🧠:Alright, let me try to work through this problem step by step. So, we have 200 cards, each sized 1x2, with +1 and -1 written on them. We need to arrange them on a 4x100 grid such that the product of the numbers in each column and each row is positive. Each card occupies two adjacent cells, either horizontally or vertically. Hmm, okay.First, let me understand the requirements. Each column and each row must have a product of +1. Since the product is positive, that means each row and each column must contain an even number of -1s. Because multiplying an even number of -1s will give +1. So, for every row and every column, the number of -1s must be even (0, 2, 4, etc.), but since each row has 100 cells and each column has 4 cells, the possible numbers of -1s in a row are 0, 2, 4, ..., 100 (even numbers), and in a column, 0, 2, 4.Now, each card is a domino covering two adjacent cells. Each domino has either two +1s, two -1s, or one +1 and one -1. Wait, but the problem states that each card has the numbers +1 and -1 written on them. Wait, does each card have both +1 and -1, or each card has either two +1s or two -1s? Let me check the problem statement again."each with the numbers +1 and -1 written on them." Hmm, the wording is a bit ambiguous. If it's "each with the numbers +1 and -1", that might mean each card has one +1 and one -1. But maybe it's possible that each card has two numbers, each of which can be +1 or -1. Wait, the problem says "the numbers +1 and -1 written on them". So maybe each card has both a +1 and a -1? So each domino has one +1 and one -1. That would make sense. So each domino is a tile with a +1 and a -1. Therefore, when placed on the grid, each domino will contribute one +1 and one -1 to the grid.But wait, maybe it's not necessarily one of each. The problem says "each with the numbers +1 and -1 written on them". So each card has two numbers, which can be either +1 or -1. So each domino can have (+1, +1), (-1, -1), (+1, -1), or (-1, +1). But the problem doesn't specify any restrictions on how the numbers are arranged on the dominoes. So perhaps we have 200 dominoes, each with two numbers, either both +1, both -1, or one of each. But the problem might actually state that each domino has one +1 and one -1. Let me check again.Wait, the problem says: "each with the numbers +1 and -1 written on them". The wording is a bit unclear. If it's "each with the numbers +1 and -1", does that mean each domino has both numbers? So one side is +1 and the other is -1? If that's the case, then each domino contributes a +1 and a -1. However, if it's possible that each domino has two numbers, each of which is either +1 or -1, then the domino could be either both +1, both -1, or mixed.But given that the problem is about arranging dominoes such that the product of each row and column is positive, the presence of dominoes with two -1s would complicate things, as they would contribute two -1s in adjacent cells. Similarly, dominoes with two +1s would be helpful for keeping the product positive, but since dominoes are 1x2, they cover two cells. So perhaps the key is that each domino has one +1 and one -1. Then, when placed horizontally or vertically, each domino contributes one +1 and one -1 to the grid. If that's the case, then the entire grid will have an equal number of +1s and -1s. Since the grid is 4x100, there are 400 cells. Therefore, 200 dominoes, each contributing one +1 and one -1, so total +1s and -1s would be 200 each. So 200 +1s and 200 -1s.But then, each row is 100 cells. So each row would have 100 numbers, half +1s and half -1s? Wait, but 100 is even, so half would be 50 each. But 200 dominoes, each contributing one +1 and one -1, so total 200 of each. So 4 rows, each with 100 cells, so each row would have 50 +1s and 50 -1s. But wait, the product of the numbers in each row is the product of 50 +1s and 50 -1s. The product would be (-1)^50 = +1, which is positive. Similarly, each column is 4 cells. If each column has an even number of -1s, then the product would be positive. But the problem is arranging dominoes such that in each row and column, the number of -1s is even.But if each domino contributes one +1 and one -1, then each domino contributes exactly one -1. Since there are 200 dominoes, there are 200 -1s. So total -1s in the grid would be 200. Since the grid is 4x100=400 cells, 200 -1s and 200 +1s. Each row would have 100 cells. If we can arrange the dominoes such that each row has an even number of -1s (i.e., 50, but 50 is even? Wait, 50 is even? 50 divided by 2 is 25, so 50 is even. Wait, 50 is an even number. Wait, 50 is even. So 50 -1s per row. So the product would be (-1)^50 = 1. So each row's product is 1. Similarly, each column is 4 cells. Each column must have an even number of -1s. So possible 0, 2, or 4 -1s per column.But since there are 200 -1s in total, and 100 columns, each column would need to have 2 -1s on average. Because 200 / 100 = 2. So each column must have exactly 2 -1s. Because the total is 200, and 100 columns, so average 2 per column. So we need each column to have exactly 2 -1s. Therefore, the problem reduces to arranging dominoes (each domino covering two adjacent cells, either horizontally or vertically), such that each row has 50 -1s and each column has 2 -1s.But dominoes are placed either horizontally or vertically. Each domino has one -1 and one +1. Wait, but if each domino is required to have one -1 and one +1, then each domino is like a +1/-1 pair. So if you place dominoes horizontally, they would occupy two cells in the same row, contributing one -1 and one +1. Similarly, vertically, they would occupy two cells in the same column, one -1 and one +1. Therefore, in a horizontal domino, the two cells in the row would have one -1 and one +1, so the row's count of -1s would be increased by 1 every two cells. Similarly, for vertical dominoes, each vertical domino in a column would contribute one -1 and one +1 to that column, so the column's count of -1s would be increased by 1 per domino. But since each domino is placed either horizontally or vertically, the way the -1s are distributed depends on the domino orientation.But wait, actually, each domino contributes one -1 and one +1, regardless of orientation. So in a horizontal domino, the two cells in the same row would have one -1 and one +1, so in that row, the count of -1s increases by 1, and in the adjacent row (if vertical), but no—wait, horizontal domino is in the same row. Wait, no, horizontal domino is in the same row, covering two columns. So in a horizontal domino in row i, columns j and j+1, there's one -1 and one +1. Similarly, vertical domino is in column j, rows i and i+1, with one -1 and one +1.Therefore, each horizontal domino contributes one -1 and one +1 to a single row, and each vertical domino contributes one -1 and one +1 to a single column. Wait, no. Wait, vertical domino spans two rows, so in column j, rows i and i+1, one -1 and one +1. So the column j gets one -1 and one +1 from that domino, but the rows i and i+1 each get either a -1 or a +1. Similarly, horizontal domino in row i, columns j and j+1: the row i gets one -1 and one +1, and the columns j and j+1 each get either a -1 or a +1.Therefore, the key is that each domino, regardless of orientation, places one -1 and one +1 in the cells it covers. Therefore, in each row, the total number of -1s would be equal to the number of dominoes that have their -1 in that row. Similarly, in each column, the number of -1s is equal to the number of dominoes that have their -1 in that column.But since dominoes can be placed both horizontally and vertically, the challenge is to arrange them such that every row has 50 -1s (so even number) and every column has 2 -1s (also even). So in total, the counts add up: 4 rows, 50 -1s each, total 200 -1s; 100 columns, 2 -1s each, total 200 -1s. So that's consistent.But how to arrange dominoes such that each row has 50 -1s and each column has 2 -1s. Let's think about tiling the 4x100 grid with dominoes, each domino covering two cells with one -1 and one +1, such that in every row, exactly 50 dominoes have their -1 in that row, and in every column, exactly 2 dominoes have their -1 in that column.Wait, but each domino contributes exactly one -1 and one +1. So for each domino, we have to decide where the -1 is placed. The problem then becomes assigning the -1s in such a way that each row has 50 -1s and each column has 2 -1s. Since dominoes can be placed horizontally or vertically, the position of the -1 affects both the row and column counts.Alternatively, maybe we can model this as a bipartite graph where one set is the rows and columns, and the other is the domino positions? Hmm, not sure. Alternatively, think of it as a matrix where each cell is either +1 or -1, such that each row has 50 -1s and each column has 2 -1s, and the -1s and +1s are arranged such that they can be tiled with dominoes, each domino covering one -1 and one +1.But tiling dominoes with one -1 and one +1 is equivalent to a perfect matching in some sense. But the problem is not just about tiling, but also about the placement of -1s and +1s such that dominoes can cover them with each domino covering one of each.Alternatively, maybe we can construct such a grid. Let's think about the structure of the grid. 4 rows, 100 columns. Each column needs 2 -1s. So in each column, 2 cells are -1 and 2 are +1. Similarly, each row needs 50 -1s. So each row has 50 -1s and 50 +1s. So the entire grid is a 4x100 matrix with exactly two -1s per column and 50 -1s per row.Moreover, the domino tiling must align such that each domino covers one -1 and one +1. That is, every pair of adjacent cells covered by a domino must consist of one -1 and one +1.Wait, but that's equivalent to the grid being a domino tiling where adjacent cells (horizontally or vertically) covered by a domino have opposite signs. So the grid must be colored like a chessboard, where adjacent cells alternate between +1 and -1. But in a chessboard coloring, each domino, regardless of orientation, would cover one +1 and one -1. However, in such a coloring, each row would have alternating +1 and -1, leading to 50 of each in a row of 100, and each column of 4 would have 2 of each. But wait, a chessboard pattern on a 4x100 grid would have each column alternating between +1 and -1. For example, column 1: +1, -1, +1, -1; column 2: -1, +1, -1, +1; and so on. Then each column would have 2 -1s and 2 +1s, and each row would have 50 of each. So that satisfies the counts. However, in such a chessboard pattern, dominoes can be placed either horizontally or vertically, each covering one +1 and one -1. Therefore, tiling the chessboard with dominoes is possible? Wait, a chessboard-colored grid can always be tiled with dominoes, since each domino will cover one black and one white square, and the total number of black and white squares is equal. But in this case, the grid is 4x100, which is even in both dimensions, so tiling is possible. However, in our case, the coloring is not chessboard but alternates signs. So if we arrange the grid in a chessboard pattern with +1 and -1, then domino tiling is possible, and each domino covers one +1 and one -1. Therefore, this would satisfy the problem's conditions: each row has 50 -1s and 50 +1s, each column has 2 -1s and 2 +1s, and the domino tiling is possible. Therefore, the answer would be yes.But wait, the problem states that the dominoes themselves have the numbers +1 and -1 written on them. Wait, hold on, earlier I thought each domino has one +1 and one -1, but maybe the dominoes can have any combination, but we need to place them such that the resulting grid meets the product conditions. The problem says: "each with the numbers +1 and -1 written on them". It's ambiguous whether each domino has both numbers (one on each square) or each domino has two numbers which can be either +1 or -1. If the dominoes can have any combination, then perhaps we can choose dominoes with two +1s, two -1s, or one of each. But the problem might require that each domino has one +1 and one -1. Let me re-read the problem.Original problem: "There are 200 cards sized 1×2, each with the numbers +1 and -1 written on them. Can these cards be arranged on a 4×100 grid of squared paper such that the product of the numbers in each column and each row of the resulting table is positive? (Each card completely occupies two adjacent cells.)"The key is "each with the numbers +1 and -1 written on them". So each card (domino) has both +1 and -1 written on them. So each domino has one +1 and one -1. Therefore, when placing a domino, one cell is +1 and the adjacent cell is -1. Therefore, the dominoes are like +1/-1 pairs. Therefore, we cannot have dominoes with two +1s or two -1s. Therefore, each domino must cover one +1 and one -1. So the entire grid must be tiled with dominoes such that each domino is placed either horizontally or vertically, covering one +1 and one -1. Therefore, the grid must be colored in such a way that adjacent cells (horizontally or vertically) alternate between +1 and -1, like a chessboard. Then dominoes can be placed to cover adjacent cells, each domino covering one +1 and one -1.In such a chessboard coloring, each row would have 50 +1s and 50 -1s, and each column (which has 4 cells) would alternate between +1 and -1, resulting in 2 +1s and 2 -1s per column. Therefore, the product of each row would be (+1)^50 * (-1)^50 = 1, and the product of each column would be (+1)^2 * (-1)^2 = 1. So this satisfies the problem's conditions.However, the question is whether such a tiling is possible with dominoes. A standard chessboard coloring of a 4x100 grid allows for a domino tiling because both dimensions are even. In fact, any grid with even area can be tiled with dominoes. But in our case, we need to not only tile the grid but also ensure that the dominoes are placed such that each domino's +1 and -1 align with the chessboard pattern. But since the dominoes themselves have fixed +1 and -1, can we arrange them accordingly?Wait, no. If the dominoes are physical cards with +1 and -1 written on them, then each domino has a specific orientation. For example, a horizontal domino could have +1 on the left and -1 on the right, or vice versa. Similarly, a vertical domino could have +1 on top and -1 on bottom, or vice versa. Therefore, to tile the chessboard pattern, we need to have dominoes that can be rotated and placed in either orientation (horizontal or vertical) with the correct +1 and -1 alignment.However, the problem does not specify any restrictions on the dominoes' orientations or the order of +1 and -1 on them. It only states that each domino has the numbers +1 and -1 written on them. Therefore, we can assume that we can rotate and flip the dominoes as needed. That is, each domino is simply a 1x2 tile with one +1 and one -1, and we can place them either horizontally or vertically, with the +1 and -1 in either order. Therefore, as long as the grid is colored in a chessboard pattern, we can place the dominoes accordingly, flipping them as needed to match the +1 and -1 of the grid.Therefore, the answer would be yes, such an arrangement is possible. We can tile the 4x100 grid in a chessboard pattern with alternating +1s and -1s, ensuring that each row has 50 of each and each column has 2 of each. Then, by tiling the grid with dominoes placed either horizontally or vertically, flipping them as necessary to align the +1 and -1 with the grid's pattern, we can satisfy the conditions.But wait, let me verify this. Suppose we color the grid in a chessboard pattern. Then, every two adjacent cells (horizontally or vertically) have opposite signs. Therefore, any domino placed on the grid, whether horizontally or vertically, will cover one +1 and one -1, which matches the domino's requirement of having one +1 and one -1. Therefore, as long as we can tile the grid with dominoes (which we can, since it's a 4x100 grid, both even dimensions), then this tiling will satisfy the product conditions.Therefore, the answer is yes, it is possible.But let me think if there's any possible issue. For example, in a 4x100 grid, the chessboard coloring would alternate each cell. So the first cell (1,1) is +1, then (1,2) is -1, (1,3) +1, etc. Similarly, row 2 would start with -1, +1, -1, etc. Then each column would alternate between +1, -1, +1, -1, or starting with -1 depending on the column. Therefore, each column has two +1s and two -1s. For example, column 1: rows 1-4: +1, -1, +1, -1. Column 2: -1, +1, -1, +1. And so on. Therefore, each column has two -1s and two +1s, so product is (+1*-1*+1*-1) = 1. Similarly, each row has alternating signs, so 50 +1s and 50 -1s, product is 1. Therefore, this satisfies the problem's conditions.Tiling this grid with dominoes is straightforward because each domino covers two adjacent cells of opposite sign, which matches the domino's requirement. Since the grid is even in both dimensions, domino tiling is possible. Therefore, the answer should be yes.But wait, let's consider that domino tiling a 4xN grid is always possible for any N. Because you can tile it with vertical dominoes for each column, but since N=100 is even, you can also tile it with horizontal dominoes. However, in our case, the coloring is fixed as a chessboard, so regardless of the tiling method (horizontal or vertical dominoes), each domino will align with the coloring.Therefore, the conclusion is that such an arrangement is possible. Thus, the answer is yes.Final Answerboxed{Yes}

✅:To determine if 200 cards sized (1 times 2) with numbers +1 and -1 can be arranged on a (4 times 100) grid such that the product of the numbers in each column and each row is positive, we need to ensure each row and column has an even number of -1s. 1. Product Requirements: Each row and column must have an even number of -1s. For a (4 times 100) grid, each row (100 cells) must have 50 -1s and each column (4 cells) must have 2 -1s.2. Domino Contribution: Each domino covers two adjacent cells with one +1 and one -1. This means the grid must have an equal number of +1s and -1s (200 each).3. Chessboard Pattern: Arranging the grid in a chessboard pattern (alternating +1 and -1) ensures each row has 50 +1s and 50 -1s, and each column has 2 +1s and 2 -1s. This satisfies the product conditions.4. Domino Tiling: A chessboard-patterned grid can be tiled with dominoes as each domino covers one +1 and one -1. The (4 times 100) grid, having even dimensions, allows for such a tiling.Thus, the answer is:[boxed{Yes}]

🔑:To determine whether it's possible to arrange the 1x2 cards such that the product of the numbers in every row and column of the (4 times 100) grid is positive, let's follow these steps.1. Understand the Problem: - We have 200 cards, each sized (1 times 2), and each card has numbers (+1) and (-1) written on it. - We need to fill a (4 times 100) grid with these cards such that each card occupies two neighboring cells. Therefore, each card can be placed either horizontally or vertically. - Our objective is to arrange these cards so that the multiplication of numbers in each row and column results in a positive number.2. Constraints Identification: - Each card must be placed to cover two adjacent cells, either horizontally within the same row or vertically within the same column. - In each row of the (4 times 100) grid, there are 4 cells. We need to consider how to structure these 4 cells. - Each column in the (4 times 100) grid will have 100 cells filled by the cards.3. Proposed Arrangement: - The key idea is to use a repetitive pattern that ensures every row and column has a product of numbers that is positive. Plan: Arrange the cards in blocks of (4 times 2) within the grid. We will use the cards in such a way that within each (4 times 2) block, every row has alternating (+1) and (-1) ensuring the overall product remains positive.4. Row-wise Arrangement: - We can arrange the cards into (50) blocks of (4 times 2) where the first column of each block will have (+1) and the second column will have (-1). This pattern will ensure within each block: [ begin{pmatrix} +1 & -1 +1 & -1 +1 & -1 +1 & -1 end{pmatrix} ]5. Combining Blocks into Grid: - Place these blocks next to each other horizontally until the (4 times 100) grid is fully covered. Each (4 times 2) block consists of two columns, and we need 50 such blocks to cover the entire grid's width of 100 columns. - This arrangement ensures that every (4 times 2) block maintains a pattern where the product of each row within the block is: [ (+1) cdot (-1) = -1 ] - To be consistent with our goal: - If each row within the block has a multiplication result of (-1), every (4 times 2) block must align in such a way that subsequent vertical columns in the entire grid will still satisfy a positive product from rearrangement because we superimpose such that +1 and -1 moves across rows column-wise too.6. Verification: - Given each card covers two cells and the grid (4 times 100) has 400 cells in total, using (200) cards is sufficient to cover the entire grid perfectly. - The planned block arrangement ensures every row and then every subsequent column multiplication indeed results in positive values. Conclusion:Thus, the correct arrangement that ensures that the product of the numbers in every row and every column of the (4 times 100) grid is positive is achievable. We strategically utilized blocks to maintain the required properties.[boxed{text{Yes, it is possible to fill the grid such that the product in each row and column is positive.}}]

🧠:Okay, let's try to figure out this logic puzzle. So, we have three friends: Ben, Dan, and Cam. Each of them makes a statement about the others' truthfulness. The goal is to determine who is lying. Let's break down each person's statement step by step.First, let me write down the statements again to keep track:- Ben says, "Exactly one of Dan and Cam is telling the truth."- Dan says, "Exactly one of Ben and Cam is telling the truth."- Cam says, "Neither Ben nor Dan is telling the truth."So, we need to check for consistency in these statements. The approach here is probably to assume each person is telling the truth and see if that leads to a contradiction, or assume someone is lying and see if the statements align.Let me start by considering each person as truthful and see if that holds.Case 1: Suppose Ben is telling the truth. Then, according to Ben, exactly one of Dan or Cam is telling the truth. That means either Dan is truthful and Cam is lying, or Cam is truthful and Dan is lying.But let's explore both possibilities under Ben's truthfulness.Subcase 1a: If Ben is truthful, and Dan is also truthful, then according to Dan's statement, exactly one of Ben and Cam is truthful. But wait, if Ben is truthful (as per our assumption here), then Dan's statement would require that exactly one of Ben and Cam is truthful. Since we're assuming Ben is truthful, then Cam must be lying for Dan's statement to hold. But wait, in this subcase, we're considering both Ben and Dan as truthful. Then, according to Dan, exactly one of Ben and Cam is truthful. But if Ben is truthful, then Cam must be lying. However, in Ben's original statement, he said exactly one of Dan and Cam is truthful. If Dan is truthful (as per this subcase), then Cam is lying, which would make Ben's statement true. So far, that's consistent. But then, Cam's statement is "Neither Ben nor Dan is telling the truth." If Cam is lying, then the opposite must be true, which is that at least one of Ben or Dan is telling the truth. But in this scenario, both Ben and Dan are telling the truth, which would make Cam's statement a lie. That works. Wait, so in this scenario, both Ben and Dan are truthful, and Cam is lying. But hold on, Ben's statement says that exactly one of Dan and Cam is truthful. If Dan is truthful and Cam is lying, then Ben's statement holds. Dan's statement says exactly one of Ben and Cam is truthful. Since Ben is truthful and Cam is lying, Dan's statement also holds. But then both Ben and Dan are telling the truth, which would mean that Ben's statement is actually false because Ben claims exactly one of Dan and Cam is truthful, but in reality, Dan is truthful and Cam is lying, so that's exactly one. Wait, so Ben's statement is correct. But Dan's statement is also correct because exactly one of Ben and Cam is truthful (Ben is truthful, Cam is lying). So, both Ben and Dan are truthful. But Cam is lying. However, if both Ben and Dan are truthful, then Cam's statement "Neither Ben nor Dan is telling the truth" is a lie, which is correct. So this seems possible. But wait, the problem is that if both Ben and Dan are telling the truth, then their statements must hold. Let's check again.Ben says exactly one of Dan and Cam is truthful. If Dan is truthful, then Cam must be lying. That's consistent. Dan says exactly one of Ben and Cam is truthful. If Ben is truthful, then Cam must be lying. That's also consistent. So in this case, both Ben and Dan are truthful, Cam is lying. But wait, that would mean there are two people telling the truth (Ben and Dan), which contradicts the statements? Wait, no. Each of their statements refers to exactly one of the other two. Let me verify:Ben's statement: Exactly one of Dan and Cam is truthful. If Dan is truthful and Cam is lying, then Ben's statement is true. So Ben is truthful.Dan's statement: Exactly one of Ben and Cam is truthful. If Ben is truthful and Cam is lying, then Dan's statement is also true. So Dan is truthful.So, both Ben and Dan can be truthful, with Cam lying. But that would mean two people are telling the truth. But the answer options don't include that. Let me check the answer options again. The options are A) Just Ben, B) Just Dan, C) Just Cam, D) Each of Ben and Cam, E) All three.Wait, if both Ben and Dan are truthful, and Cam is lying, then the answer would not be among the options given, because the options don't have "Both Ben and Dan are truthful". Wait, the options are about who is lying. So if Cam is lying, then the answer would be C) Just Cam. But wait, in this scenario, Ben and Dan are truthful. So the answer would be C. But wait, let's see if this is possible. Let me check if there's a contradiction here.Wait, if Ben and Dan are both truthful, then Cam is lying. But Cam's statement is "Neither Ben nor Dan is telling the truth." If Cam is lying, then the opposite is true: at least one of Ben or Dan is telling the truth. Since both are telling the truth, that's fine. So there's no contradiction here. So this scenario is possible: Ben and Dan truthful, Cam lying. Therefore, the answer would be C) Just Cam. But wait, let's check other possibilities because sometimes these puzzles require that only one scenario is possible.Case 2: Suppose Ben is truthful, then exactly one of Dan and Cam is truthful. Let's consider Subcase 1b where Cam is truthful and Dan is lying. So in this case, Ben is truthful, Cam is truthful, Dan is lying.But then Dan's statement is "Exactly one of Ben and Cam is truthful." If Dan is lying, then the opposite of his statement is true. The opposite of "exactly one" is either both or none. So if Dan's statement is false, then it's not the case that exactly one of Ben and Cam is truthful. Therefore, either both are truthful or both are lying. But in this subcase, we're assuming Ben and Cam are both truthful. Therefore, Dan's statement is false, which is consistent. But then Cam's statement is "Neither Ben nor Dan is telling the truth." If Cam is truthful, then both Ben and Dan are lying. But in this subcase, Ben is truthful, which contradicts Cam's statement. Therefore, this subcase leads to a contradiction. Therefore, Subcase 1b is impossible.So, when Ben is truthful, the only possible subcase is 1a, where Dan is truthful and Cam is lying. But as we saw, this leads to Ben and Dan both being truthful, which is possible. So that would mean Cam is lying. But let's check other cases.Case 3: Suppose Dan is truthful. Then, according to Dan, exactly one of Ben and Cam is truthful. So either Ben is truthful and Cam is lying, or Cam is truthful and Ben is lying.Subcase 3a: Dan is truthful, Ben is truthful, Cam is lying.But if Ben is truthful, then his statement says exactly one of Dan and Cam is truthful. If Dan is truthful and Cam is lying, then Ben's statement holds. So this is the same as Case 1a. So this is consistent, leading to Ben and Dan truthful, Cam lying.Subcase 3b: Dan is truthful, Cam is truthful, Ben is lying.If Dan is truthful and Cam is truthful, then Dan's statement requires exactly one of Ben and Cam is truthful. But in this subcase, Cam is truthful, so Dan's statement would require that exactly one of Ben and Cam is truthful. But Cam is truthful, so Ben must be lying. So that's consistent. Now, Ben is lying, so his statement "Exactly one of Dan and Cam is telling the truth" is false. Therefore, the actual case is that either both Dan and Cam are truthful or both are lying. But in this subcase, both Dan and Cam are truthful. Therefore, Ben's statement is false, which is correct. Now, Cam's statement is "Neither Ben nor Dan is telling the truth." If Cam is truthful, then both Ben and Dan are lying. But in this subcase, Dan is truthful, which contradicts Cam's statement. Therefore, this subcase leads to a contradiction. Hence, Subcase 3b is impossible.Therefore, if Dan is truthful, the only possible scenario is Subcase 3a where Ben is truthful and Cam is lying, which is the same as before.Case 4: Suppose Cam is truthful. Then Cam's statement is "Neither Ben nor Dan is telling the truth." So both Ben and Dan are lying.If Cam is truthful, then Ben and Dan are lying. Let's check Ben's statement: Ben is lying, so his statement "Exactly one of Dan and Cam is telling the truth" is false. Therefore, the actual situation is that either both Dan and Cam are truthful, or both are lying. But Cam is truthful in this case, so Dan must also be truthful for Ben's statement to be false. However, Cam's statement says Dan is lying. Therefore, there's a contradiction here: Cam says Dan is lying, but if Cam is truthful, then Dan is lying. However, Ben's lie requires that both Dan and Cam are truthful or both are lying. Since Cam is truthful, Dan must also be truthful, but Cam says Dan is lying. Therefore, contradiction. Therefore, Cam cannot be truthful because it leads to a contradiction. Hence, Cam must be lying.So, from Cam's perspective, if Cam is lying, then the negation of his statement is true. The negation of "Neither Ben nor Dan is telling the truth" is "At least one of Ben or Dan is telling the truth." So, at least one of Ben or Dan is truthful.Now, combining this with previous analysis, if Cam is lying, then at least one of Ben or Dan is truthful. Let's see the possible scenarios.If Cam is lying, then we have two possibilities: either one of Ben or Dan is truthful, or both are truthful.But let's check if both can be truthful. As in Case 1a and 3a, if both Ben and Dan are truthful, then their statements would be:- Ben: Exactly one of Dan and Cam is truthful. But if Dan is truthful, then Cam must be lying, which fits.- Dan: Exactly one of Ben and Cam is truthful. If Ben is truthful, then Cam must be lying, which also fits.So both statements hold. Therefore, this scenario is possible: Ben and Dan truthful, Cam lying.Alternatively, if only one of Ben or Dan is truthful:Subcase 4a: Cam is lying, Ben is truthful, Dan is lying.If Ben is truthful, then exactly one of Dan and Cam is truthful. Since Cam is lying, Dan must be truthful for Ben's statement to hold. But this contradicts Dan being lying. Therefore, impossible.Subcase 4b: Cam is lying, Dan is truthful, Ben is lying.If Dan is truthful, then exactly one of Ben and Cam is truthful. Since Cam is lying, Ben must be truthful for Dan's statement to hold. But this contradicts Ben being lying. Therefore, also impossible.Therefore, the only possible scenario when Cam is lying is that both Ben and Dan are truthful. But wait, in that case, both Ben and Dan's statements hold, as checked earlier. So this is possible.Thus, the only consistent scenario is Ben and Dan telling the truth, and Cam lying. Therefore, the answer is C) Just Cam.But wait, let me confirm once more. If Ben and Dan are both truthful, then:- Ben's statement: Exactly one of Dan and Cam is truthful. But Dan is truthful, Cam is lying. So exactly one (Dan) is truthful. Wait, no, "exactly one of Dan and Cam"—so Dan is truthful, Cam is lying. That's exactly one. So Ben's statement is true.- Dan's statement: Exactly one of Ben and Cam is truthful. Ben is truthful, Cam is lying. So exactly one (Ben) is truthful. Dan's statement is true.- Cam's statement: Neither Ben nor Dan is truthful. But both are, so Cam is lying.Yes, this holds. So the only liar is Cam. Therefore, the answer is C.

✅:To determine which friends are lying, we analyze each statement under different truth assumptions:1. Ben's Statement: "Exactly one of Dan and Cam is telling the truth." 2. Dan's Statement: "Exactly one of Ben and Cam is telling the truth." 3. Cam's Statement: "Neither Ben nor Dan is telling the truth." Key Analysis:- Assume Cam is truthful: If Cam is truthful, then both Ben and Dan must be lying. However: - Ben’s lie implies *both or neither* of Dan/Cam are truthful. Since Cam is truthful, Dan must also be truthful (contradicting Cam’s claim that Dan is lying). - Dan’s lie implies *both or neither* of Ben/Cam are truthful. Since Cam is truthful, Ben must also be truthful (contradicting Cam’s claim). → Cam cannot be truthful; Cam is lying.- Cam is lying: The negation of Cam’s statement is *at least one of Ben or Dan is truthful*. We test scenarios: - Case 1: Both Ben and Dan are truthful. - Ben’s claim: Exactly one of Dan/Cam is truthful. Since Dan is truthful and Cam is lying, this holds. - Dan’s claim: Exactly one of Ben/Cam is truthful. Since Ben is truthful and Cam is lying, this holds. → No contradictions. - Case 2: Only one of Ben/Dan is truthful. - If Ben is truthful and Dan is lying: - Ben’s claim requires Dan or Cam to be truthful. Since Cam is lying, Dan must be truthful (contradicts Dan lying). - If Dan is truthful and Ben is lying: - Dan’s claim requires Ben or Cam to be truthful. Since Cam is lying, Ben must be truthful (contradicts Ben lying). → Both subcases fail. Conclusion:The only consistent scenario is Ben and Dan truthful, Cam lying. Answer: boxed{C}

🔑:We need to analyze the statements made by Ben, Dan, and Cam to deduce who is lying. Let's revisit the statements:- Ben says, "Exactly one of Dan and Cam is telling the truth."- Dan says, "Exactly one of Ben and Cam is telling the truth."- Cam says, "Neither Ben nor Dan is telling the truth."Let's evaluate each scenario based on these statements.1. Assume Cam is telling the truth: - Cam says, "Neither Ben nor Dan is telling the truth." - This implies Ben and Dan are both lying. - If Ben is lying, then Ben's statement "Exactly one of Dan and Cam is telling the truth" is false. - This would mean either both Dan and Cam are lying or both are telling the truth. - If Dan is lying, then Dan's statement "Exactly one of Ben and Cam is telling the truth" is false. - This would mean either both Ben and Cam are lying or both are telling the truth. Since Cam is assumed to be telling the truth, neither Ben nor Dan can be telling the truth if Cam's statement is true. However, we have found a contradiction as the assumption leads to both Ben and Dan being both truthful and liars simultaneously. Thus, Cam cannot be telling the truth. Hence, Cam must be lying.2. Conclusion: - Since Cam is lying, let's now verify the implications on Ben and Dan's statements. - Since Cam is lying, it implies Ben and/or Dan must be telling the truth.3. Evaluate Ben's statement: - Ben says, "Exactly one of Dan and Cam is telling the truth." - Since Cam is lying (we proved this), it means Dan must be telling the truth for Ben's statement to hold4. Evaluate Dan's statement: - Dan says, "Exactly one of Ben and Cam is telling the truth." - Since Cam is lying, for Dan's statement to be true, Ben must be telling the truthBy verifying Dan's and Ben's statements under the assumption that Cam is lying, we establish no contradictions:- If Cam is lying, then Ben and Dan must be telling the truth. Therefore: - Ben's statement that exactly one of Dan and Cam is telling the truth holds, because Dan is truthful and Cam is lying. - Dan's statement that exactly one of Ben and Cam is telling the truth holds because Ben is truthful and Cam is lying. Conclusion:[boxed{text{C}}]

🧠:Alright, let's try to figure out how many pairs of integers from 1 to 100 add up to more than 100. Hmm, okay. So the problem is asking for the number of ordered pairs (a, b) where both a and b are integers between 1 and 100, inclusive, and their sum a + b is greater than 100. Repetitions are allowed, so a pair like (50, 50) is valid, right?First, I need to understand the total number of possible pairs. Since repetitions are allowed and order matters, the total number of ordered pairs is 100 * 100 = 10,000. That makes sense because for each of the 100 choices for the first number, there are 100 choices for the second number.Now, the question is how many of these 10,000 pairs have a sum greater than 100. Maybe the easiest way is to subtract the number of pairs that have a sum of 100 or less from the total. So total pairs - pairs with sum ≤ 100 = pairs with sum > 100. That seems like a good strategy because counting the complement might be simpler.Let me verify this approach. If I can find how many pairs add up to 100 or less, subtracting that from 10,000 should give me the desired number. But how do I count the pairs that add up to 100 or less?Let's consider the possible sums. The smallest possible sum is 1 + 1 = 2, and the largest is 100 + 100 = 200. We want sums from 101 to 200. But again, maybe working with sums from 2 to 100 is easier for the complement.For each possible sum s from 2 to 100, how many ordered pairs (a, b) satisfy a + b = s? Let's think about that. For a given sum s, the number of ordered pairs where a and b are positive integers adding to s is s - 1. Wait, but here a and b are from 1 to 100, so we have to consider the constraints that a ≤ 100 and b ≤ 100.Wait a second, the standard formula for the number of ordered pairs (a, b) such that a + b = s is s - 1 when a and b are positive integers. However, when s is greater than 100 + 1 = 101, this formula would give more pairs, but since our numbers can't exceed 100, we have to adjust.But in our case, for the complement, s ranges from 2 to 100. So for each s from 2 to 100, the number of ordered pairs (a, b) with a + b = s would be s - 1. Because if s is between 2 and 100, then the possible pairs are (1, s - 1), (2, s - 2), ..., (s - 1, 1). Each of these pairs has both a and b at least 1 and at most s - 1. Since s ≤ 100, s - 1 ≤ 99, so both a and b are within 1 to 100. Therefore, there's no restriction here; all those pairs are valid. Therefore, for each s from 2 to 100, the number of ordered pairs is s - 1.Therefore, the total number of ordered pairs with sum ≤ 100 is the sum from s = 2 to s = 100 of (s - 1). Let's compute that.The sum from s = 2 to 100 of (s - 1) is equal to the sum from k = 1 to 99 of k, where k = s - 1. The sum of the first n integers is n(n + 1)/2. So here, n = 99. Therefore, the sum is 99 * 100 / 2 = 4950.Therefore, the total number of ordered pairs with sum ≤ 100 is 4950. Then, the number of ordered pairs with sum > 100 is total pairs (10,000) minus 4950, which is 10,000 - 4950 = 5050.Wait, but let me check if this is correct. Let's take a small example to verify. Suppose instead of 1 to 100, we take numbers from 1 to 3. How many pairs have a sum greater than 3?Total pairs: 3*3=9. Pairs with sum ≤3: Let's list them. (1,1)=2, (1,2)=3, (2,1)=3, (2,2)=4 which is already over. Wait, no, wait. Wait, in the small case:Wait, numbers from 1 to 3. Sums greater than 3. So possible sums are 4, 5, 6. Let's count how many pairs have sum >3.Possible pairs:(1,1)=2, (1,2)=3, (1,3)=4(2,1)=3, (2,2)=4, (2,3)=5(3,1)=4, (3,2)=5, (3,3)=6So pairs with sum >3: (1,3), (2,2), (2,3), (3,1), (3,2), (3,3). That's 6 pairs.Total pairs:9. Pairs with sum <=3: 3 (sum=2:1 pair, sum=3:2 pairs). Wait, but according to our formula, the sum from s=2 to s=3 of (s -1). For s=2:1, s=3:2. Total 1+2=3. Then total pairs with sum>3 would be 9-3=6. Which matches. So that works.But in the original problem, when s is up to 100, we use the formula sum from s=2 to 100 of (s-1) = 4950, which gives us 10,000 - 4950 = 5050. So that seems correct.But wait, let me check another example to be sure. Let's take numbers from 1 to 4. How many pairs sum >4?Total pairs:16.Pairs with sum <=4:Sum=2:1Sum=3:2Sum=4:3Total:1+2+3=6. Therefore, pairs with sum >4:16-6=10.Listing them:(1,4)=5, (2,3)=5, (2,4)=6, (3,2)=5, (3,3)=6, (3,4)=7, (4,1)=5, (4,2)=6, (4,3)=7, (4,4)=8. That's 10 pairs. Correct.Therefore, the formula works here as well.Therefore, returning to the original problem, 5050 is the answer.Wait, but let me make sure that when s is up to 100, the number of pairs is indeed s-1. For example, if s=100, then pairs are (1,99), (2,98), ..., (99,1). There are 99 pairs. Similarly, for s=50, there are 49 pairs. So yes, for each s from 2 to 100, the number of pairs is s-1, and summing those gives the total number of pairs with sum <=100. Hence, 4950.Therefore, the answer should be 10,000 - 4950 = 5050. But let me think again.Alternatively, maybe there's another way to compute it directly. Let's see.Suppose we fix the first number a. For each a from 1 to 100, how many numbers b (from 1 to 100) satisfy a + b > 100?So for a given a, b must be > 100 - a. Since b is an integer, b >= 101 - a.But b has to be at least 1 and at most 100. Therefore, the number of possible b's is 100 - (101 - a) + 1 = a. Wait, let's check:If a is 1: b must be >= 100. So b=100. So only 1 number. But according to formula, a=1, so 1. Correct.If a=50: b >= 51. So b can be 51 to 100, which is 50 numbers. 100 - 51 +1 = 50. And according to the formula, a=50, so 50. Correct.If a=100: b >= 1. So all 100 numbers. But according to formula, a=100, so 100. Correct.So yes, for each a, the number of b's is a. Therefore, the total number of pairs is the sum from a=1 to a=100 of a.Wait, but that sum is 1 + 2 + ... + 100 = (100)(101)/2 = 5050. Exactly the same result as before.Therefore, both methods give the same answer, 5050. That seems solid.Therefore, the number of ordered pairs is 5050.But wait, in the first method, we considered ordered pairs. In the second method, we fixed a and counted b's, which also accounts for ordered pairs. So both methods confirm that the answer is 5050.Therefore, the answer is 5050.But just to be thorough, let's check another angle. Suppose we visualize the pairs as a grid where a is the row and b is the column. The condition a + b > 100 corresponds to the area above the line a + b = 100. In such a grid from 1 to 100 in both axes, the line a + b = 100 would go from (1,99) to (99,1). The area above this line includes all pairs where a + b > 100. Since it's a grid of 100x100 points, the number of points above the line is the number we want.In such cases, the number of points above the line is equal to the number of points below the line because the grid is symmetric. Wait, but is that true?Wait, the line a + b = 100 divides the grid. The points above the line are a + b > 100, and below are a + b < 100. The points on the line are a + b = 100. But since the total number of points is 100*100=10,000, and the number of points on the line is the number of pairs where a + b =100. For a + b =100, the number of ordered pairs is 99, as we discussed earlier (from (1,99) to (99,1)). Therefore, total points on the line:99.Therefore, total points above the line would be (10,000 - 99)/2 + points above? Wait, no. Wait, if the grid is symmetric with respect to the line a + b = 100, then the number of points above and below would be equal. However, since 10,000 is even and 99 is odd, subtracting 99 gives 9901, which is odd. Therefore, it's not possible to split 9901 into two equal integers. Hence, the symmetry idea might not hold.Wait, but let's think again. The grid is symmetric if we swap a and b. However, the line a + b =100 is symmetric in itself. However, when considering points above and below, they might not be exact mirror images because of the integer grid points. For example, consider a point (a, b) where a + b >100. Its mirror image across the line a + b =100 would be (100 - b +1, 100 - a +1) or something like that? Wait, maybe not exactly.Alternatively, perhaps there's a different symmetry. Let me check with a smaller grid. For example, 1 to 3. The line a + b =4. Points above the line: (1,4) but since 4 is beyond the grid, it's actually (3,2), (3,3), (2,3). Wait, in the 3x3 grid, the line a + b =4 would pass through (1,3) and (3,1). The points above would be (2,3), (3,2), (3,3). That's 3 points. The points below would be (1,1), (1,2), (2,1), (2,2), (3,1), (1,3). Wait, that's 6 points. So it's not symmetric. The line itself has points (1,3), (2,2), (3,1), which are 3 points. So total points above:3, below:6, on the line:3. Total 12, but the grid is 9. Wait, no, in the 3x3 grid, total is 9. The line a + b=4 passes through (1,3), (2,2), (3,1). Points above the line are those where a + b >4: (3,2), (2,3), (3,3). That's 3 points. Points below the line: a + b <4: (1,1), (1,2), (2,1). That's 3 points. Points on the line:3. So total 3+3+3=9. Hence, in this case, above and below are equal. Wait, but in this case, the sum 4 is even? Wait, but in the previous example where the numbers go up to 3, the sum for the line was 4. Wait, maybe when the maximum sum is even or odd, the symmetry changes.But in the original problem, the maximum sum is 200, which is even. The line a + b =100 is in the middle. Wait, but in the 100x100 grid, the line a + b =100 would divide the grid. The number of points above the line should equal the number of points below the line, but since the total number of points is 10,000, and the number of points on the line is 99, then (10,000 - 99)/2 = (9901)/2 = 4950.5. But that's not an integer. Therefore, this suggests that the number of points above and below can't be equal. Therefore, our previous calculation must be correct, giving 5050 points above and 4951 points below, or vice versa. Wait, but according to our previous calculations:Total pairs:10,000Pairs with sum >100:5050Pairs with sum <=100:4950But 5050 + 4950 =10,000, but according to the line of symmetry, the numbers should be different. Wait, but perhaps the line a + b =100.5 would be the actual symmetry line, but since we are dealing with integers, it's not exact.Wait, maybe the confusion comes from the fact that the line a + b =100.5 would divide the plane, but in integer grids, the counts aren't symmetric. However, according to our two methods, both give 5050 for pairs with sum >100 and 4950 for pairs with sum <=100. Since both methods agree, and they have been verified with small examples, perhaps that's the correct answer despite the symmetry intuition.Therefore, I think the answer is 5050.Another way to think about it is for each a from 1 to 100, the number of b's that satisfy b >100 - a. Since b must be at least 1 and at most 100, the number of valid b's is 100 - (100 - a) = a. Wait, but that's only if 100 - a is less than 100. Wait, no:Wait, for a given a, b must be >100 - a.So the minimum b is max(1, 101 - a). Wait, no, if 100 - a is less than 1, then b must be greater than 100 - a, which is already satisfied by all b >=1. Wait, let's think again.If a + b >100, then b >100 - a.But since b is an integer >=1, the number of b's is:If 100 - a <1, i.e., a >=100, then b can be any from 1 to100, so 100 options.If 100 - a >=1, i.e., a <=99, then b must be >=101 - a, and <=100. So the number of b's is 100 - (101 - a) +1 = a.Wait, let's verify:For a=1:b >99, so b=100. 1 option. Which is a=1, so 1.For a=50:b >50, so b=51 to100. That's 50 numbers. Which is a=50.For a=100:b >0, so b=1 to100. That's 100 numbers. Which is a=100.Therefore, indeed, for each a, the number of b's is a. Therefore, summing over a=1 to100 of a gives the total number of pairs. Sum from1 to100 is 100*101/2=5050. So that's consistent.Therefore, regardless of the approach, the answer is 5050.Just to make sure, let's take a=99:b >100 -99=1, so b>=2. Therefore, b=2 to100, which is99 numbers. Wait, but according to the formula, a=99 should give 99. But 100 - (101 -99) +1=100 -2 +1=99. Correct.Wait, but according to the formula, number of b's is a. So for a=99, it's99. But when a=99, b needs to be >=2. So from2 to100, which is99 numbers. Yes, that's correct.Therefore, the formula holds.Thus, the answer is 5050.

✅:To determine the number of ordered pairs (a, b) where both a and b are integers from 1 to 100 and their sum (a + b > 100), we can use the following systematic approach:1. Total Number of Pairs: Since repetitions are allowed and order matters, there are (100 times 100 = 10,000) possible pairs.2. Complementary Counting: Instead of directly counting pairs where (a + b > 100), we first count pairs where (a + b leq 100) and subtract this from the total.3. Counting Pairs with (a + b leq 100): For each sum (s) from 2 to 100, the number of valid ordered pairs ((a, b)) is (s - 1) (e.g., for (s = 100), there are 99 pairs: ((1, 99), (2, 98), ldots, (99, 1))). The total number of such pairs is the sum of the first 99 natural numbers: [ sum_{s=2}^{100} (s - 1) = sum_{k=1}^{99} k = frac{99 times 100}{2} = 4950. ]4. Final Calculation: Subtract the number of pairs with (a + b leq 100) from the total pairs: [ 10,000 - 4950 = 5050. ]Verification via Direct Counting: For each fixed (a) (from 1 to 100), the smallest (b) that satisfies (a + b > 100) is (b = 101 - a). The number of valid (b) values for each (a) is (a) (e.g., (a = 50) requires (b geq 51), yielding 50 choices). Summing these values gives: [sum_{a=1}^{100} a = frac{100 times 101}{2} = 5050.]Conclusion: The number of ordered pairs ((a, b)) where (a + b > 100) is (boxed{5050}).

🔑:To find the number of pairs of integers ((a, b)) from 1 to 100 such that their sum is greater than 100, we need to consider various possible pairs and count them systematically.1. Identify the Pairs: We will identify the pairs ((a, b)) such that (a + b > 100).2. Step by Step Counting: Let's start from (a = 1). When (a = 1): - We require (1 + b > 100 implies b > 99). However, (b) can go up to 100, so (b = 100). - Then, the pair is ((1,100)). There is only one such pair. When (a = 2): - We require (2 + b > 100 implies b > 98). Thus, (b) can be 99 or 100. - Then, the pairs are ((2,99)) and ((2,100)). There are two such pairs. When (a = 3): - We require (3 + b > 100 implies b > 97). Thus, (b) can be 98, 99, or 100. - Then, the pairs are ((3,98)), ((3,99)), and ((3,100)). There are three such pairs. This pattern continues until (a = 50): - We require (50 + b > 100 implies b > 50). Thus, (b) can be 51, 52, ..., 100. - Then, the pairs are ((50, 51), (50, 52), ldots, (50, 100)). There are (50) such pairs (because the number of integers from 51 to 100 inclusive is (100 - 51 + 1 = 50)). Similarly, when (a = 51): - We require (51 + b > 100 implies b > 49). Thus, (b) can be 50, 51, ..., 100. - Then, the pairs are ((51, 50), (51, 51), ldots, (51, 100)). There are (51) such pairs. And continue this pattern until (a = 99) and (a = 100) where the pair pattern becomes: - For (a = 99), (99 + b > 100 implies b > 1). There are (99) such pairs. - For (a = 100), (100 + b > 100 implies b > 0). There are (100) such pairs.3. Sum the Number of Pairs: Now, the number of pairs form an arithmetic series: - Pairs when (a = 1): 1 - Pairs when (a = 2): 2 - Continue similarly up to - Pairs when (a = 100): 100 The total number of pairs is the sum of the first 100 positive integers: [ text{Total pairs} = 1 + 2 + 3 + cdots + 100 = frac{100 cdot 101}{2} = 5050 ] Conclusion:Thus, the number of ways to pick pairs of integers from 1 to 100 such that their sum is greater than 100 is:(boxed{5050})

🧠:Okay, let's try to solve this geometry problem step by step. First, I need to understand the problem thoroughly. The problem involves triangle ABC with an incircle ω that touches the sides AB, BC, and CA at points C₁, A₁, and B₁ respectively. Then, we are to reflect A₁ over the midpoint of BC to get A₂, and similarly define B₂ and C₂. Then, A₃ is the intersection point of line AA₂ with the incircle ω that's closer to A, and similarly for B₃ and C₃. The goal is to find the ratio of the areas [A₃B₃C₃]/[ABC], given the side lengths AB = 9, BC = 10, and CA = 13.First, let me recall some properties and formulas related to triangles, incircles, and areas. The inradius of a triangle can be calculated using the formula r = A/s, where A is the area of the triangle and s is the semiperimeter. The semiperimeter s is (a + b + c)/2. Given the sides AB = 9, BC = 10, CA = 13, the semiperimeter s = (9 + 10 + 13)/2 = 32/2 = 16. Then, the area can be found using Heron's formula: √[s(s - a)(s - b)(s - c)]. Plugging the values: √[16(16 - 9)(16 - 10)(16 - 13)] = √[16*7*6*3] = √[2016]. Simplifying that, 2016 = 16*126 = 16*9*14, so √(16*9*14) = 4*3*√14 = 12√14. Therefore, the area [ABC] is 12√14, and the inradius r = 12√14 / 16 = (3√14)/4.Next, the points where the incircle touches the sides are called the touch points. The lengths from the vertices to the touch points can be calculated using the formula: if the incircle touches BC at A₁, then the length BA₁ = (AB + BC - AC)/2. Wait, let me confirm that. The formula for the touch point on BC is s - AB, right? Wait, actually, the touch point on BC is called A₁, and the lengths from B to A₁ and from C to A₁ can be calculated as s - AC and s - AB, respectively. Let me verify this.Yes, in a triangle, the length from vertex B to the touch point on BC (A₁) is equal to (AB + BC - AC)/2. Similarly, the length from C to A₁ is (BC + AC - AB)/2. Let's compute that. For BA₁: (AB + BC - AC)/2 = (9 + 10 - 13)/2 = (6)/2 = 3. Similarly, CA₁ = (BC + AC - AB)/2 = (10 + 13 - 9)/2 = 14/2 = 7. So, on BC, the touch point A₁ is 3 units from B and 7 units from C. Similarly, we can compute the touch points on the other sides.For example, the touch point B₁ on AC: the length from A to B₁ is (AB + AC - BC)/2 = (9 + 13 - 10)/2 = 12/2 = 6. So AB₁ = 6 and CB₁ = AC - AB₁ = 13 - 6 = 7. Similarly, the touch point C₁ on AB: the length from A to C₁ is (AB + AC - BC)/2 = (9 + 13 - 10)/2 = 12/2 = 6. Wait, that can't be. Wait, the formula for the touch point on AB: from A, it's (AB + AC - BC)/2, which would be (9 + 13 - 10)/2 = 12/2 = 6, so AC₁ = 6 and BC₁ = AB - AC₁ = 9 - 6 = 3. So the touch points are:- A₁ on BC: BA₁ = 3, CA₁ = 7- B₁ on AC: AB₁ = 6, CB₁ = 7- C₁ on AB: AC₁ = 6, BC₁ = 3Alright, so now we need to reflect A₁ over the midpoint of BC to get A₂. Similarly for B₂ and C₂. Let's first find the coordinates of these points. Maybe setting up coordinate axes would help here. Let me consider placing triangle ABC in a coordinate system to make calculations easier.Let me place point B at the origin (0, 0), point C at (10, 0), since BC = 10. Then, we need to find the coordinates of point A, given that AB = 9 and AC = 13. Let me denote the coordinates of A as (x, y). Then, the distance from A to B is 9, so:√(x² + y²) = 9 → x² + y² = 81.The distance from A to C is 13, so:√((x - 10)² + y²) = 13 → (x - 10)² + y² = 169.Subtracting the first equation from the second:(x - 10)² + y² - (x² + y²) = 169 - 81 → x² - 20x + 100 + y² - x² - y² = 88 → -20x + 100 = 88 → -20x = -12 → x = 12/20 = 3/5 = 0.6.Then, plugging back x = 0.6 into x² + y² = 81:(0.6)² + y² = 81 → 0.36 + y² = 81 → y² = 80.64 → y = √80.64. Let me compute that: 80.64 = 80 + 0.64. √80 ≈ 8.944, √0.64 = 0.8. But this is not linear. Let me compute √80.64 exactly. Note that 80.64 = 8064/100 = 504/6.25. Wait, maybe square of 8.98: 8.98² = 80.6404. So approximately y ≈ 8.98. Let's keep it as exact value for now.Wait, instead of approximating, maybe we can write it as a fraction. Let's compute y² = 81 - (3/5)² = 81 - 9/25 = (2025 - 9)/25 = 2016/25. Therefore, y = √(2016)/5. But 2016 factors as 16*126 = 16*9*14, so √2016 = 12√14. Therefore, y = (12√14)/5. So coordinates of A are (3/5, (12√14)/5).So to recap, coordinates:- B: (0, 0)- C: (10, 0)- A: (3/5, 12√14/5)Now, let's find the coordinates of A₁, the touch point on BC. Earlier, we found that BA₁ = 3, so since BC is from (0,0) to (10,0), moving 3 units from B gives A₁ at (3, 0).Similarly, the touch points:- A₁: (3, 0)- B₁: On AC, 6 units from A and 7 units from C. Let's compute coordinates of B₁. Since AC is from A (3/5, 12√14/5) to C (10, 0). The length of AC is 13, so moving 6 units from A towards C.Parametric equations for AC: starting at A, moving towards C. The vector from A to C is (10 - 3/5, 0 - 12√14/5) = (47/5, -12√14/5). The unit vector in that direction is (47/5)/13, (-12√14/5)/13) = (47/65, -12√14/65). Therefore, moving 6 units along this direction from A:B₁ = A + 6*(47/65, -12√14/65) = (3/5 + 6*(47/65), 12√14/5 + 6*(-12√14/65)).Compute the x-coordinate: 3/5 = 39/65, 6*(47/65) = 282/65. So 39/65 + 282/65 = 321/65 ≈ 4.938.The y-coordinate: 12√14/5 = 156√14/65, 6*(-12√14)/65 = -72√14/65. So 156√14/65 - 72√14/65 = 84√14/65. Therefore, B₁ is (321/65, 84√14/65).Similarly, touch point C₁ on AB, 6 units from A and 3 units from B. AB is from A (3/5, 12√14/5) to B (0,0). The vector from A to B is (-3/5, -12√14/5). The length of AB is 9, so moving 6 units from A towards B gives C₁.Parametric equations: unit vector from A to B is (-3/5)/9, (-12√14/5)/9) = (-1/15, -4√14/15). Moving 6 units: displacement is 6*(-1/15, -4√14/15) = (-6/15, -24√14/15) = (-2/5, -8√14/5). Therefore, C₁ = A + displacement = (3/5 - 2/5, 12√14/5 - 8√14/5) = (1/5, 4√14/5).So coordinates of touch points:- A₁: (3, 0)- B₁: (321/65, 84√14/65)- C₁: (1/5, 4√14/5)Now, the next step is to find A₂, the reflection of A₁ over the midpoint of BC. The midpoint of BC is at (5, 0). Reflecting A₁ (3, 0) over (5, 0). The reflection over a midpoint (x_m, y_m) of a point (x, y) is (2x_m - x, 2y_m - y). Therefore, reflection of A₁ over (5,0) is (2*5 - 3, 2*0 - 0) = (10 - 3, 0) = (7, 0). So A₂ is (7, 0). Similarly, we can find B₂ and C₂.Wait, B₂ is the reflection of B₁ over the midpoint of AC. The midpoint of AC is ((3/5 + 10)/2, (12√14/5 + 0)/2) = ((3/5 + 50/5)/2, 6√14/5) = (53/10, 6√14/5). So reflecting B₁ (321/65, 84√14/65) over this midpoint.Reflection formula: (2x_m - x, 2y_m - y). So x-coordinate: 2*(53/10) - 321/65. Compute 2*(53/10) = 106/10 = 53/5 = 689/65. Then 689/65 - 321/65 = (689 - 321)/65 = 368/65. Similarly, y-coordinate: 2*(6√14/5) - 84√14/65 = 12√14/5 - 84√14/65 = (156√14 - 84√14)/65 = 72√14/65. So B₂ is (368/65, 72√14/65).Similarly, C₂ is the reflection of C₁ over the midpoint of AB. The midpoint of AB is ((3/5 + 0)/2, (12√14/5 + 0)/2) = (3/10, 6√14/5). Reflecting C₁ (1/5, 4√14/5) over this midpoint. Using reflection formula:x-coordinate: 2*(3/10) - 1/5 = 6/10 - 2/10 = 4/10 = 2/5.y-coordinate: 2*(6√14/5) - 4√14/5 = 12√14/5 - 4√14/5 = 8√14/5. Therefore, C₂ is (2/5, 8√14/5).So now we have:- A₂: (7, 0)- B₂: (368/65, 72√14/65)- C₂: (2/5, 8√14/5)Next, we need to find A₃, which is the intersection of line AA₂ with the incircle ω, closer to A. Similarly for B₃ and C₃. Let's first find the equation of line AA₂.Point A is (3/5, 12√14/5), and point A₂ is (7, 0). Let's parametrize this line. Let's write parametric equations for AA₂.Vector from A to A₂: (7 - 3/5, 0 - 12√14/5) = (32/5, -12√14/5). So parametric equations:x = 3/5 + (32/5)ty = 12√14/5 - (12√14/5)twhere t ∈ [0,1] gives the segment from A to A₂. We need to find the intersection of this line with the incircle ω. The incircle ω has center at the incenter of triangle ABC. Let me find the coordinates of the incenter.The incenter coordinates can be found using the formula:I = (aA + bB + cC)/(a + b + c), where a, b, c are the lengths of sides opposite to A, B, C. Wait, actually, the formula is weighted by the lengths of the sides. Wait, in barycentric coordinates, the incenter is at (a : b : c), but in Cartesian coordinates, the coordinates are ( (aAx + bBx + cCx)/(a + b + c), (aAy + bBy + cCy)/(a + b + c) ). Wait, actually, no: the incenter coordinates are given by ( (a x_A + b x_B + c x_C ) / (a + b + c), (a y_A + b y_B + c y_C ) / (a + b + c) ), where a, b, c are the lengths of the sides opposite to angles A, B, C. Wait, but in standard notation, a is BC, b is AC, c is AB. So in our case:Given triangle ABC with sides:- BC = a = 10- AC = b = 13- AB = c = 9Therefore, incenter coordinates are:( (a x_A + b x_B + c x_C ) / (a + b + c), (a y_A + b y_B + c y_C ) / (a + b + c) )Wait, but x_A, y_A are the coordinates of A, etc. Wait, but in our coordinate system:- A is (3/5, 12√14/5)- B is (0, 0)- C is (10, 0)But in the formula, the incenter coordinates are calculated as:( (a x_A + b x_B + c x_C ) / (a + b + c), (a y_A + b y_B + c y_C ) / (a + b + c) )Wait, but actually, the formula is:If the sides opposite to A, B, C are a, b, c respectively, then the incenter is ( (a x_A + b x_B + c x_C)/(a + b + c), similarly for y).Wait, no, actually, the correct formula is that the incenter coordinates are ( (a x_A + b x_B + c x_C ) / (a + b + c ), (a y_A + b y_B + c y_C ) / (a + b + c ) ), where a, b, c are the lengths of the sides opposite to A, B, C. In our case, since in standard notation:- Side a is BC = 10, opposite to A- Side b is AC = 13, opposite to B- Side c is AB = 9, opposite to CBut in the formula, we need to use the lengths opposite to each vertex. Wait, actually, the formula is weighted by the lengths of the sides adjacent to each vertex. Wait, maybe I confused the notation. Let me check.Wait, the incenter coordinates can be calculated as:I_x = (a x_A + b x_B + c x_C) / (a + b + c)I_y = (a y_A + b y_B + c y_C) / (a + b + c)But here, a, b, c are the lengths of the sides opposite to angles A, B, C. So in standard notation:- a = BC = 10- b = AC = 13- c = AB = 9Therefore, the incenter coordinates are:I_x = (a x_A + b x_B + c x_C)/(a + b + c) = (10*(3/5) + 13*0 + 9*10)/(10 + 13 + 9) = (6 + 0 + 90)/32 = 96/32 = 3I_y = (a y_A + b y_B + c y_C)/(a + b + c) = (10*(12√14/5) + 13*0 + 9*0)/32 = (24√14 + 0 + 0)/32 = 24√14/32 = 3√14/4Therefore, the incenter I is at (3, 3√14/4), and the inradius r is 3√14/4, which matches our earlier calculation.So the incircle ω has center (3, 3√14/4) and radius 3√14/4. Now, we need to find the intersection points of line AA₂ with ω, and take the one closer to A.We have parametric equations for line AA₂:x = 3/5 + (32/5)ty = 12√14/5 - (12√14/5)tWe can substitute these into the equation of the incircle and solve for t. The equation of the incircle is:(x - 3)^2 + (y - 3√14/4)^2 = (3√14/4)^2Plugging in x and y:[ (3/5 + 32t/5 - 3)^2 + (12√14/5 - 12√14 t/5 - 3√14/4)^2 ] = (9*14)/16Simplify each term step by step.First term: (3/5 + 32t/5 - 3) = (3 + 32t - 15)/5 = (32t - 12)/5. So squared: (32t - 12)^2 / 25.Second term: Let's factor out √14:12√14/5 - 12√14 t/5 - 3√14/4 = √14*(12/5 - 12t/5 - 3/4)Compute inside the brackets:12/5 - 12t/5 - 3/4 = (12/5 - 3/4) - 12t/5 = (48/20 - 15/20) - 12t/5 = (33/20) - (12t/5) = (33 - 48t)/20Therefore, the second term squared: [√14*(33 - 48t)/20]^2 = 14*(33 - 48t)^2 / 400Therefore, putting back into the circle equation:[(32t - 12)^2 / 25] + [14*(33 - 48t)^2 / 400] = (9*14)/16Multiply both sides by 400 to eliminate denominators:16*(32t - 12)^2 + 14*(33 - 48t)^2 = 400*(9*14)/16Compute right-hand side: 400*(126)/16 = (400/16)*126 = 25*126 = 3150.Left-hand side: 16*(32t - 12)^2 + 14*(33 - 48t)^2Let me compute each term separately.First term: 16*(32t - 12)^2Let u = 32t - 12. Then, u² = (32t - 12)^2 = 1024t² - 768t + 144. Multiply by 16: 16384t² - 12288t + 2304.Second term: 14*(33 - 48t)^2Let v = 33 - 48t. Then, v² = (48t - 33)^2 = 2304t² - 3168t + 1089. Multiply by 14: 32256t² - 44352t + 15246.So total left-hand side: 16384t² -12288t + 2304 + 32256t² -44352t +15246 = (16384 + 32256)t² + (-12288 -44352)t + (2304 +15246)Calculate coefficients:t² term: 16384 + 32256 = 48640t term: -12288 -44352 = -56640constant term: 2304 +15246 = 17550Therefore, equation is:48640t² -56640t +17550 = 3150Subtract 3150:48640t² -56640t +14400 = 0Divide both sides by 160 to simplify:48640/160 = 304, 56640/160 = 354, 14400/160 = 90So equation becomes:304t² - 354t + 90 = 0Divide by 2:152t² - 177t + 45 = 0Let's solve this quadratic equation for t. Using quadratic formula:t = [177 ± √(177² - 4*152*45)] / (2*152)First, compute discriminant:177² = 313294*152*45 = 4*6840 = 27360Discriminant: 31329 - 27360 = 3969√3969 = 63Therefore, t = [177 ±63]/304Compute both solutions:t₁ = (177 + 63)/304 = 240/304 = 15/19 ≈ 0.7895t₂ = (177 - 63)/304 = 114/304 = 57/152 ≈ 0.375Now, we need to determine which of these corresponds to the intersection closer to A. Since the parameter t=0 is point A, and t=1 is point A₂, the smaller t value would be closer to A. Wait, but wait: when t increases from 0 to 1, we move from A to A₂. So the intersection points occur at t = 57/152 ≈ 0.375 and t = 15/19 ≈ 0.7895. The one closer to A is the one with smaller t, which is t = 57/152. Therefore, A₃ is the point corresponding to t = 57/152.Wait, but hold on: the parametric equations were defined from A to A₂, so t=0 is A, t=1 is A₂. Therefore, the intersection closer to A is at t = 57/152 ≈ 0.375, and the other intersection is further away, closer to A₂. Therefore, A₃ is at t = 57/152.Therefore, coordinates of A₃ are:x = 3/5 + (32/5)*(57/152) = 3/5 + (32*57)/(5*152)Simplify denominator 152 = 8*19, numerator 32*57 = 32*(57). Let's compute 32*57: 30*57=1710, 2*57=114, total=1710+114=1824. So x = 3/5 + 1824/(5*152) = 3/5 + 1824/760. Simplify 1824/760: divide numerator and denominator by 8: 228/95. Then 228/95 = 2.4. Wait, 95*2 = 190, 228-190=38, so 228/95 = 2 + 38/95 = 2 + 38/95 ≈ 2.4. Wait, but let's compute exactly:1824 ÷ 152 = 12 (since 152*12 = 1824). Therefore, 1824/(5*152) = 12/5. Therefore, x = 3/5 + 12/5 = 15/5 = 3. Wait, that seems too straightforward. Let me verify:Wait, 32/5 * 57/152 = (32*57)/(5*152). Let's factor 152 as 8*19. 32/152 = 32/(8*19) = 4/19. So 32*57/(5*152) = (4/19)*57/5 = (4*57)/(19*5) = (4*3)/5 = 12/5. Therefore, x = 3/5 + 12/5 = 15/5 = 3. Similarly, y-coordinate:y = 12√14/5 - (12√14/5)*(57/152) = 12√14/5*(1 - 57/152) = 12√14/5*(95/152) = (12√14 *95)/(5*152)Simplify 95/152 = 5/8 (since 95 = 5*19, 152=8*19, so divide numerator and denominator by 19: 5/8). Therefore, y = (12√14 *5)/(5*8) = (12√14)/8 = (3√14)/2.Therefore, A₃ is at (3, 3√14/2). Wait, but that seems suspicious. The inradius is 3√14/4, so the center is at (3, 3√14/4). The point A₃ is at (3, 3√14/2), which is vertically above the center. The distance from the center to A₃ is (3√14/2 - 3√14/4) = 3√14/4, which is equal to the radius. Therefore, this is correct.Wait, but let me confirm the calculation again. The parametric y-coordinate:12√14/5 - (12√14/5)*(57/152) = 12√14/5 [1 - 57/152] = 12√14/5 * (95/152) = 12√14/5 * (5/8) [since 95/152 = 5/8] = (12√14 *5)/(5*8) = 12√14/8 = 3√14/2. Yes, that's correct.So A₃ is at (3, 3√14/2). Interesting, so the x-coordinate is 3, which is the same as the x-coordinate of the inradius center. Therefore, A₃ lies directly above the incenter on the vertical line x=3, at a distance of 3√14/2 - 3√14/4 = 3√14/4, which is exactly the radius. So this makes sense.Similarly, we need to find B₃ and C₃. Let's attempt to do this for B₃.First, B₃ is the intersection of line BB₂ with ω closer to B. Let's find the equation of line BB₂. Point B is (0, 0), and point B₂ is (368/65, 72√14/65). Let's parametrize this line.Parametric equations:x = 0 + (368/65)ty = 0 + (72√14/65)twhere t ∈ [0,1] gives the segment from B to B₂.We need to find the intersection of this line with the incircle ω. The incircle has center (3, 3√14/4) and radius 3√14/4. So the equation is (x -3)^2 + (y - 3√14/4)^2 = (3√14/4)^2.Substitute x = 368t/65, y = 72√14 t /65 into the equation:(368t/65 - 3)^2 + (72√14 t /65 - 3√14/4)^2 = (9*14)/16Simplify each term.First term: 368t/65 - 3 = (368t - 195)/65. Squared: (368t - 195)^2 / (65^2).Second term: factor out √14:√14*(72t/65 - 3/4) = √14*(288t/260 - 195/260) = √14*(288t - 195)/260. Therefore, squared: 14*(288t - 195)^2 / (260^2).Therefore, the equation becomes:[(368t - 195)^2 + 14*(288t - 195)^2] / (65^2) = 126/16Multiply both sides by 65^2:(368t - 195)^2 + 14*(288t - 195)^2 = (126/16)*4225Compute the right-hand side: (126/16)*4225 = (126*4225)/16. Let's compute 126*4225:First, note that 4225 = 65^2. 126*65 = 8190. Then, 8190*65 = 532,350. Therefore, 126*4225 = 532,350. Then, divide by 16: 532350 /16 = 33,271.875.So equation:(368t - 195)^2 +14*(288t -195)^2 = 33,271.875This seems complicated. Let me see if there's a better way. Alternatively, perhaps we can compute the quadratic equation.Alternatively, substitute t as parameter and expand.First, expand (368t -195)^2:= (368t)^2 - 2*368*195 t + 195^2= 135,424t² - 143,520t + 38,025Then expand 14*(288t -195)^2:First compute (288t -195)^2 = 82,944t² - 112,320t + 38,025. Multiply by 14:1,161,216t² - 1,572,480t + 532,350.Therefore, total left-hand side:135,424t² -143,520t +38,025 +1,161,216t² -1,572,480t +532,350 =(135,424 +1,161,216)t² + (-143,520 -1,572,480)t + (38,025 +532,350)Compute coefficients:t²: 1,296,640t: -1,716,000constant: 570,375So equation:1,296,640t² -1,716,000t +570,375 = 33,271.875Subtract 33,271.875:1,296,640t² -1,716,000t +537,103.125 = 0This is a bit unwieldy. Maybe we can divide by some factor. Let me check if all coefficients are divisible by, say, 12.5? Let's see:1,296,640 ÷ 12.5 = 103,731.2, which is not integer. Maybe there's an error in my calculations. Let me double-check the expansions.Wait, perhaps there is a miscalculation in expanding (288t -195)^2. Let's verify:(288t -195)^2 = (288t)^2 - 2*288t*195 + 195^2 = 82,944t² - 112,320t + 38,025. Multiply by 14:82,944*14 = 1,161,216t²-112,320*14 = -1,572,480t38,025*14 = 532,350. That's correct.Then, adding (368t -195)^2: 135,424t² -143,520t +38,025. Adding to the previous gives:135,424 +1,161,216 = 1,296,640t²-143,520 -1,572,480 = -1,716,000t38,025 +532,350 = 570,375. Correct.Then, subtract 33,271.875: 570,375 -33,271.875 = 537,103.125. So equation is correct.This seems messy. Perhaps instead of expanding, we can use parametric substitution.Alternatively, perhaps notice that the line BB₂ passes through the incenter or has some symmetry. Alternatively, maybe use the parametric line and solve for t.Alternatively, since the inradius is 3√14/4, and center at (3, 3√14/4), the equation is (x - 3)^2 + (y - 3√14/4)^2 = (3√14/4)^2.Substitute x = 368t/65, y = 72√14 t/65 into the equation:(368t/65 - 3)^2 + (72√14 t/65 - 3√14/4)^2 = (9*14)/16Let me compute each term separately.First term: (368t/65 - 3) = (368t - 195)/65. Squared: (368t -195)^2 / 4225.Second term: (72√14 t/65 - 3√14/4) = √14*(72t/65 - 3/4) = √14*(288t - 195)/260. Squared: 14*(288t -195)^2 / 67600.Therefore, the equation is:[(368t -195)^2)/4225 + (14*(288t -195)^2)/67600] = 126/16Multiply both sides by 67600 (which is 4225*16) to eliminate denominators:16*(368t -195)^2 +14*(288t -195)^2 = 67600*(126)/16Compute RHS: 67600*126/16. Let's compute 67600/16 = 4225. Then, 4225*126 = 4225*(100 + 20 +6) = 422500 + 84500 +25350 = 422500 + 84500 = 507,000 +25,350 = 532,350.Therefore, the equation is:16*(368t -195)^2 +14*(288t -195)^2 = 532,350But this is the same equation as before. So we have to solve this quadratic equation. Given the complexity, maybe use substitution.Let me denote u = t. Let’s denote:Let’s compute 16*(368u -195)^2 +14*(288u -195)^2 = 532,350.Let me expand the terms:First term: 16*(368u -195)^2 =16*(135,424u² -143,520u +38,025) = 2,166,784u² -2,296,320u +608,400.Second term:14*(288u -195)^2 =14*(82,944u² -112,320u +38,025) =1,161,216u² -1,572,480u +532,350.Adding both terms:2,166,784u² +1,161,216u² = 3,328,000u²-2,296,320u -1,572,480u = -3,868,800u608,400 +532,350 = 1,140,750Therefore, equation:3,328,000u² -3,868,800u +1,140,750 = 532,350Subtract 532,350:3,328,000u² -3,868,800u +608,400 = 0Divide by 800:3,328,000 /800 = 4,160-3,868,800 /800 = -4,836608,400 /800 = 760.5So equation: 4,160u² -4,836u +760.5 = 0Multiply by 2 to eliminate the decimal:8,320u² -9,672u +1,521 = 0Now, let's apply the quadratic formula:u = [9,672 ±√(9,672² -4*8,320*1,521)]/(2*8,320)First compute discriminant:9,672² = let's compute 9,672*9,672:First, 9,000² = 81,000,000Then, 672² = 451,584Then, cross term: 2*9,000*672 = 12,096,000So total: 81,000,000 +12,096,000 +451,584 =93,547,584Now, 4*8,320*1,521 =4*8,320=33,280; 33,280*1,521.Compute 33,280*1,500=49,920,000; 33,280*21=698, 880. So total 49,920,000 +698,880 =50,618,880Therefore, discriminant D =93,547,584 -50,618,880 =42,928,704√D = √42,928,704. Let's check: 6,552² = (6,500 +52)^2 =6,500² + 2*6,500*52 +52²=42,250,000 + 676,000 +2,704=42,928,704. Yes! Therefore, √D =6,552.Therefore, u = [9,672 ±6,552]/16,640Compute both solutions:u₁ = (9,672 +6,552)/16,640 =16,224/16,640 = 16224 ÷ 16,640. Divide numerator and denominator by 16: 1014/1040 = 507/520 = 0.975u₂ = (9,672 -6,552)/16,640 =3,120/16,640 = 3120 ÷16,640 = 312/1664 = 39/208 ≈0.1875Therefore, the two intersection points are at t ≈0.975 and t≈0.1875. The one closer to B is t≈0.1875, so B₃ is at t=39/208.Therefore, coordinates of B₃ are:x = 368/65 *39/208, y =72√14/65 *39/208Simplify:First, simplify 368/65 *39/208:368 = 16*23, 39=3*13, 208=16*13, 65=5*13.Therefore, (16*23)/ (5*13) * (3*13)/(16*13) )= (23/5*13) * (3*13)/(16*13) Wait, better to cancel factors:368/65 *39/208 = (368*39)/(65*208)Factor numerator and denominator:368 = 16*2339 = 3*1365 =5*13208=16*13Therefore,(16*23 *3*13)/(5*13 *16*13) ) = (23*3)/(5*13) = 69/65 ≈1.0615Wait, but let me verify:368*39 = (16*23)*(3*13) =16*3*23*1365*208 = (5*13)*(16*13) =5*16*13*13Therefore, (16*3*23*13)/(5*16*13*13) )= (3*23)/(5*13) = 69/65 ≈1.0615Similarly, y-coordinate:72√14/65 *39/208 = (72*39)/(65*208) *√14Same denominator and numerator:72=8*9, 39=3*13, 65=5*13, 208=16*13Therefore,(8*9 *3*13)/(5*13 *16*13) ) = (8*9*3)/(5*16*13) )= (216)/(1040) =54/260 =27/130 ≈0.2077Therefore, y =27√14/130Therefore, coordinates of B₃ are (69/65, 27√14/130). Let's simplify:69/65 = 1 + 4/65 = approx 1.061527/130 = approx 0.2077So B₃ is at (69/65, 27√14/130).Similarly, we need to find C₃ by reflecting C₁ over midpoint of AB and finding intersection. However, this process seems very tedious. Given the complexity of these coordinates, maybe there's a better approach.Alternatively, perhaps there is symmetry or homothety involved. Given that the problem is asking for the ratio of areas, perhaps the triangle A₃B₃C₃ is similar to ABC or has a specific ratio.Alternatively, given that all three points A₃, B₃, C₃ lie on the incircle, maybe triangle A₃B₃C₃ is the incentral triangle or something similar, and the area ratio can be found using properties of the incircle.Alternatively, since each of the points A₃, B₃, C₃ are intersections of lines from the vertices through the reflected touch points, maybe there's a homothety or inversion that maps ABC to A₃B₃C₃.But since the coordinates are very specific and messy, perhaps another approach is needed. Alternatively, use barycentric coordinates with respect to triangle ABC.But barycentric coordinates might also be complicated. Alternatively, note that A₃ is the point where AA₂ meets the incircle closer to A. Since A₂ is the reflection of A₁ over the midpoint of BC, which is point (5,0) in our coordinate system. So A₂ is (7,0). The line AA₂ goes from A (3/5, 12√14/5) to (7,0). We found A₃ at (3, 3√14/2).Similarly, B₃ is at (69/65, 27√14/130) and C₃ would be found similarly. Then, once we have coordinates of all three points, we can compute the area [A₃B₃C₃] using determinant formula and compare it to [ABC] =12√14.But computing coordinates for C₃ would be similarly tedious. Let's attempt it.C₃ is the intersection of line CC₂ with the incircle closer to C. Point C is (10,0). Point C₂ is (2/5, 8√14/5). Let's parametrize line CC₂.Parametric equations:x = 10 + (2/5 -10)t =10 + (-48/5)ty =0 + (8√14/5 -0)t =8√14 t/5Where t ∈[0,1] gives the segment from C to C₂.We need to find the intersection of this line with the incircle ω: (x -3)^2 + (y -3√14/4)^2 = (3√14/4)^2.Substitute x =10 -48t/5, y=8√14 t/5 into the equation.Compute (10 -48t/5 -3)^2 + (8√14 t/5 -3√14/4)^2 = (9*14)/16Simplify:First term: (7 -48t/5) = (35 -48t)/5. Squared: (35 -48t)^2 /25.Second term: factor out √14:√14*(8t/5 -3/4) = √14*(32t/20 -15/20) = √14*(32t -15)/20. Squared:14*(32t -15)^2 /400.Therefore, equation becomes:[(35 -48t)^2 /25] + [14*(32t -15)^2 /400] = 126/16Multiply both sides by 400:16*(35 -48t)^2 +14*(32t -15)^2 = 400*(126)/16 = 25*126 =3150Compute each term:First term:16*(35 -48t)^2 =16*(1225 - 3360t +2304t²) =19,600 -53,760t +36,864t²Second term:14*(32t -15)^2 =14*(1024t² -960t +225)=14,336t² -13,440t +3,150Sum:36,864t² +14,336t² =51,200t²-53,760t -13,440t =-67,200t19,600 +3,150 =22,750Equation:51,200t² -67,200t +22,750 =3,150Subtract 3,150:51,200t² -67,200t +19,600 =0Divide by 800:51,200 /800 =64, 67,200 /800=84, 19,600 /800=24.5Equation:64t² -84t +24.5 =0Multiply by 2 to eliminate decimal:128t² -168t +49 =0Apply quadratic formula:t = [168 ±√(168² -4*128*49)] / (2*128)Compute discriminant:168² =28,2244*128*49 =4*6,272 =25,088Discriminant D=28,224 -25,088 =3,136√D=56Therefore, t=(168 ±56)/256Solutions:t₁=(168+56)/256=224/256=7/8=0.875t₂=(168-56)/256=112/256=7/16≈0.4375Since we need the intersection closer to C, which is at t=0. Therefore, the smaller t is closer to C. Wait, but the parametrization is from C (t=0) to C₂ (t=1). The smaller t corresponds to closer to C. Therefore, the intersection closer to C is t=7/16. Therefore, C₃ is at t=7/16.Compute coordinates:x =10 -48t/5 =10 -48*(7/16)/5 =10 - (48/16)*(7/5) =10 -3*(7/5) =10 -21/5= (50 -21)/5=29/5=5.8y=8√14 t/5=8√14*(7/16)/5= (56√14)/80=7√14/10.Therefore, C₃ is at (29/5, 7√14/10).Now, we have coordinates for A₃, B₃, C₃:- A₃: (3, 3√14/2)- B₃: (69/65, 27√14/130)- C₃: (29/5, 7√14/10)Now, we need to compute the area of triangle A₃B₃C₃. Using the shoelace formula:Area = (1/2)|x₁(y₂ - y₃) +x₂(y₃ - y₁) +x₃(y₁ - y₂)|Let’s denote:A₃: (x₁, y₁) = (3, 3√14/2)B₃: (x₂, y₂) = (69/65, 27√14/130)C₃: (x₃, y₃) = (29/5, 7√14/10)Plug into the formula:Area = (1/2)| 3*(27√14/130 -7√14/10) + (69/65)*(7√14/10 -3√14/2) + (29/5)*(3√14/2 -27√14/130) |Let's compute each term step by step.First term: 3*(27√14/130 -7√14/10)Convert to common denominator:7√14/10 =91√14/13027√14/130 -91√14/130 = -64√14/130 = -32√14/65Multiply by 3: 3*(-32√14/65) = -96√14/65Second term: (69/65)*(7√14/10 -3√14/2)Convert to common denominator:7√14/10 =7√14/103√14/2 =15√14/107√14/10 -15√14/10 = -8√14/10 = -4√14/5Multiply by (69/65): (69/65)*(-4√14/5) = -276√14/325Third term: (29/5)*(3√14/2 -27√14/130)Convert to common denominator:3√14/2 =195√14/130195√14/130 -27√14/130 =168√14/130 =84√14/65Multiply by (29/5): (29/5)*(84√14/65) = (29*84√14)/(325) = (2436√14)/325Now, sum all three terms:-96√14/65 -276√14/325 +2436√14/325Convert to common denominator 325:-96√14/65 = -480√14/325-276√14/325 +2436√14/325 =2160√14/325Therefore, total sum:-480√14/325 +2160√14/325 =1680√14/325 = 336√14/65Take absolute value and multiply by 1/2:Area = (1/2)*|336√14/65| =168√14/65Therefore, [A₃B₃C₃] =168√14/65The area of triangle ABC is12√14, so the ratio is:(168√14/65)/12√14 = (168/65)/12 =168/(65*12) =14/65 ≈0.21538Simplify 14/65: GCD of 14 and 65 is 1, so it's 14/65. But wait, 14 and 65: 14 divides by 7 and 2, 65 divides by 5 and 13. No common factors. So ratio is 14/65.But wait, let me check the calculations again, as 168 divided by 12 is 14, and 65 stays as 65. Yes, so the ratio is14/65.But let me verify the steps once more to ensure no arithmetic errors.First term computation:3*(27√14/130 -7√14/10):Convert 7/10 to 91/130: 7*13=91, 10*13=130.27/130 -91/130 = -64/130 = -32/65.Multiply by 3: -96/65√14. That's correct.Second term:(69/65)*(7√14/10 -3√14/2):7/10 -3/2 =7/10 -15/10= -8/10= -4/5. Multiply by 69/65: -4/5 *69/65= -276/325. Correct.Third term:(29/5)*(3√14/2 -27√14/130):Convert 3/2 to 195/130, so 195/130 -27/130=168/130=84/65. Multiply by 29/5: 84/65 *29/5=2436/325. Correct.Sum:-96/65√14 -276/325√14 +2436/325√14Convert -96/65 to -480/325:-480/325 -276/325 +2436/325 = ( -480 -276 +2436 )/325 = (2436 -756)/325 =1680/325=336/65. Then multiply by √14 and 1/2 gives168/65√14. Yes, correct.Therefore, ratio is (168√14/65)/(12√14) )= (168/65)/12=14/65.Thus, the final answer is 14/65.Final Answerboxed{dfrac{14}{65}}

✅:To solve the problem, we start by calculating the semiperimeter ( s ) and the area of triangle ( ABC ) using Heron's formula. Given ( AB = 9 ), ( BC = 10 ), and ( CA = 13 ):[s = frac{9 + 10 + 13}{2} = 16][[ABC] = sqrt{s(s - a)(s - b)(s - c)} = sqrt{16 times 7 times 6 times 3} = 12sqrt{14}]The inradius ( r ) is calculated as:[r = frac{[ABC]}{s} = frac{12sqrt{14}}{16} = frac{3sqrt{14}}{4}]Next, we determine the coordinates of the touch points ( A_1 ), ( B_1 ), and ( C_1 ) of the incircle with the sides of the triangle. Using the touch point formulas, we find:- ( A_1 ) on ( BC ): ( BA_1 = 3 ), ( CA_1 = 7 )- ( B_1 ) on ( AC ): ( AB_1 = 6 ), ( CB_1 = 7 )- ( C_1 ) on ( AB ): ( AC_1 = 6 ), ( BC_1 = 3 )We then reflect these touch points over the midpoints of their respective sides to find ( A_2 ), ( B_2 ), and ( C_2 ). The coordinates of these reflections are:- ( A_2 ) is the reflection of ( A_1 ) over the midpoint of ( BC ): ( (7, 0) )- ( B_2 ) is the reflection of ( B_1 ) over the midpoint of ( AC ): ( left( frac{368}{65}, frac{72sqrt{14}}{65} right) )- ( C_2 ) is the reflection of ( C_1 ) over the midpoint of ( AB ): ( left( frac{2}{5}, frac{8sqrt{14}}{5} right) )Next, we find the coordinates of ( A_3 ), ( B_3 ), and ( C_3 ) by intersecting lines ( AA_2 ), ( BB_2 ), and ( CC_2 ) with the incircle. Using parametric equations and solving for intersections, we find:- ( A_3 ) is at ( (3, frac{3sqrt{14}}{2}) )- ( B_3 ) is at ( left( frac{69}{65}, frac{27sqrt{14}}{130} right) )- ( C_3 ) is at ( left( frac{29}{5}, frac{7sqrt{14}}{10} right) )Finally, we calculate the area of triangle ( A_3B_3C_3 ) using the shoelace formula and find the ratio:[left[ A_3B_3C_3 right] = frac{168sqrt{14}}{65}]The ratio of the areas is:[frac{left[ A_3B_3C_3 right]}{left[ ABC right]} = frac{frac{168sqrt{14}}{65}}{12sqrt{14}} = frac{14}{65}]Thus, the final answer is:[boxed{dfrac{14}{65}}]

🔑:Let's solve the problem step-by-step.Step 1: Understanding the configurationGiven triangle ( ABC ) with side lengths ( AB = 9 ), ( BC = 10 ), and ( CA = 13 ), its incircle ( omega ) touches ( AB ), ( BC ), and ( CA ) at points ( C_1 ), ( A_1 ), and ( B_1 ) respectively. Points ( A_2 ), ( B_2 ), and ( C_2 ) are reflections of ( A_1 ), ( B_1 ), and ( C_1 ) over the midpoints of ( BC ), ( CA ), and ( AB ) respectively. Points ( A_3 ), ( B_3 ), and ( C_3 ) are intersection points of ( AA_2 ), ( BB_2 ), and ( CC_2 ) with ( omega ), closer to ( A ), ( B ), and ( C ) respectively.Step 2: Homothety and ReflectionNotice that ( A_2 ) is the point of tangency of the excircle opposite to ( A ) with ( BC ). By considering the homothety centered at ( A ) mapping the excircle to the incircle, ( A_3 ) is the intersection of the incircle ( omega ) with the tangent line parallel to ( BC ). Thus, ( A_3 ), ( B_3 ), and ( C_3 ) form a triangle congruent to ( A_1B_1C_1 ) reflected through the center of ( omega ).Step 3: Ratio of AreasWe need to find the ratio (frac{[A_3B_3C_3]}{[ABC]}). First, we calculate (frac{[A_1B_1C_1]}{[ABC]}).Step 4: Calculate areas using tangency pointsThe lengths from the tangency points to the vertices of ( ABC ) are required. Using the lengths of the sides of triangle ( ABC ):- The semiperimeter ( s ) of ( ABC ) is: [ s = frac{AB + BC + CA}{2} = frac{9 + 10 + 13}{2} = 16 ]- The lengths of the segments from vertices to the points of tangency are: [ s - a = 16 - 13 = 3; quad s - b = 16 - 9 = 7; quad s - c = 16 - 10 = 6 ]- By calculating the areas of the smaller triangles ( Delta A_1BC_1 ), ( Delta B_1CA_1 ), and ( Delta C_1AB_1 ): [ frac{[A_1 BC_1]}{[ABC]} = frac{(s - a)^2}{BC cdot AB} = frac{3^2}{10 cdot 9} = frac{9}{90} = frac{1}{10} ] [ frac{[B_1 CA_1]}{[ABC]} = frac{(s - b)^2}{CA cdot BC} = frac{7^2}{13 cdot 10} = frac{49}{130} ] [ frac{[C_1 AB_1]}{[ABC]} = frac{(s - c)^2}{AB cdot CA} = frac{6^2}{9 cdot 13} = frac{36}{117} = frac{4}{13} ]- The total area of (Delta A_1B_1C_1 ) is: [ [ABC] - ( left[ Delta A_1 BC_1 right] + left[ Delta B_1 CA_1 right] + left[ Delta C_1 AB_1 right] ) ] So, [ frac{[A_1B_1C_1]}{[ABC]} = 1 - left( frac{1}{10} + frac{49}{130} + frac{4}{13} right) ]Step 5: Simplify the final ratio [ 1 - left( frac{1}{10} + frac{49}{130} + frac{4}{13} right) ] Convert all fractions to a common denominator: [ 1 = frac{130}{130}, quad frac{1}{10} = frac{13}{130}, quad frac{49}{130} = frac{49}{130}, quad frac{4}{13} = frac{40}{130} ]Sum: [ frac{13 + 49 + 40}{130} = frac{102}{130} = frac{51}{65} ]Subtract: [ 1 - frac{51}{65} = frac{65 - 51}{65} = frac{14}{65} ]Conclusion:[boxed{frac{14}{65}}] Hence, the ratio of the area of triangle (Delta A_3B_3C_3) to the area of triangle (Delta ABC) is (frac{14}{65}).

🧠:Okay, so I need to solve this problem about two parabolas with distinct vertices. Each vertex lies on the other parabola. The leading coefficients are p and q, and I need to find possible values of p + q. Hmm, let me think through this step by step.First, let me recall that a quadratic function in standard form is f(x) = ax² + bx + c, and its vertex is at (-b/(2a), f(-b/(2a))). Alternatively, in vertex form, it's f(x) = a(x - h)² + k, where (h, k) is the vertex. Since the problem mentions leading coefficients p and q, I can assume that the two parabolas are f(x) = p x² + b1 x + c1 and g(x) = q x² + b2 x + c2. Their vertices will be at (-b1/(2p), f(-b1/(2p))) and (-b2/(2q), g(-b2/(2q))) respectively.But the problem states that the vertex of each parabola lies on the other parabola. So, the vertex of f lies on g, and the vertex of g lies on f. That gives us two equations. Let me formalize that.Let’s denote the vertex of f as (h1, k1) and the vertex of g as (h2, k2). Then:1. (h1, k1) lies on g, so k1 = q h1² + b2 h1 + c22. (h2, k2) lies on f, so k2 = p h2² + b1 h2 + c1But also, since (h1, k1) is the vertex of f, h1 = -b1/(2p), and k1 = f(h1) = p h1² + b1 h1 + c1. Similarly, h2 = -b2/(2q), and k2 = g(h2) = q h2² + b2 h2 + c2.Wait, maybe it's better to express the quadratics in vertex form. Let me try that.Let’s write f(x) = p(x - h1)² + k1 and g(x) = q(x - h2)² + k2. Then the vertex of f is (h1, k1) and the vertex of g is (h2, k2). According to the problem, (h1, k1) lies on g, so substituting x = h1 into g(x) gives k1 = q(h1 - h2)² + k2. Similarly, (h2, k2) lies on f, so substituting x = h2 into f(x) gives k2 = p(h2 - h1)² + k1.So now we have the system of equations:1. k1 = q(h1 - h2)² + k22. k2 = p(h2 - h1)² + k1Let me note that (h1 - h2)² is the same as (h2 - h1)², so let's denote that as D = (h1 - h2)². Then equations become:1. k1 = q D + k22. k2 = p D + k1Now, substituting equation 2 into equation 1:k1 = q D + (p D + k1)Which simplifies to:k1 = q D + p D + k1Subtract k1 from both sides:0 = (p + q) DSince D is the square of (h1 - h2), it is non-negative. Moreover, the problem states that the vertices are distinct, so h1 ≠ h2, which means D > 0. Therefore, for the product (p + q) D to be zero, since D is positive, we must have p + q = 0.Wait, that seems straightforward. So p + q must be zero? Let me check my steps again to make sure.Starting from the two equations:k1 = q D + k2k2 = p D + k1If I substitute the second equation into the first:k1 = q D + p D + k1Then subtract k1 from both sides: 0 = (p + q) DSince D ≠ 0 (vertices are distinct), then p + q must be zero. So p + q = 0. Therefore, the only possible value is 0. Is this correct?Wait, but the question says "What could be the value of p + q?" So it's possible that 0 is the answer. Let me confirm with another approach.Alternatively, let's use the standard form. Let me denote f(x) = p x² + b x + c, vertex at (-b/(2p), f(-b/(2p))). Similarly, g(x) = q x² + d x + e, vertex at (-d/(2q), g(-d/(2q))).Then the vertex of f lies on g, so:f(-b/(2p)) = g(-b/(2p))Similarly, the vertex of g lies on f:g(-d/(2q)) = f(-d/(2q))Let me compute f(-b/(2p)) first. Since f(x) = p x² + b x + c, substituting x = -b/(2p):f(-b/(2p)) = p*(b²/(4p²)) + b*(-b/(2p)) + c = (b²)/(4p) - (b²)/(2p) + c = (-b²)/(4p) + cSimilarly, the vertex of f is (-b/(2p), (-b²)/(4p) + c). This point lies on g(x) = q x² + d x + e, so:(-b²)/(4p) + c = q*(-b/(2p))² + d*(-b/(2p)) + eCompute the right-hand side:q*(b²/(4p²)) - d*(b/(2p)) + eSo equation 1:(-b²)/(4p) + c = (q b²)/(4p²) - (d b)/(2p) + eSimilarly, the vertex of g is (-d/(2q), (-d²)/(4q) + e). This lies on f(x), so:(-d²)/(4q) + e = p*(-d/(2q))² + b*(-d/(2q)) + cCompute the right-hand side:p*(d²)/(4q²) - (b d)/(2q) + cSo equation 2:(-d²)/(4q) + e = (p d²)/(4q²) - (b d)/(2q) + cNow we have two equations:1. (-b²)/(4p) + c = (q b²)/(4p²) - (d b)/(2p) + e2. (-d²)/(4q) + e = (p d²)/(4q²) - (b d)/(2q) + cLet me rearrange equation 1 to:c - e = (q b²)/(4p²) - (d b)/(2p) + (b²)/(4p)Similarly, equation 2:e - c = (p d²)/(4q²) - (b d)/(2q) + (d²)/(4q)Adding equations 1 and 2 together:(c - e) + (e - c) = [ (q b²)/(4p²) - (d b)/(2p) + (b²)/(4p) ] + [ (p d²)/(4q²) - (b d)/(2q) + (d²)/(4q) ]Left-hand side is 0. Right-hand side:= (q b²)/(4p²) + (b²)/(4p) - (d b)/(2p) - (b d)/(2q) + (p d²)/(4q²) + (d²)/(4q)This seems complicated, but maybe there's a symmetry here. Let's see if we can find relationships between the coefficients.Alternatively, perhaps we can assume that the two parabolas are symmetric in some way. Let me try setting p = -q. Then p + q = 0. Let me see if that works.If p + q = 0, then q = -p. Let's substitute q = -p into the equations.First, let's look back at the vertex form equations. If p + q = 0, then the previous conclusion was that p + q must be zero. So maybe that's the answer.But let me verify with the standard form equations. Let's suppose q = -p.Then equation 1 becomes:(-b²)/(4p) + c = (-p b²)/(4p²) - (d b)/(2p) + eSimplify:Left side: (-b²)/(4p) + cRight side: (-b²)/(4p) - (d b)/(2p) + eSet them equal:(-b²)/(4p) + c = (-b²)/(4p) - (d b)/(2p) + eSubtract (-b²)/(4p) from both sides:c = - (d b)/(2p) + eSimilarly, equation 2 with q = -p:(-d²)/(4*(-p)) + e = (p d²)/(4*(-p)^2) - (b d)/(2*(-p)) + cSimplify:Left side: (d²)/(4p) + eRight side: (p d²)/(4p²) + (b d)/(2p) + c = (d²)/(4p) + (b d)/(2p) + cSet equal:(d²)/(4p) + e = (d²)/(4p) + (b d)/(2p) + cSubtract (d²)/(4p) from both sides:e = (b d)/(2p) + cNow from equation 1, we had c = - (d b)/(2p) + e. Let's substitute e from equation 2 into equation 1.From equation 2: e = (b d)/(2p) + cSubstitute into equation 1:c = - (d b)/(2p) + (b d)/(2p) + cSimplify right side: - (d b)/(2p) + (d b)/(2p) + c = 0 + cSo c = c, which is an identity.Therefore, if p + q = 0, the equations are satisfied as long as e = (b d)/(2p) + c and c = - (d b)/(2p) + e. But substituting e into c gives an identity, so there's no contradiction. That means that with p + q = 0, there exist coefficients b, d, c, e that satisfy the conditions. Therefore, p + q = 0 is possible.But does this mean that p + q must necessarily be zero, or could there be other solutions?In the vertex form approach earlier, we deduced that (p + q) D = 0, with D > 0, hence p + q = 0. Therefore, that seems to be the only solution. So p + q must be zero. Therefore, the value could be 0.Wait, but let me check if there's any other possibility. Suppose D = 0, but the vertices are distinct, so D = (h1 - h2)^2 > 0. Hence, D cannot be zero, so the only way (p + q) D = 0 is if p + q = 0. Therefore, p + q must be zero.So the answer is 0.But let me think if there's a different approach where p + q might not be zero, but the problem states "could be the value", so maybe 0 is the only possible value. Let me see.Suppose we take specific examples. Let's construct two parabolas satisfying the given conditions with p + q = 0.Let’s take p = 1, then q = -1.Let’s define f(x) = x² + bx + c and g(x) = -x² + dx + e.The vertex of f is at (-b/2, f(-b/2)). Let's compute f(-b/2):f(-b/2) = (-b/2)^2 + b*(-b/2) + c = b²/4 - b²/2 + c = -b²/4 + cSimilarly, the vertex of g is at (-d/(2*(-1))) = d/2, and g(d/2) = - (d/2)^2 + d*(d/2) + e = -d²/4 + d²/2 + e = d²/4 + eNow, the vertex of f, which is (-b/2, -b²/4 + c), lies on g(x). So substituting x = -b/2 into g(x):g(-b/2) = - (-b/2)^2 + d*(-b/2) + e = -b²/4 - (d b)/2 + eThis must equal the y-coordinate of the vertex of f, which is -b²/4 + c. Therefore:- b²/4 - (d b)/2 + e = -b²/4 + cSimplify: - (d b)/2 + e = c => e = c + (d b)/2Similarly, the vertex of g is (d/2, d²/4 + e), which lies on f(x). Substituting x = d/2 into f(x):f(d/2) = (d/2)^2 + b*(d/2) + c = d²/4 + (b d)/2 + cThis must equal the y-coordinate of the vertex of g, which is d²/4 + e. Therefore:d²/4 + (b d)/2 + c = d²/4 + eSimplify: (b d)/2 + c = eWhich is the same as the previous equation: e = c + (b d)/2So these conditions are consistent. Therefore, as long as we choose e = c + (b d)/2, the vertices will lie on each other's parabolas. So such parabolas exist. For example, let’s choose some values.Let’s take p = 1, q = -1. Let’s set b = 2, d = 4. Then e = c + (2*4)/2 = c + 4.Let’s choose c = 0, then e = 4.So f(x) = x² + 2x + 0, vertex at (-1, -1 + 0) = (-1, -1). Check if (-1, -1) lies on g(x) = -x² + 4x + 4.Compute g(-1) = -1 + (-4) + 4 = -1. Yes, it does.Now, the vertex of g(x) is at d/2 = 4/2 = 2. Compute g(2) = -4 + 8 + 4 = 8. So the vertex of g is (2, 8). Check if (2, 8) lies on f(x) = x² + 2x. Compute f(2) = 4 + 4 = 8. Yes, it does. So this works.Therefore, p = 1, q = -1, so p + q = 0. This example confirms that p + q = 0 is possible.Are there other possibilities? Let me see. Suppose p + q ≠ 0. Then from the vertex form equations, we would have (p + q) D = 0, but D ≠ 0, so p + q must be zero. Therefore, there are no other possibilities. Hence, the only possible value of p + q is 0.Therefore, the answer is boxed{0}.

✅:To determine the value of ( p + q ) for two parabolas with distinct vertices where each vertex lies on the other parabola, we start by expressing the parabolas in vertex form. Let the parabolas be ( f(x) = p(x - h_1)^2 + k_1 ) and ( g(x) = q(x - h_2)^2 + k_2 ), with vertices ((h_1, k_1)) and ((h_2, k_2)) respectively.Given that the vertex of each parabola lies on the other, we have the following equations:1. ( k_1 = q(h_1 - h_2)^2 + k_2 )2. ( k_2 = p(h_2 - h_1)^2 + k_1 )Let ( D = (h_1 - h_2)^2 ). Substituting ( D ) into the equations, we get:1. ( k_1 = qD + k_2 )2. ( k_2 = pD + k_1 )Substituting the second equation into the first:[ k_1 = qD + (pD + k_1) ][ k_1 = qD + pD + k_1 ]Subtracting ( k_1 ) from both sides:[ 0 = (p + q)D ]Since the vertices are distinct, ( D neq 0 ). Therefore, ( p + q ) must be zero. To verify, consider specific examples where ( p = -q ). For instance, if ( p = 1 ) and ( q = -1 ), construct parabolas ( f(x) = x^2 + 2x ) and ( g(x) = -x^2 + 4x + 4 ). Their vertices are ((-1, -1)) and ((2, 8)), which lie on each other's parabolas, confirming the solution.Thus, the only possible value of ( p + q ) is (boxed{0}).

🔑:1. Consider the first parabola with its vertex at the origin ((0, 0)). Its equation can be written as: [ y = px^2 ] where ( p ) is the leading coefficient of the quadratic term.2. Let the vertex of the second parabola be at ( (a, b) ). Given that this vertex lies on the first parabola, we have: [ b = pa^2 ]3. The equation of the second parabola, which opens upwards with vertex at ((a, b)), is given by: [ y = q(x - a)^2 + b ] where ( q ) is the leading coefficient of the quadratic term for the second parabola.4. Since the vertex of the second parabola lies on the first parabola, we substitute ( a ) and ( b ) into the first parabola’s equation for the vertex condition: [ y = q(x - a)^2 + pa^2 ]5. Additionally, note that the vertex of the parabola ( y = px^2 ) (which is ((0, 0))) must lie on the second parabola. Thus, substituting the vertex ((0,0)) into the equation of the second parabola: [ 0 = q(0 - a)^2 + b ] Simplifying, we get: [ 0 = qa^2 + b ]6. Using ( b = pa^2 ) from step 2, we substitute into the equation obtained in step 5: [ 0 = qa^2 + pa^2 ]7. Factor out ( a^2 ): [ 0 = a^2 (p + q) ]8. Since (a^2 neq 0) (the vertices are different, implying ( a neq 0 )), we must have: [ p + q = 0 ] Conclusion:[boxed{0}]

🧠:Okay, so I need to prove that for any positive integer m in the set {r+1, r+2, ..., 2r+1}, the sum Σ a_j x_j^m is congruent to 0 modulo m. The given conditions are that Σ a_j x_j^k = 0 for k = 1, 2, ..., r. All the a_j and x_j are integers, n is at least 2, and r is at least 2. Hmm. Let me try to break this down.First, the problem involves sums of powers of integers multiplied by coefficients, and these sums vanish for exponents from 1 up to r. Then, I need to show that for exponents m from r+1 up to 2r+1, the same sum modulo m is zero. So, modulo m, the sum is a multiple of m. Let me recall some algebraic concepts that might be helpful here. Since we have multiple equations where linear combinations of powers of x_j are zero, maybe there's a connection to polynomial equations or linear algebra. The fact that the sums are zero for k = 1 to r suggests that the coefficients a_j are in some way orthogonal to the vectors (x_j^k) for these exponents. Maybe generating functions or recurrence relations could be involved.Alternatively, considering the problem modulo m, perhaps Fermat's little theorem applies? But m is not necessarily prime here. However, if m is composite, maybe we can use Euler's theorem, which generalizes Fermat's, but that requires the base to be coprime to m. But x_j might not be coprime to m. Hmm.Wait, the problem states that m is in {r+1, ..., 2r+1}, and r is at least 2. So m ranges from r+1 up to 2r+1. For example, if r=2, m would be 3,4,5. Since r is at least 2, m is at least 3. Not sure if that's relevant yet.Another thought: the equations given are linear in the coefficients a_j. If we write the system of equations for k=1 to r, we have:a_0 x_0 + a_1 x_1 + ... + a_n x_n = 0a_0 x_0^2 + a_1 x_1^2 + ... + a_n x_n^2 = 0...a_0 x_0^r + a_1 x_1^r + ... + a_n x_n^r = 0So these are r equations with n+1 variables a_j. But since the a_j are integers, perhaps there's some structure here. But how does that help for higher exponents m?Maybe we can consider generating a polynomial from the coefficients a_j. Let me think. If I define a polynomial P(t) = Σ_{j=0}^n a_j t^{d_j} for some exponents d_j. Wait, but the x_j are different variables here. Alternatively, if we think of the generating function Σ_{k=0}^infty (Σ_{j=0}^n a_j x_j^k) t^k. Then, the given conditions say that the first r coefficients (for t^1 to t^r) are zero. Maybe this generating function can be expressed as a rational function, given that the higher coefficients satisfy some linear recurrence?Alternatively, if I consider the generating function G(t) = Σ_{k=0}^infty (Σ_{j=0}^n a_j x_j^k) t^k. Then, this is equal to Σ_{j=0}^n a_j Σ_{k=0}^infty (x_j t)^k = Σ_{j=0}^n a_j / (1 - x_j t), assuming |x_j t| < 1. But since we are dealing with formal power series, maybe we can use this expression. Then, if the first r coefficients of G(t) are zero, that would mean that Σ_{j=0}^n a_j / (1 - x_j t) has a numerator that is a polynomial of degree at most n, but the first r terms in the expansion are zero. Hmm, perhaps not the most straightforward approach.Wait, but the problem is about congruence modulo m for specific m. So maybe we can work modulo m directly. Let me fix an m in {r+1, ..., 2r+1}, and try to show that Σ a_j x_j^m ≡ 0 mod m.Given that m is between r+1 and 2r+1, perhaps m can be expressed as a combination of the previous exponents 1 through r. For example, if m is r+1, maybe there's a way to relate x_j^{r+1} to lower powers of x_j, which are already known to make the sum zero. But how?Alternatively, maybe we can use the fact that x_j^m ≡ x_j^{m mod (something)} modulo m. For example, if we can find an exponent e such that x_j^e ≡ 1 mod m for all j, then x_j^m ≡ x_j^{m mod e} mod m. But since x_j might not be coprime to m, Euler's theorem might not apply. However, if we can use Carmichael's theorem, which gives the minimal exponent such that x^λ(m) ≡ 1 mod m for x coprime to m. But again, x_j may not be coprime to m.Alternatively, consider the polynomial f(t) = Σ_{j=0}^n a_j t^{x_j}. Wait, but x_j are integers, not exponents. Hmm, not sure.Wait, maybe using the given conditions, the sums of the a_j times x_j^k are zero for k=1 to r. So if I consider the vector (a_0, a_1, ..., a_n) and the vectors (x_0^k, x_1^k, ..., x_n^k) for k=1 to r, then the dot product is zero. So the vector a is orthogonal to each of these r vectors. Then, perhaps the vector a is in the orthogonal complement of the space spanned by these vectors.But how does that relate to higher exponents m? If the space spanned by the vectors (x_j^k) for k=1 to r is of dimension r, then maybe the higher exponents can be expressed in terms of these? But unless there's a linear recurrence relation or something.Wait, suppose that for each x_j, the sequence x_j^k satisfies a linear recurrence of order r. For example, if x_j is a root of a polynomial of degree r, then x_j^k can be expressed in terms of previous terms. But since x_j are integers, unless they all satisfy some common recurrence relation, which might not be the case.Alternatively, since the equations hold for all k from 1 to r, maybe we can use induction for higher exponents. Suppose that for some k >= r, the sum Σ a_j x_j^k = 0. Then, perhaps multiply by x_j and sum again? But not sure.Alternatively, using generating functions again. Let’s define S(k) = Σ a_j x_j^k. We know that S(k) = 0 for k=1,...,r. We need to show that S(m) ≡ 0 mod m for m = r+1,...,2r+1.If we can express S(m) as a linear combination of S(1), ..., S(r), then since each S(k) is zero, S(m) would be zero. But m is larger than r, so how?Alternatively, think of the generating function for S(k). Let’s denote G(t) = Σ_{k=0}^infty S(k) t^k. Then, G(t) = Σ_{j=0}^n a_j Σ_{k=0}^infty (x_j t)^k = Σ_{j=0}^n a_j / (1 - x_j t). But since S(k) = 0 for k=1,...,r, this means that G(t) = S(0) + Σ_{k=r+1}^infty S(k) t^k. So, maybe this generating function can be written as a rational function where the numerator is a polynomial of degree at most n, but given that the first r coefficients after S(0) are zero, perhaps there's a relationship here. However, since the x_j are integers, and a_j are integers, the generating function is a sum of reciprocals of linear terms. But I don't see immediately how this helps.Alternatively, let's think about the problem modulo m. Let me fix m and work modulo m. Then, we need to show that Σ a_j x_j^m ≡ 0 mod m. Let me note that x_j^m ≡ x_j^{m - phi(m)} mod m if we can apply Euler's theorem, but phi(m) is Euler's totient function. However, since m can be composite, and x_j and m might not be coprime, this might not hold for all x_j. But maybe we can decompose m into prime factors and use the Chinese Remainder Theorem.Alternatively, consider that for each prime power p^e dividing m, we can show that Σ a_j x_j^m ≡ 0 mod p^e. Then, by the Chinese Remainder Theorem, it would hold modulo m. So maybe it's sufficient to prove the congruence for prime powers dividing m.But how does the given condition (sums zero for k=1,...,r) relate to prime powers in m? Hmm. Let's think about the case when m is a prime p. Then, perhaps Fermat's little theorem applies: x^{p-1} ≡ 1 mod p if x not divisible by p. So x^p ≡ x mod p. Then, if m = p, x^m ≡ x mod p. Therefore, Σ a_j x_j^m ≡ Σ a_j x_j ≡ 0 mod p. But from the given condition, Σ a_j x_j = 0, so this would imply Σ a_j x_j^m ≡ 0 mod p. Then, for higher prime powers, maybe use lifting the exponent?Wait, if m is a prime power p^e, then perhaps we can use Hensel's lemma or the lifting the exponent lemma (LTE). But LTE usually applies to specific cases, like when x ≡ y mod p but not mod p^2, etc. Not sure if directly applicable here.But let's test this idea for m being a prime. Suppose m is a prime number in {r+1, ..., 2r+1}. Then, by Fermat's little theorem, x_j^m ≡ x_j mod m. Therefore, Σ a_j x_j^m ≡ Σ a_j x_j ≡ 0 mod m. Since we are given that Σ a_j x_j = 0, this gives the result. So for prime m, this works. What about composite m?But m can be composite, like 4, 6, etc., depending on r. For example, if r=2, m can be 3,4,5. If r=3, m can be 4,5,6,7. So m can be composite. For composite m, Fermat's little theorem doesn't hold as such, but maybe there's a generalization. Euler's theorem says x^phi(m) ≡ 1 mod m if x and m are coprime. So for x coprime to m, x^phi(m) ≡ 1 mod m. Therefore, x^m ≡ x^{m mod phi(m)} mod m. But unless m and phi(m) have some relationship.Alternatively, if we can write m as a product of primes, but I think this approach might get complicated. Maybe there's another way.Wait, another angle: the set of exponents m is from r+1 to 2r+1. The number of exponents here is r+1. Maybe using the fact that m <= 2r+1, and the given sums up to r. Maybe there's a relation through polynomial identities. For example, if we have a polynomial of degree r that annihilates the x_j in some way.Wait, let's consider the linear system. Let me define the vectors v_k = (x_0^k, x_1^k, ..., x_n^k) for k=1,...,r. The given conditions are that the vector a = (a_0, a_1, ..., a_n) is orthogonal to each v_k. Therefore, a is in the orthogonal complement of the span of {v_1, ..., v_r}. If the dimension of the span of {v_1, ..., v_r} is r, then the orthogonal complement has dimension n+1 - r. But since n >= 2 and r >= 2, n+1 - r could be greater than 0. But how does this relate to the higher exponents?Alternatively, if we consider the matrix M with rows v_1, v_2, ..., v_r, then the vector a is in the nullspace of M. Then, perhaps for higher exponents m, the vector v_m = (x_0^m, ..., x_n^m) can be expressed as a linear combination of the rows of M, which would imply that a · v_m = 0. But if m > r, is v_m a linear combination of v_1, ..., v_r?This would depend on the x_j. If the x_j are roots of a polynomial of degree r, then their powers can be expressed in terms of lower powers. But the x_j are arbitrary integers, so unless they satisfy such a condition, we can't assume that. Hmm.Wait, but even if the x_j don't satisfy a common linear recurrence, maybe we can use the fact that the sums Σ a_j x_j^k = 0 for k=1,...,r. So, suppose we take any linear combination of these equations. For example, if we multiply the equation for k=1 by some coefficient and add to the equation for k=2, etc. But how does that help with higher exponents?Alternatively, consider that the equations Σ a_j x_j^k = 0 for k=1,...,r can be seen as the evaluations of the polynomial P(t) = Σ a_j x_j t at t=1, but not exactly. Wait, no. Alternatively, if we have a polynomial Q(t) = Σ_{k=1}^r c_k t^k, then Σ a_j Q(x_j) = Σ_{k=1}^r c_k Σ a_j x_j^k = 0. So Q(x_j) is a polynomial that, when evaluated at each x_j and combined with coefficients a_j, gives zero. But this is for any coefficients c_k. Therefore, the vector a is orthogonal to all polynomials of degree up to r evaluated at the x_j. Maybe using the concept of duality here.But how can we connect this to x_j^m? If m is between r+1 and 2r+1, maybe x_j^m can be expressed as a linear combination of lower-degree terms modulo m. For example, using some modular relation. But how?Wait, here's an idea inspired by the previous thought about primes. For any integer x_j, by Fermat's little theorem, if m is prime, x_j^m ≡ x_j mod m. So then Σ a_j x_j^m ≡ Σ a_j x_j ≡ 0 mod m. But this only works if m is prime. However, the problem allows m to be composite. But maybe we can generalize this idea. If m is composite, maybe for each prime p dividing m, we can relate x_j^m mod p to something. Wait, but m could be a power of a prime, or have multiple prime factors.Alternatively, consider that for each x_j, x_j^m ≡ x_j^{m mod (p-1)} mod p by Fermat's little theorem, if p is prime and doesn't divide x_j. If p divides x_j, then x_j^m ≡ 0 mod p. So maybe decompose the sum into primes dividing m and handle each prime separately.Let me try this approach. Let m be in {r+1, ..., 2r+1}, and let p be a prime dividing m. Let’s write m = p^e * q where q is coprime to p. Then, by the Chinese Remainder Theorem, it's enough to show that Σ a_j x_j^m ≡ 0 mod p^e and mod q, then combine the results.First, consider modulo p^e. If p divides x_j, then x_j^m ≡ 0 mod p^e, since x_j is divisible by p and m >= p^e (if e >= 1). Wait, but m could be equal to p^e * q, but unless q is 1, but m can be composite. Hmm, perhaps this is getting too complicated.Wait, let's take a step back. Suppose m is a prime power, say p^e. If p divides x_j, then x_j^m is divisible by p^m, which is certainly divisible by p^e, since m >= p^e (as m >= r+1 >= 3, and p is a prime divisor of m, so p <= m). Therefore, x_j^m ≡ 0 mod p^e. For x_j not divisible by p, Euler's theorem tells us that x_j^{phi(p^e)} ≡ 1 mod p^e, where phi(p^e) = p^e - p^{e-1} = p^{e-1}(p - 1). Therefore, x_j^{phi(p^e)} ≡ 1 mod p^e. Therefore, x_j^m ≡ x_j^{m mod phi(p^e)} mod p^e.But m is between r+1 and 2r+1. Let me note that phi(p^e) = p^{e-1}(p - 1). So m mod phi(p^e) is some value. However, unless m is related to phi(p^e), this might not help. But let's see. If we can write m = k * phi(p^e) + s, then x_j^m ≡ x_j^s mod p^e. However, since m is in {r+1, ..., 2r+1}, and phi(p^e) could be larger or smaller.Alternatively, suppose that m <= 2r + 1 and we have the given equations up to k = r. Maybe m - phi(p^e) <= r? If that were the case, then x_j^m ≡ x_j^{m - phi(p^e)} mod p^e, and m - phi(p^e) <= r, so we could use the given condition that Σ a_j x_j^{m - phi(p^e)} = 0. But I need to verify if m - phi(p^e) is in the range 1 to r.Let’s take an example. Let m be 6 (so r could be 2 or 3). If p = 3, e = 1, then phi(3) = 2. Then m - phi(p^e) = 6 - 2 = 4, which is greater than r=2 or 3. So that might not help. Hmm.Alternatively, maybe use induction on m. Suppose that for some m in {r+1, ..., 2r+1}, all smaller exponents in that set have the property. But I'm not sure how the induction step would work.Wait, another idea: using Newton's identities or symmetric sums. The given equations relate power sums to zero. Maybe the coefficients a_j can be considered as part of a generating function whose power sums vanish up to degree r. If the first r power sums are zero, maybe the generating function is a polynomial of degree r, but I need to think more carefully.Suppose we have a polynomial Q(t) = Π_{j=0}^n (t - x_j)^{a_j}. Wait, but a_j are coefficients, not exponents. Alternatively, perhaps something like Σ a_j / (t - x_j) = 0, which would relate to the given generating function. If we have Σ a_j / (t - x_j) = 0, then combining these fractions would give a rational function where the numerator is a polynomial of degree n - 1. But I don't see the direct connection here.Wait, going back to the generating function G(t) = Σ_{k=0}^infty S(k) t^k = Σ_{j=0}^n a_j / (1 - x_j t). If S(1) = S(2) = ... = S(r) = 0, then G(t) = S(0) + Σ_{k=r+1}^infty S(k) t^k. On the other hand, if we write G(t) as a rational function with denominator Π_{j=0}^n (1 - x_j t), then the numerator would be a polynomial of degree at most n. But if the first r coefficients after S(0) are zero, perhaps the numerator is divisible by t^{r+1}, which would require that the numerator is a polynomial of degree n multiplied by t^{r+1}. But unless n <= r, which is not necessarily the case since n >= 2 and r >= 2. Hmm, this seems complicated.Wait, but if we consider the given that S(k) = 0 for k=1,...,r, then the generating function G(t) - S(0) = Σ_{k=1}^infty S(k) t^k = Σ_{k=r+1}^infty S(k) t^k. So, this is a power series starting from t^{r+1}. Therefore, G(t) - S(0) is divisible by t^{r+1}. Therefore, G(t) = S(0) + t^{r+1} H(t) for some power series H(t). But G(t) is also equal to Σ_{j=0}^n a_j / (1 - x_j t). Therefore, Σ_{j=0}^n a_j / (1 - x_j t) = S(0) + t^{r+1} H(t). Maybe manipulate this equation to find a relation?Multiply both sides by Π_{j=0}^n (1 - x_j t):Σ_{j=0}^n a_j Π_{i≠j} (1 - x_i t) = [S(0) + t^{r+1} H(t)] Π_{j=0}^n (1 - x_j t).The left-hand side (LHS) is a polynomial of degree n, since each term Π_{i≠j} (1 - x_i t) is degree n. The right-hand side (RHS) is [S(0) + t^{r+1} H(t)] times a polynomial of degree n+1, which would be a polynomial of degree n+1 + something. But this seems messy. Maybe this approach is not helpful.Alternatively, recall that in linear algebra, if we have a vector orthogonal to several vectors, then it's orthogonal to any linear combination of those vectors. But here, the higher power terms might be linear combinations of the lower ones if there's a dependency. But since the x_j are arbitrary integers, such a dependency may not exist. However, modulo m, dependencies might arise because of periodicity in exponents.Wait, here's a crucial observation: For any integer x_j, x_j^m ≡ x_j^{m - (p-1)} mod p if p is a prime not dividing x_j. So if m - (p-1) is in the range 1 to r, then we can relate x_j^m to x_j^{m - (p-1)}, which is covered by the given conditions. But m is between r+1 and 2r+1. Let's see: m - (p-1) <= m - 2 (since p >=2, so p-1 >=1). Since m <= 2r+1, m - (p-1) >= (r+1) - (p-1). But since p divides m, and m is between r+1 and 2r+1, the primes p are <= m <= 2r+1. So, if p divides m, then p <= m <= 2r+1. Therefore, p-1 <= 2r. Therefore, m - (p-1) >= (r+1) - (2r) = -r +1. Not sure if helpful.Alternatively, consider that for each prime p dividing m, we can write m = kp + s, but this might not directly help.Wait, perhaps using the fact that m <= 2r +1. So, for each prime p dividing m, p <= m <= 2r +1. Then, p -1 <= 2r. Therefore, m can be written as m = (p -1) + t, where t = m - (p -1). Since m >= r+1 and p <= m, t >= r +1 - (p -1) >= r +1 - (m -1) >= r +1 - (2r +1 -1) = r +1 - 2r = -r +1. Hmm, not helpful.Wait, let's try specific examples to see a pattern.Example 1: Let r=2. Then m can be 3,4,5.Take m=3. Need to show Σ a_j x_j^3 ≡ 0 mod 3. Since 3 is prime, by Fermat's little theorem, x_j^3 ≡ x_j mod 3. Therefore, Σ a_j x_j^3 ≡ Σ a_j x_j ≡ 0 mod 3. Which is given (k=1). So this works.For m=4. Now, 4 is composite. Let's take mod 4. For even x_j, x_j^4 ≡ 0 mod 4. For odd x_j, x_j^4 ≡ 1 mod 4. So Σ a_j x_j^4 ≡ Σ_{x_j odd} a_j *1 + Σ_{x_j even} a_j *0 ≡ Σ_{x_j odd} a_j mod 4. Need this to be 0 mod 4. But how is this related to the given conditions? The given sums for k=1 and 2 are zero. Let's see:Σ a_j x_j = 0 (given for k=1)Σ a_j x_j^2 = 0 (given for k=2)But how to relate Σ a_j (x_j^4 mod 4). Let's compute x_j^4 mod 4. If x_j is even, x_j ≡ 0 or 2 mod 4. Then x_j^4 ≡ 0 mod 4. If x_j is odd, x_j ≡ 1 or 3 mod 4, so x_j^4 ≡ 1 mod 4. Therefore, Σ a_j x_j^4 ≡ Σ_{x_j odd} a_j mod 4.But how do we know this sum is zero mod 4? Let's think. From the given conditions:Σ a_j x_j = 0. Let's separate even and odd x_j.Let E be the set of j where x_j is even, and O be the set where x_j is odd. Then:Σ_{j∈E} a_j x_j + Σ_{j∈O} a_j x_j = 0.Similarly, Σ_{j∈E} a_j x_j^2 + Σ_{j∈O} a_j x_j^2 = 0.But x_j even: x_j ≡ 0 mod 2, so x_j^2 ≡ 0 mod 4.x_j odd: x_j ≡ 1 mod 2, so x_j^2 ≡ 1 mod 4. Therefore, Σ a_j x_j^2 ≡ Σ_{j∈O} a_j mod 4. But given that Σ a_j x_j^2 = 0, which implies Σ_{j∈O} a_j ≡ 0 mod 4. Therefore, Σ a_j x_j^4 ≡ Σ_{j∈O} a_j ≡ 0 mod 4. Hence, Σ a_j x_j^4 ≡ 0 mod 4.Similarly, for m=4, it works. For m=5, which is prime, similar to m=3: x_j^5 ≡ x_j mod 5, so Σ a_j x_j^5 ≡ Σ a_j x_j ≡ 0 mod 5.So for m=3,4,5 with r=2, it works. So in these cases, the result holds because either m is prime (using Fermat's little theorem) or m is a power of 2 and the sum modulo 4 can be related to the given sum for k=2.Another example: take r=3, so m=4,5,6,7.For m=7 (prime): x_j^7 ≡ x_j mod 7, so Σ a_j x_j^7 ≡ Σ a_j x_j ≡ 0 mod 7.For m=6: composite. Let's compute modulo 6. Using Chinese Remainder Theorem, check modulo 2 and modulo 3.Modulo 2: x_j^6 ≡ (x_j mod 2)^6. If x_j even, 0; if odd, 1. So Σ a_j x_j^6 ≡ Σ_{j odd} a_j mod 2. But from given k=1: Σ a_j x_j ≡ Σ_{j odd} a_j ≡ 0 mod 2. Hence, Σ a_j x_j^6 ≡ 0 mod 2.Modulo 3: x_j^6 ≡ (x_j^{3})^2 ≡ x_j^2 mod 3 (since x_j^3 ≡ x_j mod 3 by Fermat). So Σ a_j x_j^6 ≡ Σ a_j x_j^2 ≡ 0 mod 3 (given for k=2). Therefore, by Chinese Remainder Theorem, Σ a_j x_j^6 ≡ 0 mod 6.Similarly, for m=4, as before. For m=5, which is prime: x_j^5 ≡ x_j mod 5. So Σ a_j x_j^5 ≡ 0 mod 5.For m=6: done above.Another example: m=6, r=3. Wait, m=6 is in {4,5,6,7} when r=3. Wait, but for r=3, m ranges up to 2*3 +1=7. So m=4,5,6,7.So for m=6, using modulo 2 and 3, we saw it works. For m=4, similar to previous case. For m=5 and 7, primes.Therefore, the pattern seems to be that for primes, use Fermat's theorem, and for composites, factor into primes and use previous given conditions. So perhaps the general approach is:For each prime p dividing m, note that m is in {r+1, ..., 2r+1}. So p <= m <= 2r+1. Then, depending on p:- If p is a prime, then for any x_j not divisible by p, x_j^{p-1} ≡ 1 mod p. So x_j^m ≡ x_j^{m mod (p-1)} mod p. If m mod (p-1) is in {1, ..., r}, then we can use the given condition that Σ a_j x_j^{m mod (p-1)} = 0. But m mod (p-1) can be written as m - k*(p-1). Since m <= 2r+1 and p >=2 (since p divides m >= r+1 >=3), p-1 >=1. So k*(p-1) <= m <= 2r+1. If we can write m = s + t*(p-1), where s <= r, then x_j^m ≡ x_j^s mod p. Then, Σ a_j x_j^m ≡ Σ a_j x_j^s ≡ 0 mod p. But m <= 2r+1 and p divides m. Let's see. If p divides m, then p <= m <=2r+1. So p-1 <= 2r. So m = (p-1)*k + s, where 0 <= s < p-1. But since m <= 2r+1 and p-1 >=1 (since p>=2), k can be up to floor(m/(p-1)). But since m <=2r+1 and p-1 could be up to m-1 (if p= m), then s = m - (p-1)*k. If p= m, then p-1 = m-1, so k=1, s= m - (m-1)*1=1. So x_j^m ≡ x_j mod p (since x_j^{p-1} ≡1 mod p for x_j not divisible by p). Therefore, Σ a_j x_j^m ≡ Σ a_j x_j ≡0 mod p. For p < m, suppose p divides m, then m = p * t. Then, p <= m <=2r+1. But m is in {r+1, ...,2r+1}, so p is between 2 and 2r+1. Then, for x_j not divisible by p, x_j^{p-1} ≡1 mod p. So x_j^m =x_j^{p*t} ≡ (x_j^{p-1})^{t} *x_j^{t} ≡1^t *x_j^t ≡x_j^t mod p. So x_j^m ≡x_j^t mod p. Now, t = m/p. Since m <=2r+1 and p >=2, t <= (2r+1)/2 <r+1. So t <= r. Therefore, x_j^m ≡x_j^{m/p} mod p. Since m/p <=r (because m <=2r+1 and p >=2, so m/p <= (2r+1)/2 <r+1). Since m/p is an integer (because p divides m), then m/p <=r. Thus, m/p is in {1,...,r}. Therefore, Σ a_j x_j^m ≡ Σ a_j x_j^{m/p} ≡0 mod p, since the given conditions include k=m/p.But m/p is an integer between 1 and r because p divides m and m is between r+1 and 2r+1. Therefore, m/p is at least (r+1)/p >= (r+1)/(2r+1) (since p >=2). Wait, but m/p is integer. For example, if m=4, p=2, then m/p=2, which is <=r=2. If m=6, p=2, m/p=3, which is <=r=3. Similarly, m=6, p=3, m/p=2 <=r=3. So in general, if m is composite, say m=p*q, then m/p=q, which is in {1,...,r} because m <=2r+1 and p>=2. Therefore, q =m/p <= (2r+1)/2 <r+1. So q <=r. Hence, m/p is an integer <=r. Therefore, x_j^m ≡x_j^{m/p} mod p, and Σ a_j x_j^{m/p} =0, so Σ a_j x_j^m ≡0 mod p. For primes p dividing m, this works.For primes p dividing m and x_j divisible by p, x_j^m ≡0 mod p. So the sum Σ a_j x_j^m mod p is Σ_{p|x_j} a_j*0 + Σ_{p∤x_j} a_j x_j^{m/p} ≡ Σ_{p∤x_j} a_j x_j^{m/p} ≡0 mod p, since Σ a_j x_j^{m/p}=0. Therefore, for each prime p dividing m, Σ a_j x_j^m ≡0 mod p. By Chinese Remainder Theorem, since this holds for all prime divisors of m, it holds modulo m. Hence, Σ a_j x_j^m ≡0 mod m.This seems to be the key argument. Let me formalize it:Let m be in {r+1, ...,2r+1}. Let p be a prime divisor of m. Write m =p *k, where k is an integer (since p divides m). Then, k =m/p. Since m <=2r+1 and p >=2, k <= (2r+1)/2 <r+1. Hence, k <=r. Therefore, k is an integer in {1,...,r}. By the given condition, Σ a_j x_j^k =0. Now, for each x_j, if p divides x_j, then x_j^m = (x_j)^{p*k} is divisible by p^{p*k}, hence x_j^m ≡0 mod p. If p does not divide x_j, then by Fermat's little theorem, x_j^{p-1} ≡1 mod p. Therefore, x_j^{p*k} = (x_j^{p-1})^{k} *x_j^{k} ≡1^k *x_j^{k} ≡x_j^{k} mod p. Therefore, x_j^{m} ≡x_j^{k} mod p. Therefore, Σ a_j x_j^{m} ≡Σ a_j x_j^{k} mod p. But Σ a_j x_j^{k}=0 by the given condition. Therefore, Σ a_j x_j^{m} ≡0 mod p. Since this holds for every prime p dividing m, by the Chinese Remainder Theorem, Σ a_j x_j^{m} ≡0 mod m. Therefore, the result is proven.

✅:To prove that for any positive integer ( m in {r+1, r+2, cdots, 2r+1} ), the sum ( sum_{j=0}^{n} a_{j} x_{j}^{m} equiv 0 pmod{m} ), we proceed as follows:Key Steps:1. Prime Divisor Analysis: Let ( m ) be in the given range and consider any prime ( p ) dividing ( m ). Write ( m = p cdot k ), where ( k = m/p ). Since ( m leq 2r+1 ) and ( p geq 2 ), it follows that ( k leq (2r+1)/2 < r+1 ), so ( k in {1, 2, ldots, r} ).2. Fermat's Little Theorem Application: For each ( x_j ): - If ( p mid x_j ), then ( x_j^m equiv 0 pmod{p} ). - If ( p nmid x_j ), Fermat's theorem gives ( x_j^{p-1} equiv 1 pmod{p} ). Hence, ( x_j^{m} = x_j^{p cdot k} equiv (x_j^{p-1})^k cdot x_j^k equiv 1^k cdot x_j^k equiv x_j^k pmod{p} ).3. Sum Modulo Prime ( p ): The sum becomes: [ sum_{j=0}^{n} a_j x_j^m equiv sum_{p mid x_j} a_j cdot 0 + sum_{p nmid x_j} a_j x_j^k equiv sum_{j=0}^{n} a_j x_j^k pmod{p}. ] Since ( k leq r ), the given condition ( sum_{j=0}^{n} a_j x_j^k = 0 ) implies ( sum_{j=0}^{n} a_j x_j^m equiv 0 pmod{p} ).4. Chinese Remainder Theorem: As the congruence holds for all prime divisors ( p ) of ( m ), combining them via the Chinese Remainder Theorem yields ( sum_{j=0}^{n} a_j x_j^m equiv 0 pmod{m} ).Conclusion: For each ( m in {r+1, ldots, 2r+1} ), the sum ( sum_{j=0}^{n} a_j x_j^m ) is divisible by ( m ), hence the result holds.[boxed{sum_{j=0}^{n} a_{j} x_{j}^{m} equiv 0 pmod{m} text{ for all } m in {r+1, r+2, ldots, 2r+1}}]

🔑:To prove that for any positive integer ( m in {r+1, r+2, ldots, 2r+1} ), we have:[sum_{j=0}^{n} a_{j} x_{j}^{m} equiv 0 pmod{m},]given (sum_{j=0}^{n} a_{j} x_{j}^{k} = 0 ) for ( k=1,2, ldots, r).1. Purpose: We aim to show that (sum_{j=0}^{n} a_{j} x_{j}^{m} equiv 0 pmod{m}) for any positive integer ( m ) in the given range. 2. Factorize (m): Let ( p^alpha | m ), where ( p ) is a prime number, and ( alpha geq 1 ). By Euler's totient function, we have: [ varphi(p^alpha) = p^{alpha-1}(p-1). ]3. Euler's Totient Function Property: Since ( varphi(p^alpha) leftlvert left(m-frac{m}{p}right) right. ): [ varphi(p^alpha) mid left( m - frac{m}{p} right), ] which implies ( p^{alpha-1}(p-1) leftlvert left( m - frac{m}{p} right) right. ).4. Bound (frac{m}{p}): Given ( m leq 2r+1 ): [ frac{m}{p} leq frac{2r+1}{p} < r+1. ] Hence, we have: [ frac{m}{p} leq r. ]5. Estimate (alpha): Since ( r geq frac{m}{p} geq p^{alpha-1} geq alpha ): [ r geq p^{alpha-1} geq alpha. ]6. Evaluate ( x_j^m ): For any ( x_j ), if ( p mid x_j ), then using ( m > frac{m}{p} geq alpha ): [ x_j^m equiv x_j^{frac{m}{p}} pmod{p^alpha}. ] If ( p nmid x_j ), from ( varphi(p^alpha) mid left(m - frac{m}{p}right) ) and Euler's theorem, we get: [ x_j^{m-frac{m}{p}} equiv 1 pmod{p^alpha}. ] Thus: [ x_j^m equiv x_j^{frac{m}{p}} pmod{p^alpha}. ]7. Conclusion for Sum: Because (frac{m}{p} leq r), we obtain: [ sum_{j=0}^{n} a_{j} x_{j}^{m} equiv sum_{j=0}^{n} a_{j} x_{j}^{frac{m}{p}} equiv 0 pmod{p^alpha}. ]8. Final Assertion: Since the previous result holds for any ( p^alpha | m ): [ sum_{j=0}^{n} a_{j} x_{j}^{m} equiv 0 pmod{m}. ]We have thus proved the required statement.[boxed{sum_{j=0}^{n} a_{j} x_{j}^{m} equiv 0 pmod{m}.}]