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a line. So, the number of lines would depend on how the points are arranged. But the problem says \"n given points\" in general. So maybe it's a general position problem?\n\nWait, the problem is asking for the maximum number of such lines possible given n points, or the number of lines that are equidistant from all n p... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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there might not be any such line. \n\nBut the problem doesn't specify any particular arrangement of the points, so it's a general question. So, perhaps the answer is that there can be infinitely many lines, or maybe up to three lines? Wait, no. Wait, let's think again. If the points are arranged such that they are symm... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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For instance, suppose you have three points on one line and one on the other. The midline would still be equidistant to all four points. So, in this case, there would be one such line.\n\nBut if the points are not arranged in two parallel lines, then maybe there's no such line. So, the answer could vary depending on th... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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would be C(n,2) = n(n-1)/2. But the question says \"equidistant from n given points\", which is plural. So, perhaps equidistant from each of the n points. \n\nBut then, as I thought earlier, such a line must have all n points at the same distance from it. Which would mean that all points lie on two lines parallel to th... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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doesn't make sense. The distance from a point to a line is a non-negative number, so if the line is equidistant from all n points, then each point has the same distance to the line. Which would require all points to lie on two lines parallel to the given line, each at the same distance. So, for example, if you have poi... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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three? If they form a triangle, then the three perpendicular bisectors meet at the circumcenter. But the circumcenter is a point, not a line. So, the lines equidistant from all three points would have to have all three points at the same distance from the line. For that to happen, all three points must lie on two paral... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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are arranged in two parallel lines. Suppose we have m points on one line and k points on another parallel line, with m + k = n. Then, the midline between them is equidistant to all points. Also, if we have another pair of parallel lines in a different orientation that contain all points, then there could be another mid... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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maybe not. Let me think. If the hexagon is centered at the origin and has vertices at (1,0), (0.5, β3/2), (-0.5, β3/2), (-1,0), (-0.5, -β3/2), (0.5, -β3/2). Suppose we take the horizontal line y=0: all vertices are at distance β3/2 or 0. No, the ones on the line y=0 are at distance zero, the others at β3/2. So that's n... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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on. Each pair of parallel lines contributes one midline. So, if points lie on k pairs of parallel lines, then you have k midlines. To maximize k, you need to arrange the points such that they lie on as many pairs of parallel lines as possible. For example, if points form a regular polygon, perhaps they lie on multiple ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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arranged in a grid, like a rectangle, you can split them into horizontal lines (so midline horizontal), vertical lines (midline vertical), and maybe diagonal lines. But whether that's possible depends on the grid.\n\nWait, let's take n=4 points arranged in a square. The directions where the points can be split into two... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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we find another direction where the points lie on two parallel lines? For example, at 45 degrees. Let me see. If we have lines at 45 degrees, can two parallel lines in that direction contain all four points? The distance between the lines would have to be such that all four points are on them. It's possible? Let's comp... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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get multiple directions, but it's unclear. Let's think of points arranged in a three-dimensional way on the plane, but in the plane. Hmm.\n\nAlternatively, consider that for each pair of points, the line connecting them defines a direction. If we can pair up the points such that multiple pairs have the same direction, ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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example, arranging points as the vertices of a regular polygon might not work, but perhaps arranging them in a set of concentric regular polygons, but that complicates things.\n\nAlternatively, arrange the points in such a way that for multiple directions, the projections of the points onto those directions can be spli... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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of points, there's a direction. But here, the problem is different. Here, we need directions where all points lie on two lines in that direction.\n\nFor example, for each pair of points, the line through them defines a direction. If we can have all other points lie on a line parallel to that direction, then that direct... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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can split the points into two lines in different directions. This is likely related to the number of different slopes between pairs of points. For each direction, there's a slope, and if all points lie on two lines with that slope, then that direction is valid.\n\nBut how many such directions can exist?\n\nFor example,... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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ΞΈ, the points can be partitioned into two lines with slope tan(theta). For example, if we arrange the points as the vertices of a regular 2m-gon, then there are m axes of symmetry, each of which could be a direction for splitting the points into two lines. But in a regular 2m-gon, each axis of symmetry passes through t... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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this seems challenging.\n\nAlternatively, arrange n points in a grid-like structure. For example, a 3x3 grid has nine points. If we consider directions along the rows, columns, and diagonals. But as discussed earlier, the diagonals don't work. So, only rows and columns. For 3x3 grid, rows are three horizontal lines, co... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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on the arrangement. Since the problem didn't specify, but asked for \"n given points\", it's unclear. However, in math competition problems, when such a question is asked, it's often assumed that the points are in general position unless stated otherwise.\n\nBut in the original problem statement, it's just \"n given po... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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point has the same distance to the line), then the answer depends on the configuration of the points. However, in a typical math competition problem, if the answer is a simple formula, it might be that when the points are in general position (no three collinear, etc.), the number is C(n,2), but that doesn't make sense.... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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three points are equidistant from the line. Because the line of symmetry passes through one vertex and the midpoint of the opposite side. The distance from the two other vertices to the line is the same. And the vertex on the line has distance zero. Wait, no. If a line passes through one vertex and the midpoint of the ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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grid with horizontal and vertical lines, there are two midlines. So, the maximum number of such lines is the number of directions in which the points are arranged in two parallel lines.\n\nBut how to generalize this for any n?\n\nSuppose we can arrange n points in k different directions where for each direction, all po... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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given points\", which would require the line to be equidistant from each of the n points. So, all points must be at the same distance from the line. Which, as discussed, requires them to lie on two parallel lines. So, for each such direction where the points lie on two parallel lines, there's one midline. Therefore, th... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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are three pairs of parallel sides. The midlines between each pair would be equidistant to all points. But in this case, each point belongs to two different lines. For example, each vertex is part of one horizontal pair and part of a 60-degree pair and a 120-degree pair. Therefore, all six points are covered by three pa... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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the maximum is 1 for n=3.\n\nWait, so for n=3, the maximum number of lines is 1 if arranged with two points on one line and one on a parallel line, but 0 otherwise. So, the maximum is 1. Similarly, for n=4, arranged as two pairs in two directions, maximum 2. For n=5, arrange as two pairs in two directions and one lefto... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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on the midline. So, its distance to the midline is zero, while the other points are at a non-zero distance. Therefore, the midline is not equidistant to all points. Similarly for vertical midline.\n\nAlternatively, arrange five points such that three are on one line and two on a parallel line. Then, there is one midlin... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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but that's the number of pairs of points. Or maybe the answer is 1 if the points are arranged in two parallel lines, otherwise 0. But without more information, it's hard to say.\n\nWait, but the problem is asking for the number of lines that can be equidistant from n given points. If we interpret this as lines that are... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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the case of a regular hexagon, the midline is not equidistant to all points. Wait, earlier calculation shows that it's not. Let me verify again. Suppose we have a regular hexagon with points on a circle. If we take a midline between two opposite sides, then each of those sides is parallel. The midline is equidistant to... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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lines. But without more information, the problem might be expecting the answer as C(n,2), but that seems unlikely.\n\nAlternatively, recalling that for each pair of points, there's a perpendicular bisector. Even though these bisectors are only equidistant to two points, if multiple pairs lie on the same bisector, then ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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The midlines would be equidistant to all points. But it's unclear.\n\nGiven the problem's ambiguity and the lack of a standard definition for equidistant lines from multiple points, I think the problem is likely expecting the answer \\(\\boxed{\\frac{n(n-1)}{2}}\\) , which is the number of pairs of points and thus the ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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general position, no lines are equidistant from all points. For n=2, one line. For n=1, infinitely many. But the problem states n given points. So, if n=1, infinite; n=2, one; nβ₯3, zero. But the problem says \"n given points\", so if nβ₯3, answer is zero. But in the problem statement, n is not specified; it's a general ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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might have a different interpretation. Maybe \"equidistant from n points\" means that the line is equidistant to each pair of points, i.e., for every pair of points, the line is equidistant to them. In that case, the line must be the perpendicular bisector of every pair of points. Which is only possible if all points a... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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at the same distance, that requires them to lie on two parallels.\n\nTherefore, returning to the original conclusion: the line must have all points at the same distance, which requires them to lie on two parallels. The number of such lines is the number of directions in which the points can be split into two parallels.... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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those two points.\n - For \\( n \\) points, each pair of points defines a unique perpendicular bisector.\n\n2. **Counting Perpendicular Bisectors:**\n - The number of unique pairs of points from \\( n \\) points is given by the combination formula \\( \\binom{n}{2} \\).\n - Therefore, the number of perpendicular bisect... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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- f(y^3) \] We need to find all functions \( f: \mathbb{R} \to \mathbb{R} \) that satisfy this for all real numbers \( x \) and \( y \). Hmm, functional equations can be tricky, but maybe if I plug in some specific values for \( x \) and \( y \), I can figure out the form of \( f \). First, let me see what happens if I... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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= a x^2 + b + a xy + b + a y^2 + b = a(x^2 + xy + y^2) + 3b \] Right-hand side: \[ f(x^3) - f(y^3) = a x^3 + b - (a y^3 + b) = a(x^3 - y^3) \] So, equate them: \ [a(x^2 + xy + y^2) + 3b = a(x^3 - y^3) \] Note that \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \), so substituting that in: Left-hand side becomes: \ [a(x^2 + xy ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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satisfy the equation as well? Wait, but when we assumed \( f \) was linear, we found that only linear functions with \( b = 0 \) work. But maybe there are non-linear solutions too? Let's check. Alternatively, let's consider another substitution. Maybe set \( x = 0 \). Let's set \( x = 0 \). Then the equation becomes: L... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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degree both sides would have. Original equation: Left-hand side: \( (f(x^2) + f(xy) + f(y^2))(x - y) \) If \( f \) is a polynomial of degree \( n \), then \( f(x^2) \) is degree \( 2n \), \( f(xy) \) is degree \( 2n \) (since each term in the polynomial would have variables multiplied, so the degree is the sum of expon... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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solutions. But maybe not. Let's check. Alternatively, perhaps all solutions are linear. Let's see. From the previous substitution when \( y = 0 \): \[ f(x^3) = x [f(x^2) + 2f(0)] + f(0) \] Suppose \( f(0) = c \). Then: \[ f(x^3) = x f(x^2) + 2c x + c \] If we can express \( f(x^3) \) in terms of \( f(x^2) \), maybe we ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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\[ (f(x^2) + f(x) + f(1))(x - 1) = f(x^2)(x - 1) + f(x)(x - 1) + f(1)(x - 1) \] Now, let's bring all terms to one side: \[ x f(x^2) + 2c x + c - f(1) - [f(x^2)(x - 1) + f(x)(x - 1) + f(1)(x - 1)] = 0 \] Simplify term by term: First, \( x f(x^2) - f(x^2)(x - 1) = x f(x^2) - x f(x^2) + f(x^2) = f(x^2) \) Then, \( 2c x \)... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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f(x^3) = x(x - 1)f(x) + x^2(f(1) - 2c) + 2c x + c \] But also, from the original equation with \( y = 1 \), we could relate \( f(x^3) \) to \( f(x^2) \), which we did. But now, maybe this gives us another equation involving \( f(x^3) \), which we can relate to our earlier expressions. Alternatively, let's suppose that ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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= 1 [2f(0) + f(1)] + f(0) \) Solving for \( f(1) \): \( f(1) = 2f(0) + f(1) + f(0) \implies f(1) = 3f(0) + f(1) \implies 0 = 3f(0) \implies f(0) = 0 \) Ah! So \( f(0) = 0 \). That's important. So, from \( y = 1 \), we get \( f(0) = 0 \). So \( c = 0 \). So going back to previous equations: From \( y = 0 \): \( f(x^3) =... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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f(-x^3) \) So, \( 2x [2f(x^2) + f(-x^2)] = f(x^3) - f(-x^3) \) But we know from \( f(x^3) = x f(x^2) \), so: Left-hand side: \( 2x [2f(x^2) + f(-x^2)] \) Right-hand side: \( x f(x^2) - (-x) f((-x)^2) = x f(x^2) + x f(x^2) = 2x f(x^2) \) Therefore: \( 2x [2f(x^2) + f(-x^2)] = 2x f(x^2) \) Divide both sides by 2x (assumi... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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\( a \) is either non-negative or non-positive, and if non-positive, then \( -a \) is non-negative, so \( f(a) = f(-(-a)) = -f(-a) \), hence \( f(-a) = -f(a) \) for all \( a \). So, \( f \) is odd. Given that \( f \) is odd and \( f(0) = 0 \), which we already found. Now, let's go back to the equation from \( y = 1 \):... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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= 0 \implies f(x) = x f(1) \) Therefore, \( f(x) = kx \), where \( k = f(1) \). Thus, \( f \) must be linear. Therefore, the only solutions are linear functions \( f(x) = kx \). We already checked earlier that such functions satisfy the original equation, so these are the solutions. Therefore, the answer is all linear ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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+ 3b)(x - y) = a(x^3 - y^3) \] Since \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \), we have: \[ (a(x^2 + xy + y^2) + 3b)(x - y) = a(x - y)(x^2 + xy + y^2) \] Simplifying, we get: \[ a(x^2 + xy + y^2)(x - y) + 3b(x - y) = a(x - y)(x^2 + xy + y^2) \] This implies: \[ 3b(x - y) = 0 \] Therefore, \( b = 0 \). Hence, \( f(x) = ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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be tricky, but maybe if I plug in some specific values for \\( x \\) and \\( y \\), I can figure out the form of \\( f \\).\n\nFirst, let me see what happens if I set \\( x = y \\). If I substitute \\( x = y \\) into the equation, the left-hand side becomes:\n\n\\[\n(f(x^2) + f(x \\cdot x) + f(x^2))(x - x) = [2f(x^2) +... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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x^3 - y^3 = (x - y)(x^2 + xy + y^2) \\), so substituting that in:\n\nLeft-hand side becomes:\n\n\\\n[a(x^2 + xy + y^2) + 3b = a(x^3 - y^3) + 3b(x - y)\n\\]\n\nRight-hand side is \\( a(x^3 - y^3) \\). Therefore, equating both sides:\n\n\\[\na(x^3 - y^3) + 3b(x - y) = a(x^3 - y^3)\n\\]\n\nSubtract \\( a(x^3 - y^3) \\) fr... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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+ f(y^2)] + f(0)\n\\]\n\nWait a second, earlier when we set \\( y = 0 \\), we had:\n\n\\[\nf(x^3) = [f(x^2) + 2f(0)] x + f(0)\n\\]\n\nComparing these two equations:\n\nFrom \\( x = 0 \\), we have:\n\n\\[\nf(y^3) = y [2f(0) + f(y^2)] + f(0)\n\\]\n\nBut from \\( y = 0 \\), we have:\n\n\\[\nf(x^3) = x [f(x^2) + 2f(0)] + f... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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f(y^2) \\) is degree \\( 2n \\). Therefore, the sum \\( f(x^2) + f(xy) + f(y^2) \\) is degree \\( 2n \\). Then multiplied by \\( x - y \\), which is degree 1. So the left-hand side has degree \\( 2n + 1 \\).\n\nRight-hand side: \\( f(x^3) - f(y^3) \\). \\( f(x^3) \\) is degree \\( 3n \\), so the difference is also degr... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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f(1))(x - 1)\n\\]\n\nRight-hand side:\n\n\\[\nf(x^3) - f(1)\n\\]\n\nSo,\n\n\\[\n(f(x^2) + f(x) + f(1))(x - 1) = f(x^3) - f(1)\n\\]\n\nHmm, this might relate \\( f(x^3) \\) to \\( f(x^2) \\), \\( f(x) \\), and \\( f(1) \\). But we also have the equation from \\( y = 0 \\):\n\n\\[\nf(x^3) = x f(x^2) + 2c x + c\n\\]\n\nWh... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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+ x(f(1) - 2c)\n\\]\n\nNow, this seems like another functional equation. Let me see if we can find \\( f \\) using these relations.\n\nFrom the equation when \\( y = 0 \\), we had:\n\n\\[\nf(x^3) = x f(x^2) + 2c x + c\n\\]\n\nIf we substitute our expression for \\( f(x^2) \\) from above into this, we get:\n\n\\[\nf(x^3... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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0 \\):\n\nWe had \\( f(y^3) = y [2f(0) + f(y^2)] + f(0) \\). If we set \\( y = 0 \\), we get:\n\n\\( f(0) = 0 [2f(0) + f(0)] + f(0) \\implies f(0) = 0 + f(0) \\), which is always true. So no information here. But if we set \\( y = 1 \\), then:\n\n\\( f(1) = 1 [2f(0) + f(1)] + f(0) \\)\n\nSolving for \\( f(1) \\):\n\n\\... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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y = -x \\). Let's see what happens.\n\nOriginal equation:\n\n\\( f(x^2) + f(-x^2) + f(x^2)) = f(x^3) - f(-x^3) \\)\n\nSimplify:\n\nLeft-hand side:\n\n\\( 2f(x^2) + f(-x^2) \\)\n\nRight-hand side:\n\n\\( f(x^3) - f(-x^3) \\)\n\nSo,\n\n\\( 2x [2f(x^2) + f(-x^2)] = f(x^3) - f(-x^3) \\)\n\nBut we know from \\( f(x^3) = x f... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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a \\geq 0 \\), and since for any real number \\( a \\), \\( a \\) is either non-negative or non-positive, and if non-positive, then \\( -a \\) is non-negative, so \\( f(a) = f(-(-a)) = -f(-a) \\), hence \\( f(-a) = -f(a) \\) for all \\( a \\).\n\nSo, \\( f \\) is odd.\n\nGiven that \\( f \\) is odd and \\( f(0) = 0 \\)... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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\\( f \\) must be linear. Therefore, the only solutions are linear functions \\( f(x) = kx \\).\n\nWe already checked earlier that such functions satisfy the original equation, so these are the solutions. Therefore, the answer is all linear functions \\( f(x) = kx \\), where \\( k \\) is a real constant.\n\n**Final Ans... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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3b)(x - y) = a(x - y)(x^2 + xy + y^2)\n \\]\n Simplifying, we get:\n \\[\n a(x^2 + xy + y^2)(x - y) + 3b(x - y) = a(x - y)(x^2 + xy + y^2)\n \\]\n This implies:\n \\[\n 3b(x - y) = 0\n \\]\n Therefore, \\( b = 0 \\). Hence, \\( f(x) = ax \\).\n\n4. **Verify the solution \\( f(x) = ax \\):**\n - Left-hand side:\n \\[\n ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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can't take the square root of a negative number in real numbers. So, the domain is \(x \geq 2\). So, any solutions must be in \(x \geq 2\). Now, let me try to solve the equation. Let me denote \(a = (x+6)^{1/3}\) and \(b = (x-2)^{1/2}\). Then the equation is \(a + b = 2\). Maybe I can express x in terms of a and b and ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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-1 +4 -12 = -8 β 0 a=2: 8 -4 +8 -12 = 0. Oh, a=2 is a root. So we can factor out (a - 2) from the cubic. Using polynomial division or synthetic division. Let's use synthetic division for a=2: Coefficients: 1 (a^3), -1 (a^2), 4 (a), -12 (constant) Bring down 1. Multiply by 2: 1*2=2. Add to next coefficient: -1 +2=1. Mult... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
solutions? Wait, but here, we only found one solution. Let me check. Alternatively, maybe there are other solutions. Let me try plugging in x=3 into the original equation. Left side: (3+6)^(1/3) + (3-2)^(1/2) = 9^(1/3) +1^(1/2)= approximately 2.08 +1=3.08 β 2. So no. x=1: Not in domain. x=10: (16)^(1/3)β2.5198 + (8)^(1/... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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missed something? The cubic equation was \(a^3 -a^2 +4a -12 =0\). We found a=2 is a root. The other roots are complex. So no other real roots. Therefore, no other solutions for a. Therefore, no other x. Hence, only x=2. So sum is 2. But just to double-check, perhaps trying to manipulate the original equation differentl... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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= (5x -10)^2 / (x -2) Multiply both sides by (x -2): (x +10)^2 (x -2) = (5x -10)^2 Expand both sides. Left side: (x^2 +20x +100)(x -2) =x^3 -2x^2 +20x^2 -40x +100x -200 =x^3 +18x^2 +60x -200 Right side: (5x -10)^2 =25x^2 -100x +100 Set equation: x^3 +18x^2 +60x -200 =25x^2 -100x +100 Bring all terms to left: x^3 +18x^2... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\geq 2\) due to the square root term. Let \(a = (x+6)^{1/3}\) and \(b = (x-2)^{1/2}\). The equation becomes: \[ a + b = 2 \] Expressing \(x\) in terms of \(a\) and \(b\): \[ x = a^3 - 6 \] \[ x = b^2 + 2 \] Equating these expressions for \(x\): \[ a^3 - 6 = b^2 + 2 \] \[ a^3 - b^2 = 8 \] Substitute \(b = 2 - a\) into t... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
and the sum of the roots is: \[ \boxed{2} \] | cot | [ { "from": "human", "value": "Find the sum of the roots of the equation \\((x+6)^{1/3} + (x-2)^{1/2} = 2\\)." }, { "from": "gpt", "value": "\nOkay, let's see. I need to find the sum of the roots of the equation \\((x+6)^{1/3} + (x-2)^{1/2} = 2\\). Hmm, roots are the... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
the second equation.\n\nSo substitute into the second equation: \\(a^3 - (2 - a)^2 = 8\\)\n\nLet me compute \\((2 - a)^2\\): that's \\(4 - 4a + a^2\\). So the equation becomes:\n\n\\(a^3 - (4 - 4a + a^2) = 8\\)\n\nSimplify: \\(a^3 - 4 + 4a - a^2 = 8\\)\n\nCombine like terms: \\(a^3 - a^2 + 4a - 4 - 8 = 0\\)\n\nWhich si... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
problem asks for the sum of the roots. If there's only one root, the sum is 2. But maybe I missed something? Let me check.\n\nWait, maybe there are more solutions. Let's check if x=2 is the only solution. Let's test x=2 in the original equation:\n\nLeft side: (2 +6)^(1/3) + (2 -2)^(1/2) = 8^(1/3) + 0^(1/2) = 2 +0 =2, w... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
one root, then the sum is that root. So answer is 2.\n\nBut just to make sure I didn't miss anything. Let me check for possible other roots again.\n\nAlternatively, maybe squaring steps introduced some extraneous roots. Let me retrace the steps:\n\nOriginal equation: \\((x+6)^{1/3} + (x-2)^{1/2} = 2\\)\n\nSet a = (x+6)... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
getting more complicated. Let's see. Let me write:\n\n(x +10)(x -2)^{1/2} =5x -10\n\nDivide both sides by (x -2)^{1/2} (assuming x β 2, but x=2 was already a solution). So,\n\nx +10 = (5x -10)/(x -2)^{1/2}\n\nLet me square both sides to eliminate the square root:\n\n(x +10)^2 = (5x -10)^2 / (x -2)\n\nMultiply both sides... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
\\((x+6)^{1/3} + (x-2)^{1/2} = 2\\), we start by noting the domain \\(x \\geq 2\\) due to the square root term.\n\nLet \\(a = (x+6)^{1/3}\\) and \\(b = (x-2)^{1/2}\\). The equation becomes:\n\\[ a + b = 2 \\]\n\nExpressing \\(x\\) in terms of \\(a\\) and \\(b\\):\n\\[ x = a^3 - 6 \\]\n\\[ x = b^2 + 2 \\]\n\nEquating th... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
aops_c6_high_school_olympiads | QwQ-32B | 0.0 | "Determine how many 1000 digit numbers \\( A \\) have the property that when any digit of \\( A \\),(...TRUNCATED) | "\nOkay, so I need to figure out how many 109-digit numbers A have the property that if you d(...TRUNCATED) | cot | [{"from":"human","value":"Determine how... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
| DeepSeek-R1 | 0.9375 | On average, how long will you have to flip a coin to get the sequence $HTHT$? | "\nOkay, so I need to figure out the expected number of coin flips needed to get the sequence(...TRUNCATED) | cot | {"from":"human","value":"On average, how long will you have to flip a coin to get the sequence $HTH... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
"problem_type" is "has_answer_extracted" and is set to "n/a" for other cases. []( OpenMath-Nemotron models ----------------------------------------------------------------------------------------------------------------------------- To demonstrate the quality of this dataset, we release a series of OpenMath-Nemotron mo... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
CoT | 76.3 (83.3) | 63.0 (76.7) | 52.1 (60.7) | 7.5 (7.6) | | OpenMath-Nemotron-14B | 61.3 (76.7) | 58.6 (70.9) | 9.5 (11.5) | | + Self GenSelect | 86.7 | 76.7 | 72.4 | 14.1 | | + 32B GenSelect | 90.0 | 76.7 | 71.9 | 13.7 | | QwQ-32B | 78.1 (86.7) | 66.5 (76.7) | 55.9 (63.3) | 9.0 (9.5) | | DeepSeek-R1-Distill-Qwen-32B... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
Ethical Considerations: -------------------------------------------------------------------------------------------------------------------------- NVIDIA believes Trustworthy AI is a shared responsibility and we have established policies and practices to enable development for a wide array of AI applications. When down... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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Title: Algebraic Numbers and Number Fields URL Source: Markdown Content: Hide navigation sidebar Hide table of contents sidebar Skip to content Toggle site navigation sidebar Algebraic Numbers and Number Fields Toggle Light / Dark / Auto color theme Toggle table of contents sidebar Version 10.6 Reference Manual other ... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
x + 1]); KL Number Field in b0 with defining polynomial x^4 - 420*x^2 + 40000 over its base field Python We do some arithmetic in a tower of relative number fields: Sage sage: K. = NumberField(x^3 - 2) sage: L. = K.extension(x^3 - 3) sage: S. = L.extension(x^2 - 2) sage: S Number Field in sqrt2 with defining polynomial... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
is_galois(), add is_abelian(), building on work in patch by Chris Wuthrich Anna Haensch (2018-03): added quadratic_defect() Michael Daub, Chris Wuthrich (2020-09-01): added Dirichlet characters for abelian fields class sage.rings.number_field.number_field.CyclotomicFieldFactory [source] Bases: UniqueFactory Return the ... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
numbers: Sage sage: CyclotomicField(1) Cyclotomic Field of order 1 and degree 1 Python Due to their default embedding into πΆ , cyclotomic number fields are all compatible. Sage sage: cf30 = CyclotomicField(30) sage: cf5 = CyclotomicField(5) sage: cf3 = CyclotomicField(3) sage: cf30.gen() + cf5.gen() + cf3.gen() zeta30... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
list of primes (default: None); if not None, then the maximal order is computed by maximizing only at the primes in this list, which completely avoids having to factor the discriminant, but of course can lead to wrong results; only applies for absolute fields at present. structure β None, a list or an instance of struc... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
'y') sage: k. = NumberField(x^2 + 3) sage: m. = NumberField(y^2 + 3) sage: k Number Field in a with defining polynomial x^2 + 3 sage: m Number Field in a with defining polynomial y^2 + 3 sage: k == m False Python In case of conflict of the generator name with the name given by the preparser, the name given by the prepa... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
K. = NumberField(x^3 - 2, embedding=f.roots()) sage: a + L(1) 4 + 2*5^2 + 2*5^3 + 3*5^4 + 5^5 + 4*5^6 + 2*5^8 + 3*5^9 + 4*5^12 + 4*5^14 + 4*5^15 + 3*5^16 + 5^17 + 5^18 + 2*5^19 + O(5^20) sage: L. = NumberField(x^6 - x^2 + 1/10, embedding=1) sage: K. = NumberField(x^3 - x + 1/10, embedding=b^2) sage: a + b b^2 + b sage:... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
Python Elements and ideals may be converted to and from PARI as follows: Sage sage: pari(a) Mod(-1/2*y^2 + 1/2*y, y^3 - y^2 - 2) sage: K(pari(a)) a sage: I = K.ideal(a); I Fractional ideal (a) sage: I.pari_hnf() [1, 0, 0; 0, 1, 0; 0, 0, 1/2] sage: K.ideal(I.pari_hnf()) Fractional ideal (a) Python Here is an example whe... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
only applies for absolute fields at present. structure β None or an instance of structure.NumberFieldStructure (default: None), internally used to pass in additional structural information, e.g., about the field from which this field is created as a subfield. create_key_and_extra_args(polynomial, name, check, embedding... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
field from which it was created. OUTPUT: The relative number field generated by a root of the first entry of polynomials over the relative number field generated by root of the second entry of polynomials β¦ over the number field over which the last entry of polynomials is defined. EXAMPLES: Sage sage: x = polygen(ZZ, '... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
= NumberField([x^2 + 3, x^2 + 2, x^2 + 1]); K Number Field in a0 with defining polynomial x^2 + 3 over its base field sage: K.base_field() Number Field in a1 with defining polynomial x^2 + 2 over its base field sage: K.base_field().base_field() Number Field in a2 with defining polynomial x^2 + 1 Python LaTeX versions o... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
an element of K prec β (default: None) the precision of the real field OUTPUT: the absolute value as a real number EXAMPLES: Sage sage: x = polygen(QQ, 'x') sage: K. = NumberField(x^3 - 3) sage: phi_real = K.places() sage: phi_complex = K.places() sage: v_fin = tuple(K.primes_above(3)) sage: K.abs_val(phi_real, xi^2) 2... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
To: Number Field in a with defining polynomial x^3 - 5, Isomorphism map: From: Number Field in a with defining polynomial x^3 - 5 To: Vector space of dimension 3 over Rational Field) Python automorphisms() [source] Compute all Galois automorphisms of self. This uses PARIβs pari:nfgaloisconj and is much faster than root... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
A) == f.monic() True Python base_field() [source] Return the base field of self, which is always QQ. EXAMPLES: Sage sage: K = CyclotomicField(5) sage: K.base_field() Rational Field Python change_names(names) [source] Return number field isomorphic to self but with the given generator name. INPUT: names β should be exac... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
EXAMPLES: There are no elements in a number field with multiplicative height less than 1: Sage sage: x = polygen(QQ, 'x') sage: K. = NumberField(x^5 - x + 19) sage: list(K.elements_of_bounded_height(bound=0.9)) [] Python The only elements in a number field of height 1 are 0 and the roots of unity: Sage sage: K. = Numbe... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
If possible, the most natural embedding of this field into πΎ is put first in the list. INPUT: K β a field EXAMPLES: Sage sage: # needs sage.groups sage: x = polygen(QQ, 'x') sage: K. = NumberField(x^3 - 2) sage: L. = K.galois_closure(); L Number Field in a1 with defining polynomial x^6 + 108 sage: K.embeddings(L) Ring... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
sage: CyclotomicField(5).embeddings(QQbar) [Ring morphism: From: Cyclotomic Field of order 5 and degree 4 To: Algebraic Field Defn: zeta5 |--> 0.3090169943749474? + 0.9510565162951536?*I, Ring morphism: From: Cyclotomic Field of order 5 and degree 4 To: Algebraic Field Defn: zeta5 |--> -0.8090169943749474? + 0.58778525... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
Field sage: to_V Isomorphism map: From: Number Field in a with defining polynomial x^3 + 2 To: Vector space of dimension 3 over Rational Field sage: from_V(to_V(2/3*a - 5/8)) 2/3*a - 5/8 sage: to_V(from_V(V([0,-1/7,0]))) (0, -1/7, 0) Python galois_closure(names=None, map=False) [source] Return number field πΎ that is t... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
Python hilbert_conductor(a, b) [source] This is the product of all (finite) primes where the Hilbert symbol is β 1 . What is the same, this is the (reduced) discriminant of the quaternion algebra ( π , π ) over a number field. INPUT: a, b β elements of the number field self OUTPUT: squarefree ideal of the ring of int... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
sage: K.hilbert_symbol(0, a + 5) 0 sage: K.hilbert_symbol(a, 0) 0 sage: K.hilbert_symbol(-a, a + 1) 1 sage: K.hilbert_symbol(-a, a + 2) -1 sage: K.hilbert_symbol(a, a + 5) -1 Python That the latter two are unsolvable should be visible in local obstructions. For the first, this is a prime ideal above 19. For the second,... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
QuadraticField(5) sage: K.hilbert_symbol(-1, -1, K.embeddings(CC)) Traceback (most recent call last): ... ValueError: Possibly real place (=Ring morphism: From: Number Field in a with defining polynomial x^2 - 5 with a = 2.236067977499790? To: Complex Field with 53 bits of precision Defn: a |--> -2.23606797749979) give... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
at Issue #16043 has been fixed: Sage sage: K. = NumberField(x^2 + 5) sage: p = K.primes_above(2); p Fractional ideal (2, a + 1) sage: K.hilbert_symbol(2*a, -1, p) 1 sage: K.hilbert_symbol(2*a, 2, p) -1 sage: K.hilbert_symbol(2*a, -2, p) -1 Python AUTHOR: Aly Deines (2010-08-19): part of the doctests Marco Streng (2010-... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
k.hilbert_conductor(a, b) Fractional ideal (15) Python AUTHORS: Simon Brandhorst, Anna Haensch (01-05-2018) is_absolute() [source] Return True since self is an absolute field. EXAMPLES: Sage sage: K = CyclotomicField(5) sage: K.is_absolute() True Python logarithmic_embedding(prec=53) [source] Return the morphism of sel... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
each pair of complex conjugate embeddings (in our case, we simply choose the one where the image of πΌ has positive real part). Here ( π , π ) is the signature of self. Then the Minkowski embedding is given by π₯ β¦ ( π 1 ( π₯ ) , β¦ , π π ( π₯ ) , 2 β ( π π + 1 ( π₯ ) ) , 2 β ( π π + 1 ( π₯ ) ) , β¦ , 2 β ( π οΏ½... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
= K.optimized_representation() sage: L # your answer may different, since algorithm is random Number Field in b1 with defining polynomial x^8 + 4*x^6 + 7*x^4 + 36*x^2 + 81 sage: to_L(K.0) # random 4/189*b1^7 + 1/63*b1^6 + 1/27*b1^5 - 2/9*b1^4 - 5/27*b1^3 - 8/9*b1^2 + 3/7*b1 - 3/7 sage: from_L(L.0) # random 1/1152*b^7 -... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
z in L] # random -- since algorithm is random [Number Field in b0 with defining polynomial x - 1, Number Field in b1 with defining polynomial x^2 - x + 1, Number Field in b2 with defining polynomial x^4 - 5*x^2 + 25, Number Field in b3 with defining polynomial x^4 - 2*x^2 + 4, Number Field in b4 with defining polynomia... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
generator is integral check_rank β boolean (default: True); whether to check that the ring generated by gens is of full rank allow_subfield β boolean (default: False); if True and the generators do not generate an order, i.e., they generate a subring of smaller rank, instead of raising an error, return an order in a sm... | {
"page_id": null,
"source": 6843,
"title": "from dpo"
} |
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