2S implies S = 32]From equation 2:[E + S = D + 27 implies E + 32 = D + 27 implies E = D - 5]From the total weight:[D + E = 55 implies D + (D - 5) = 55 implies 2D - 5 = 55 implies 2D = 60 implies D = 30]Thus, ( E = 30 - 5 = 25 ).Verifying with the third equation:[S + D = 32 + 30 = 62 quad text{and} quad E + 37 = 25 + 37 = 62]All equations are satisfied. Therefore, the combined weight of Da Mao, Er Mao, and San Mao is (boxed{87}) kilograms.]

🔑:87

🧠:[<think>Okay, so I need to solve this geometry problem where a sphere is inscribed in a truncated cone (which is also called a frustum). The key information given is that the sum of the lengths of the diameters of the upper and lower bases of the cone is five times the radius of the sphere. I need to find the angle between the generatrix (which is the slant side) of the cone and the plane of the base.First, let me visualize the problem. A truncated cone is like a cone with the top cut off, so it has two circular bases of different radii. A sphere inscribed in this frustum would touch both the top and bottom bases, as well as the lateral surface. The generatrix is the slant height of the frustum, and the angle they're asking for is the angle between this slant height and the base of the cone. That angle would be related to the slope of the lateral surface.Let me denote the following variables:- Let R be the radius of the lower base of the frustum.- Let r be the radius of the upper base.- Let h be the vertical height of the frustum (the distance between the two bases).- Let l be the slant height (generatrix) of the frustum.- Let s be the radius of the inscribed sphere.The problem states that the sum of the diameters of the upper and lower bases is five times the radius of the sphere. Since the diameters are 2R and 2r, their sum is 2R + 2r. This is equal to 5s. So:2R + 2r = 5s ⇒ R + r = (5/2)s. (Equation 1)Now, since the sphere is inscribed in the frustum, it must be tangent to both the top and bottom bases, as well as the lateral surface. The sphere will touch the top and bottom bases, so the vertical distance between the bases (h) should be equal to the diameter of the sphere, right? Wait, but actually, in a frustum with an inscribed sphere, the vertical height h is equal to twice the radius of the sphere. Because the sphere has to fit snugly between the two bases. So h = 2s. Is that correct?Wait, maybe I need to verify that. Let me think. If the sphere is touching both the top and bottom, then the distance between the two bases would be equal to the diameter of the sphere. Because the sphere has radius s, so diameter 2s. So yes, h = 2s. That seems correct.But let me confirm. In a frustum, if you inscribe a sphere, the sphere must be tangent to both circular bases, so the vertical distance between the two bases must be equal to the diameter of the sphere. So h = 2s. Okay, so that gives me h = 2s. (Equation 2)Next, the sphere is also tangent to the lateral surface of the frustum. To find the relation between R, r, h, l, and s, I need to use the geometry of the frustum and the sphere.Let me recall that in a frustum, the slant height l can be found using the Pythagorean theorem:l = sqrt((R - r)^2 + h^2). (Equation 3)But we know h = 2s, so l = sqrt((R - r)^2 + (2s)^2).Alternatively, maybe there's another way to relate R, r, and s using the tangent condition of the sphere to the lateral surface.In a truncated cone with an inscribed sphere, there's a relationship between the radii of the bases, the slant height, and the radius of the sphere. Let me try to derive that.Imagine the original cone before truncation. The frustum is part of this cone. Let me denote H as the height of the original cone, and L as its slant height. The radius of the base of the original cone would be R, and when we cut off the top to form the frustum, we remove a smaller cone with height H - h and radius r.So the original cone has height H, base radius R, and slant height L = sqrt(R^2 + H^2). The smaller cone that's cut off has height H - h and radius r, so similar triangles give us:r / R = (H - h) / H ⇒ H - h = (r / R) H ⇒ h = H - (r / R) H = H (1 - r / R). (Equation 4)But maybe this is complicating things. Alternatively, in the frustum, the slant height l is related to R, r, and h by Equation 3. Since we have h = 2s, we can write l = sqrt((R - r)^2 + (2s)^2).Now, the sphere inscribed in the frustum touches the lateral side. To find the relation, perhaps we can consider the cross-section of the frustum and sphere through its axis. The cross-section would be a trapezoid with the two parallel sides of lengths 2R and 2r (the diameters of the bases) and the lateral sides (the slant heights) of length l. The inscribed sphere would appear as a circle with radius s, tangent to both the top, bottom, and the slant sides of the trapezoid.Alternatively, if we consider the cross-section of the original cone, it would be a triangle. The frustum is a trapezoid, and the inscribed circle (representing the sphere) would be tangent to both bases and the sides. The center of the circle would lie along the axis of the frustum.In this cross-sectional view, the center of the circle is at a distance s from both the top and bottom bases (since the radius is s). The circle is also tangent to the slant sides. Let's consider one of the slant sides.Let me set up coordinate axes. Let me place the origin at the center of the sphere (which is also the center along the axis of the frustum). The bottom base is at y = -s, and the top base is at y = s. The axis of the frustum is the y-axis. The slant side can be represented as a line in this coordinate system.The left slant side goes from the point (R, -s) to (r, s). Similarly, the right slant side goes from (-R, -s) to (-r, s). Let's focus on the right slant side. The equation of the line connecting (-R, -s) to (-r, s).Wait, actually, if the center of the sphere is at the origin, then the bottom base is at y = -s and the top base at y = s. The radius of the bottom base is R, so the bottom base circle is in the plane y = -s with radius R, centered at (0, -s). Similarly, the top base is in the plane y = s with radius r, centered at (0, s).So the cross-sectional trapezoid has vertices at (R, -s), (-R, -s), (-r, s), (r, s). The sides of the trapezoid are the lines connecting (R, -s) to (r, s) and (-R, -s) to (-r, s).The sphere in cross-section is a circle of radius s centered at the origin (0,0). The circle must be tangent to the slant sides. Let's find the condition for the circle x² + y² = s² to be tangent to the line connecting (R, -s) to (r, s).First, find the equation of the line connecting (R, -s) to (r, s). Let's compute the slope:Slope m = (s - (-s)) / (r - R) = (2s)/(r - R).So the equation of the line is:y - (-s) = m(x - R) ⇒ y + s = (2s)/(r - R)(x - R).Simplifying:y = (2s)/(r - R)(x - R) - s.We can write this as:y = (2s)/(r - R)x - (2sR)/(r - R) - s.Combine the constants:= (2s)/(r - R)x - s(2R/(r - R) + 1).= (2s)/(r - R)x - s( (2R + (r - R)) / (r - R) )= (2s)/(r - R)x - s( (R + r) / (r - R) )Therefore, the equation is:y = [2s/(r - R)]x - [s(R + r)/(r - R)]Now, the distance from the center of the circle (0,0) to this line must be equal to the radius of the circle, which is s.The formula for the distance from a point (x₀, y₀) to the line ax + by + c = 0 is |ax₀ + by₀ + c| / sqrt(a² + b²). So let's write the line equation in ax + by + c = 0 form.Starting from:y = [2s/(r - R)]x - [s(R + r)/(r - R)]Bring all terms to the left:[2s/(r - R)]x - y - [s(R + r)/(r - R)] = 0Multiply numerator and denominator to make it clearer. Let's denote the line as:(2s)x - (r - R)y - s(R + r) = 0Wait, actually, multiplying both sides by (r - R):2s x - (r - R)y - s(R + r) = 0So the coefficients are:a = 2sb = -(r - R)c = -s(R + r)Now, the distance from (0,0) to this line is:|a*0 + b*0 + c| / sqrt(a² + b²) = |c| / sqrt(a² + b²) = | -s(R + r) | / sqrt( (2s)^2 + ( -(r - R) )^2 )Simplify:= s(R + r) / sqrt(4s² + (R - r)^2 )This distance must be equal to the radius s:s(R + r) / sqrt(4s² + (R - r)^2 ) = sDivide both sides by s (assuming s ≠ 0, which it is a sphere):(R + r) / sqrt(4s² + (R - r)^2 ) = 1Multiply both sides by the denominator:(R + r) = sqrt(4s² + (R - r)^2 )Square both sides:(R + r)^2 = 4s² + (R - r)^2Expand both sides:Left side: R² + 2Rr + r²Right side: 4s² + R² - 2Rr + r²Set equal:R² + 2Rr + r² = 4s² + R² - 2Rr + r²Subtract R² + r² from both sides:2Rr = 4s² - 2RrBring terms together:2Rr + 2Rr = 4s² ⇒ 4Rr = 4s² ⇒ Rr = s². (Equation 5)So we have another equation: Rr = s².Earlier, from Equation 1: R + r = (5/2)sSo now, we have two equations:1) R + r = (5/2)s2) Rr = s²This is a system of equations in R and r. Let me solve for R and r in terms of s.Let me denote S = R + r = (5/2)s, and P = Rr = s².These are the sum and product of R and r. So, the quadratic equation x² - Sx + P = 0 would have roots R and r.So, x² - (5/2)s x + s² = 0Solving this quadratic:x = [ (5/2)s ± sqrt( (5/2 s)^2 - 4*1*s² ) ] / 2Compute discriminant:(25/4)s² - 4s² = (25/4 - 16/4)s² = (9/4)s²Therefore,x = [ (5/2)s ± (3/2)s ] / 2Compute both roots:First root:[ (5/2 + 3/2 )s ] / 2 = (8/2 s)/2 = (4s)/2 = 2sSecond root:[ (5/2 - 3/2 )s ] / 2 = (2/2 s)/2 = (s)/2Therefore, R and r are 2s and s/2. Since R is the radius of the lower base and r is the upper base, typically R > r in a frustum, so R = 2s and r = s/2.So, R = 2s, r = s/2.Now, with R and r in terms of s, let's recall that h = 2s (Equation 2). Then, the slant height l can be computed from Equation 3:l = sqrt( (R - r)^2 + h^2 ) = sqrt( (2s - s/2)^2 + (2s)^2 ) = sqrt( (3s/2)^2 + (2s)^2 ) = sqrt( 9s²/4 + 4s² ) = sqrt( 9s²/4 + 16s²/4 ) = sqrt(25s²/4 ) = (5s)/2.So, the slant height l is (5s)/2.Now, the problem asks for the angle between the generatrix (slant height) and the plane of the base. Let's denote this angle as θ. In the cross-sectional view, this angle is the angle between the slant side (generatrix) and the horizontal base.In the right triangle formed by the slant height, the difference in radii (R - r), and the height h, the angle θ is adjacent to the height h and opposite to the radius difference (R - r). Alternatively, in the cross-section, the generatrix makes an angle θ with the horizontal base. So, the angle θ can be found using trigonometric ratios.In the right triangle with sides h (height = 2s), horizontal leg (R - r) = 3s/2, and hypotenuse l = 5s/2. The angle θ is between the hypotenuse (l) and the horizontal leg (R - r). Wait, no. Let's clarify.Wait, in the cross-sectional trapezoid, the generatrix is the slant side. The angle between the generatrix and the base is the angle between the slant side and the horizontal. So, in the right triangle formed when you extend the frustum to form the full cone, the slant height makes an angle with the base.But perhaps in the frustum itself, the angle θ is the angle between the generatrix and the vertical axis. Wait, but the problem says "the angle between the generatrix of the cone and the plane of the base." The plane of the base is horizontal, so the angle between the generatrix (which is a line) and the plane is the complement of the angle between the generatrix and the normal to the plane (which is the vertical axis). Wait, perhaps I need to recall the definition.The angle between a line and a plane is the complement of the angle between the line and the normal to the plane. So, if φ is the angle between the generatrix and the normal (vertical axis), then the angle θ between the generatrix and the plane is 90° - φ.But in any case, in the right triangle, the angle between the generatrix (hypotenuse) and the horizontal base is θ. To find θ, we can use tanθ = opposite/adjacent = h / (R - r).Wait, in the right triangle formed by the height h, the horizontal difference (R - r), and the generatrix l. So, tan(theta) = h / (R - r) ?Wait, no. If θ is the angle between the generatrix and the base (i.e., the horizontal), then in the right triangle, θ is adjacent to the horizontal leg (R - r) and opposite to the vertical leg h. Therefore, tan(theta) = h / (R - r). So θ = arctan( h / (R - r) ).Alternatively, maybe I got the angle wrong. Let me draw a mental picture. If you have a right triangle with vertical side h and horizontal side (R - r), then the angle between the hypotenuse (generatrix) and the horizontal side is θ. So tan(theta) = h / (R - r). Therefore, theta = arctan( h / (R - r) ).Given that h = 2s, and R - r = 2s - s/2 = 3s/2.So tan(theta) = (2s) / (3s/2) = (2s) * (2)/(3s) ) = 4/3.Therefore, theta = arctan(4/3). So the angle between the generatrix and the plane of the base is arctan(4/3). The problem might want the answer in degrees or as a trigonometric function. Since the problem says "angle between the generatrix...", likely expects an exact value, perhaps in terms of arctan, or maybe a specific angle whose tangent is 4/3.Alternatively, maybe we can express this angle in another way. Let's check.Wait, 4/3 is the tangent of the angle. To get the angle itself, we can compute arctan(4/3). However, 4/3 is a common ratio. For example, in a 3-4-5 triangle, which is what we have here. Wait, in our case, the sides adjacent and opposite are 3s/2 and 2s, which when scaled by 2/s gives 3 and 4. So the triangle is similar to a 3-4-5 triangle. Therefore, the angle theta has tan(theta) = 4/3, so theta is arctan(4/3). This is a well-known angle, approximately 53.13 degrees, but since the problem doesn't specify, we can leave it in terms of arctan(4/3). However, perhaps we need to express it as arcsin or arccos for some reason.Alternatively, since the problem might expect the angle between the generatrix and the base, which is theta, but maybe it's more standard to give the angle between the generatrix and the axis, which would be complementary. Wait, let me check the problem statement again."Find the angle between the generatrix of the cone and the plane of the base."The angle between a line and a plane is defined as the angle between the line and its orthogonal projection onto the plane. Alternatively, it's the complement of the angle between the line and the normal to the plane. So if phi is the angle between the generatrix and the normal (the vertical axis), then the angle theta between the generatrix and the plane is 90° - phi.But in our case, if we already have theta as arctan(4/3), which is the angle between the generatrix and the horizontal, then is that the angle with the plane? Wait, yes. Because the plane is horizontal, and the angle between the line and the plane is measured as the smallest angle between the line and any line in the plane. That angle is equal to theta as we calculated, arctan(4/3). Alternatively, if we consider the angle between the generatrix and the normal to the plane (the vertical axis), then that angle phi is 90° - theta, so sin(phi) = 4/5, cos(phi) = 3/5, because in the 3-4-5 triangle.But according to standard definitions, the angle between a line and a plane is the angle between the line and its projection onto the plane, which is the same as the complement of the angle between the line and the normal to the plane. So, if phi is the angle between the generatrix and the normal (vertical axis), then theta = 90° - phi is the angle between the generatrix and the plane. Therefore, if we have tan(theta) = 4/3, then theta = arctan(4/3). Alternatively, if we have phi = arcsin(4/5), then theta = 90° - arcsin(4/5) = arccos(4/5). Wait, that may be another way to express it.Wait, let's see:In the right triangle with legs h = 2s and (R - r) = 3s/2, hypotenuse l = 5s/2.If phi is the angle between the generatrix and the vertical axis (normal to the base), then:sin(phi) = (R - r)/l = (3s/2)/(5s/2) = 3/5cos(phi) = h/l = (2s)/(5s/2) = 4/5Therefore, phi = arcsin(3/5) or arccos(4/5). Then, the angle theta between the generatrix and the plane is 90° - phi.But sin(phi) = 3/5 ⇒ phi = arcsin(3/5), so theta = 90° - arcsin(3/5) = arccos(3/5), since sin(phi) = 3/5 ⇒ cos(theta) = 3/5.Alternatively, since theta = arctan(4/3), and arctan(4/3) is equal to arcsin(4/5) or arccos(3/5). Because in a right triangle with opposite side 4, adjacent 3, hypotenuse 5. So, yes, theta = arcsin(4/5) = arccos(3/5) = arctan(4/3). All equivalent.Therefore, the angle theta can be expressed as arctan(4/3), arcsin(4/5), or arccos(3/5). The problem may expect one of these forms. Since it's an Olympiad problem, the answer might be expected in terms of arctan(4/3), but sometimes they prefer arcsin or arccos if it's a standard ratio. Since 3-4-5 is a Pythagorean triplet, arcsin(4/5) or arccos(3/5) might be acceptable. Let me check the problem statement again. It says, "Find the angle...". It doesn't specify the form, so likely either is acceptable, but perhaps in the boxed answer, they expect an expression like arctan(4/3), or maybe a numerical value. But since it's exact, we should present it in terms of inverse trigonometric functions.But let me see if there's another way. Wait, in the problem statement, it's a cone, so maybe there's another approach where the angle is related to the original cone's apex angle, but since it's a frustum, we derived everything properly.Alternatively, check the answer. If we have theta = arctan(4/3), that's approximately 53.13 degrees. Alternatively, if they want the answer in radians, but the problem doesn't specify. However, in Olympiad problems, unless stated, angles are usually expected in degrees with inverse trigonometric functions or exact expressions.But let me confirm once again:We have R = 2s, r = s/2, h = 2s. The slant height l = 5s/2.In the right triangle, the legs are h = 2s and (R - r) = 3s/2. The angle theta between the slant height and the base (horizontal) is given by tan(theta) = h / (R - r) = (2s)/(3s/2) = 4/3. So theta = arctan(4/3). Therefore, the answer is arctan(4/3). Alternatively, as mentioned, since the triangle is 3-4-5, theta is arcsin(4/5) or arccos(3/5). However, arctan(4/3) is the direct answer from the right triangle.But in some textbooks, the angle between the generatrix and the base is considered as the angle whose sine is opposite over hypotenuse, which would be (R - r)/l = 3/5. Wait, no. If we consider theta as the angle between the generatrix and the base, then the opposite side is h, and the adjacent side is (R - r). Therefore, tan(theta) = h / (R - r) = 4/3. So theta is arctan(4/3). Therefore, I think arctan(4/3) is the correct answer here.However, let me check with another approach. Since the sphere is inscribed, there might be a formula related to the radius of the sphere in a frustum. Let me recall that in a frustum with radii R and r, height h, the radius s of an inscribed sphere is given by s = (h R r) / (sqrt(R r) (R + r))). Wait, but I might be misremembering.Alternatively, I can check the formula for the radius of a sphere inscribed in a frustum. The formula is s = (h R r) / (sqrt(R r) (R + r)) ). Wait, not sure. Alternatively, the formula might come from similar triangles.Wait, in the cross-section, the center of the sphere is at the midpoint between the two bases, since the vertical distance is h = 2s, so the center is at h/2 = s from each base. Then, the sphere is tangent to the lateral side. The point of tangency can be found by the distance from the center to the lateral side, which is equal to the radius s. We already used this to derive Rr = s² and R + r = (5/2)s. So, the previous steps are correct.Therefore, the angle theta is arctan(4/3). So the answer is arctan(4/3), which can be written as boxed{arctan frac{4}{3}}.Alternatively, if they want it in degrees, but I think in mathematical problems, especially when it's a ratio of integers, we leave it as arctan(4/3). So that's the answer.Final AnswerThe angle between the generatrix of the cone and the plane of the base is boxed{arctan frac{4}{3}}.</think>Given a sphere inscribed in a truncated cone (frustum), we need to find the angle between the generatrix (slant height) of the cone and the plane of the base. The sum of the diameters of the upper and lower bases is five times the radius of the sphere.1. Define Variables: - ( R ): Radius of the lower base. - ( r ): Radius of the upper base. - ( h ): Vertical height of the frustum. - ( l ): Slant height (generatrix) of the frustum. - ( s ): Radius of the inscribed sphere.2. Given Conditions: - The sum of the diameters of the bases is five times the radius of the sphere: [ 2R + 2r = 5s implies R + r = frac{5}{2}s ] - The vertical height ( h ) of the frustum is twice the radius of the sphere: [ h = 2s ]3. Slant Height Calculation: - Using the Pythagorean theorem for the slant height ( l ): [ l = sqrt{(R - r)^2 + h^2} ]4. Tangent Condition: - The sphere is tangent to the lateral surface, leading to the relationship: [ Rr = s^2 ]5. Solving for ( R ) and ( r ): - Using the system of equations: [ R + r = frac{5}{2}s quad text{and} quad Rr = s^2 ] - Solving the quadratic equation ( x^2 - frac{5}{2}sx + s^2 = 0 ), we find: [ R = 2s quad text{and} quad r = frac{s}{2} ]6. Calculating Slant Height ( l ): - Substituting ( R = 2s ), ( r = frac{s}{2} ), and ( h = 2s ): [ l = sqrt{left(2s - frac{s}{2}right)^2 + (2s)^2} = sqrt{left(frac{3s}{2}right)^2 + (2s)^2} = sqrt{frac{25s^2}{4}} = frac{5s}{2} ]7. Finding the Angle: - The angle ( theta ) between the generatrix and the base is found using the tangent ratio in the right triangle formed by ( h ), ( R - r ), and ( l ): [ tan(theta) = frac{h}{R - r} = frac{2s}{frac{3s}{2}} = frac{4}{3} ] - Therefore, the angle is: [ theta = arctanleft(frac{4}{3}right) ]The angle between the generatrix of the cone and the plane of the base is boxed{arctan frac{4}{3}}.,<think>Okay, so I have this geometry problem here. A sphere is inscribed in a truncated cone, which I think is also called a frustum. The problem says that the sum of the lengths of the diameters of the upper and lower bases of the cone is five times the length of the radius of the sphere. I need to find the angle between the generatrix of the cone and the plane of the base. Alright, let's break this down step by step.First, let me visualize the situation. A truncated cone is a cone with the top cut off by a plane parallel to the base. So, the frustum has two circular bases, one larger (the original base) and one smaller (the cut-off part). The sphere inscribed in this frustum would touch both the top and bottom bases, as well as the lateral surface of the frustum. That gives me a starting point.The problem states that the sum of the diameters of the upper and lower bases is five times the radius of the sphere. Let me denote some variables here. Let’s say:- Let R be the radius of the lower base (the larger one).- Let r be the radius of the upper base (the smaller one).- Let’s denote the radius of the sphere as a.- The problem says 2R + 2r = 5a. Wait, sum of diameters: diameter of lower base is 2R, diameter of upper base is 2r. So, 2R + 2r = 5a. That simplifies to R + r = (5/2)a. Okay, so that's one equation.Now, I need to relate this to the angle between the generatrix (which is the slant side of the frustum) and the plane of the base. Let me recall that the angle θ we're supposed to find is the angle between the generatrix and the base. So, in a right triangle formed by the height of the frustum, the difference in radii (R - r), and the generatrix (slant height), the angle θ would be between the generatrix and the base, which is adjacent to the height and opposite to the (R - r) side.Wait, but the sphere is inscribed in the frustum. So, there must be a relationship between the sphere's radius and the dimensions of the frustum.Let me recall that in a frustum with an inscribed sphere (which makes it a "truncated cone" with a sphere tangent to both bases and the lateral surface), there are certain conditions that must be satisfied. Specifically, the sphere must fit perfectly within the frustum, touching all surfaces. So, the sphere's diameter would be related to the height of the frustum, but also the radii R and r.But how exactly? Let me think. If the sphere is tangent to both the top and bottom bases, then the distance between the two bases (the height h of the frustum) must be equal to the diameter of the sphere, which is 2a. Wait, is that correct? Hmm. If the sphere is touching both the top and bottom, then the height h would indeed be equal to twice the radius of the sphere, so h = 2a. Is that right?Wait, hold on. If the sphere is sitting inside the frustum, touching both the top and bottom, then yes, the vertical distance between the two bases must be equal to the diameter of the sphere. Because the sphere's center would be halfway between the two bases, so from the bottom base to the center is a, and from the center to the top base is another a, so total height h = 2a. Okay, that seems reasonable. So, h = 2a.But there's also the lateral tangent. The sphere must also be tangent to the lateral side of the frustum. So, the sphere touches the slant side. This is going to relate the radius of the sphere to the slant height and the radii R and r.Let me recall that in such problems, similar triangles come into play. Because the frustum is part of a larger cone. Let me consider the original cone before truncation. Let’s denote H as the height of the original cone, L as its slant height, and R as its base radius. Then, when we cut off the top to form the frustum, we remove a smaller cone similar to the original one. The smaller cone has height H - h and radius r.Since the cones are similar, the ratios are equal. So, (H - h)/H = r/R. Therefore, H - h = (r/R) * H, which gives H = (R/(R - r)) * h. But in our case, h is the height of the frustum, which we already determined as h = 2a. So, H = (R/(R - r)) * 2a.But I need to relate this to the sphere inscribed in the frustum. Let me think. The sphere touches the lateral side, so there might be a right triangle involving the sphere's radius, the slant height, and the difference in radii.Alternatively, maybe I can use the formula for the radius of an inscribed sphere in a frustum. I don't remember the exact formula, but maybe I can derive it.Let’s denote the slant height of the frustum as l. Then, l can be calculated using the Pythagorean theorem: l = sqrt((R - r)^2 + h^2). But we already have h = 2a, so l = sqrt((R - r)^2 + (2a)^2).Now, the sphere inscribed in the frustum will have its center lying along the axis of the frustum. The radius of the sphere is a, so the center is at a distance a from both the top and bottom bases. The sphere is also tangent to the lateral face. So, the distance from the center of the sphere to the lateral face must be equal to a.To find this distance, perhaps we can use the formula for the distance from a point to a plane. Let me consider the lateral face as part of the original cone. The generatrix (slant height) of the frustum is part of the lateral surface of the cone.Alternatively, maybe we can use similar triangles here. Let me imagine the cross-section of the frustum and sphere through its axis. The cross-section would be a trapezoid with the inscribed circle (the sphere in cross-section). The trapezoid has the two parallel sides of lengths 2R and 2r, and the height h = 2a. The non-parallel sides are the slant heights l.In this cross-section, the inscribed circle (radius a) is tangent to both the top, bottom, and the slant sides. So, maybe in this trapezoid, the radius of the inscribed circle is a. Let me recall that in a trapezoid, the radius of an inscribed circle (if it exists) is equal to half the height, but here the height is 2a, so that would make the radius a, which matches. Wait, but not all trapezoids can have an inscribed circle. A trapezoid can have an inscribed circle only if the sum of the lengths of the two bases equals the sum of the lengths of the two legs. So, in this case, the trapezoid is tangential, meaning it has an inscribed circle.So, the condition for a trapezoid to be tangential is that the sum of the lengths of the two bases equals the sum of the lengths of the two legs. So, in our case, the two bases are 2R and 2r, and the two legs are both l (since it's a trapezoid formed by truncating a cone, the two non-parallel sides are equal in length). Therefore, the condition is 2R + 2r = 2l, so R + r = l. Wait, but according to the problem statement, R + r = (5/2)a. So, l = (5/2)a.But we also have l = sqrt((R - r)^2 + (2a)^2). So, substituting l = (5/2)a into that equation:(5/2 a)^2 = (R - r)^2 + (2a)^2Let me compute the left side: (25/4)a²Right side: (R - r)^2 + 4a²Therefore:25/4 a² = (R - r)^2 + 4a²Subtract 4a² from both sides:25/4 a² - 16/4 a² = (R - r)^2Which simplifies to:9/4 a² = (R - r)^2Take square roots:(R - r) = (3/2)aSo, we have R - r = (3/2)aEarlier, we had R + r = (5/2)aSo now we have two equations:1. R + r = (5/2)a2. R - r = (3/2)aWe can solve these equations for R and r.Adding the two equations:(R + r) + (R - r) = (5/2)a + (3/2)a2R = 4aTherefore, R = 2aSubtracting the two equations:(R + r) - (R - r) = (5/2)a - (3/2)a2r = aTherefore, r = (1/2)aSo, R = 2a, r = a/2Now, the angle θ between the generatrix (slant height) and the base. In the cross-section, the generatrix is the leg of the trapezoid, which we found to be l = (5/2)a. The height of the trapezoid is h = 2a, and the difference in radii is R - r = (3/2)a.Wait, but in the right triangle formed by the slant height l, the height h, and the horizontal difference (R - r), the angle between the generatrix and the base is adjacent to the height h and opposite to the (R - r) side.Therefore, in that right triangle, tan(theta) = (R - r)/h = [(3/2)a]/[2a] = (3/2)/2 = 3/4Therefore, theta = arctan(3/4)Wait, but the angle between the generatrix and the base is theta, which is the angle whose tangent is (R - r)/h. So yes, that gives theta = arctan(3/4). Let me confirm.In the right triangle, the height is h = 2a, the horizontal leg is R - r = (3/2)a. So, the angle at the base (between the generatrix and the base) would have opposite side (R - r) and adjacent side h. So, tan(theta) = (R - r)/h = (3/2 a)/(2a) = 3/4. Therefore, theta = arctan(3/4). To get this angle, we can compute arctan(3/4). But the problem asks for the angle, so unless they want it in degrees, which they might. But since the problem doesn't specify, maybe we can leave it in terms of arctan(3/4). But let me check the problem again.Wait, the problem is given in a Russian textbook problem (12.238), and usually, such problems expect an exact answer, perhaps in terms of a known angle or a fraction with π. But arctan(3/4) is not a standard angle. Wait, maybe there's another approach. Wait, perhaps the angle can be expressed as arcsin(3/5) or arccos(4/5). Let me check.If tan(theta) = 3/4, then in a right triangle with opposite side 3 and adjacent side 4, the hypotenuse is 5. Therefore, sin(theta) = 3/5 and cos(theta) = 4/5. Therefore, theta can be expressed as arcsin(3/5) or arccos(4/5). Either of these would be acceptable. But let me check if this matches the problem's expectation.Alternatively, maybe they want the answer in degrees, but since it's a competition-style problem, likely expecting an exact expression. Since 3-4-5 is a Pythagorean triple, we can express the angle as arcsin(3/5) or arccos(4/5). Either is acceptable, but perhaps the answer expects it in one form or the other.Looking back at the problem statement: "Find the angle between the generatrix of the cone and the plane of the base." The angle between a line and a plane is defined as the angle between the line and its orthogonal projection onto the plane. Wait, but in this case, the generatrix is already on the plane's surface? Wait, no. The generatrix is the slant side of the frustum. The plane of the base is the base of the frustum.Wait, the angle between a line and a plane is the complement of the angle between the line and the normal to the plane. Wait, actually, the angle between a line and a plane is defined as the angle between the line and its projection onto the plane, which is the same as the complement of the angle between the line and the normal to the plane.But in our case, the generatrix is inclined at an angle theta from the vertical axis (the axis of the cone). Wait, maybe I need to clarify this.Wait, in the cross-sectional view, the axis of the cone is vertical, and the generatrix makes an angle theta with the horizontal base. Wait, no. Wait, in the right triangle we considered earlier, the angle theta is between the generatrix (hypotenuse) and the vertical height. Wait, no, actually, in the right triangle with vertical leg h = 2a and horizontal leg (R - r) = 3/2 a, the angle between the generatrix and the vertical axis is arctan( (R - r)/h ) = arctan(3/4), but the angle between the generatrix and the base would be the angle between the generatrix and the horizontal plane.Wait, the angle between a line and a plane is the smallest angle between the line and any line in the plane. In this case, the smallest angle would be between the generatrix and its projection onto the base. Since the generatrix is already inclined relative to the base, its projection onto the base is along the horizontal component (R - r). Therefore, the angle between the generatrix and the base is the same as theta, which is arctan( (R - r)/h ) = arctan(3/4). But actually, that's the angle between the generatrix and the vertical axis. Wait, confusion arises here.Wait, let's clarify. The generatrix is the slant side. The axis of the cone is vertical. The angle between the generatrix and the axis is theta, such that tan(theta) = (R - r)/h = 3/4. Therefore, theta is arctan(3/4). But the angle between the generatrix and the base would be the complement of this angle. Because the angle with the axis is theta, so the angle with the base is 90° - theta. Alternatively, since the generatrix and the base form a triangle where the angle at the base is between the generatrix and the horizontal.Wait, perhaps we need to think in terms of the angle between the generatrix and the base. Let me consider the cross-section again. The generatrix is the hypotenuse of a right triangle with vertical side h = 2a and horizontal side (R - r) = 3/2 a. The angle between the generatrix and the horizontal base is phi, such that tan(phi) = h / (R - r) = (2a) / (3/2 a) = (2)/(3/2) = 4/3. Therefore, phi = arctan(4/3). But is this the angle between the generatrix and the base?Wait, actually, the angle between a line and a plane is defined as the angle between the line and its orthogonal projection onto the plane. In this case, the projection of the generatrix onto the base would be a horizontal line from the base of the generatrix to the point directly below the upper end. That horizontal distance is (R - r) = 3/2 a. The length of the generatrix is l = 5/2 a. So, the projection of the generatrix onto the base is (R - r) = 3/2 a, and the angle between the generatrix and its projection is phi, such that cos(phi) = (projection length)/(generatrix length) = (3/2 a)/(5/2 a) = 3/5. Therefore, phi = arccos(3/5). Alternatively, sin(phi) = h/l = (2a)/(5/2 a) = 4/5, so phi = arcsin(4/5). Wait, but arcsin(4/5) is the same as arccos(3/5), because sin^2(phi) + cos^2(phi) = 1. So, either way, the angle between the generatrix and the base is arcsin(4/5) or arccos(3/5). But which is it?Wait, the angle between the generatrix and the plane of the base is the angle phi, which is the angle between the generatrix and its projection onto the base. The projection is the horizontal component (R - r), so the angle phi is given by:cos(phi) = adjacent/hypotenuse = (projection length)/l = (3/2 a)/(5/2 a) = 3/5Therefore, phi = arccos(3/5). Alternatively, sin(phi) = opposite/hypotenuse = h/l = (2a)/(5/2 a) = 4/5, so phi = arcsin(4/5). Both are correct, but arccos(3/5) is more direct here since we have the adjacent side.But the problem states "the angle between the generatrix of the cone and the plane of the base". The angle between a line and a plane is typically defined as the complement of the angle between the line and the normal to the plane. Since the normal to the base is vertical (aligned with the axis), the angle between the generatrix and the normal is theta = arctan(3/4), as we had before. Therefore, the angle between the generatrix and the base would be 90° - theta. So, 90° - arctan(3/4). But let's compute this.If theta is arctan(3/4), then 90° - theta is equal to arctan(4/3), since tan(90° - theta) = cot(theta) = 4/3. Wait, tan(arctan(3/4)) = 3/4, so cot(theta) = 4/3, so tan(90° - theta) = 4/3. Therefore, 90° - theta = arctan(4/3). But arctan(4/3) is equal to arcsin(4/5) or arccos(3/5). Therefore, the angle between the generatrix and the base is arctan(4/3), which is equivalent to arcsin(4/5) or arccos(3/5). However, typically, the angle between a line and a plane is measured as the smallest angle between the line and any line in the plane, which would be the angle phi = arcsin(4/5), since it's the angle between the generatrix and its projection onto the plane.Wait, but definitions can vary. Let me check the precise definition.According to standard definitions, the angle between a line and a plane is the angle between the line and its orthogonal projection onto the plane, which is the smallest angle between the line and any line in the plane. So, in this case, the angle phi between the generatrix and the base is equal to arcsin(sin(theta')), where theta' is the angle between the generatrix and the normal to the plane.Wait, perhaps it's better to compute it using the formula:sin(phi) = |n ⋅ v| / (|n||v|)But here, n is the normal vector to the plane, and v is the direction vector of the generatrix. Let's model this.Assuming the base is the xy-plane, and the generatrix goes from (0,0,0) to ( (R - r), 0, h ). Wait, in our case, in cross-section, the generatrix goes from (0,0) to ( (R - r), h ) in a 2D coordinate system. The direction vector of the generatrix is ( (R - r), h ). The normal vector to the base (which is the x-axis in 2D, but in 3D, the base is the xy-plane, so normal vector is along the z-axis). Wait, perhaps in 3D, the generatrix has a direction vector with components (R - r, 0, h) if we consider movement in x and z directions (assuming circular symmetry around the z-axis). The normal vector to the base (xy-plane) is along the z-axis, (0,0,1). The angle between the generatrix and the normal vector is given by:cos(theta') = (v ⋅ n)/( |v| |n| ) = h / |v| = h / lTherefore, theta' = arccos(h/l) = arccos(2a / (5/2 a)) = arccos(4/5). Therefore, the angle between the generatrix and the normal is arccos(4/5). Then, the angle between the generatrix and the plane is 90° - theta' = arcsin(4/5). So, since sin(theta') = sqrt(1 - (4/5)^2 ) = 3/5, but wait, no. Wait, theta' is arccos(4/5), so sin(theta') = 3/5. Therefore, the angle between the generatrix and the plane is defined as phi = 90° - theta' = arcsin(3/5). Wait, this is getting confusing.Wait, according to the formula, the angle between a line and a plane is equal to the complement of the angle between the line and the normal to the plane. Therefore:phi = 90° - theta'Where theta' is the angle between the line and the normal. So, if theta' = arccos(4/5), then phi = 90° - arccos(4/5) = arcsin(4/5). Alternatively, since sin(phi) = 4/5, so phi = arcsin(4/5). But we can check this with the direction vector.Alternatively, using the direction vector (R - r, h) which is (3/2 a, 2a). The projection onto the base would be (3/2 a, 0). The angle between the generatrix and its projection is phi. Then, cos(phi) = |projection| / |generatrix| = (3/2 a) / (5/2 a) = 3/5. Therefore, phi = arccos(3/5). Therefore, the angle between the generatrix and the plane of the base is arccos(3/5). Wait, but now we have conflicting results. Which one is correct?This is the problem with different definitions. Let me refer to a standard source. According to the definition, the angle between a line and a plane is the angle between the line and its orthogonal projection onto the plane, which is the smallest angle between the line and any line in the plane. In this case, the projection of the generatrix onto the base is a horizontal line of length (R - r) = 3/2 a, and the generatrix itself is of length l = 5/2 a. Therefore, the angle phi between them satisfies cos(phi) = (3/2 a)/(5/2 a) = 3/5, so phi = arccos(3/5). Therefore, the angle is arccos(3/5).But another way to compute it is using the angle between the generatrix and the normal. The angle between the generatrix and the normal is theta' = arccos(4/5), so the angle between the generatrix and the plane is 90° - theta', which is arcsin(4/5). However, arcsin(4/5) is equal to arccos(3/5), since sin(phi) = 4/5 implies cos(phi) = 3/5. Therefore, phi = arcsin(4/5) = arccos(3/5). So both approaches give the same result, just expressed differently. Therefore, the angle is arccos(3/5) or arcsin(4/5). Either is acceptable, but perhaps the answer is expected in one form.Looking back at the problem, in the original problem statement, they mention "the angle between the generatrix of the cone and the plane of the base". In standard geometric terms, this angle is measured as the smallest angle between the generatrix and the plane, which is arcsin(4/5). However, depending on the convention, sometimes it's expressed as arccos(3/5). But to verify, let's compute both.Given that:arccos(3/5) ≈ 53.13 degreesarcsin(4/5) ≈ 53.13 degreesWait, actually, arcsin(4/5) and arccos(3/5) are the same angle, since sin(phi) = 4/5 and cos(phi) = 3/5. So, phi ≈ 53.13 degrees, which is the angle between the generatrix and the plane. Therefore, the answer is arccos(3/5) or arcsin(4/5). However, since the problem is likely expecting an exact answer, perhaps in terms of arctangent. Let me check.Wait, earlier, we found that tan(theta) = (R - r)/h = 3/4, where theta is the angle between the generatrix and the vertical axis. Therefore, theta = arctan(3/4). Then, the angle between the generatrix and the base is 90° - theta = arctan(4/3). Wait, arctan(4/3) is approximately 53.13 degrees, which is the same as arcsin(4/5). So, arctan(4/3) is equivalent to arcsin(4/5). Let me confirm:If tan(phi) = 4/3, then sin(phi) = 4/5 and cos(phi) = 3/5. Therefore, phi = arctan(4/3) = arcsin(4/5) = arccos(3/5). All equivalent expressions.Therefore, the answer can be expressed as arctan(4/3), arcsin(4/5), or arccos(3/5). However, in the context of the problem, which expression is most appropriate?The problem asks for the angle between the generatrix and the plane of the base. In the right triangle we considered, with legs (R - r) = 3/2 a and h = 2a, the generatrix is the hypotenuse. The angle between the generatrix and the base can be found by looking at the angle adjacent to the (R - r) side and opposite to the h side. Wait, no. Wait, if we consider the triangle, the angle at the lower base between the generatrix and the horizontal (the base) is phi. The opposite side to phi is the vertical side h, and the adjacent side is (R - r). Therefore, tan(phi) = h / (R - r) = (2a) / (3/2 a) = 4/3. Therefore, phi = arctan(4/3). Therefore, the angle is arctan(4/3). But since tan(phi) = 4/3, this is equivalent to the previous expressions. So, depending on the method, we can write it in different forms.But in the problem statement, they might expect the answer as arctan(4/3). Alternatively, since we found that the generatrix length l = 5/2 a, and the height h = 2a, then sin(phi) = h / l = (2a)/(5/2 a) = 4/5, so phi = arcsin(4/5). So both are valid. However, in terms of the angle between the generatrix and the plane, which is defined as the complement of the angle between the generatrix and the normal, which is arcsin(4/5). But given that the problem is from a Russian textbook, which often expresses angles in terms of arctangent, maybe they expect arctan(4/3).Wait, let me check my previous steps again. When I considered the right triangle with legs (R - r) and h, the generatrix is the hypotenuse. The angle between the generatrix and the base is phi, which is the angle at the base of the triangle. Therefore, tan(phi) = h / (R - r) = 2a / (3/2 a) = 4/3. Therefore, phi = arctan(4/3). Therefore, the angle is arctan(4/3). But also, using the definition of the angle between a line and a plane, it's arcsin(4/5). Wait, but these are the same angle, just expressed differently. For example, arctan(4/3) is equal to arcsin(4/5), because in a 3-4-5 triangle, opposite side 4, adjacent 3, hypotenuse 5, so sin(phi) = 4/5, cos(phi) = 3/5, tan(phi) = 4/3. Therefore, all these expressions are equivalent.Therefore, the answer can be written as arctan(4/3), arcsin(4/5), or arccos(3/5). Since the problem asks for the angle, and it's common to present it in terms of arctangent when the sides are known, perhaps arctan(4/3) is the expected answer. Alternatively, if the answer is expected in a specific form, but given the problem comes from a Russian source, they might prefer arcsin or arccos. However, without more context, any of these forms is correct.But let me cross-verify with another approach. Let's consider the original cone before truncation. The sphere inscribed in the frustum would also be related to the original cone. Let me recall that in a cone, the radius of an inscribed sphere can be found using certain formulas. However, since this is a frustum, perhaps we can consider it as a portion of the cone.But maybe another way is to use the fact that the sphere touches the lateral side. In the cross-section, the sphere touches the slant side. The center of the sphere is at a distance a from both the top and bottom bases, so it's located at the midpoint of the height h = 2a. Let me denote the center as point O, located at ( (R - r)/2, a ) in the cross-sectional coordinate system, where the bottom base is at (0,0) and the top base is at (R - r, 2a). Wait, actually, in the cross-section, the trapezoid has the bottom base from (-R, 0) to (R, 0), and the top base from (-r, h) to (r, h). But since the sphere is inscribed symmetrically, the center is at (0, a). The sphere equation in cross-section is x² + (y - a)² = a².The slant side of the trapezoid is a line connecting (R, 0) to (r, h). Let me find the equation of this line. The left slant side connects (-R, 0) to (-r, h), but due to symmetry, we can just consider the right slant side from (R, 0) to (r, h).The slope of this line is (h - 0)/(r - R) = (2a)/( (a/2) - 2a ) = (2a)/( -3a/2 ) = -4/3. So, the equation of the line is y = (-4/3)(x - R). Let me verify this. At x = R, y = 0, which is correct. At x = r = a/2, y = (-4/3)( (a/2) - 2a ) = (-4/3)( -3a/2 ) = 2a, which is correct. So, the equation of the right slant side is y = (-4/3)(x - R). But R = 2a, so substituting, y = (-4/3)(x - 2a).The distance from the center of the sphere (0, a) to this line must be equal to the radius of the sphere, which is a. Let's compute the distance from point (0, a) to the line y = (-4/3)(x - 2a).First, let's rewrite the line equation in standard form: 4x + 3y - 8a = 0.Wait, starting from y = (-4/3)(x - 2a). Multiply both sides by 3: 3y = -4x + 8a. Then, 4x + 3y - 8a = 0. So, the standard form is 4x + 3y - 8a = 0.The distance from (0, a) to this line is |4*0 + 3*a - 8a| / sqrt(4² + 3²) = | -5a | / 5 = 5a / 5 = a. Which matches the radius, as expected. So, this confirms that our previous calculations are correct.Now, relating back to the angle. The slope of the slant side is -4/3, so the angle phi between the slant side and the horizontal axis is arctan(|slope|) = arctan(4/3). However, since the slope is negative, the angle with respect to the positive x-axis is 180° - arctan(4/3), but since we're talking about the angle between the generatrix and the base (which is the horizontal axis), it's the acute angle, which is arctan(4/3). Therefore, the angle between the generatrix and the plane of the base is arctan(4/3).Thus, after verifying through multiple approaches—using the properties of the trapezoid, calculating the distance from the center to the slant side, and analyzing the slope—it seems that the angle is indeed arctan(4/3). Since arctan(4/3) is the angle whose tangent is 4/3, this is the angle between the generatrix and the base.Therefore, the answer should be arctan(4/3), which can be written as boxed{arctan frac{4}{3}}.Final Answerboxed{arctan frac{4}{3}}</think>Given a sphere inscribed in a truncated cone (frustum), where the sum of the diameters of the upper and lower bases is five times the radius of the sphere, we need to find the angle between the generatrix of the cone and the plane of the base.1. Define Variables: - Let ( R ) be the radius of the lower base. - Let ( r ) be the radius of the upper base. - Let ( a ) be the radius of the sphere. - The given condition translates to ( 2R + 2r = 5a ), simplifying to ( R + r = frac{5}{2}a ).2. Height of the Frustum: - The height ( h ) of the frustum is equal to the diameter of the sphere, so ( h = 2a ).3. Slant Height of the Frustum: - Using the Pythagorean theorem, the slant height ( l ) is given by ( l = sqrt{(R - r)^2 + h^2} ).4. Condition for Inscribed Sphere: - The trapezoidal cross-section of the frustum must be tangential, leading to the condition ( R + r = l ).5. Solve for ( R ) and ( r ): - From ( R + r = frac{5}{2}a ) and solving the equations derived from the slant height and height, we find: - ( R - r = frac{3}{2}a ) - Solving these, we get ( R = 2a ) and ( r = frac{a}{2} ).6. Calculate the Angle: - The angle between the generatrix and the base is found using the right triangle formed by the height ( h ), the difference in radii ( R - r ), and the slant height ( l ). - The tangent of the angle ( theta ) is given by ( tan(theta) = frac{h}{R - r} = frac{2a}{frac{3}{2}a} = frac{4}{3} ). - Therefore, the angle is ( arctanleft(frac{4}{3}right) ).The final answer is (boxed{arctan frac{4}{3}}).]

🔑:## Solution.Let the isosceles trapezoid ABCD be the axial section of the given truncated cone, E and F be the centers of its lower and upper bases, O be the center of the inscribed sphere, EF be its diameter and the height of the trapezoid ABCD (Fig. 12.105). If R is the radius of the sphere, then EF = 2R, AD + BC = 5R.BC + AD = AB + CD = 2CD Rightarrow CD = frac{5R}{2}.CK is the height of the trapezoid ABCD, CK = EF = 2R.angle CDK is the angle between the generatrix CD of the cone and the plane of the base.Fig. 12.105From triangle CKD left(angle CKD = 90^circright): sin angle CDK = frac{CK}{CD} = frac{4}{5}; angle CDK = arcsin frac{4}{5}.Answer: arcsin frac{4}{5}.

🧠:[<think>Okay, so I need to figure out the value of m that makes the polynomial x⁴ - 5x² + 4x - m divisible by (2x + 1). Hmm, let's see. I remember that if a polynomial is divisible by a binomial like (ax + b), then the remainder when you divide the polynomial by (ax + b) is zero. Also, there's the Remainder Theorem which states that the remainder of dividing a polynomial f(x) by (x - c) is f(c). But in this case, the divisor is (2x + 1), which isn't exactly in the form (x - c), so maybe I need to adjust it a bit.Wait, right. If I have a divisor like (2x + 1), I can factor out the 2 to get 2(x + 1/2). So, (2x + 1) = 2(x - (-1/2)). But since we're talking about divisibility, the factor of 2 in front might not matter because if the polynomial is divisible by (2x + 1), it's also divisible by (x + 1/2), except for some scaling factor. But maybe the Remainder Theorem can still be applied here. Let me think.According to the Remainder Theorem, if I divide f(x) by (x - c), the remainder is f(c). So, if I want the remainder when dividing by (2x + 1) to be zero, maybe I need to set c such that 2x + 1 = 0, which gives x = -1/2. Then, plugging x = -1/2 into the polynomial should give zero. That might be the key here.Let me write that down. If (2x + 1) is a factor of the polynomial f(x) = x⁴ - 5x² + 4x - m, then f(-1/2) must equal zero. So, I can substitute x = -1/2 into the polynomial and set it equal to zero, then solve for m. Let's try that.First, compute each term step by step:1. x⁴: (-1/2)⁴. Let's calculate that. (-1/2)⁴ is the same as (1/2)⁴, which is 1/16.2. -5x²: -5 * (-1/2)². First, square (-1/2): (-1/2)² = 1/4. Then multiply by -5: -5*(1/4) = -5/4.3. 4x: 4*(-1/2) = -2.4. -m: Just -m.So, adding all these up: 1/16 - 5/4 - 2 - m = 0.Now, let's compute the constants step by step. Let's convert all terms to sixteenths to add them up:1/16 - 5/4 is 1/16 - 20/16 = -19/16.Then, -19/16 - 2 is the same as -19/16 - 32/16 = -51/16.So, -51/16 - m = 0. Therefore, -m = 51/16, so m = -51/16.Wait, but that seems a bit strange. Let me check the calculations again to be sure.First term: (-1/2)^4 = (1/2)^4 = 1/16. Correct.Second term: -5*(-1/2)^2. The square is positive, so (-1/2)^2 = 1/4, then multiplied by -5: -5*(1/4) = -5/4. Correct.Third term: 4*(-1/2) = -2. Correct.Fourth term: -m.So, adding them: 1/16 - 5/4 - 2 - m.Convert all to sixteenths:1/16 - 20/16 - 32/16 - m = (1 - 20 - 32)/16 - m = (-51)/16 - m.Set equal to zero: (-51/16) - m = 0 → m = -51/16. Wait, but that's a negative value for m. Is that possible? Let me check once more.Alternatively, maybe I messed up the sign when plugging in x = -1/2. Let's verify:Original polynomial: x⁴ -5x² +4x -m.At x = -1/2:(-1/2)^4 -5*(-1/2)^2 +4*(-1/2) -m.Compute each term:(-1/2)^4 = 1/16. Correct.-5*(-1/2)^2: (-1/2)^2 = 1/4, so -5*(1/4) = -5/4. Correct.4*(-1/2) = -2. Correct.Then -m. So total is 1/16 -5/4 -2 -m.Convert all to sixteenths:1/16 -20/16 -32/16 -m = (1 -20 -32)/16 -m = (-51)/16 -m.Set to zero: -51/16 -m =0 → m= -51/16. So that seems correct.But is the answer supposed to be positive? Hmm. Maybe I made an error in the Remainder Theorem application. Let me recall: If a polynomial f(x) is divisible by (ax + b), then f(-b/a) = 0. In this case, (2x +1) corresponds to a = 2, b =1. Therefore, root at x = -b/a = -1/2. So substituting x= -1/2 into f(x) gives zero. So that part is correct.So, if substituting x= -1/2 into the polynomial gives zero, then m must be -51/16. But let's confirm that m is indeed supposed to be negative here. The problem didn't specify any constraints on m, so it's possible.Alternatively, perhaps there's a miscalculation in converting the fractions. Let me compute again:1/16 -5/4 -2 = 1/16 - (5/4 + 2). 5/4 is 1.25, 2 is 2.0, so total is 1.25 + 2.0 = 3.25. Then, 1/16 is 0.0625. So 0.0625 -3.25 = -3.1875. Then, -3.1875 -m =0 → m= -3.1875. But 51/16 is 3.1875, so m= -51/16. So that's correct.Alternatively, maybe I need to use polynomial division? Let's try that to confirm.Alternatively, using synthetic division. But synthetic division is typically for divisors like (x - c). However, since our divisor is (2x +1), which is equivalent to 2(x +1/2). So, perhaps if I factor out the 2, I can use synthetic division with x = -1/2, but the result will need to be scaled accordingly.Alternatively, perform polynomial long division. Let's attempt dividing x⁴ -5x² +4x -m by (2x +1).But polynomial division can be tedious. Let me set it up.Divide x⁴ -5x² +4x -m by 2x +1.First term: x⁴ divided by 2x gives (x³)/2. Multiply (2x +1) by (x³)/2: x⁴ + (x³)/2. Subtract from the original polynomial:(x⁴ -5x² +4x -m) - (x⁴ + (x³)/2) = - (x³)/2 -5x² +4x -m.Next term: -(x³)/2 divided by 2x gives - (x²)/4. Multiply (2x +1) by -x²/4: - (2x³)/4 - x²/4 = -x³/2 -x²/4. Subtract this from the current dividend:[- (x³)/2 -5x² +4x -m] - [ -x³/2 -x²/4 ] = (-x³/2 + x³/2) + (-5x² + x²/4) +4x -m = (- (5 - 1/4)x² ) +4x -m = (-19/4 x²) +4x -m.Next term: -19/4 x² divided by 2x gives -19/8 x. Multiply (2x +1) by -19/8 x: -38/8 x² -19/8 x = -19/4 x² -19/8 x. Subtract this from the current dividend:(-19/4 x² +4x -m) - (-19/4 x² -19/8 x ) = ( -19/4 x² +19/4 x² ) + (4x +19/8 x ) -m = (0) + ( (32/8 +19/8)x ) -m = (51/8 x ) -m.Next term: 51/8 x divided by 2x gives 51/16. Multiply (2x +1) by 51/16: 102/16 x +51/16 = 51/8 x +51/16. Subtract this from the current dividend:(51/8 x -m ) - (51/8 x +51/16 ) = (51/8 x -51/8 x ) + (-m -51/16 ) = -m -51/16.So, the remainder is -m -51/16. For the polynomial to be divisible by (2x +1), the remainder must be zero. Therefore:-m -51/16 =0 → -m =51/16 → m= -51/16.So, that matches the previous result. Therefore, m is indeed -51/16.Alternatively, if I use the Remainder Factor Theorem properly, since (2x +1) is a factor, then substituting x= -1/2 into the polynomial gives zero. Which is exactly what I did first, leading to the same answer. So, both methods confirm that m= -51/16.Therefore, the value of m is -51/16.Final AnswerThe value of ( m ) is boxed{-dfrac{51}{16}}.</think>To determine ( m ) such that the polynomial ( x^4 - 5x^2 + 4x - m ) is divisible by ( (2x + 1) ), we use the Remainder Theorem. If ( (2x + 1) ) is a factor, then substituting ( x = -frac{1}{2} ) into the polynomial should yield zero.Substituting ( x = -frac{1}{2} ) into the polynomial:[left( -frac{1}{2} right)^4 - 5 left( -frac{1}{2} right)^2 + 4 left( -frac{1}{2} right) - m]Calculating each term step by step:1. ( left( -frac{1}{2} right)^4 = frac{1}{16} )2. ( -5 left( -frac{1}{2} right)^2 = -5 left( frac{1}{4} right) = -frac{5}{4} )3. ( 4 left( -frac{1}{2} right) = -2 )4. ( -m )Combining these terms:[frac{1}{16} - frac{5}{4} - 2 - m = 0]Converting all terms to sixteenths for easier addition:[frac{1}{16} - frac{20}{16} - frac{32}{16} - m = 0]Simplifying the constants:[frac{1 - 20 - 32}{16} - m = 0 implies frac{-51}{16} - m = 0]Solving for ( m ):[-m = frac{51}{16} implies m = -frac{51}{16}]Thus, the value of ( m ) is boxed{-dfrac{51}{16}}.,<think>Okay, let me try to figure out how to solve this problem. So, the question is asking for the value of ( m ) such that the polynomial ( x^{4} - 5x^{2} + 4x - m ) is divisible by ( 2x + 1 ). Hmm, right. Divisible means that when we divide the polynomial by ( 2x + 1 ), there should be no remainder. So, I need to find ( m ) that makes the remainder zero. I remember something about the Remainder Theorem. Let me recall. The Remainder Theorem states that if a polynomial ( f(x) ) is divided by ( x - c ), the remainder is ( f(c) ). So, if we want the remainder to be zero, ( c ) must be a root of the polynomial, right? So, ( f(c) = 0 ). But here, the divisor is ( 2x + 1 ), not ( x - c ). Let me see. If I can express ( 2x + 1 ) in the form ( x - c ), then maybe I can apply the Remainder Theorem. Let's solve for ( x ) in ( 2x + 1 = 0 ). Subtracting 1 from both sides gives ( 2x = -1 ), then dividing by 2, we get ( x = -frac{1}{2} ). So, the root here is ( x = -frac{1}{2} ). Therefore, if the polynomial is divisible by ( 2x + 1 ), then substituting ( x = -frac{1}{2} ) into the polynomial should give zero. That makes sense. So, I need to substitute ( x = -frac{1}{2} ) into ( x^{4} - 5x^{2} + 4x - m ) and set it equal to zero, then solve for ( m ). Let me write that out:( left(-frac{1}{2}right)^4 - 5left(-frac{1}{2}right)^2 + 4left(-frac{1}{2}right) - m = 0 ).Now, let's compute each term step by step. First term: ( left(-frac{1}{2}right)^4 ). Since the exponent is even, the negative becomes positive. ( left(frac{1}{2}right)^4 = frac{1}{16} ).Second term: ( -5left(-frac{1}{2}right)^2 ). Again, squaring a negative gives a positive. ( left(-frac{1}{2}right)^2 = frac{1}{4} ). Multiply that by -5: ( -5 times frac{1}{4} = -frac{5}{4} ).Third term: ( 4left(-frac{1}{2}right) ). That's straightforward: ( 4 times -frac{1}{2} = -2 ).Fourth term: ( -m ).Putting all these together:( frac{1}{16} - frac{5}{4} - 2 - m = 0 ).Now, let's combine the constants. Let me convert everything to sixteenths to make it easier. ( frac{1}{16} - frac{5}{4} - 2 ) becomes:First term is already in sixteenths: ( frac{1}{16} ).Second term: ( frac{5}{4} = frac{20}{16} ), so ( -frac{20}{16} ).Third term: ( 2 = frac{32}{16} ), so ( -frac{32}{16} ).Adding these together:( frac{1}{16} - frac{20}{16} - frac{32}{16} = frac{1 - 20 - 32}{16} = frac{-51}{16} ).So, the equation becomes:( -frac{51}{16} - m = 0 ).Solving for ( m ):( -m = frac{51}{16} )Multiply both sides by -1:( m = -frac{51}{16} ).Wait, that seems a bit odd. Let me check my calculations again to be sure. First term: ( (-frac{1}{2})^4 = frac{1}{16} ). Correct.Second term: ( -5 times (-frac{1}{2})^2 ). Wait, hold on. Is it ( (-5) times (text{something}) )?Yes, the term is -5 multiplied by ( (-frac{1}{2})^2 ). Let's re-examine this:( (-5) times (-frac{1}{2})^2 ). The square of ( -frac{1}{2} ) is ( frac{1}{4} ), so ( -5 times frac{1}{4} = -frac{5}{4} ). That's correct.Third term: ( 4 times (-frac{1}{2}) = -2 ). Correct.So, adding those three terms: ( frac{1}{16} - frac{5}{4} - 2 ).Converting all terms to sixteenths:( frac{1}{16} - frac{20}{16} - frac{32}{16} ). 1 - 20 is -19, then -19 - 32 is -51. So, ( -51/16 ). Then subtract ( m ), so:( -51/16 - m = 0 ), so ( m = -51/16 ). That's correct.But let me confirm if I applied the Remainder Theorem properly. The divisor is ( 2x + 1 ), which is equal to 2*(x + 1/2). So, the root is at x = -1/2. So, according to the Remainder Theorem, substituting x = -1/2 into the polynomial should give the remainder, which we need to set to zero. Therefore, yes, the calculation is correct. Alternatively, maybe I should use polynomial long division to verify. Let me try dividing ( x^4 - 5x^2 + 4x - m ) by ( 2x + 1 ) and see what remainder I get, then set that remainder to zero and solve for ( m ). That might be a good way to cross-check.So, let's perform the division:Divide ( x^4 - 5x^2 + 4x - m ) by ( 2x + 1 ).First, arrange the dividend in descending powers. The dividend is already arranged: ( x^4 + 0x^3 - 5x^2 + 4x - m ).The divisor is ( 2x + 1 ).Start with the leading term ( x^4 ) divided by the leading term of the divisor, which is ( 2x ). So, ( x^4 ÷ 2x = (1/2)x^3 ). That's the first term of the quotient.Multiply the divisor ( 2x + 1 ) by ( (1/2)x^3 ):( (2x + 1) * (1/2)x^3 = x^4 + (1/2)x^3 ).Subtract this from the dividend:( (x^4 + 0x^3 - 5x^2 + 4x - m) - (x^4 + (1/2)x^3) = - (1/2)x^3 -5x^2 + 4x - m ).Bring down the next term, but since all terms are already there, proceed.Next term: ( - (1/2)x^3 ÷ 2x = - (1/4)x^2 ).Multiply divisor by - (1/4)x^2:( (2x + 1) * (-1/4)x^2 = - (1/2)x^3 - (1/4)x^2 ).Subtract this from the current dividend:( (- (1/2)x^3 -5x^2 + 4x - m) - (- (1/2)x^3 - (1/4)x^2 ) = (-5x^2 + (1/4)x^2) + 4x - m ).Simplify:( (-5 + 1/4)x^2 + 4x - m = (-19/4)x^2 + 4x - m ).Next term: ( (-19/4)x^2 ÷ 2x = (-19/8)x ).Multiply divisor by (-19/8)x:( (2x + 1)*(-19/8)x = (-38/8)x^2 - (19/8)x = (-19/4)x^2 - (19/8)x ).Subtract this from the current dividend:( (-19/4)x^2 + 4x - m - [ (-19/4)x^2 -19/8 x ] = (4x + 19/8 x) - m ).Convert 4x to 32/8 x for common denominators:32/8 x + 19/8 x = 51/8 x. So, the remainder is 51/8 x - m.Next term: Divide 51/8 x by 2x, which is 51/16.Multiply divisor by 51/16:(2x +1)*(51/16) = (51/8)x + 51/16.Subtract this from the current remainder:(51/8 x - m) - (51/8 x + 51/16) = -m -51/16.So, the remainder is -m -51/16.Since the polynomial is divisible by 2x +1, the remainder must be zero. Therefore: -m -51/16 =0.Solving for m: -m =51/16Multiply both sides by -1:m= -51/16.Okay, so that matches the earlier result. Therefore, the value of m is -51/16.But let me check once more. Sometimes when doing polynomial division, signs can be tricky. Let me verify with substitution again. If m = -51/16, then the polynomial is x^4 -5x^2 +4x - (-51/16) =x^4 -5x^2 +4x +51/16. Plugging x= -1/2 into this:(-1/2)^4 -5*(-1/2)^2 +4*(-1/2) +51/16.Compute each term:(-1/2)^4 = 1/16.-5*(-1/2)^2 = -5*(1/4) = -5/4.4*(-1/2) = -2.Adding these: 1/16 -5/4 -2 +51/16.Convert all to sixteenths:1/16 -20/16 -32/16 +51/16.Adding numerators:1 -20 -32 +51 = (1 +51) + (-20 -32) =52 -52 =0.So, yes, it gives zero. Therefore, m = -51/16 is correct.Hence, the answer is m = -51/16.Final AnswerThe value of ( m ) is boxed{-dfrac{51}{16}}.</think>To determine ( m ) such that ( x^4 - 5x^2 + 4x - m ) is divisible by ( 2x + 1 ), we use the Remainder Theorem. The root of ( 2x + 1 = 0 ) is ( x = -frac{1}{2} ). Substituting ( x = -frac{1}{2} ) into the polynomial and setting it equal to zero gives:[left(-frac{1}{2}right)^4 - 5left(-frac{1}{2}right)^2 + 4left(-frac{1}{2}right) - m = 0]Calculating each term:1. (left(-frac{1}{2}right)^4 = frac{1}{16})2. (-5left(-frac{1}{2}right)^2 = -5 times frac{1}{4} = -frac{5}{4})3. (4left(-frac{1}{2}right) = -2)Combining these terms:[frac{1}{16} - frac{5}{4} - 2 - m = 0]Converting to a common denominator (sixteenths):[frac{1}{16} - frac{20}{16} - frac{32}{16} - m = 0]Simplifying the fractions:[frac{1 - 20 - 32}{16} - m = 0 implies frac{-51}{16} - m = 0]Solving for ( m ):[-m = frac{51}{16} implies m = -frac{51}{16}]Thus, the value of ( m ) is (boxed{-dfrac{51}{16}}).]

🔑:If we divide the expression x^{4}-5 x^{2}+4 x-m by (2 x+1), we obtainx^{4}-5 x^{2}+4 x-m=(2 x+1)left(frac{x^{3}}{2}-frac{x^{2}}{4}-frac{19 x}{8}+frac{51}{16}right)-left(frac{51}{16}+mright)Therefore, our expression is divisible by (2 x+1) iffrac{51}{16}+m=0which means ifm=-frac{51}{16}=-3 frac{3}{16}(Borota Braniszláv, Szeged.)

🧠:[<think>Okay, let's see. The problem is about finding the maximum value of the product of distances from any point P on the ellipse to the two foci, F₁ and F₂. The ellipse equation is given as x²/25 + y²/9 = 1. The options are A: 9, B: 16, C: 25, D: 25/2. Hmm, I need to figure out which one is the correct answer.First, I recall that for an ellipse, the standard form is x²/a² + y²/b² = 1, where a is the semi-major axis and b is the semi-minor axis. Comparing that with the given equation, a² = 25 so a = 5, and b² = 9 so b = 3. Therefore, this is an ellipse centered at the origin, stretched along the x-axis because a > b.Next, the foci of the ellipse are located at (±c, 0), where c is the distance from the center to each focus, and c² = a² - b². Let me compute c. So, c² = 25 - 9 = 16, which means c = 4. Therefore, the foci F₁ and F₂ are at (-4, 0) and (4, 0) respectively.Now, the question is to find the maximum value of |PF₁|·|PF₂|, where P is any point on the ellipse. Hmm, how do I approach this?I remember that on an ellipse, the sum of the distances from any point to the two foci is constant and equal to 2a. So, |PF₁| + |PF₂| = 2a = 10 in this case. But here, we need the product of the distances instead of the sum. The problem is asking for the maximum of this product.Let me denote |PF₁| as d₁ and |PF₂| as d₂. Then, we have d₁ + d₂ = 10, and we need to maximize d₁·d₂.This seems like a problem that can be solved using the arithmetic mean-geometric mean inequality (AM-GM), but wait, AM-GM gives a maximum or minimum under certain conditions. However, since d₁ and d₂ are not independent variables—they are related by the equation d₁ + d₂ = 10—it's a constrained optimization problem. Alternatively, since we have one equation and two variables, maybe we can express one variable in terms of the other and then find the maximum of the product.Alternatively, maybe there is a geometric interpretation here. Let me think.Alternatively, since the ellipse is symmetric with respect to both axes, perhaps the maximum occurs at one of the vertices or co-vertices. Let's check that.First, let's consider the vertices. The ellipse has major vertices at (±a, 0) = (±5, 0) and minor vertices at (0, ±b) = (0, ±3).If P is at (5, 0), then |PF₁| is the distance from (5, 0) to (-4, 0), which is |5 - (-4)| = 9 units. Similarly, |PF₂| is the distance from (5, 0) to (4, 0), which is |5 - 4| = 1 unit. Therefore, the product is 9*1 = 9. That's option A. But wait, if we take the other vertex at (-5, 0), the distances would be | -5 - (-4)| = 1 and | -5 - 4| = 9, so the product is still 1*9=9. So at the major vertices, the product is 9.Now, check the minor vertices. Let's take P as (0, 3). Then, compute |PF₁| and |PF₂|. The distance from (0, 3) to (-4, 0) is sqrt((0 - (-4))² + (3 - 0)²) = sqrt(16 + 9) = sqrt(25) = 5. Similarly, the distance to (4, 0) is sqrt((0 - 4)² + (3 - 0)²) = sqrt(16 + 9) = sqrt(25) = 5. Therefore, the product is 5*5=25. That's option C. Hmm, so 25 is bigger than 9. So maybe the maximum is 25. But wait, let me verify if there are other points where the product could be even larger.Wait, so at the minor vertices, we get the product as 25, which is higher than at the major vertices. But is 25 the maximum?Wait, let me think. Maybe there's another point on the ellipse where the product is larger. Let's suppose a general point P(x, y) on the ellipse. Let's express |PF₁|·|PF₂| in terms of coordinates.Given F₁ is (-4, 0) and F₂ is (4, 0). Then, |PF₁| = sqrt((x + 4)^2 + y^2), and |PF₂| = sqrt((x - 4)^2 + y^2). Then, the product is sqrt[(x + 4)^2 + y^2] * sqrt[(x - 4)^2 + y^2]. Hmm, that looks complicated. Maybe we can simplify this expression.Let me square the product to make it easier: [sqrt((x + 4)^2 + y^2) * sqrt((x - 4)^2 + y^2)]² = [(x + 4)^2 + y^2][(x - 4)^2 + y^2]. Let's compute this product.First, expand each term:(x + 4)^2 + y^2 = x² + 8x + 16 + y²(x - 4)^2 + y^2 = x² - 8x + 16 + y²Multiply them together:(x² + 8x + 16 + y²)(x² - 8x + 16 + y²)Let me denote A = x² + y² + 16, and B = 8x. Then, the expression becomes (A + B)(A - B) = A² - B².Therefore, the product is (x² + y² + 16)² - (8x)^2Compute that:= (x² + y² + 16)² - 64x²Now, let's expand (x² + y² + 16)^2:= x^4 + y^4 + 256 + 2x²y² + 32x² + 32y²Then subtract 64x²:= x^4 + y^4 + 256 + 2x²y² + 32x² + 32y² - 64x²Simplify:= x^4 + y^4 + 256 + 2x²y² - 32x² + 32y²Hmm, this seems messy. Maybe there's a better way. Let's recall that P(x, y) is on the ellipse, so x²/25 + y²/9 = 1. So, we can express y² in terms of x²: y² = 9(1 - x²/25) = 9 - (9/25)x².Therefore, substitute y² into the expression for the squared product:First, compute x² + y² + 16:= x² + [9 - (9/25)x²] + 16= x² + 9 - (9/25)x² + 16= (1 - 9/25)x² + 25= (16/25)x² + 25Similarly, (8x)^2 = 64x²Therefore, the squared product is [(16/25)x² + 25]^2 - 64x²Let me compute this:Let me denote u = (16/25)x². Then, the expression becomes (u + 25)^2 - 64x².But x² can be expressed in terms of u: x² = (25/16)u.Wait, perhaps expanding directly is better.First, expand [(16/25)x² + 25]^2:= (16/25 x²)^2 + 2*(16/25 x²)*25 + 25^2= (256/625)x^4 + 2*(16/25 x²)*25 + 625Simplify:= (256/625)x^4 + 32x² + 625Then subtract 64x²:= (256/625)x^4 + 32x² + 625 - 64x²= (256/625)x^4 - 32x² + 625So, the squared product is (256/625)x^4 - 32x² + 625. To find the maximum of the product, we need to maximize this expression with respect to x, given that P(x, y) is on the ellipse, so x² ≤ 25 (since x²/25 ≤ 1).Let me set f(x) = (256/625)x^4 - 32x² + 625. To find the maximum of f(x), take the derivative f'(x) and set it equal to zero.First, compute derivative:f'(x) = 4*(256/625)x^3 - 64xSimplify:= (1024/625)x^3 - 64xSet equal to zero:(1024/625)x^3 - 64x = 0Factor out x:x[(1024/625)x² - 64] = 0So, either x = 0 or [(1024/625)x² - 64] = 0Case 1: x = 0If x = 0, then we are at the minor vertices (0, ±3). Then, f(0) = 0 - 0 + 625 = 625. Therefore, the squared product is 625, so the product is sqrt(625) = 25. Which matches what we found earlier.Case 2: (1024/625)x² - 64 = 0Solve for x²:(1024/625)x² = 64Multiply both sides by 625/1024:x² = 64*(625/1024)Simplify:64 and 1024: 64 is 2^6, 1024 is 2^10, so 64/1024 = 1/16Similarly, 625 is 25², so:x² = 64*(625)/1024 = (64/1024)*625 = (1/16)*625 = 625/16Therefore, x = ±√(625/16) = ±25/4 = ±6.25But wait, our ellipse has x² ≤ 25, so 625/16 is 39.0625, which is larger than 25. Wait, that's not possible. So, this solution is outside the ellipse. Therefore, the critical points at x = ±25/4 are not on the ellipse, so they are invalid. Therefore, the only critical points within the ellipse are at x = 0. Hence, the maximum of f(x) is at x = 0, giving the squared product as 625, so the product is 25.But wait, let me check again. If x = 25/4 is 6.25, but the ellipse only goes up to x = 5. So, indeed, x² = 625/16 is 39.0625, which is more than 25. Hence, invalid. So the only valid critical point is at x = 0, which gives the maximum.Therefore, the maximum value of |PF₁|·|PF₂| is 25, which is achieved at the minor vertices (0, ±3). So the answer should be C: 25.But let me cross-verify this with another approach. Let's use parametric equations of the ellipse. The parametric equations for the ellipse x²/a² + y²/b² = 1 are x = a cos θ, y = b sin θ, where θ is the parameter.So, for our ellipse, x = 5 cos θ, y = 3 sin θ.Then, the distances to F₁ and F₂ can be written as:|PF₁| = sqrt[(5 cos θ + 4)^2 + (3 sin θ)^2]|PF₂| = sqrt[(5 cos θ - 4)^2 + (3 sin θ)^2]Multiply these two:sqrt[(5 cos θ + 4)^2 + 9 sin² θ] * sqrt[(5 cos θ - 4)^2 + 9 sin² θ]Again, squaring the product gives:[(5 cos θ + 4)^2 + 9 sin² θ][(5 cos θ - 4)^2 + 9 sin² θ]Let me compute each bracket first.First bracket: (5 cos θ + 4)^2 + 9 sin² θExpand (5 cos θ + 4)^2:= 25 cos² θ + 40 cos θ + 16Add 9 sin² θ:= 25 cos² θ + 40 cos θ + 16 + 9 sin² θSimilarly, second bracket: (5 cos θ - 4)^2 + 9 sin² θExpand (5 cos θ - 4)^2:= 25 cos² θ - 40 cos θ + 16Add 9 sin² θ:= 25 cos² θ - 40 cos θ + 16 + 9 sin² θNow, the product becomes:[25 cos² θ + 40 cos θ + 16 + 9 sin² θ][25 cos² θ - 40 cos θ + 16 + 9 sin² θ]Again, this looks similar to (A + B)(A - B) where A = 25 cos² θ + 16 + 9 sin² θ and B = 40 cos θ. So,= (A + B)(A - B) = A² - B²Compute A²:A = 25 cos² θ + 9 sin² θ + 16So, A² = (25 cos² θ + 9 sin² θ + 16)^2B² = (40 cos θ)^2 = 1600 cos² θTherefore, the squared product is A² - B² = [25 cos² θ + 9 sin² θ + 16]^2 - 1600 cos² θLet me simplify A first. Let's combine the terms:25 cos² θ + 9 sin² θ = 25 cos² θ + 9(1 - cos² θ) = 25 cos² θ + 9 - 9 cos² θ = 16 cos² θ + 9Therefore, A = 16 cos² θ + 9 + 16 = 16 cos² θ + 25So, A = 16 cos² θ + 25Therefore, A² = (16 cos² θ + 25)^2 = 256 cos^4 θ + 800 cos² θ + 625Then, A² - B² = 256 cos^4 θ + 800 cos² θ + 625 - 1600 cos² θSimplify:= 256 cos^4 θ - 800 cos² θ + 625So, squared product = 256 cos^4 θ - 800 cos² θ + 625Let me denote t = cos² θ. Then, since cos² θ ranges from 0 to 1, t ∈ [0, 1].Then, squared product becomes:256 t² - 800 t + 625Now, we can consider this as a quadratic function in terms of t, but actually, it's a quartic in cos θ, but since we substituted t = cos² θ, it's a quadratic in t.So, let's write f(t) = 256 t² - 800 t + 625, where t ∈ [0, 1]To find the maximum of f(t), we can take the derivative with respect to t and find critical points.Compute derivative:f’(t) = 512 t - 800Set derivative to zero:512 t - 800 = 0Solve for t:t = 800 / 512 = 25 / 16 ≈ 1.5625But t is in [0, 1], so t = 25/16 is outside the domain. Therefore, the maximum of f(t) on [0,1] occurs at one of the endpoints.Compute f(0) = 0 - 0 + 625 = 625Compute f(1) = 256(1) - 800(1) + 625 = 256 - 800 + 625 = (256 + 625) - 800 = 881 - 800 = 81Therefore, the maximum value of f(t) is 625 at t = 0, which corresponds to cos² θ = 0, so θ = π/2 or 3π/2, which are the points (0, ±3), the minor vertices. Therefore, the squared product is 625, so the product is sqrt(625) = 25. This confirms our earlier result.Therefore, the maximum value of |PF₁|·|PF₂| is 25, which is option C.But just to be thorough, let me check another point. Suppose θ = π/4. Then cos θ = √2/2, sin θ = √2/2.Compute |PF₁|:sqrt[(5*(√2/2) + 4)^2 + (3*(√2/2))^2]= sqrt[( (5√2/2) + 4 )^2 + ( (3√2/2 )^2 )]Compute (5√2/2 + 4)^2:First, convert to decimals for easier estimation:5√2 ≈ 5*1.4142 ≈ 7.071, divided by 2 ≈ 3.5355, plus 4 ≈ 7.5355So squared ≈ (7.5355)^2 ≈ 56.78(3√2/2)^2 ≈ (4.2426/2)^2 ≈ (2.1213)^2 ≈ 4.5So |PF₁| ≈ sqrt(56.78 + 4.5) ≈ sqrt(61.28) ≈ 7.828Similarly, |PF₂| would be sqrt[(5*(√2/2) - 4)^2 + (3*(√2/2))^2]Compute (5√2/2 - 4):≈ 3.5355 - 4 ≈ -0.4645, squared ≈ 0.2158Add (3√2/2)^2 ≈ 4.5 as beforeSo |PF₂| ≈ sqrt(0.2158 + 4.5) ≈ sqrt(4.7158) ≈ 2.171Multiply the two distances: 7.828 * 2.171 ≈ 17.0, which is about 17. So the product here is approximately 17, which is less than 25. Hence, the maximum seems to indeed be at the minor vertices.Alternatively, if I take a point on the ellipse where x is something else. For example, let's take x = 3. Then, y² = 9*(1 - 9/25) = 9*(16/25) = 144/25, so y = ±12/5 = ±2.4.Compute |PF₁|: distance from (3, 2.4) to (-4,0):sqrt((3 + 4)^2 + (2.4 - 0)^2) = sqrt(49 + 5.76) = sqrt(54.76) ≈ 7.4|PF₂|: distance from (3, 2.4) to (4,0):sqrt((3 - 4)^2 + (2.4 - 0)^2) = sqrt(1 + 5.76) = sqrt(6.76) ≈ 2.6Product ≈ 7.4 * 2.6 ≈ 19.24, which is still less than 25.Therefore, all these checks suggest that the maximum product is indeed 25, achieved at the minor vertices.Another way to think about this problem is by using the property of an ellipse and the concept of the geometric mean. Since the sum of the distances is constant, the product would be maximized when the distances are as equal as possible. In the case of the minor vertices, both distances are equal (both 5), so that's when their product is maximized. Whereas at the major vertices, one distance is maximum and the other is minimum, leading to a lower product.This is similar to the AM-GM inequality, where for a fixed sum, the product is maximized when the numbers are equal. However, in this case, the numbers are distances from a point on the ellipse to the foci, which are not independent variables. But the result aligns with the AM-GM intuition: when d₁ + d₂ = 10, the product d₁d₂ is maximized when d₁ = d₂ = 5. However, in reality, not all points on the ellipse have d₁ = d₂. The only points where d₁ = d₂ are the minor vertices. Therefore, in this case, equality is achieved there, and hence the maximum product is 5*5 = 25.Therefore, the correct answer should be C: 25.Final Answerboxed{C}</think>Given the ellipse (frac{x^2}{25} + frac{y^2}{9} = 1), we need to find the maximum value of (|PF_1| cdot |PF_2|) where (F_1) and (F_2) are the foci of the ellipse and (P) is any point on the ellipse.1. Identify the semi-major and semi-minor axes: - The semi-major axis (a = 5). - The semi-minor axis (b = 3).2. Calculate the distance to the foci: - The distance (c) from the center to each focus is given by (c^2 = a^2 - b^2). - (c^2 = 25 - 9 = 16) hence (c = 4). - The foci are located at ((-4, 0)) and ((4, 0)).3. Key properties of the ellipse: - The sum of distances from any point on the ellipse to the foci is (2a = 10).4. Evaluate the product of distances at key points: - Vertices on the major axis: At ((5, 0)) and ((-5, 0)), the product (|PF_1| cdot |PF_2| = 9 cdot 1 = 9). - Vertices on the minor axis: At ((0, 3)) and ((0, -3)), the distances to both foci are 5, giving the product (|PF_1| cdot |PF_2| = 5 cdot 5 = 25).5. General parametric form and optimization: - Using parametric equations (x = 5 cos theta) and (y = 3 sin theta), the product of distances was derived and maximized. - The maximum value of the product was found to occur at the minor vertices, confirming the result.6. Conclusion: - The maximum value of (|PF_1| cdot |PF_2|) is achieved at the minor vertices ((0, pm 3)), resulting in the product (25).Thus, the maximum value of (|PF_1| cdot |PF_2|) is (boxed{C}).,<think>Okay, let's see. The problem is about an ellipse given by the equation x²/25 + y²/9 = 1. We need to find the maximum value of the product |PF₁| · |PF₂| where P is any point on the ellipse, and F₁, F₂ are the foci of the ellipse. The answer choices are A through D, with C being 25. Hmm, interesting.First, let me recall some properties of ellipses. The standard form of an ellipse is x²/a² + y²/b² = 1, where a is the semi-major axis and b is the semi-minor axis. The foci are located at (±c, 0) where c² = a² - b². In this case, the given equation is x²/25 + y²/9 = 1, so a² = 25 and b² = 9. Therefore, a = 5 and b = 3. Then c² = 25 - 9 = 16, so c = 4. Therefore, the foci F₁ and F₂ are at (-4, 0) and (4, 0) respectively.Now, the question is to find the maximum value of |PF₁| · |PF₂| as P moves along the ellipse. Let me think about how to approach this. One possible method is to parametrize the ellipse and then express |PF₁| and |PF₂| in terms of the parameter, multiply them, and then find the maximum. Alternatively, maybe there's a geometric property or an algebraic identity that could simplify this product. Let me explore both options.First, parametrization. For an ellipse, a common parametrization is using the parametric equations x = a cosθ, y = b sinθ. So here, x = 5 cosθ, y = 3 sinθ. Then, any point P on the ellipse can be written as (5 cosθ, 3 sinθ).Then, let's compute |PF₁| and |PF₂|. The distance from P to F₁ and F₂ can be calculated using the distance formula.F₁ is at (-4, 0), so |PF₁| = sqrt[(5 cosθ + 4)² + (3 sinθ - 0)²]Similarly, F₂ is at (4, 0), so |PF₂| = sqrt[(5 cosθ - 4)² + (3 sinθ - 0)²]Therefore, the product |PF₁| · |PF₂| is sqrt[(5 cosθ + 4)² + 9 sin²θ] · sqrt[(5 cosθ - 4)² + 9 sin²θ]This looks complicated, but maybe we can simplify it. Let me square the product to make it easier, find the maximum of the square, and then take the square root at the end. Since the square of the product is the product of the squares, that might be manageable.So, [ |PF₁| · |PF₂| ]² = [ (5 cosθ + 4)² + 9 sin²θ ] · [ (5 cosθ - 4)² + 9 sin²θ ]Let me compute each term inside the product first.First term: (5 cosθ + 4)² + 9 sin²θ.Expanding (5 cosθ + 4)²: 25 cos²θ + 40 cosθ + 16.Adding 9 sin²θ: 25 cos²θ + 40 cosθ + 16 + 9 sin²θ.Similarly, second term: (5 cosθ - 4)² + 9 sin²θ.Expanding (5 cosθ - 4)²: 25 cos²θ - 40 cosθ + 16.Adding 9 sin²θ: 25 cos²θ - 40 cosθ + 16 + 9 sin²θ.So now, the product becomes:[25 cos²θ + 40 cosθ + 16 + 9 sin²θ] · [25 cos²θ - 40 cosθ + 16 + 9 sin²θ]Notice that both terms have 25 cos²θ + 16 + 9 sin²θ, with the middle term ±40 cosθ. Let me denote A = 25 cos²θ + 16 + 9 sin²θ, then the product is (A + 40 cosθ)(A - 40 cosθ) = A² - (40 cosθ)².So, expanding this, we have:A² - 1600 cos²θNow compute A:A = 25 cos²θ + 16 + 9 sin²θ = (25 cos²θ + 9 sin²θ) + 16Let me compute 25 cos²θ + 9 sin²θ. Factor terms:25 cos²θ + 9 sin²θ = 9(cos²θ + sin²θ) + 16 cos²θ = 9(1) + 16 cos²θ = 9 + 16 cos²θSo A = 9 + 16 cos²θ + 16 = 25 + 16 cos²θTherefore, A = 25 + 16 cos²θSo, A² = (25 + 16 cos²θ)^2 = 625 + 800 cos²θ + 256 cos⁴θThen, subtract 1600 cos²θ:A² - 1600 cos²θ = 625 + 800 cos²θ + 256 cos⁴θ - 1600 cos²θ = 625 - 800 cos²θ + 256 cos⁴θTherefore, [ |PF₁| · |PF₂| ]² = 625 - 800 cos²θ + 256 cos⁴θHmm, that's a quartic in cosθ. Let me see if this can be simplified or written as a quadratic in cos²θ. Let’s set u = cos²θ. Then, the expression becomes:256 u² - 800 u + 625Which is a quadratic in u. Let's denote f(u) = 256 u² - 800 u + 625.To find the maximum of [ |PF₁| · |PF₂| ]², we need to analyze f(u). Since u = cos²θ, u ranges between 0 and 1. Therefore, u ∈ [0, 1]. But since the quadratic f(u) is 256 u² - 800 u + 625, which is a parabola opening upwards (since coefficient of u² is positive). Therefore, the minimum of f(u) is at the vertex, and the maximum occurs at the endpoints of the interval [0,1].Wait, but we need the maximum of f(u) over u ∈ [0,1]. Since the parabola opens upwards, the maximum will be at either u=0 or u=1.Compute f(0) = 0 - 0 + 625 = 625.Compute f(1) = 256(1) - 800(1) + 625 = 256 - 800 + 625 = (256 + 625) - 800 = 881 - 800 = 81.Therefore, the maximum of f(u) on [0,1] is 625, occurring at u=0. Therefore, the maximum of [ |PF₁| · |PF₂| ]² is 625, so the maximum of |PF₁| · |PF₂| is sqrt(625) = 25. Wait, but 25 is option C. However, the answer options include 25, which is option C. But let me check if that's correct.Wait, but if u=0, that corresponds to cosθ=0, so θ=π/2 or 3π/2, so points P at (0, 3) or (0, -3). Let's compute |PF₁| · |PF₂| at P=(0,3).Compute |PF₁|: distance from (0,3) to (-4,0): sqrt( (0 +4)^2 + (3 -0)^2 ) = sqrt(16 + 9) = sqrt(25) = 5.Similarly, |PF₂|: distance from (0,3) to (4,0): same thing, sqrt(16 +9)=5. Therefore, the product is 5*5=25. So that's 25. If θ=0, which is u=1, then cosθ=1, sinθ=0, so P=(5,0). Let's compute |PF₁| · |PF₂|. |PF₁| is distance from (5,0) to (-4,0): 5 - (-4) = 9. |PF₂| is distance from (5,0) to (4,0): 5 -4 =1. Product is 9*1=9. Which is option A. So the product here is 9.Wait, but according to the previous calculation, the maximum is 25 and the minimum is 9? But the problem is asking for the maximum value. So according to this, the answer is 25, which is option C. However, I need to confirm this, because sometimes when squaring and taking square roots, there can be confusion, but in this case, since distances are positive, it's okay.Wait a second, but let me think again. If P is at (0,3), then both distances to F₁ and F₂ are 5, so the product is 25. If P is at (5,0), then the product is 9*1=9, and if P is at (-5,0), the product would be distance to F₁: |-5 - (-4)| = 1, distance to F₂: |-5 -4| =9, product 9*1=9. So the product ranges from 9 to 25? That seems counterintuitive. Wait, but according to the calculation, when u=0 (i.e., cosθ=0), which is at the top and bottom of the ellipse, the product is maximized. So 25 is the maximum, and 9 is the minimum. Therefore, the answer is 25, which is C.But wait, in the problem statement, the options are A:9, B:16, C:25, D:25/2. So 25 is there, so the answer should be C.But let me see another way to approach this problem to verify.Another method: recall that in an ellipse, the sum |PF₁| + |PF₂| is constant and equal to 2a, which is 10 in this case. However, here we are dealing with the product, not the sum. The problem is to maximize the product of distances from a point on the ellipse to the two foci.Alternatively, we can use coordinates. Let me suppose that P is (x,y) on the ellipse. Then |PF₁| = sqrt( (x +4)^2 + y^2 ), |PF₂| = sqrt( (x -4)^2 + y^2 ). The product is sqrt( (x +4)^2 + y^2 ) * sqrt( (x -4)^2 + y^2 ). Let's square this product to make it easier:[(x +4)^2 + y^2][(x -4)^2 + y^2]Let me expand this product. First, note that [(x +4)^2 + y^2][(x -4)^2 + y^2] can be written as [(x^2 +8x +16 + y^2)(x^2 -8x +16 + y^2)]Let’s denote A = x^2 + y^2 +16, then the expression becomes (A +8x)(A -8x) = A² - (8x)^2 = (x^2 + y^2 +16)^2 - 64x².Now, expanding (x^2 + y^2 +16)^2:= x^4 + 2x^2 y^2 + y^4 + 32x^2 + 32y^2 + 256Then subtract 64x²:= x^4 + 2x^2 y^2 + y^4 +32x^2 +32y^2 +256 -64x²Simplify:= x^4 + 2x^2 y^2 + y^4 -32x² +32y^2 +256Hmm, not sure if this helps. Let me recall that the ellipse equation is x²/25 + y²/9 =1, so x² =25(1 - y²/9). Maybe substitute x² into the expression.Alternatively, perhaps use the ellipse equation to express y² in terms of x²: y² =9(1 -x²/25).So substitute y² into the product squared:[(x +4)^2 + y^2][(x -4)^2 + y^2] = [(x +4)^2 +9(1 -x²/25)][(x -4)^2 +9(1 -x²/25)]Let me compute each bracket separately.First, (x +4)^2 +9(1 -x²/25):= x² +8x +16 +9 - (9x²)/25= x² - (9x²)/25 +8x +25= (25x² -9x²)/25 +8x +25= (16x²)/25 +8x +25Similarly, (x -4)^2 +9(1 -x²/25):= x² -8x +16 +9 - (9x²)/25= x² - (9x²)/25 -8x +25= (16x²)/25 -8x +25Therefore, the product is:[ (16x²/25 +8x +25) ] * [ (16x²/25 -8x +25) ]This is similar to the previous approach. Let me denote A =16x²/25 +25, then the product becomes (A +8x)(A -8x) = A² - (8x)^2Compute A² = (16x²/25 +25)^2 = (16x²/25)^2 + 2*(16x²/25)*25 +25² = 256x⁴/625 + 32x² +625Then subtract (8x)^2 =64x²:A² -64x² =256x⁴/625 +32x² +625 -64x² =256x⁴/625 -32x² +625Therefore, the squared product is 256x⁴/625 -32x² +625. To find the maximum of this expression, we can treat it as a function of x². Let me let t =x². Since x is on the ellipse, x² ≤25, so t ∈ [0,25].So the expression becomes 256t²/625 -32t +625.Let me write this as (256/625)t² -32t +625.To find the maximum of this quadratic in t. Since the coefficient of t² is positive (256/625 >0), the parabola opens upwards, so the minimum is at the vertex, and the maximum occurs at the endpoints of t's interval [0,25].Compute the value at t=0:(256/625)*0 -32*0 +625 =625.At t=25:(256/625)*(625) -32*25 +625 =256 -800 +625 = (256 +625) -800 =881 -800=81.Therefore, the maximum of the squared product is 625, hence the maximum product is sqrt(625)=25, same as before. Therefore, the maximum value is 25, which is option C.Alternatively, maybe there's a smarter way using properties of ellipses. Let me think. For a point P on the ellipse, the product |PF₁|*|PF₂|. Since we know that the sum |PF₁| + |PF₂| =2a=10. If we can express the product in terms of the sum and something else, perhaps.We know that for two numbers, the product is maximized when the numbers are equal, given a fixed sum. However, here we have a product of distances which are constrained by the ellipse. Wait, but in our case, when P is at (0,3), the distances are both 5, so the product is 25, which is a maximum. When P is at (5,0), the distances are 9 and 1, product 9, which is a minimum. So, in this case, when the two distances are equal, the product is maximized, which aligns with the idea that for a fixed sum, the product is maximized when the numbers are equal.But here, the sum is fixed? Wait, but in the ellipse, the sum |PF₁| + |PF₂| is fixed at 2a=10. Therefore, given that the sum is fixed, the product is maximized when the two numbers are equal. Therefore, when |PF₁|=|PF₂|=5, which occurs when P is at the top or bottom of the ellipse, i.e., (0, ±3). Then, the product is 5*5=25, which is the maximum. Therefore, the answer is 25, option C.Therefore, this makes sense. So, even without parametrizing or expanding, we can use the fact that for fixed sum, the product is maximized when the terms are equal. Therefore, maximum product is (10/2)^2=25. Wait, actually, if S = |PF₁| + |PF₂| =10, then the maximum of |PF₁|*|PF₂| is when |PF₁|=|PF₂|=5, so the product is 25. So this is a straightforward approach.So, in this case, the maximum value is indeed 25, which is option C. Therefore, the correct answer is C.Final Answerboxed{C}</think>Given the ellipse (frac{x^2}{25} + frac{y^2}{9} = 1), we need to find the maximum value of (|PF_1| cdot |PF_2|) where (P) is any point on the ellipse and (F_1) and (F_2) are the foci of the ellipse.1. Identify the semi-major axis, semi-minor axis, and foci: - The semi-major axis (a = 5). - The semi-minor axis (b = 3). - The distance to the foci (c) is given by (c^2 = a^2 - b^2 = 25 - 9 = 16), so (c = 4). - The foci are at ((-4, 0)) and ((4, 0)).2. Parametrize the ellipse: - Any point (P) on the ellipse can be parametrized as (P(5 cos theta, 3 sin theta)).3. Calculate distances (|PF_1|) and (|PF_2|): - (|PF_1| = sqrt{(5 cos theta + 4)^2 + (3 sin theta)^2}) - (|PF_2| = sqrt{(5 cos theta - 4)^2 + (3 sin theta)^2})4. Square the product of distances: - ([|PF_1| cdot |PF_2|]^2 = [(5 cos theta + 4)^2 + 9 sin^2 theta] cdot [(5 cos theta - 4)^2 + 9 sin^2 theta])5. Simplify the expression: - Expand and simplify the product to find it is a quadratic in (cos^2 theta). - The expression simplifies to (256 cos^4 theta - 800 cos^2 theta + 625).6. Maximize the quadratic expression: - By analyzing the quadratic expression, we find the maximum value occurs when (cos theta = 0) (i.e., at points ((0, pm 3))).7. Verify the maximum product: - At points ((0, pm 3)), the distances to both foci are 5, resulting in the product (5 cdot 5 = 25).8. Conclusion: - The maximum value of (|PF_1| cdot |PF_2|) is 25, which corresponds to option C.(boxed{C})]

🔑:Let P(x_0, y_0), |PF_1| = 5 + frac{4}{5}x_0, and |PF_2| = 5 - frac{4}{5}x_0.Thus, |PF_1| cdot |PF_2| = 25 - frac{16}{25}x_0^2.Hence, the maximum value of |PF_1| cdot |PF_2| is 25.Therefore, the answer is boxed{text{C}}.Let P(x_0, y_0), |PF_1| = 5 + frac{4}{5}x_0, |PF_2| = 5 - frac{4}{5}x_0, and |PF_1| cdot |PF_2| = 25 - frac{16}{25}x_0^2. From this, we can find the maximum value of |PF_1| cdot |PF_2|.This problem tests our understanding of the focal radii of an ellipse. While solving, be sure to carefully read the question and solve it meticulously to avoid errors.