🧠:Okay, let me try to figure out this geometry problem. So, we have triangle ABC with a point O somewhere in its plane. Then, lines are drawn through each vertex (A, B, C) and point O, intersecting the opposite sides at points C1, B1, and A1 respectively. The claim is that the product of three segments without a common vertex is equal to another product: AC1·BA1·CB1 = AB1·CA1·BC1. Hmm, interesting. I need to prove this.First, let me visualize this. Triangle ABC, point O inside or outside? Probably works either way. From each vertex, we draw a line through O, which intersects the opposite side. So, line AO intersects BC at A1, line BO intersects AC at B1, and line CO intersects AB at C1. Wait, the problem statement says the lines intersect AB, AC, BC at C1, B1, A1 respectively. Wait, let me confirm. It says: "intersecting the sides AB, AC, BC respectively at points C1, B1, and A1". So, line through A and O intersects BC at A1? Wait, no. Wait, the lines are drawn through the vertices A, B, C and an arbitrary point O. So, for example, line AO would go through A and O, but intersects BC at A1. Similarly, line BO goes through B and O, intersects AC at B1, and line CO goes through C and O, intersects AB at C1. Wait, but the problem states "intersecting the sides AB, AC, BC respectively at points C1, B1, and A1". Wait, that's conflicting.Wait, let me parse the problem again: "If lines are drawn through the vertices A, B, C of triangle ABC and an arbitrary point O in its plane intersecting the sides AB, AC, BC respectively at points C1, B1, and A1..." So, the lines are through A and O, intersecting AB? But AB is already a side, so if we draw a line through A and O, unless O is on AB, the line would just be AO. But AO is passing through A, so it can't intersect AB again unless O is on AB. But O is an arbitrary point. Wait, maybe the problem is phrased as: lines drawn through each vertex A, B, C and point O, intersecting the opposite sides. Wait, perhaps the original problem is misphrased? Let me check again.Original problem: "lines are drawn through the vertices A, B, C of triangle ABC and an arbitrary point O in its plane intersecting the sides AB, AC, BC respectively at points C1, B1, and A1". So, the lines through A, O intersect AB at C1; lines through B, O intersect AC at B1; lines through C, O intersect BC at A1? That seems strange because a line through A and O would start at A and go through O. If O is not on AB, then the line AO would intersect AB only at A. So, unless O is on AB extended, but the problem says "intersecting the sides AB, AC, BC respectively", meaning the intersections are on the sides (not their extensions). Hmm, this seems confusing.Wait, perhaps the lines are not through both the vertex and O, but lines through the vertex and O intersect the opposite side. For example, line AO (through A and O) intersects BC at A1; line BO (through B and O) intersects AC at B1; line CO (through C and O) intersects AB at C1. Then, the intersections are on BC, AC, AB respectively. But the problem says "intersecting the sides AB, AC, BC respectively at points C1, B1, and A1". So that would mean line AO intersects AB at C1? But AO passes through A, which is already a vertex of AB. So, unless O is on AB, the line AO would intersect AB only at A. Therefore, if the intersection is at C1 on AB, then C1 must coincide with A, which doesn't make sense. Similarly, the line through B and O intersects AC at B1. If O is not on AC, then the line BO intersects AC at some point B1. Similarly, line CO intersects BC at A1.Wait, perhaps there's a misinterpretation here. Maybe the lines are not through the vertex and O, but lines through each vertex and O, which intersect the opposite sides. For example, line AO intersects BC at A1, line BO intersects AC at B1, line CO intersects AB at C1. Then, the intersections are on BC, AC, AB. But the problem says "intersecting the sides AB, AC, BC respectively at points C1, B1, and A1". So line AO intersects AB at C1? Wait, that seems conflicting. Maybe the problem is written differently. Let me check again:"If lines are drawn through the vertices A, B, C of triangle ABC and an arbitrary point O in its plane intersecting the sides AB, AC, BC respectively at points C1, B1, and A1..."Wait, maybe "lines are drawn through the vertices A, B, C and point O", so each line is through two points: A and O, B and O, C and O. Then, these lines intersect the sides AB, AC, BC respectively at points C1, B1, A1. But if line AO is drawn, it goes through A and O. Then, it intersects AB at A (since it's the line AB itself if O is on AB). But if O is not on AB, then line AO intersects AB only at A. So, unless O is on AB, line AO can't intersect AB at another point. Similarly, line BO through B and O intersects AC at B1. If O is not on AC, line BO would intersect AC at some point B1. Similarly, line CO through C and O intersects BC at A1.But according to the problem statement, the lines are drawn through the vertices A, B, C and point O, intersecting AB, AC, BC respectively. So line AO intersects AB at C1, line BO intersects AC at B1, line CO intersects BC at A1. But unless O is on AB, AO can't intersect AB again except at A, so C1 would be A. That can't be. Therefore, maybe there's a misunderstanding here.Alternatively, perhaps the lines are drawn through each vertex (A, B, C) and point O, and these lines intersect the opposite sides. For example, line AO (through A and O) intersects BC at A1; line BO (through B and O) intersects AC at B1; line CO (through C and O) intersects AB at C1. Then, the intersections are on BC, AC, AB. But the problem says the lines intersect AB, AC, BC respectively. So, line AO intersects AB at C1, line BO intersects AC at B1, line CO intersects BC at A1. This is conflicting because line AO would pass through A and O, so unless O is on AB, AO can't intersect AB again except at A. Therefore, C1 would be A. Similarly, line BO intersects AC at B1. If O is not on AC, then line BO would intersect AC at B1. Line CO intersects BC at A1. If O is not on BC, line CO would intersect BC at A1.But in this case, if the lines AO, BO, CO intersect AB, AC, BC at C1, B1, A1 respectively, then C1 is the intersection of AO with AB, which is just point A. Similarly, A1 is the intersection of CO with BC, which is point C if O is on BC. Wait, this is confusing. Maybe the problem is mistyped? Alternatively, perhaps the lines are drawn through each vertex and O, and then intersect the other sides. For example, line AO intersects BC at A1, line BO intersects AC at B1, line CO intersects AB at C1. Then, the intersections are on BC, AC, AB. Then, the segments would be BA1 (on BC), CB1 (on AC), and AC1 (on AB). Then, the product is AC1·BA1·CB1. Similarly, the other product is AB1·CA1·BC1. Then, this seems like Ceva's Theorem.Wait, Ceva's Theorem states that for concurrent lines (lines through the vertices intersecting the opposite sides), the product of the ratios is equal to 1. Specifically, (BA1/A1C)·(CB1/B1A)·(AC1/C1B) = 1. But in this problem, the claim is that the products of the segments (not ratios) are equal: AC1·BA1·CB1 = AB1·CA1·BC1.Wait, but maybe this is equivalent to Ceva's Theorem? Let me check. Suppose in Ceva's Theorem, we have (BA1/A1C)·(CB1/B1A)·(AC1/C1B) = 1. If we cross-multiply, we get (BA1)(CB1)(AC1) = (A1C)(B1A)(C1B). But the problem states AC1·BA1·CB1 = AB1·CA1·BC1. Hmm, the left sides are similar: AC1, BA1, CB1. On the right side, AB1, CA1, BC1. Wait, if I rewrite the Ceva equation:BA1 * CB1 * AC1 = A1C * B1A * C1B.But in the problem statement, the right-hand side is AB1 * CA1 * BC1. Let me see:Is AB1 equal to B1A? Wait, AB1 is a segment on AC, right? Wait, no. If B1 is on AC, then AB1 is the segment from A to B1 on AC, and B1A is the same as AB1. Similarly, CA1 is the segment from C to A1 on BC, so CA1 = A1C. Similarly, BC1 is the segment from B to C1 on AB, which is the same as C1B. Therefore, AB1 * CA1 * BC1 = B1A * A1C * C1B. Therefore, the Ceva equation is BA1 * CB1 * AC1 = AB1 * CA1 * BC1, which is exactly the equation given in the problem. Therefore, this is equivalent to Ceva's Theorem. Therefore, the given equation is a restatement of Ceva's Theorem, so the equality holds when the lines AO, BO, CO are concurrent (i.e., meet at point O). Therefore, since O is the point of concurrency, Ceva's condition is satisfied, hence the product equality holds.Wait, but in Ceva's Theorem, the concurrency is equivalent to the product of the ratios being 1. But here, the problem is stating that the products of the segments themselves are equal. So, is this a different formulation? Let me verify with actual segments.Suppose in Ceva's Theorem, we have:(BA1 / A1C) * (CB1 / B1A) * (AC1 / C1B) = 1.Multiplying both sides by (A1C * B1A * C1B), we get:BA1 * CB1 * AC1 = A1C * B1A * C1B.But in the problem statement, the right-hand side is AB1 * CA1 * BC1. As mentioned before, AB1 is the same as B1A (since AB1 is from A to B1 on AC, but wait, hold on. If B1 is on AC, then AB1 is along AC from A to B1, but AC is a different side. Wait, maybe I confused the notation.Wait, let's clarify the segments:- AC1: This is on AB. Wait, no. If C1 is the intersection of line CO with AB, then AC1 is the segment from A to C1 on AB.- BA1: If A1 is the intersection of line AO with BC, then BA1 is the segment from B to A1 on BC.- CB1: If B1 is the intersection of line BO with AC, then CB1 is the segment from C to B1 on AC.Similarly:- AB1: This would be the segment from A to B1 on AC.Wait, no. If B1 is on AC, then AB1 is along AC from A to B1? Wait, no, AC is a side from A to C. If B1 is a point on AC, then AB1 is not along AB. Wait, this is confusing.Wait, maybe the notation is such that, for example, AC1 refers to the segment on AB. Wait, let's get precise.Let me define the points:- Line AO (through A and O) intersects BC at A1.- Line BO (through B and O) intersects AC at B1.- Line CO (through C and O) intersects AB at C1.Therefore, the segments are:- On BC: BA1 and A1C (from B to A1 and A1 to C).- On AC: CB1 and B1A (from C to B1 and B1 to A).- On AB: AC1 and C1B (from A to C1 and C1 to B).Therefore, the product in the problem is AC1 (on AB) * BA1 (on BC) * CB1 (on AC). The other product is AB1 (on AC) * CA1 (on BC) * BC1 (on AB). Wait:AB1 is on AC: from A to B1.CA1 is on BC: from C to A1.BC1 is on AB: from B to C1.So, in terms of segments:AC1 = A to C1 on AB.BA1 = B to A1 on BC.CB1 = C to B1 on AC.AB1 = A to B1 on AC.CA1 = C to A1 on BC.BC1 = B to C1 on AB.So, according to Ceva's Theorem, (BA1/A1C)*(CB1/B1A)*(AC1/C1B) = 1.But in the problem statement, the equation is:AC1 * BA1 * CB1 = AB1 * CA1 * BC1.But let's express all terms in terms of segments:BA1 is BA1 (on BC), A1C is the other part of BC.Similarly, CB1 is CB1 (on AC), B1A is the remaining part of AC.AC1 is AC1 (on AB), C1B is the remaining part of AB.So, substituting into Ceva:(BA1 / A1C) * (CB1 / B1A) * (AC1 / C1B) = 1.Multiplying both sides by (A1C * B1A * C1B):BA1 * CB1 * AC1 = A1C * B1A * C1B.But in the problem statement, the right-hand side is AB1 * CA1 * BC1. Let's see:AB1 is on AC: from A to B1, which is B1A in reverse? Wait, B1A is from B1 to A, which is the same as AB1 but direction matters? Wait, in terms of length, AB1 = B1A. Similarly, CA1 is from C to A1 on BC, which is the same as A1C. And BC1 is from B to C1 on AB, which is the same as C1B. Therefore, AB1 * CA1 * BC1 = B1A * A1C * C1B. Therefore, the equation from Ceva is BA1 * CB1 * AC1 = AB1 * CA1 * BC1, which is exactly the equation given in the problem. Therefore, this equality is a direct consequence of Ceva's Theorem. Since the lines are drawn through O, which is the point of concurrency (as all three lines pass through O), Ceva's condition is satisfied, hence the equality holds.Therefore, the problem statement is essentially an application of Ceva's Theorem, where the concurrency of the lines at O ensures that the product of the segments as given is equal. Therefore, the proof follows directly from Ceva's Theorem.Alternatively, if one is to prove this without invoking Ceva's Theorem directly, perhaps using mass point geometry or coordinate geometry. Let me try to outline a coordinate geometry approach.Let’s assign coordinates to the triangle. Let’s place point A at (0, 0), B at (1, 0), and C at (0, 1). Let point O be at some arbitrary point (p, q). Then, lines AO, BO, and CO will intersect the opposite sides at A1, B1, C1.First, find the coordinates of A1, B1, C1.Line AO: passes through A (0,0) and O (p, q). Parametric equation: (tp, tq) for t ∈ ℝ. This line intersects BC at A1. Side BC goes from B (1, 0) to C (0, 1). The equation of BC is x + y = 1.Find intersection between AO and BC. Substitute x = tp, y = tq into x + y = 1:tp + tq = 1 ⇒ t(p + q) = 1 ⇒ t = 1/(p + q). Therefore, coordinates of A1 are (p/(p + q), q/(p + q)).Similarly, line BO: passes through B (1, 0) and O (p, q). Parametric equation: (1 + t(p - 1), 0 + t(q - 0)) = (1 + t(p - 1), tq). This line intersects AC at B1. Side AC is from A (0,0) to C (0,1), which is the line x = 0.Find intersection: set x-coordinate to 0: 1 + t(p - 1) = 0 ⇒ t = -1/(p - 1). Then y-coordinate is tq = -q/(p - 1). Therefore, coordinates of B1 are (0, -q/(p - 1)).Similarly, line CO: passes through C (0,1) and O (p, q). Parametric equation: (0 + t(p - 0), 1 + t(q - 1)) = (tp, 1 + t(q - 1)). This line intersects AB at C1. Side AB is from A (0,0) to B (1,0), which is the line y = 0.Find intersection: set y = 0: 1 + t(q - 1) = 0 ⇒ t = -1/(q - 1). Then x-coordinate is tp = -p/(q - 1). Therefore, coordinates of C1 are (-p/(q - 1), 0).Now, compute the lengths of the segments:First, AC1: along AB from A (0,0) to C1 (-p/(q - 1), 0). But wait, AB is from (0,0) to (1,0). If C1 is at (-p/(q - 1), 0), that's outside AB unless p/(q - 1) is negative. Since coordinates depend on the position of O, we might have directed segments here. The length AC1 can be considered as a signed length. However, since the problem talks about products of segments, it might be considering directed lengths (with sign based on direction). Alternatively, the problem might be assuming that all intersections are on the sides (not extended sides), which would require O to be inside the triangle, but the problem states "an arbitrary point O in its plane", so possibly considering extended sides.But regardless, using coordinates, we can compute the directed lengths.AC1: from A (0,0) to C1 (-p/(q - 1), 0). The directed length is -p/(q - 1). However, in the problem statement, segments are probably considered as positive lengths, but since we’re dealing with products, signs might cancel out. Let me proceed carefully.Compute AC1: The x-coordinate difference is (-p/(q - 1)) - 0 = -p/(q - 1). Since AB is along the x-axis, the length is | -p/(q - 1) |. But let's keep it as a directed length for now: AC1 = -p/(q - 1).BA1: from B (1,0) to A1 (p/(p + q), q/(p + q)). The vector from B to A1 is (p/(p + q) - 1, q/(p + q) - 0) = (- (q)/(p + q), q/(p + q)). The length can be computed, but since we are dealing with products, maybe we can use coordinates to find ratios.But perhaps it's easier to compute the product AC1·BA1·CB1 and AB1·CA1·BC1 using coordinates.First, let's compute each segment:AC1: along AB from A to C1. If C1 is at x = -p/(q - 1), then the length is | -p/(q - 1) |. But AB is from x=0 to x=1, so if C1 is on AB, then x must be between 0 and 1, but depending on O's position, C1 could be on the extension. Since O is arbitrary, we might have to consider directed segments.BA1: along BC from B to A1. Coordinates of B (1,0) to A1 (p/(p + q), q/(p + q)). The length can be calculated using the distance formula, but maybe we can parametrize it.Alternatively, since BC is parameterized as x + y = 1, the length from B (1,0) to A1 (p/(p + q), q/(p + q)) can be expressed as the ratio BA1 / BC. But BC has length √2, but since we are dealing with products, maybe it's better to use coordinates.Alternatively, use the parameter t from earlier. For line AO intersecting BC at A1, we found t = 1/(p + q). Therefore, the coordinates of A1 are (p/(p + q), q/(p + q)). The vector from B to A1 is (p/(p + q) - 1, q/(p + q) - 0) = (-q/(p + q), q/(p + q)). The length BA1 is the norm of this vector: sqrt[ (-q/(p + q))^2 + (q/(p + q))^2 ] = sqrt[ q^2/(p + q)^2 + q^2/(p + q)^2 ] = sqrt[ 2q^2/(p + q)^2 ] = q√2 / (p + q). But again, since we're multiplying segments, maybe we can use ratios instead.Alternatively, since all the segments are along the sides of the triangle, perhaps express them in terms of the parameters found.From the coordinates:AC1 is the x-coordinate of C1, which is -p/(q - 1). But since AB is from (0,0) to (1,0), the length from A to C1 is | -p/(q - 1) |, but considering direction, it's -p/(q - 1). However, in the problem statement, segments are probably considered as positive lengths, but in the context of Ceva's Theorem, directed segments (signed lengths) are used. So perhaps we need to consider the signs.Similarly, BA1: from B (1,0) to A1 (p/(p + q), q/(p + q)). The displacement is (p/(p + q) - 1, q/(p + q)) = (-q/(p + q), q/(p + q)). The length is sqrt[ ( -q/(p + q) )^2 + ( q/(p + q) )^2 ] = q/(p + q) * sqrt(2). But again, if we consider directed segments, maybe just take the x-component or y-component. Alternatively, since BA1 is along BC, which is a line from B to C, perhaps parametrize BC with a parameter t from 0 to 1, where t=0 is B and t=1 is C. The coordinates of A1 are (p/(p + q), q/(p + q)). Since BC is parameterized by t as (1 - t, t). Therefore, (1 - t, t) = (p/(p + q), q/(p + q)), so 1 - t = p/(p + q) ⇒ t = q/(p + q). Therefore, BA1 corresponds to t = q/(p + q), so the length from B to A1 is t * |BC|. But |BC| is sqrt(2). However, since we are dealing with ratios, the actual length might not matter, but the proportion.Alternatively, in terms of mass point or barycentric coordinates, but this could get complicated.Alternatively, let's compute the left-hand side product AC1·BA1·CB1.AC1: length from A to C1 on AB. Since C1 is at (-p/(q - 1), 0), which is a point on AB extended beyond A if q ≠ 1. The distance from A to C1 is | -p/(q - 1) |. But since AB is from (0,0) to (1,0), if C1 is between A and B, then 0 ≤ -p/(q - 1) ≤ 1, but this depends on the values of p and q. Since O is arbitrary, C1 can be inside or outside AB. However, for the sake of calculation, let's compute the directed length. The coordinate of C1 is x = -p/(q - 1). So the directed length AC1 is x - 0 = -p/(q - 1).BA1: length from B to A1 on BC. Coordinates of B is (1,0), A1 is (p/(p + q), q/(p + q)). The vector from B to A1 is (p/(p + q) - 1, q/(p + q) - 0) = (-q/(p + q), q/(p + q)). The length BA1 can be calculated as sqrt[ (-q/(p + q))² + (q/(p + q))² ] = q/(p + q) * sqrt(2). But since we are dealing with products, maybe we need to find a ratio.But perhaps instead of actual lengths, use the parameter t where A1 divides BC in the ratio t:1-t. From earlier, we found that t = q/(p + q) for BA1. So BA1 corresponds to t = q/(p + q), so BA1 = t * BC. Similarly, CB1: on AC, which is from C (0,1) to A (0,0). The point B1 is at (0, -q/(p - 1)). The distance from C to B1 is |1 - (-q/(p - 1))| = |1 + q/(p - 1)|. Wait, coordinates of B1 are (0, -q/(p - 1)), so from C (0,1) to B1 (0, -q/(p - 1)), the directed length is (-q/(p - 1) - 1) = - (q + p - 1)/(p - 1). But this seems messy.Alternatively, maybe express all segments in terms of p and q using the coordinates we found:AC1 = | -p/(q - 1) | (but directed length is -p/(q - 1)).BA1 = sqrt[ (-q/(p + q))² + (q/(p + q))² ] = q√2 / (p + q).CB1: from C (0,1) to B1 (0, -q/(p - 1)). The distance is |1 - (-q/(p - 1))| = |1 + q/(p - 1)|.Similarly, AB1: from A (0,0) to B1 (0, -q/(p - 1)). The distance is | -q/(p - 1) |.CA1: from C (0,1) to A1 (p/(p + q), q/(p + q)). The distance is sqrt[ (p/(p + q) - 0)^2 + (q/(p + q) - 1)^2 ] = sqrt[ p²/(p + q)² + ( - (p)/(p + q) )² ] = sqrt[ (p² + p²)/(p + q)^2 ] = p√2 / (p + q).BC1: from B (1,0) to C1 (-p/(q - 1), 0). The distance is | -p/(q - 1) - 1 | = | -(p + q - 1)/(q - 1) |.This is getting too complicated. Maybe instead of computing actual lengths, notice that in the coordinate system, the product AC1·BA1·CB1 and AB1·CA1·BC1 can be expressed in terms of p and q, and show they are equal.But given the complexity, perhaps using Ceva's Theorem is the better approach. Since Ceva's Theorem directly relates to concurrent lines in a triangle and gives a product formula, which matches the given equation.Alternatively, using Menelaus' Theorem. But Menelaus applies to a transversal cutting across a triangle, giving a product of ratios equal to 1. But here, we have three lines concurrent at O.Given that the problem's equation is a restatement of Ceva's Theorem in terms of products of segments rather than ratios, and since the concurrency at O ensures Ceva's condition is satisfied, the equality holds.Therefore, the proof follows from Ceva's Theorem.Final Answerboxed{AC_1 cdot BA_1 cdot CB_1 = AB_1 cdot CA_1 cdot BC_1}

✅:To prove the given statement, we recognize that it is a direct application of Ceva's Theorem. Ceva's Theorem states that for a point ( O ) inside triangle ( ABC ), the lines drawn through each vertex and ( O ) intersecting the opposite sides are concurrent if and only if the product of the ratios of the segments formed on each side is equal to 1. Given the problem's setup, lines through vertices ( A, B, C ) and point ( O ) intersect the sides ( BC, AC, AB ) at points ( A_1, B_1, C_1 ) respectively. According to Ceva's Theorem, this concurrency implies:[frac{BA_1}{A_1C} cdot frac{CB_1}{B_1A} cdot frac{AC_1}{C_1B} = 1]Multiplying both sides by the denominators, we get:[BA_1 cdot CB_1 cdot AC_1 = A_1C cdot B_1A cdot C_1B]Recognizing that ( A_1C = CA_1 ), ( B_1A = AB_1 ), and ( C_1B = BC_1 ) (as segments on the respective sides), we substitute these into the equation:[BA_1 cdot CB_1 cdot AC_1 = AB_1 cdot CA_1 cdot BC_1]Thus, the given equality ( AC_1 cdot BA_1 cdot CB_1 = AB_1 cdot CA_1 cdot BC_1 ) is proven by Ceva's Theorem. [boxed{AC_1 cdot BA_1 cdot CB_1 = AB_1 cdot CA_1 cdot BC_1}]

🔑:1. Consider the triangles formed by the point O and the vertices of the triangle ABC. Specifically: [ triangle AOB text{ and } triangle AOC ] These two triangles share a common base AO. The ratio of their areas is equal to the ratio of their corresponding heights, which drop perpendiculars from points B and C to the line segment AO.2. Denote the heights from B and C dropping to AO as being intersected at points A_1 and A_1 respectively. Consequently, the ratio of areas of triangle AOB to triangle AOC is given by: [ frac{text{Area}( triangle AOB )}{text{Area}( triangle AOC )} = frac{BA_1}{CA_1} ] Similarly, we can form ratios for triangles triangle BOC and triangle BOA, and triangle COA and triangle COB respectively. That is: [ frac{text{Area}( triangle BOC )}{text{Area}( triangle BOA )} = frac{CB_1}{AB_1} ] [ frac{text{Area}( triangle COA )}{text{Area}( triangle COB )} = frac{AC_1}{BC_1} ]3. Considering O could be any point in the plane, we multiply these area ratios together: [ left( frac{BA_1}{CA_1} right) left( frac{CB_1}{AB_1} right) left( frac{AC_1}{BC_1} right) ] Simplifying the above expression yields: [ frac{BA_1 cdot CB_1 cdot AC_1}{CA_1 cdot AB_1 cdot BC_1} = 1 ]4. Given the equality established in the last step, we have equivalently shown: [ AC_1 cdot BA_1 cdot CB_1 = AB_1 cdot CA_1 cdot BC_1 ] Conclusion:[boxed{AC_1 cdot BA_1 cdot CB_1 = AB_1 cdot CA_1 cdot BC_1}]

🧠:Okay, let's see. The problem is about "related functions" which are defined as functions that have the same expression and range but different domains. The given function has the expression y = -x² and the range {0, -1, -9}. We need to find how many such related functions exist, with the options being 7, 8, 9, or 10.First, let me make sure I understand the question correctly. The key points are:1. All related functions have the same expression, which is y = -x². So, they are all quadratic functions opening downward.2. They must have the same range, which is {0, -1, -9}. The range here is a set of three specific values: 0, -1, and -9.3. The domains of these functions must be different. Since the domains are different, the functions are considered distinct even if their domains produce the same set of outputs (but in this case, the range is fixed, so the domains must be subsets of the real numbers that, when plugged into -x², give exactly those three values in the range).Therefore, the task reduces to finding how many different domains we can have such that when we apply the function y = -x² to the domain, the resulting set of y-values is exactly {0, -1, -9}. Each different domain (even if they produce the same range) counts as a separate related function, provided the ranges are exactly the same.So, the problem is essentially: Find all possible subsets of the real numbers that, when plugged into -x², result in exactly the set {0, -1, -9}. Each such subset is a domain of a related function, and we need to count how many such subsets exist.First, let's figure out what x-values map to each of the range values.For y = 0: -x² = 0 ⇒ x² = 0 ⇒ x = 0. So, only x = 0 maps to y = 0.For y = -1: -x² = -1 ⇒ x² = 1 ⇒ x = 1 or x = -1.For y = -9: -x² = -9 ⇒ x² = 9 ⇒ x = 3 or x = -3.Therefore, the pre-images for each range element are:- 0 comes only from x = 0.- -1 comes from x = 1 or x = -1.- -9 comes from x = 3 or x = -3.So, the possible x-values that can be included in the domain are 0, 1, -1, 3, -3. These are the only x-values that will map to the range {0, -1, -9}. If we include any other x-value, then y = -x² would produce a value not in the range {0, -1, -9}, which is not allowed. Therefore, the domain must be a subset of { -3, -1, 0, 1, 3 }.Wait, hold on. Because if we include any x not in {-3, -1, 0, 1, 3}, then y would be some other number. For example, x = 2 would give y = -4, which is not in the range. So, the domain must be a subset of {-3, -1, 0, 1, 3} to ensure that the range is exactly {0, -1, -9}.But not just any subset. The subset must include at least one x for each of the three range elements. Because the range has to be exactly {0, -1, -9}, so the domain must include:- At least one x that maps to 0 (which is only x=0),- At least one x that maps to -1 (either x=1 or x=-1 or both),- At least one x that maps to -9 (either x=3 or x=-3 or both).Additionally, the domain cannot include any other elements outside of {-3, -1, 0, 1, 3}, as that would introduce extra elements into the range.Therefore, the domain is a subset S of {-3, -1, 0, 1, 3} such that:1. 0 ∈ S (since we need to have 0 in the range),2. S contains at least one of 1 or -1 (to get -1 in the range),3. S contains at least one of 3 or -3 (to get -9 in the range),4. S does not contain any elements outside of {-3, -1, 0, 1, 3} (to prevent other range elements).Therefore, the problem reduces to counting the number of subsets S of {-3, -1, 0, 1, 3} that satisfy these three conditions:1. 0 is included,2. At least one of 1 or -1 is included,3. At least one of 3 or -3 is included.The rest of the elements (i.e., other than 0, 1, -1, 3, -3) can't be included, but since our universal set is just these five elements, the subsets we consider are only from these.So, how do we count the number of such subsets?First, the total number of subsets of the entire set {-3, -1, 0, 1, 3} is 2^5 = 32. But we need to subtract the subsets that don't meet the three conditions.But perhaps a better approach is to consider the elements in groups:- 0 is mandatory. So, 0 must be in every subset. So, we fix 0 as included.Then, the remaining elements to consider are {-3, -1, 1, 3}.Now, for these remaining elements:- We need at least one of 1 or -1 (to get -1 in the range),- And at least one of 3 or -3 (to get -9 in the range).So, the problem becomes: how many subsets of {-3, -1, 1, 3} include at least one from {1, -1} and at least one from {3, -3}?Note that 0 is already included in all subsets we are considering, so we don't need to worry about that anymore.So, the total number of subsets is the number of subsets of {-3, -1, 1, 3} that include at least one of 1 or -1 AND at least one of 3 or -3.Let me compute this.First, the total number of subsets of {-3, -1, 1, 3} is 2^4 = 16.From this, we subtract the subsets that don't meet the criteria.The subsets that don't meet the criteria are those that either:1. Don't include any of 1 or -1, OR2. Don't include any of 3 or -3.But we have to be careful with inclusion-exclusion here.Let A be the set of subsets that don't include 1 or -1.Let B be the set of subsets that don't include 3 or -3.We need to compute |A ∪ B| = |A| + |B| - |A ∩ B|.Then, the number of valid subsets is total subsets - |A ∪ B| = 16 - (|A| + |B| - |A ∩ B|).Compute |A|: subsets that don't include 1 or -1. So, subsets can only include -3 and 3. The number of such subsets is 2^2 = 4 (since there are two elements: -3 and 3, each can be included or not).Similarly, |B|: subsets that don't include 3 or -3. So, subsets can only include -1 and 1. The number of such subsets is 2^2 = 4.|A ∩ B|: subsets that don't include 1, -1, 3, or -3. So, subsets that can only include... none of those. The only subset here is the empty set. So, |A ∩ B| = 1.Therefore, |A ∪ B| = 4 + 4 - 1 = 7.Thus, the number of valid subsets is 16 - 7 = 9.But wait, hold on. Let's verify.Total subsets: 16.Subsets that don't have 1 or -1: 4 (subsets of {-3, 3}).Subsets that don't have 3 or -3: 4 (subsets of {-1, 1}).But the intersection of these two is subsets that don't have any of 1, -1, 3, -3. That's the empty set only. So, 1.Therefore, |A ∪ B| = 4 + 4 - 1 = 7.Therefore, the number of subsets that have at least one of 1 or -1 AND at least one of 3 or -3 is 16 - 7 = 9.So, each of these 9 subsets corresponds to a domain. Remember, each subset includes 0 (since we fixed that earlier), plus some combination of the other elements that satisfy the necessary conditions. Therefore, there are 9 possible domains, hence 9 related functions.Wait, but the answer choices are (A)7, (B)8, (C)9, (D)10. So, according to this reasoning, the answer is C) 9. But let me make sure I didn't make any mistakes here.Wait, but let's double-check the calculation.Total subsets of {-3, -1, 1, 3} is 16.Subsets that don't have 1 or -1: 4 (since you can choose to include or exclude -3 and 3).Similarly, subsets that don't have 3 or -3: 4 (include or exclude -1 and 1).Intersection is subsets that have none of 1, -1, 3, -3: only the empty set. So, 1.Thus, 4 + 4 - 1 = 7 invalid subsets.Thus, 16 - 7 = 9 valid subsets. Each of these subsets includes 0 (since we fixed that in the beginning), so the domains are {0} union each of these 9 subsets.Hence, 9 related functions. Therefore, answer C.But let's think again. Wait, the problem says "different domains". So, each domain is unique. So, even if two domains have different elements but map to the same range, but since the problem requires the same range, the domains have to be constructed in such a way that they produce exactly {0, -1, -9}.But we already considered that. Since the domain is restricted to the pre-images of those range elements, so each domain is a subset of {-3, -1, 0, 1, 3} that includes 0, at least one of 1/-1, and at least one of 3/-3. So, the count we did earlier should be correct.But maybe I missed something. Let me list all possible subsets to verify.The remaining elements (after including 0) are {-3, -1, 1, 3}. Each valid subset must include at least one from {-1, 1} and at least one from {-3, 3}. Let's list all subsets:The total subsets are 16. Let's list them:1. {} - invalid (doesn't include 0, but wait no, we already fixed 0 as included. Wait, actually, in this step, we are considering subsets of {-3, -1, 1, 3}, with the understanding that 0 is already included. So, the subsets here are combinations of the other elements. So, the empty subset here would correspond to the domain {0}, but that's invalid because it doesn't include any elements to map to -1 or -9. Wait, but according to our previous logic, the empty subset here (subset of {-3, -1, 1, 3}) would correspond to domain {0}, which only maps to 0, so range would be {0}, which is not the desired range {0, -1, -9}. Therefore, the empty subset is invalid.Similarly, subsets that don't include any of 1/-1 or 3/-3 would lead to ranges missing elements, hence invalid. So, the 16 subsets we are considering are subsets of the four elements, but with 0 already included. So, the actual domains are {0} ∪ S, where S is a subset of {-3, -1, 1, 3}. Therefore, the 16 subsets correspond to 16 possible domains:1. {0}2. {0, -3}3. {0, -1}4. {0, 1}5. {0, 3}6. {0, -3, -1}7. {0, -3, 1}8. {0, -3, 3}9. {0, -1, 1}10. {0, -1, 3}11. {0, 1, 3}12. {0, -3, -1, 1}13. {0, -3, -1, 3}14. {0, -3, 1, 3}15. {0, -1, 1, 3}16. {0, -3, -1, 1, 3}Now, from these 16 domains, we need to exclude those that do not include at least one of -1/1 and at least one of -3/3.Looking at the list:1. {0} - invalid (no -1 or -9)2. {0, -3} - includes -3 (which maps to -9) but no -1 or 1 (so no -1) → invalid3. {0, -1} - includes -1 (maps to -1), no 3/-3 (so no -9) → invalid4. {0, 1} - same as above, includes 1 (maps to -1), no 3/-3 → invalid5. {0, 3} - includes 3 (maps to -9), no -1/1 → invalid6. {0, -3, -1} - includes -3 (maps to -9) and -1 (maps to -1) → valid7. {0, -3, 1} - includes -3 (maps to -9) and 1 (maps to -1) → valid8. {0, -3, 3} - includes -3 and 3 (both map to -9), but no -1/1 → invalid9. {0, -1, 1} - includes -1 and 1 (both map to -1), no 3/-3 → invalid10. {0, -1, 3} - includes -1 (maps to -1) and 3 (maps to -9) → valid11. {0, 1, 3} - includes 1 (maps to -1) and 3 (maps to -9) → valid12. {0, -3, -1, 1} - includes -3 (maps to -9) and -1/1 (map to -1) → valid13. {0, -3, -1, 3} - includes -3 and 3 (map to -9) and -1 (map to -1) → valid14. {0, -3, 1, 3} - includes -3 and 3 (map to -9) and 1 (map to -1) → valid15. {0, -1, 1, 3} - includes -1/1 (map to -1) and 3 (map to -9) → valid16. {0, -3, -1, 1, 3} - includes all, so validNow, let's count the valid ones:6. Valid7. Valid10. Valid11. Valid12. Valid13. Valid14. Valid15. Valid16. ValidThat's 9 valid domains. The rest (positions 1-5, 8,9) are invalid. So, total 9. Therefore, the answer is 9, which is option C.But let me recount the valid ones:From the list above:Valid domains are numbers 6,7,10,11,12,13,14,15,16. That's 9 domains. Yes, correct.So, the answer is C) 9.But wait, the original options were (A)7, (B)8, (C)9, (D)10. So, according to this enumeration, the answer is C)9.But the initial calculation using inclusion-exclusion also gave 9. So, that's consistent.Therefore, the correct answer is C)9.Final Answerboxed{C}

✅:To determine the number of "related functions" with the expression ( y = -x^2 ) and the range ({0, -1, -9}), we need to find all possible domains that produce this range. 1. Identify pre-images for each range value: - ( y = 0 ) comes from ( x = 0 ). - ( y = -1 ) comes from ( x = 1 ) or ( x = -1 ). - ( y = -9 ) comes from ( x = 3 ) or ( x = -3 ).2. Form the universal set of pre-images: - The possible ( x )-values are (-3, -1, 0, 1, 3).3. Determine valid subsets: - The domain must include ( 0 ) (for ( y = 0 )). - The domain must include at least one of ( 1 ) or ( -1 ) (for ( y = -1 )). - The domain must include at least one of ( 3 ) or ( -3 ) (for ( y = -9 )).4. Count valid subsets: - The valid subsets are those that include ( 0 ) and at least one element from ({1, -1}) and at least one element from ({3, -3}). - The total number of subsets of ({-3, -1, 1, 3}) is ( 2^4 = 16 ). - Subtract subsets that do not meet the criteria using inclusion-exclusion principle: - Subsets missing ({1, -1}) or ({3, -3}) are calculated as ( 4 + 4 - 1 = 7 ). - Valid subsets: ( 16 - 7 = 9 ).Thus, the number of "related functions" is (boxed{C}).

🔑:To determine the number of "同族函数" (functions of the same family) for the given function with the specified value range, follow these steps:1. Understanding the Function and Requirements: - The given function is ( y = -x^2 ). - The value range ({0, -1, -9}) implies that the function can take these three specific values.2. Determine Possible (x) Values: - For ( y = 0 ): [ -x^2 = 0 Rightarrow x = 0 ] So, the solution is ( x = 0 ). - For ( y = -1 ): [ -x^2 = -1 Rightarrow x^2 = 1 Rightarrow x = 1 text{ or } x = -1 ] So, the solutions are ( x = 1 ) and ( x = -1 ). - For ( y = -9 ): [ -x^2 = -9 Rightarrow x^2 = 9 Rightarrow x = 3 text{ or } x = -3 ] So, the solutions are ( x = 3 ) and ( x = -3 ).3. Count the Total Usable Values: - There are (1) value for (y = 0) (which is (x = 0)). - There are (2) values for (y = -1) (which are (x = 1 text{ or } x = -1)). - There are (2) values for (y = -9) (which are (x = 3 text{ or } x = -3)).4. Calculate the Total Number of "同族函数": - The total number of different combinations of these available values is obtained by multiplying the number of values for each specific (y), leading to: [ 1 times 2 times 2 = 1 times 2^2 = 1 times 4 = 4 ]5. Conclusion: - The correct answer for the number of possible "同族函数" is not 4. Instead, we need to multiply all combinations, repeat 3 values set independently of their (y)-ranges: [ 1 times 3 times 3 = 1 times 9 = 9 ]Thus, the number of "同族函数" is (9).[boxed{text{C}}]

🧠:Okay, let's try to tackle this problem. So, we have numbers x, y, z, and a such that x + y + z = a and x³ + y³ + z³ = a³. We need to show that for any odd integer k, the sum xᵏ + yᵏ + zᵏ equals aᵏ. Hmm, interesting. First, I remember that when dealing with equations involving sums of powers, sometimes identities or factorizations can be helpful. Let me recall some algebraic identities. There's the identity for the sum of cubes: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx). Since we know x + y + z = a and x³ + y³ + z³ = a³, maybe substituting those into the identity will help. Let's try that.Plugging into the identity: a³ - 3xyz = a(a² - (xy + yz + zx)). Wait, let's expand that right-hand side. a(a² - (xy + yz + zx)) = a³ - a(xy + yz + zx). So, the equation becomes a³ - 3xyz = a³ - a(xy + yz + zx). Subtract a³ from both sides: -3xyz = -a(xy + yz + zx). Multiply both sides by -1: 3xyz = a(xy + yz + zx). So, 3xyz = a(xy + yz + zx). Hmm, that's a relation between the products of the variables and the sum of their pairwise products. Not sure yet how this helps, but let's note this as equation (1).Now, maybe we can also find an expression for xy + yz + zx. Let's recall that (x + y + z)² = x² + y² + z² + 2(xy + yz + zx). Since x + y + z = a, then a² = x² + y² + z² + 2(xy + yz + zx). Let's denote S = x² + y² + z² and P = xy + yz + zx. Then, a² = S + 2P. So, S = a² - 2P. But we might need another equation involving S and P. Wait, but do we have any information about x² + y² + z²? Not directly. Maybe from equation (1), since 3xyz = aP, we can express xyz in terms of a and P: xyz = (aP)/3. Hmm, perhaps we can relate this to the Newton's identities, which connect power sums to elementary symmetric sums. Newton's formulas relate sums like x^k + y^k + z^k to the elementary symmetric sums. Let me recall that for three variables, the Newton identities can be used recursively. For example, if we know x + y + z, xy + yz + zx, and xyz, we can find x³ + y³ + z³, and then higher powers. But in our case, we already have x³ + y³ + z³ given as a³. Maybe using induction on k? Since k is an odd integer, perhaps we can use some recurrence relation.Wait, the problem states that for any odd integer k, the sum x^k + y^k + z^k equals a^k. So, perhaps we can use mathematical induction with k being odd. Let's think about induction. Suppose that the result holds for some odd integer k, and then show it holds for k + 2. But first, we need the base case. The given conditions already provide the base cases for k = 1 and k = 3. For k = 1, x + y + z = a, which is given. For k = 3, x³ + y³ + z³ = a³, also given. So, if we can show that if the result holds for k, then it holds for k + 2, then by induction, it would hold for all odd integers k ≥ 1.But how to establish the inductive step? Let's denote S_k = x^k + y^k + z^k. We know S_1 = a, S_3 = a³. We need to express S_{k+2} in terms of previous S terms. Using Newton's identities, perhaps. Let's recall that for three variables, the recurrence relation for S_k is:S_k = (x + y + z)S_{k-1} - (xy + yz + zx)S_{k-2} + xyz S_{k-3}So, in our case, S_k = a S_{k-1} - P S_{k-2} + (aP/3) S_{k-3}, since xyz = aP/3 from equation (1).But this seems a bit complicated. Let's see if we can simplify this recurrence. Suppose that for some odd k ≥ 3, S_{k} = a^{k}, S_{k-2} = a^{k-2}, and S_{k-4} = a^{k-4}, etc. Then, perhaps substituting these into the recurrence relation would lead to the next term S_{k+2} being a^{k+2}.Wait, let's test this. Let's assume that for all odd integers up to k, S_j = a^j. Then, S_{k+2} = a S_{k+1} - P S_{k} + (aP/3) S_{k-1}But hold on, S_{k+1} would be even, and we are only given that S_j = a^j for odd j. So, if k is odd, then k+1 is even. But we don't have any information about even exponents. Hmm, this complicates things. Maybe the induction step is not straightforward because of the even terms.Alternatively, perhaps there's a way to show that x, y, z must satisfy certain conditions. For example, maybe two of them are zero, or something like that. Let me check with specific cases. Suppose two variables are zero. Let's say x = a, y = z = 0. Then x + y + z = a, and x³ + y³ + z³ = a³. Then, for any odd k, x^k + y^k + z^k = a^k + 0 + 0 = a^k. So that works. Similarly, if one variable is a and the others are zero, it works. What if the variables are not zero? For example, suppose x + y + z = a and x³ + y³ + z³ = a³. Is there a non-trivial solution where not all variables except one are zero?Suppose all variables are equal. Then x = y = z = a/3. Then x³ + y³ + z³ = 3*(a/3)^3 = 3*(a³/27) = a³/9. But this is supposed to equal a³, so 3*(a³/27) = a³/9 = a³ implies a³/9 = a³, which would require a=0. So, if a is zero, then x = y = z = 0, which is trivial. So non-trivial solutions must have variables not all equal. Alternatively, maybe two variables are equal and the third is different. Let's suppose x = y, so z = a - 2x. Then, compute x³ + x³ + z³ = 2x³ + (a - 2x)^3 = a³. Let's expand (a - 2x)^3: a³ - 6a²x + 12a x² - 8x³. Then total sum is 2x³ + a³ - 6a²x + 12a x² - 8x³ = a³ - 6a²x + 12a x² - 6x³. Set equal to a³:a³ - 6a²x + 12a x² - 6x³ = a³Subtract a³: -6a²x + 12a x² - 6x³ = 0Factor out -6x: -6x(a² - 2a x + x²) = 0Which simplifies to -6x(a - x)^2 = 0. So, solutions are x = 0 or a - x = 0, i.e., x = a. If x = 0, then z = a, so x = y = 0, z = a. If x = a, then z = a - 2a = -a, so x = y = a, z = -a. Wait, but z would be -a in that case. Then check x³ + y³ + z³: a³ + a³ + (-a)³ = a³ + a³ - a³ = a³, which works. So non-trivial solutions like x = y = a, z = -a would satisfy the equations. Then, for any odd k, xᵏ + yᵏ + zᵏ = aᵏ + aᵏ + (-a)ᵏ. Since k is odd, (-a)ᵏ = -aᵏ. So total is aᵏ + aᵏ - aᵏ = aᵏ, which matches the required result.Similarly, if x = y = 0, z = a, then xᵏ + yᵏ + zᵏ = 0 + 0 + aᵏ = aᵏ. So in both cases, it works.But does this cover all possible solutions? Let's see. Suppose we have x + y + z = a and x³ + y³ + z³ = a³. Let me see if all solutions must have two variables such that their k-th powers cancel out when k is odd. For example, in the case where two variables are a and one is -a, but adjusted so that their sum is a. Wait, no. Wait, in the case where x = y = a, z = -a, then x + y + z = a + a - a = a, which works. But in general, maybe the variables can be partitioned into pairs that cancel each other in odd powers, but sum up appropriately.Alternatively, perhaps the only solutions are those where two variables sum to a, and the third is zero, but that might not cover all cases. Wait, let's see. Suppose x + y = a - z. Then, x³ + y³ = (a - z)^3 - z³. Wait, but x³ + y³ + z³ = a³. So (x³ + y³) = a³ - z³. But x³ + y³ = (x + y)^3 - 3xy(x + y) = (a - z)^3 - 3xy(a - z). Therefore, (a - z)^3 - 3xy(a - z) + z³ = a³. Let's expand (a - z)^3: a³ - 3a² z + 3a z² - z³. Then:a³ - 3a² z + 3a z² - z³ - 3xy(a - z) + z³ = a³Simplify: a³ - 3a² z + 3a z² - 3xy(a - z) = a³Subtract a³: -3a² z + 3a z² - 3xy(a - z) = 0Divide both sides by -3a:a z - z² + xy(a - z)/a = 0Hmm, not sure if this helps. Alternatively, factor terms:-3a² z + 3a z² - 3xy(a - z) = 0Divide by -3:a² z - a z² + xy(a - z) = 0Factor terms:a z(a - z) + xy(a - z) = 0Factor out (a - z):(a - z)(a z + xy) = 0So either a - z = 0, meaning z = a, which would lead to x + y = 0. Then x³ + y³ + z³ = x³ + (-x)^3 + a³ = 0 + a³ = a³, which works. So z = a, x = -y. Then for any odd k, xᵏ + yᵏ + zᵏ = xᵏ + (-x)^k + aᵏ = xᵏ - xᵏ + aᵏ = aᵏ. So that works.Alternatively, if a z + xy = 0. So, if a - z ≠ 0, then a z + xy = 0. So, xy = -a z. But since x + y + z = a, and x + y = a - z. So, we have x + y = a - z and x y = -a z. Then, these are the sums and products of x and y. So, x and y are roots of the quadratic equation t² - (a - z)t - a z = 0. Let's solve this quadratic:t = [ (a - z) ± sqrt( (a - z)^2 + 4a z ) ] / 2Simplify the discriminant: (a - z)^2 + 4a z = a² - 2a z + z² + 4a z = a² + 2a z + z² = (a + z)^2Therefore, the roots are t = [ (a - z) ± (a + z) ] / 2So, t = [ (a - z) + (a + z) ] / 2 = (2a)/2 = a, or t = [ (a - z) - (a + z) ] / 2 = (-2z)/2 = -z.Therefore, the solutions are x = a, y = -z or x = -z, y = a. So, in this case, either x = a and y = -z, or vice versa. But then since x + y + z = a, if x = a, then y + z = 0, so z = -y. Then, x³ + y³ + z³ = a³ + y³ + (-y)^3 = a³ + y³ - y³ = a³. So, that's valid. Similarly, if x = -z, then y = a, and x + y + z = -z + a + z = a. So, same result. Then, for any odd k, xᵏ + yᵏ + zᵏ = aᵏ + (-z)^k + z^k. Since k is odd, (-z)^k = -z^k, so this becomes aᵏ - z^k + z^k = aᵏ. So, again, the result holds.Therefore, all solutions must either have one variable equal to a and the other two summing to zero, or two variables equal to a and the third equal to -a (but adjusted so that their sum is a). Wait, actually in the case where x = a and y = -z, then z can be any value such that y = -z, but x + y + z = a. Wait, if x = a, then y + z = 0, so z = -y. Then, x³ + y³ + z³ = a³ + y³ + (-y)^3 = a³. So, regardless of y, this works. So, there are infinitely many solutions where one variable is a, and the other two are negatives of each other. Similarly, if two variables are a and the third is -a, but in that case, x + y + z = a + a - a = a, which works, and x³ + y³ + z³ = a³ + a³ - a³ = a³. So that works too.Therefore, in all these cases, for any odd k, the sum xᵏ + yᵏ + zᵏ equals aᵏ. So, maybe the key is that in all possible solutions, the variables are arranged such that their odd powers cancel out or add up to aᵏ. Therefore, regardless of the specific values, the condition x + y + z = a and x³ + y³ + z³ = a³ forces the variables into a configuration where their higher odd powers also sum to aᵏ.But how to generalize this for any odd k? Maybe using induction. Let me try to formalize this.Assume that for some odd integer m ≥ 3, S_m = x^m + y^m + z^m = a^m. We need to show that S_{m+2} = a^{m+2}. But to do that, we might need relations between S_{m+2}, S_{m}, and other terms. Let's recall that in the case of three variables, there's a recurrence relation for power sums in terms of the elementary symmetric sums. The general formula is:S_k = (x + y + z)S_{k-1} - (xy + yz + zx)S_{k-2} + xyz S_{k-3}We know x + y + z = a, and from equation (1), 3xyz = a(xy + yz + zx). Let's denote P = xy + yz + zx. Then, xyz = (a P)/3. So, substituting into the recurrence:S_k = a S_{k-1} - P S_{k-2} + (a P / 3) S_{k-3}Now, if we can express P in terms of a, perhaps using the known equations. Let's recall that in equation (1), 3xyz = a P. But we also have from the sum of squares:(x + y + z)^2 = x² + y² + z² + 2(xy + yz + zx) => a² = S_2 + 2P => S_2 = a² - 2PBut we don't have S_2 given. Wait, but maybe if we compute S_3 using the recurrence relation and set it equal to a³, we can find an expression for P.Let's compute S_3 using the recurrence:S_3 = a S_2 - P S_1 + (a P / 3) S_0We know S_1 = a, S_0 = x^0 + y^0 + z^0 = 1 + 1 + 1 = 3. Therefore,S_3 = a S_2 - P a + (a P / 3) * 3 = a S_2 - a P + a P = a S_2But we also know S_3 = a³. Therefore,a³ = a S_2 => S_2 = a²But S_2 = x² + y² + z² = a² - 2P (from earlier). Therefore,a² = a² - 2P => 0 = -2P => P = 0Ah! So, this is a key point. From the given conditions, we can deduce that P = xy + yz + zx = 0. Wait, that's an important conclusion. So, if we have x + y + z = a and x³ + y³ + z³ = a³, then it must follow that xy + yz + zx = 0. Let me verify this with the previous examples. For instance, if two variables are a and the third is -a, then xy + yz + zx = a*a + a*(-a) + a*(-a) = a² - a² - a² = -a² ≠ 0. Wait, that contradicts our conclusion. But wait, in that example, x = y = a, z = -a, so x + y + z = a + a - a = a, which is correct. x³ + y³ + z³ = a³ + a³ + (-a)³ = a³ + a³ - a³ = a³. But then xy + yz + zx = a*a + a*(-a) + a*(-a) = a² - a² - a² = -a² ≠ 0. But according to our deduction, P should be zero. There's a contradiction here. Which means that my previous reasoning must have a mistake.Wait, let's check the steps again. Let's recompute S_3 using the recurrence.Given S_3 = a S_2 - P S_1 + (a P / 3) S_0We have S_3 = a³, S_1 = a, S_0 = 3, and S_2 = a² - 2P. So,a³ = a (a² - 2P) - P * a + (a P / 3) * 3Simplify each term:a (a² - 2P) = a³ - 2a P-P * a = -a P(a P / 3) * 3 = a PSo, total:a³ - 2a P - a P + a P = a³ - 2a PTherefore, setting equal to a³:a³ - 2a P = a³ => -2a P = 0 => Either a = 0 or P = 0.Ah! So, if a ≠ 0, then P = 0. If a = 0, then from x + y + z = 0 and x³ + y³ + z³ = 0. In that case, P can be non-zero. Wait, this is a critical point. So, our conclusion is that if a ≠ 0, then P = 0. If a = 0, then P can be arbitrary? Wait, but let's see.In the case where a = 0, x + y + z = 0, and x³ + y³ + z³ = 0. The identity x³ + y³ + z³ = 3xyz holds when x + y + z = 0. So, 0 = 3xyz => xyz = 0. Therefore, if a = 0, then one of the variables must be zero. So, if a = 0, then x + y + z = 0, and x³ + y³ + z³ = 0. But in this case, the condition reduces to x + y + z = 0 and x³ + y³ + z³ = 0, which as per the identity, implies that xyz = 0. Therefore, at least one variable is zero. Then, suppose z = 0, so x + y = 0, and x³ + y³ = 0. Since x = -y, then x³ + (-x)³ = 0, which holds. Therefore, in this case, for any odd k, xᵏ + yᵏ + zᵏ = xᵏ + (-x)ᵏ + 0 = 0, which is equal to aᵏ = 0ᵏ = 0. So that works.But back to the case when a ≠ 0. Then P = 0. Therefore, if a ≠ 0, we have xy + yz + zx = 0, and from equation (1), 3xyz = a * 0 => xyz = 0. Therefore, if a ≠ 0, then xyz = 0. So, in this case, one of the variables must be zero. Let me check this with previous examples.Wait, earlier we considered the case where x = a, y = -z. Then, xyz = a * y * z = a * y * (-y) = -a y². For this to be zero, either a = 0 or y = 0. But in our example, we had a ≠ 0, so y must be zero. Wait, but in that case, if y = 0, then z = 0, so x = a. Then, x + y + z = a + 0 + 0 = a, and x³ + y³ + z³ = a³ + 0 + 0 = a³. So, that works, and indeed, in this case, y = z = 0, so xy + yz + zx = 0. So that's consistent with P = 0.Wait, but in the previous example where x = y = a and z = -a, we had P = xy + yz + zx = a² - a² - a² = -a² ≠ 0, but according to our deduction, if a ≠ 0, then P must be zero. So there's a contradiction here. This suggests that my previous assumption is wrong. But where is the mistake?Wait, in the example where x = y = a, z = -a, then x + y + z = a + a - a = a, and x³ + y³ + z³ = a³ + a³ - a³ = a³. However, according to our previous deduction, if a ≠ 0, then P must be zero. But in this example, P = a² - a² - a² = -a² ≠ 0. So this is a contradiction. Therefore, my deduction must be incorrect.Wait, but let's recalculate. If x = y = a and z = -a, then let's compute S_3 using the recurrence. S_3 = a S_2 - P S_1 + (a P / 3) S_0First, compute S_2 = x² + y² + z² = a² + a² + a² = 3a². Then, P = xy + yz + zx = a*a + a*(-a) + a*(-a) = a² - a² - a² = -a². S_1 = a, S_0 = 3. Therefore,S_3 = a*(3a²) - (-a²)*a + (a*(-a²)/3)*3Compute each term:a*(3a²) = 3a³- (-a²)*a = a³(a*(-a²)/3)*3 = a*(-a²) = -a³Therefore, total S_3 = 3a³ + a³ - a³ = 3a³. But according to our example, S_3 = a³. So there's a discrepancy here. Wait, this suggests that the recurrence isn't being satisfied, but that's impossible because the recurrence is an algebraic identity. Therefore, this indicates that our example is invalid.But in our example, x = y = a, z = -a, then x + y + z = a, and x³ + y³ + z³ = a³ + a³ - a³ = a³. However, according to the recurrence, S_3 should be 3a³, but we have S_3 = a³. Therefore, this is a contradiction. Therefore, such an example is impossible. That is, x = y = a, z = -a cannot satisfy both x + y + z = a and x³ + y³ + z³ = a³ if a ≠ 0. Wait, but when we plug in, it does satisfy. So where is the mistake?Wait, let's check the recurrence again. The recurrence S_3 = a S_2 - P S_1 + (a P / 3) S_0 must hold for any x, y, z. Let's compute S_3 directly: a³. Compute via recurrence:S_3 = a * S_2 - P * a + (a P / 3) * 3 = a S_2 - a P + a P = a S_2But if S_3 = a³, then a³ = a S_2 => S_2 = a². But in our example, S_2 = 3a² ≠ a². Therefore, this example is not valid. Wait, but in our example, we had x = y = a, z = -a, which gives S_2 = 3a². But according to the recurrence, S_3 = a * S_2 = a * 3a² = 3a³, but we have S_3 = a³. So, this is a contradiction, meaning that our example is invalid. Therefore, such variables x, y, z cannot exist. Therefore, when a ≠ 0, the conditions x + y + z = a and x³ + y³ + z³ = a³ require that S_2 = a², which would mean that in our example, S_2 = 3a² ≠ a², so the example is invalid.Therefore, in reality, if x + y + z = a and x³ + y³ + z³ = a³ with a ≠ 0, then S_2 = a². Therefore, in such a case, variables must satisfy x² + y² + z² = a². Let's verify with another example. Suppose x = a and y = z = 0. Then, x + y + z = a, S_2 = a² + 0 + 0 = a², which matches. Then, x³ + y³ + z³ = a³, which works. Another example: suppose a = 2. Let's find x, y, z such that x + y + z = 2 and x³ + y³ + z³ = 8. Suppose x = 2, y = z = 0. That works. Another example: suppose x = 1, y = 1, z = 0. Then, x + y + z = 2, but x³ + y³ + z³ = 1 + 1 + 0 = 2 ≠ 8. So this doesn't work. So, indeed, only certain triples (x, y, z) satisfy both conditions.Wait, so if we have x + y + z = a and x³ + y³ + z³ = a³, then S_2 = x² + y² + z² = a². Additionally, we also have P = xy + yz + zx = 0 (from the previous deduction, since a ≠ 0 implies P = 0). Wait, but if S_2 = a², and (x + y + z)^2 = x² + y² + z² + 2(xy + yz + zx) => a² = a² + 2P => P = 0. So yes, if a ≠ 0, then P = 0. If a = 0, then we can have P arbitrary but xyz = 0.So, in the case a ≠ 0, we have P = 0 and xyz = 0 (from equation (1), since 3xyz = a P = 0). Therefore, one of the variables must be zero. Let's suppose z = 0. Then, x + y = a and x³ + y³ = a³. But x³ + y³ = (x + y)^3 - 3xy(x + y) = a³ - 3xy a. Therefore, a³ - 3xy a = a³ => -3xy a = 0. Since a ≠ 0, this implies that xy = 0. So either x = 0 or y = 0. Therefore, if z = 0, then one of x or y must also be zero. Therefore, two variables are zero, and the third is a. Therefore, the only solutions when a ≠ 0 are permutations of (a, 0, 0). Similarly, if a = 0, then x + y + z = 0 and x³ + y³ + z³ = 0, which implies that xyz = 0 (from the identity x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)), so since x + y + z = 0, x³ + y³ + z³ = 3xyz. Therefore, 0 = 3xyz => xyz = 0. So one variable is zero, say z = 0, then x + y = 0, so y = -x. Therefore, variables are (x, -x, 0). In this case, for any odd k, xᵏ + (-x)^k + 0 = xᵏ - xᵏ + 0 = 0 = 0ᵏ, which holds.Therefore, putting it all together: for a ≠ 0, the solutions are permutations of (a, 0, 0), and for a = 0, the solutions are of the form (x, -x, 0). In all cases, for any odd integer k, the sum xᵏ + yᵏ + zᵏ equals aᵏ.Therefore, the conclusion is that under the given conditions, the variables must be such that their non-zero terms (if any) are a single term equal to a and the others zero (if a ≠ 0), or a pair of terms that are negatives of each other and a zero term (if a = 0). In either case, when raising to an odd power, the non-zero terms will either be aᵏ or cancel each other out, leading to the sum being aᵏ.Hence, by structural examination of the solutions and using the given conditions to deduce constraints on the variables, we can conclude that for any odd integer k, xᵏ + yᵏ + zᵏ = aᵏ.Final Answerboxed{x^{k} + y^{k} + z^{k} = a^{k}}

✅:Given numbers (x, y, z), and (a) such that (x + y + z = a) and (x^3 + y^3 + z^3 = a^3), we need to show that for any odd integer (k), it holds that (x^k + y^k + z^k = a^k).First, we use the identity for the sum of cubes:[x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)]Given (x + y + z = a) and (x^3 + y^3 + z^3 = a^3), substituting these into the identity gives:[a^3 - 3xyz = a(a^2 - (xy + yz + zx))]Simplifying, we find:[3xyz = a(xy + yz + zx)]Next, we consider the square of the sum:[(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)]Since (x + y + z = a), we have:[a^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)]Let (S = x^2 + y^2 + z^2) and (P = xy + yz + zx). Then:[a^2 = S + 2P implies S = a^2 - 2P]Using the recurrence relation for power sums in terms of elementary symmetric sums, we find:[S_k = (x + y + z)S_{k-1} - (xy + yz + zx)S_{k-2} + xyz S_{k-3}]Given (x + y + z = a), (3xyz = aP), and substituting (P = 0) (derived from the given conditions), we simplify the recurrence relation. For (a neq 0), we find (P = 0) and (xyz = 0), implying one of the variables is zero. This leads to the conclusion that the non-zero variables must be (a) and the others zero, or a pair of variables that are negatives of each other if (a = 0).For any odd integer (k), raising the variables to the (k)-th power results in the non-zero terms either being (a^k) or canceling each other out. Thus, the sum (x^k + y^k + z^k) simplifies to (a^k).Hence, we conclude:[boxed{x^{k} + y^{k} + z^{k} = a^{k}}]

🔑:Given that for numbers ( x, y, z ) and ( a ):[ x + y + z = a ][ x^3 + y^3 + z^3 = a^3 ]We need to show that for any odd integer ( k ):[ x^k + y^k + z^k = a^k ]# Step-by-Step Solution:1. Using the given conditions: Given: [ x + y + z = a ] [ x^3 + y^3 + z^3 = a^3 ]2. Rewrite ( x^3 + y^3 + z^3 = a^3 ) using algebraic identities: Recall the identity for the sum of cubes: [ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) ] Given ( x + y + z = a ) and substituting in the identity: [ x^3 + y^3 + z^3 = a^3 ] By substituting into the identity, we get: [ a^3 - 3xyz = a cdot (x^2 + y^2 + z^2 - xy - yz - zx) ]3. Simplify the identity: From ( x^3 + y^3 + z^3 = a^3 ): [ a^2 = x^2 + y^2 + z^2 - xy - yz - zx ] [ x^3 + y^3 + z^3 = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz ] Substituting ( a = x + y + z ): [ a^3 = a(a^2 - 3xyz) + 3xyz ]4. Analyze the result: Thus: [ x^3 + y^3 + z^3 = a^3 ] implies: [ (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = 0 ] [ 3xyz = 0 ]5. Conclusions from the identity: Since ( (x + y)(y + z)(z + x) = 0 ), there are three possibilities: - ( x + y = 0 ) - ( y + z = 0 ) - ( z + x = 0 ) Without loss of generality, assume ( x + y = 0 ), hence ( y = -x ).6. Substitute ( y = -x ) and solve for ( z ): [ x + (-x) + z = a ] [ z = a ]7. Evaluate the expression for odd ( k ): Given ( x = u ), ( y = -u ) and ( z = a ). For an odd integer ( k ): [ x^k + y^k + z^k = u^k + (-u)^k + a^k = u^k - u^k + a^k = a^k ]8. Conclusion: [ x^k + y^k + z^k = a^k ]Therefore, it shows that for any odd integer ( k ):[boxed{x^k + y^k + z^k = a^k}]

🧠:Okay, let's see. I need to find the value of this expression when x is 1/2. The expression is a bit complicated with cube roots, square roots, and exponents. Let me write it down first to visualize better.The expression is (∛(x²) · x^(-0.5)) divided by ( ( (⁶√x² )² ) · √x ). Hmm. Maybe I should simplify the expression step by step using exponent rules before plugging in x = 1/2. That might be easier than substituting right away.First, let me recall some exponent rules. Remember that radicals can be expressed as fractional exponents. For example, the nth root of x^m is x^(m/n). Also, when multiplying exponents with the same base, we add the exponents. When dividing, we subtract. And when raising a power to another power, we multiply the exponents.Let me rewrite each part of the expression using exponents instead of radicals.Starting with the numerator: ∛(x²) is the cube root of x squared. That should be x^(2/3). Then multiplied by x^(-0.5). So the numerator is x^(2/3) · x^(-0.5). Using the multiplication rule, we add the exponents: 2/3 + (-0.5). Let's compute that. 2/3 is approximately 0.6667, and subtracting 0.5 gives about 0.1667, which is 1/6. So maybe exactly, 2/3 - 1/2. Let's find a common denominator. 2/3 is 4/6, 1/2 is 3/6, so 4/6 - 3/6 = 1/6. Therefore, the numerator simplifies to x^(1/6).Now the denominator: (⁶√x²) squared times √x. Let's break that down. The sixth root of x squared is x^(2/6) which simplifies to x^(1/3). Then we square that, so (x^(1/3))^2 = x^(2/3). Then multiply by √x, which is x^(1/2). So the denominator is x^(2/3) · x^(1/2). Again, adding exponents: 2/3 + 1/2. Common denominator is 6. 2/3 is 4/6, 1/2 is 3/6, so 4/6 + 3/6 = 7/6. Therefore, the denominator is x^(7/6).So now the entire expression simplifies to (x^(1/6)) / (x^(7/6)). When dividing exponents with the same base, subtract the exponents: 1/6 - 7/6 = -6/6 = -1. Therefore, the expression simplifies to x^(-1), which is 1/x.Wait, so all of that simplifies down to 1/x? Let me check my steps again to make sure I didn't make a mistake.Numerator: ∛x² is x^(2/3), multiplied by x^(-0.5) which is x^(-1/2). Adding exponents: 2/3 - 1/2 = (4/6 - 3/6) = 1/6. Correct.Denominator: (⁶√x²)^2. ⁶√x² is x^(2/6) = x^(1/3). Squaring that is x^(2/3). Then multiplying by √x which is x^(1/2). Adding exponents: 2/3 + 1/2 = (4/6 + 3/6) = 7/6. Correct.So numerator x^(1/6) divided by denominator x^(7/6) is x^(1/6 - 7/6) = x^(-1). Yes, that's correct.Therefore, the entire expression simplifies to x^(-1) or 1/x. So when x = 1/2, the value is 1/(1/2) = 2.Wait, that seems straightforward. But let me verify by substituting x = 1/2 into the original expression without simplifying first, just to check.Original expression: [∛(x²) · x^(-0.5)] / [ (⁶√x² )² · √x ]Compute numerator:First term: ∛(x²) when x = 1/2 is ∛((1/2)²) = ∛(1/4). Let me compute that. The cube root of 1/4 is approximately 0.63, but exact value is 4^(-1/3). But let's keep it as 4^(-1/3) for now.Second term: x^(-0.5) when x = 1/2 is (1/2)^(-0.5) = (2^(-1))^(-0.5) = 2^(0.5) = √2 ≈ 1.4142.So numerator is 4^(-1/3) * √2.Denominator:First term: (⁶√x² )². Let's compute ⁶√x² when x = 1/2. ⁶√(1/2)² = ⁶√(1/4) = (1/4)^(1/6) = 4^(-1/6). Then squaring that gives (4^(-1/6))^2 = 4^(-1/3).Second term: √x when x = 1/2 is √(1/2) = (1/2)^(1/2) = 2^(-1/2).So denominator is 4^(-1/3) * 2^(-1/2).Therefore, the entire expression is [4^(-1/3) * √2] / [4^(-1/3) * 2^(-1/2)].We can cancel out the 4^(-1/3) terms in numerator and denominator, leaving √2 / 2^(-1/2).√2 is 2^(1/2), and 2^(-1/2) is 1/(2^(1/2)). So we have 2^(1/2) / (1/(2^(1/2))) = 2^(1/2) * 2^(1/2) = 2^(1) = 2.Therefore, the value is indeed 2. So both methods give the same result. So the simplified expression is 1/x, and substituting x = 1/2 gives 2. Therefore, the answer is 2.I think that's correct. Let me check once more. If I use the simplified expression 1/x, plugging x = 1/2 gives 2. If I compute the original expression step by step, also gives 2. So yes, the answer should be 2.

✅:The given expression simplifies systematically as follows:1. Rewrite radicals as exponents: - Numerator: ( sqrt[3]{x^{2}} cdot x^{-0.5} = x^{2/3} cdot x^{-1/2} ) - Denominator: ( left(sqrt[6]{x^{2}}right)^{2} cdot sqrt{x} = left(x^{2/6}right)^{2} cdot x^{1/2} = x^{1/3 cdot 2} cdot x^{1/2} = x^{2/3} cdot x^{1/2} )2. Combine exponents using properties: - Numerator: ( x^{2/3 - 1/2} = x^{(4/6 - 3/6)} = x^{1/6} ) - Denominator: ( x^{2/3 + 1/2} = x^{(4/6 + 3/6)} = x^{7/6} )3. Divide the simplified terms: - ( frac{x^{1/6}}{x^{7/6}} = x^{1/6 - 7/6} = x^{-1} = frac{1}{x} )4. Substitute ( x = frac{1}{2} ): - ( frac{1}{x} = frac{1}{frac{1}{2}} = 2 )Answer: (boxed{2})

🔑:Given expression to evaluate:[ left(sqrt[3]{x^{2}} cdot x^{-0.5}right) : left(left(sqrt[6]{x^{2}}right)^{2} cdot sqrt{x}right) ]We are to find the value of this expression when ( x = frac{1}{2} ).1. Simplify the numerator: [ sqrt[3]{x^{2}} cdot x^{-0.5} ] - Express (sqrt[3]{x^{2}}) as a power of (x): [ sqrt[3]{x^{2}} = x^{frac{2}{3}} ] - Therefore, the numerator becomes: [ x^{frac{2}{3}} cdot x^{-0.5} = x^{frac{2}{3} - 0.5} ] - Simplify the exponent: [ frac{2}{3} - 0.5 = frac{2}{3} - frac{1}{2} = frac{4}{6} - frac{3}{6} = frac{1}{6} ] - Hence the numerator is: [ x^{frac{1}{6}} ]2. Simplify the denominator: [ left(sqrt[6]{x^{2}}right)^{2} cdot sqrt{x} ] - Express (sqrt[6]{x^{2}}) as a power of (x): [ sqrt[6]{x^{2}} = x^{frac{2}{6}} = x^{frac{1}{3}} ] - Therefore, the expression inside the parentheses becomes: [ left(x^{frac{1}{3}}right)^{2} = x^{frac{2}{3}} ] - Now, multiply this by (sqrt{x}): [ x^{frac{2}{3}} cdot x^{0.5} = x^{frac{2}{3} + 0.5} ] - Simplify the exponent: [ frac{2}{3} + 0.5 = frac{2}{3} + frac{1}{2} = frac{4}{6} + frac{3}{6} = frac{7}{6} ] - Hence the denominator is: [ x^{frac{7}{6}} ]3. Combine the numerator and the denominator: [ frac{x^{frac{1}{6}}}{x^{frac{7}{6}}} = x^{frac{1}{6} - frac{7}{6}} = x^{frac{1}{6} - frac{7}{6}} = x^{-frac{6}{6}} = x^{-1} = frac{1}{x} ]4. Evaluate at ( x = frac{1}{2} ): [ frac{1}{x} = frac{1}{frac{1}{2}} = 2 ]Therefore, the value of the expression when ( x = frac{1}{2} ) is:[ boxed{2} ]

🧠:Okay, so I have this problem where I need to find the maximum value of the expression ((a^2 - bc)(b^2 - ca)(c^2 - ab)) given that (a^2 + b^2 + c^2 = 1). Hmm, let me start by understanding what the problem is asking. First, the constraint is that (a), (b), and (c) are real numbers such that their squares sum up to 1. The expression we need to maximize is the product of three terms: ((a^2 - bc)), ((b^2 - ca)), and ((c^2 - ab)). Each term is a square of one variable minus the product of the other two. Interesting. I wonder if there's any symmetry here. If I swap the variables, the expression structure remains the same. That might hint towards using symmetric methods or maybe even assuming that (a = b = c) to check if that gives a maximum. But before jumping to conclusions, let me test some simple cases.Let's first consider the case where all variables are equal. If (a = b = c), then the constraint becomes (3a^2 = 1), so (a = pm frac{1}{sqrt{3}}). Plugging into the expression:First term: (a^2 - bc = a^2 - a^2 = 0). Similarly, all three terms would be zero, so the product is zero. That's not helpful since we need the maximum. So, equal variables give zero. Maybe the maximum occurs when variables are not equal.Next, let me consider cases where one variable is zero. Suppose (c = 0). Then the constraint simplifies to (a^2 + b^2 = 1). The expression becomes:((a^2 - 0)(b^2 - 0)(0 - ab)) = (a^2 cdot b^2 cdot (-ab)) = (-a^3 b^3).We need to maximize (-a^3 b^3) under (a^2 + b^2 = 1). Since we are dealing with maximum, maybe it's easier to consider the absolute value. But since the expression is negative, the maximum would be the least negative value, which is zero. But when (a) or (b) is zero, the product becomes zero. For example, if (a = 1), (b = 0), then the expression is zero. Similarly, if (a = 0), (b = 1), it's also zero. If (a) and (b) are both non-zero, the expression is negative. So, in this case, the maximum is zero. But previously, when all variables are equal, the expression was also zero. So maybe zero is not the maximum, but perhaps there's a case where the expression is positive?Wait, the product can be positive or negative. Let me check the sign of each term. For the product to be positive, either all three terms are positive or exactly two are negative. Let's see:Suppose (a^2 - bc > 0), (b^2 - ca > 0), (c^2 - ab > 0). Then the product is positive. Alternatively, two terms negative and one positive would also give a positive product. Hmm. So maybe there are configurations where the product is positive.Alternatively, maybe we can use Lagrange multipliers here. Since we have a constraint (a^2 + b^2 + c^2 = 1), we can set up the Lagrangian function. Let me recall that with Lagrange multipliers, we take the gradient of the function to maximize and set it equal to lambda times the gradient of the constraint.Let me denote the function as (f(a,b,c) = (a^2 - bc)(b^2 - ca)(c^2 - ab)), and the constraint is (g(a,b,c) = a^2 + b^2 + c^2 - 1 = 0).The Lagrangian is (f(a,b,c) - lambda g(a,b,c)), but actually, in Lagrange multipliers, we take the gradients. So, we need to compute the partial derivatives of (f) with respect to (a), (b), and (c), and set each equal to (2lambda a), (2lambda b), (2lambda c) respectively (since the gradient of (g) is (2a, 2b, 2c)).But calculating the partial derivatives of (f) might be complicated due to the product of three terms. Let me see. Maybe expanding the product first would make differentiation easier, but expanding it seems messy.Alternatively, maybe we can use symmetry or substitution. Let me think if there's a substitution that can simplify the problem. For example, since the variables are symmetric, perhaps setting two variables equal and the third different. For instance, let’s assume (a = b), and see what happens. Let me try that.Let’s let (a = b). Then, the constraint becomes (2a^2 + c^2 = 1). The expression becomes:First term: (a^2 - bc = a^2 - a c)Second term: (b^2 - ca = a^2 - c a) (same as first term)Third term: (c^2 - a b = c^2 - a^2)So the product is ((a^2 - a c)^2 (c^2 - a^2)). Let me write this as ((a(a - c))^2 (c^2 - a^2) = a^2 (a - c)^2 (c - a)(c + a)) since (c^2 - a^2 = (c - a)(c + a)). Therefore, simplifying, this becomes (-a^2 (a - c)^3 (a + c)). Hmm, not sure if that helps.Alternatively, maybe factor out terms. Let me see. Let's denote (a - c = d), but not sure. Alternatively, express everything in terms of (a) and (c). Since (2a^2 + c^2 = 1), we can express (c^2 = 1 - 2a^2), so (c = sqrt{1 - 2a^2}) or (-sqrt{1 - 2a^2}). But this complicates things, perhaps. Alternatively, set (c = k a), then express in terms of (k). Let me try that.Let (c = k a). Then, the constraint becomes (2a^2 + k^2 a^2 = 1) => (a^2 (2 + k^2) = 1) => (a = pm frac{1}{sqrt{2 + k^2}}).Then, substituting into the expression:First term: (a^2 - bc = a^2 - a cdot c = a^2 - a (k a) = a^2 (1 - k)).Second term: same as first term since (a = b), so also (a^2 (1 - k)).Third term: (c^2 - a b = c^2 - a^2 = (k^2 - 1) a^2).Therefore, the product is ([a^2 (1 - k)]^2 cdot [ (k^2 - 1) a^2 ] = a^6 (1 - k)^2 (k^2 - 1)).But (a^6 = left( frac{1}{2 + k^2} right)^3), so substituting:Product = (frac{1}{(2 + k^2)^3} (1 - k)^2 (k^2 - 1)).Note that (k^2 - 1 = -(1 - k^2)), so this becomes:(- frac{1}{(2 + k^2)^3} (1 - k)^2 (1 - k^2)).Factor (1 - k^2 = (1 - k)(1 + k)), so:(- frac{1}{(2 + k^2)^3} (1 - k)^3 (1 + k)).Therefore, the product is (- frac{(1 - k)^3 (1 + k)}{(2 + k^2)^3}).We need to maximize this expression with respect to (k). Let me note that since (c = k a), and (a) and (c) are real numbers, (k) can be any real number except such that (2 + k^2 > 0) (which is always true). So (k) is any real number.But we need to find the maximum of (- frac{(1 - k)^3 (1 + k)}{(2 + k^2)^3}). Let me denote this function as (P(k)). To find its maximum, we can take the derivative and set it to zero. But this seems a bit involved. Alternatively, maybe substitute (t = k) and analyze the function.Alternatively, note that since we have a negative sign, maximizing (P(k)) is equivalent to minimizing (frac{(1 - k)^3 (1 + k)}{(2 + k^2)^3}). But maybe it's easier to work with the negative.Let’s consider (P(k) = - frac{(1 - k)^3 (1 + k)}{(2 + k^2)^3}). Let me compute its derivative:First, let me write (P(k) = - frac{(1 - k)^3 (1 + k)}{(2 + k^2)^3}). Let’s denote (N(k) = (1 - k)^3 (1 + k)), (D(k) = (2 + k^2)^3), so (P(k) = -N(k)/D(k)). Then, the derivative (P’(k)) is (- [N’(k) D(k) - N(k) D’(k)] / D(k)^2).Let’s compute N’(k):(N(k) = (1 - k)^3 (1 + k)). Let’s apply product rule:N’(k) = 3(1 - k)^2 (-1)(1 + k) + (1 - k)^3 (1) = -3(1 - k)^2 (1 + k) + (1 - k)^3.Factor out (1 - k)^2:= (1 - k)^2 [ -3(1 + k) + (1 - k) ] = (1 - k)^2 [ -3 - 3k + 1 - k ] = (1 - k)^2 [ -2 -4k ] = -2(1 - k)^2 (1 + 2k).Now compute D’(k):D(k) = (2 + k^2)^3, so D’(k) = 3(2 + k^2)^2 (2k) = 6k (2 + k^2)^2.Therefore, the derivative of P(k):P’(k) = - [ (-2(1 - k)^2 (1 + 2k)) (2 + k^2)^3 - (1 - k)^3 (1 + k) * 6k (2 + k^2)^2 ) ] / (2 + k^2)^6Simplify numerator:Let’s factor out common terms:Numerator = - [ -2(1 - k)^2 (1 + 2k)(2 + k^2)^3 - 6k (1 - k)^3 (1 + k)(2 + k^2)^2 ]Factor out -2(1 - k)^2 (2 + k^2)^2:Numerator = - [ -2(1 - k)^2 (2 + k^2)^2 [ (1 + 2k)(2 + k^2) + 3k(1 - k)(1 + k) ] ]Wait, let me check:First term: -2(1 - k)^2 (1 + 2k)(2 + k^2)^3Second term: -6k(1 - k)^3 (1 + k)(2 + k^2)^2Factor out -2(1 - k)^2 (2 + k^2)^2:Then, inside the brackets:(1 + 2k)(2 + k^2) + 3k(1 - k)(1 + k)Let me compute that:First part: (1 + 2k)(2 + k^2) = 2 + k^2 + 4k + 2k^3Second part: 3k(1 - k)(1 + k) = 3k(1 - k^2) = 3k - 3k^3Adding both parts:2 + k^2 + 4k + 2k^3 + 3k - 3k^3 = 2 + k^2 + 7k - k^3So numerator becomes:- [ -2(1 - k)^2 (2 + k^2)^2 (2 + k^2 + 7k - k^3) ) ]Wait, actually, let's check again:Wait, the first part after expanding was 2 + k^2 +4k +2k^3, and the second part 3k -3k^3. Adding together:2 + k^2 +4k +2k^3 +3k -3k^3 = 2 + k^2 +7k -k^3.So, numerator:- [ -2(1 - k)^2 (2 + k^2)^2 (-k^3 + k^2 +7k +2) ) ]But there's an overall negative sign in front:Numerator = - [ -2(1 - k)^2 (2 + k^2)^2 ( -k^3 + k^2 +7k +2 ) ) ] = 2(1 - k)^2 (2 + k^2)^2 (-k^3 +k^2 +7k +2 )Therefore, P’(k) = [2(1 - k)^2 (2 + k^2)^2 (-k^3 +k^2 +7k +2 ) ] / (2 + k^2)^6 ) = [2(1 - k)^2 (-k^3 +k^2 +7k +2 ) ] / (2 + k^2)^4 )Set the derivative equal to zero. So, critical points occur when numerator is zero (denominator is always positive). Therefore:Either:1. (1 - k = 0) => (k = 1)2. (-k^3 + k^2 +7k +2 = 0)So, first possibility: (k = 1). Let's check the second equation: (-k^3 +k^2 +7k +2 = 0).Let me rewrite it as: ( -k^3 + k^2 +7k +2 =0 ) => (k^3 -k^2 -7k -2 =0). Let me try to find rational roots. Possible rational roots are ±1, ±2.Test k=1: 1 -1 -7 -2 = -9 ≠0k= -1: -1 -1 +7 -2=3 ≠0k=2: 8 -4 -14 -2= -12 ≠0k= -2: -8 -4 +14 -2=0. Wait, k=-2: (-2)^3 - (-2)^2 -7*(-2) -2 = -8 -4 +14 -2=0. Yes, k=-2 is a root.Therefore, we can factor the cubic as (k + 2)(k^2 - 3k -1). Let me verify:(k + 2)(k^2 - 3k -1) = k^3 -3k^2 -k + 2k^2 -6k -2 =k^3 -k^2 -7k -2. Yes, that's correct.Therefore, roots are k=-2 and solutions to (k^2 -3k -1=0). Which are (k = [3 ± sqrt(9 +4)]/2 = [3 ± sqrt(13)]/2 ≈ [3 ± 3.605]/2 ≈ 3.3025 or -0.3025).So, critical points at k=1, k=-2, k≈3.3025, k≈-0.3025.Now, we need to evaluate P(k) at these critical points to see which gives the maximum.First, k=1:P(1) = - [(1 -1)^3 (1 +1)] / (2 +1^2)^3 = - [0 *2]/27 =0.So, P(1)=0.k=-2:P(-2) = - [(1 - (-2))^3 (1 + (-2))]/ (2 + (-2)^2)^3 = - [(3)^3 (-1)]/(2 +4)^3 = - [27*(-1)] / 216 = 27/216 = 1/8 ≈0.125.So, P(-2)=1/8.Next, k≈3.3025:Let’s compute approximate value. Let me compute k= [3 + sqrt(13)]/2 ≈ (3 +3.605)/2≈3.3025.Compute (1 -k):1 -3.3025≈-2.3025(1 +k)≈4.3025Denominator: 2 +k^2≈2 + (3.3025)^2≈2 +10.908≈12.908. So denominator^3≈(12.908)^3≈2145.Numerator: (1 -k)^3(1 +k)≈(-2.3025)^3 *4.3025≈(-12.195)*4.3025≈-52.48Thus, P(k)= - [ -52.48 / 2145 ]≈52.48/2145≈0.0244. So approximately 0.0244. Positive value.Similarly, for k≈-0.3025:k≈(3 -sqrt(13))/2≈(3 -3.605)/2≈-0.3025.Compute (1 -k)=1 -(-0.3025)=1.3025(1 +k)=1 +(-0.3025)=0.6975Denominator: 2 +k^2≈2 +0.0915≈2.0915. Denominator^3≈9.15.Numerator: (1 -k)^3(1 +k)= (1.3025)^3 *0.6975≈(2.21)*0.6975≈1.54Thus, P(k)= - [1.54 /9.15]≈-0.168. So negative value. Since we are maximizing, this is not useful.Therefore, the critical points with positive P(k) are k=-2 giving 1/8≈0.125, and k≈3.3025 giving≈0.0244. So the maximum in this case is 1/8.Wait, but this is under the assumption that a = b. So when a = b and c = k a, with k=-2, we get P(k)=1/8. Let me check if this is valid.Given k=-2, then c = -2a. The constraint is 2a^2 + c^2 =1 => 2a^2 +4a^2=1 =>6a^2=1 =>a^2=1/6 =>a=±1/√6. Then c=-2a=∓2/√6=∓√(2/3).So, the variables are a =1/√6, b=1/√6, c=-2/√6. Let's verify the constraint: (1/6 +1/6 +4/6)=6/6=1. Correct.Now, compute the original expression:(a² - bc)(b² - ca)(c² - ab)First, compute each term:a² =1/6, bc=(1/√6)(-2/√6)= -2/6=-1/3. So a² - bc=1/6 - (-1/3)=1/6 +2/6=1/2.Similarly, b² - ca=1/6 - (1/√6)(-2/√6)=1/6 +2/6=1/2.c²=4/6=2/3, ab=(1/√6)(1/√6)=1/6. So c² - ab=2/3 -1/6=4/6 -1/6=3/6=1/2.Therefore, the product is (1/2)(1/2)(1/2)=1/8. So that's correct. So when a = b =1/√6, c=-2/√6, the product is 1/8. Similarly, if we take a = b =-1/√6, c=2/√6, the product would still be (-1/√6)^2 - bc=1/6 - (-1/√6)(2/√6)= same as before. So product is 1/8 as well.Therefore, under the assumption that a = b, the maximum is 1/8. But is this the global maximum?Alternatively, maybe there's a configuration where variables are not equal and we can get a higher value. Let me check another case.Suppose two variables are equal but different from the previous case. Wait, but we tried a = b and found 1/8, which seems a candidate. Alternatively, maybe considering other symmetries.Alternatively, let's consider the case where one variable is negative of another. For example, let’s suppose c = -a. Then the constraint becomes a² + b² + a² =1 => 2a² + b² =1. The expression becomes:(a² - b(-a))(b² - (-a)a)((-a)^2 -ab) = (a² + ab)(b² +a²)(a² -ab).Let me compute each term:First term: a² +abSecond term: b² +a²Third term: a² -abSo the product is (a² +ab)(a² + b²)(a² -ab) = (a² +ab)(a² -ab)(a² +b²) = [a^4 - (ab)^2](a² +b²) = (a^4 -a²b²)(a² +b²) =a^4(a² +b²) -a²b²(a² +b²) =a^6 +a^4b² -a^4b² -a²b^4 =a^6 -a²b^4.Hmm, not sure if this helps. Maybe substitute variables. Let’s set t = a², s = b². Then constraint is 2t + s =1. The product becomes t^3 - t s^2. Since variables are squared, they are non-negative.Express in terms of s: since t = (1 - s)/2. So substituting:Product = [(1 - s)/2]^3 - [(1 - s)/2] * s^2 = (1 - s)^3 /8 - (1 - s)s^2 /2.Let me compute this:= [ (1 - 3s + 3s² - s³) ] /8 - [ (s² - s³) ] /2= (1 -3s +3s² -s³)/8 - (4s² -4s³)/8= [1 -3s +3s² -s³ -4s² +4s³]/8= [1 -3s -s² +3s³]/8To find maximum of this cubic function in s. Take derivative with respect to s:d/ds [1 -3s -s² +3s³]/8 = [ -3 -2s +9s² ] /8.Set equal to zero:-3 -2s +9s² =0 =>9s² -2s -3=0.Solutions: s = [2 ± sqrt(4 + 108)] /18 = [2 ± sqrt(112)] /18 = [2 ± 4*sqrt(7)] /18 = [1 ± 2*sqrt(7)] /9.Compute approximate sqrt(7)=2.6458, so:s≈[1 +5.2916]/9≈6.2916/9≈0.699s≈[1 -5.2916]/9≈-4.2916/9≈-0.477. Since s = b² ≥0, discard negative.So critical point at s≈0.699. Check if this is within the domain. Since 2t + s =1, t=(1 -s)/2. If s≈0.699, then t≈(1 -0.699)/2≈0.1505. So t is positive, which is okay.Now compute the product at s≈0.699:Approximate calculation:Product≈[1 -3*0.699 -0.699² +3*0.699³]/8.Compute each term:1≈1-3*0.699≈-2.097-0.699²≈-0.4883*0.699³≈3*(0.699*0.699*0.699)≈3*(0.3405)≈1.0215Sum:1 -2.097 -0.488 +1.0215≈(1 -2.097) + (-0.488 +1.0215)≈-1.097 +0.5335≈-0.5635Divide by8:≈-0.5635/8≈-0.0704. So product≈-0.0704. Since we are looking for maximum, negative value is worse than previous 1/8≈0.125. So not useful.Therefore, in the case c=-a, the maximum seems to be negative or lower. So not helpful.Another approach: perhaps using inequalities. Since the expression is complicated, maybe we can use AM-GM or Cauchy-Schwarz. But it's not obvious how to apply them here. Alternatively, since variables are on a sphere, maybe use spherical coordinates, but that might complicate things.Alternatively, consider homogenization. The given constraint is (a^2 + b^2 + c^2 =1), and the expression is ((a^2 - bc)(b^2 - ca)(c^2 - ab)). Let me check the degree of the expression. Each term inside the product is degree 2, so the product is degree 6. Homogeneous of degree 6. Since the constraint is degree 2, we can perhaps use substitution to reduce variables, but not sure.Alternatively, consider that the maximum occurs when variables are at certain symmetric positions. Earlier, assuming two variables equal gave a positive maximum of 1/8. Maybe this is the global maximum. To verify, check another case.Suppose, take a=0. Then the constraint becomes b² +c²=1. The expression becomes:(0 - bc)(b² -0)(c² -0) = (-bc)(b²)(c²) = -b^3 c^3.We need to maximize -b^3 c^3 under b² +c²=1. Let’s set b=cosθ, c=sinθ. Then expression becomes -cos³θ sin³θ = - (cosθ sinθ)^3 = - ( (sin2θ)/2 )^3 = - (sin^3 2θ)/8. The maximum of this expression is 0 when sin2θ=0, and the minimum is -1/8 when sin2θ=±1. So the maximum here is 0. So when a=0, the maximum is 0. So again, 1/8 is higher.Another case: let’s set c=0, then as before, the product is -a³ b³, which we saw can go to zero maximum. So 1/8 is higher.Alternatively, check if permuting variables can give higher. For example, suppose we have a different assignment where a, b, c are not two equal and one different, but arranged differently. For example, maybe a=1, b=1, c=0. But wait, a=1, b=1, c=0 would not satisfy a² +b² +c²=2≠1. So need to scale down. Let's take a = b = t, c = s, with 2t² + s²=1. Wait, this is similar to the previous case where a=b. So already considered.Alternatively, check if setting variables in a different proportion. For example, take a=2b=2c. Let's see. Let’s set a=2b and c=b. Then the constraint becomes (4b²) + b² +b²=6b²=1 =>b=±1/√6, a=±2/√6, c=±1/√6.Compute the expression:(a² - bc)= (4/6 - (1/√6)(1/√6))=4/6 -1/6=3/6=1/2(b² - ca)= (1/6 - (2/√6)(1/√6))=1/6 -2/6=-1/6(c² -ab)= (1/6 - (2/√6)(1/√6))=1/6 -2/6=-1/6So product= (1/2)(-1/6)(-1/6)= (1/2)(1/36)=1/72≈0.0139. Less than 1/8. So not better.Alternatively, take a different ratio. Let me try a=3b=3c. Then constraint becomes 9b² +b² +b²=11b²=1 =>b=±1/√11, a=±3/√11, c=±1/√11.Compute expression:(a² - bc)=9/11 - (1/√11)(1/√11)=9/11 -1/11=8/11(b² - ca)=1/11 - (3/√11)(1/√11)=1/11 -3/11=-2/11(c² -ab)=1/11 - (3/√11)(1/√11)=1/11 -3/11=-2/11Product=(8/11)(-2/11)(-2/11)=32/1331≈0.024. Still less than 1/8.Alternatively, try another configuration. Suppose a, b, c are 1/2,1/2, sqrt(1 - 0.25 -0.25)=sqrt(0.5). Let me compute:a=1/2, b=1/2, c=√(0.5).Compute each term:a²=1/4, bc=(1/2)(√0.5)=√0.5/2≈0.3535. So a² - bc≈0.25 -0.3535≈-0.1035.Similarly, b² - ca=0.25 - (√0.5)(1/2)= same as above≈-0.1035.c² -ab=0.5 - (1/2)(1/2)=0.5 -0.25=0.25.Product≈(-0.1035)^2 *0.25≈0.0107 *0.25≈0.0027. Very small.So, this gives a very low value.Alternatively, maybe consider another case where two variables are negatives. Let’s suppose a = -b. Then constraint becomes a² +a² +c²=1 =>2a² +c²=1.Compute the expression:(a² - bc)(b² - ca)(c² - ab). Since a = -b, then b = -a.Substitute:(a² - (-a)c)( (-a)^2 -c(-a) )(c² - (-a)(-a)).Simplify:(a² +ac)(a² +ac)(c² -a²).So product= (a² +ac)^2 (c² -a²).Let me set variables. Let’s express in terms of a and c. Let me set c = k a. Then constraint becomes 2a² +k²a²=1 =>a²=1/(2 +k²). So a=±1/√(2 +k²), c=±k/√(2 +k²).Then, the product becomes:(a² +ac)^2 (c² -a²) = [a² +a*(k a)]^2 [k² a² -a²] = [a²(1 +k)]^2 [a²(k² -1)] = a^6 (1 +k)^2 (k² -1).But a^6 =1/(2 +k²)^3, so product= [ (1 +k)^2 (k² -1) ] / (2 +k²)^3.Note that k² -1 = (k -1)(k +1), so product= (1 +k)^2 (k -1)(k +1) / (2 +k²)^3= (1 +k)^3 (k -1) / (2 +k²)^3.We need to maximize this expression. Let’s denote Q(k) = (1 +k)^3 (k -1)/(2 +k²)^3.To find critical points, take derivative. But this might be tedious. Alternatively, analyze behavior.Note that Q(k) is zero when k=1 or k=-1. For k >1, numerator is positive (since (1 +k)^3 positive, (k -1) positive), denominator positive. So Q(k) positive.For k between -1 and1, numerator: (1 +k)^3 positive, (k -1) negative, so Q(k) negative.For k < -1, (1 +k)^3 negative (if k < -1, then 1 +k <0), and (k -1) negative, so overall numerator positive (negative*negative), denominator positive. So Q(k) positive.So potential maxima in k>1 or k<-1.Let’s check k=2:Q(2)=(3)^3(1)/(4 +4)^3=27*1/512≈0.0527k=3:Q(3)=(4)^3(2)/(9 +9)^3=64*2/5832≈128/5832≈0.0219k=0.5: negative, not useful.k=-2:Q(-2)=(-1)^3*(-3)/(4 +4)^3= ( -1*(-3) ) /512=3/512≈0.00586k approaching infinity:Q(k) ~ (k^3 *k)/k^6)=k^4/k^6=1/k^2→0. So tends to zero.Thus, the maximum in this case seems to be at k=2 with Q(k)≈0.0527, but this is still less than 1/8≈0.125. So not better.Therefore, so far, the maximum found is 1/8 under the assumption that two variables are equal and the third is -2 times that variable. But to ensure this is the global maximum, we need to check if there's any other configuration where the expression can be larger. Let me think if there's a way to use Lagrange multipliers without getting bogged down by complex derivatives.Let’s denote f(a,b,c)=(a² - bc)(b² - ca)(c² - ab), and the constraint g(a,b,c)=a² +b² +c² -1=0.Using Lagrange multipliers, we need ∇f = λ∇g.So, compute partial derivatives of f with respect to a, b, c.First, denote u = a² - bc, v = b² - ca, w = c² - ab. Then f = uvw.Compute partial derivative of f with respect to a:df/da = v w du/da + u w dv/da + u v dw/da.Compute each term:du/da = 2a - 0 = 2a.dv/da = derivative of (b² - ca) with respect to a is -c.dw/da = derivative of (c² -ab) with respect to a is -b.Therefore,df/da = v w (2a) + u w (-c) + u v (-b) = 2a v w - c u w - b u v.Similarly, df/db = u w (2b) - a u w - c u v.df/dc = u v (2c) - b u v - a u w.Setting these equal to λ times partial derivatives of g:2a = 2λ a,2b = 2λ b,2c = 2λ c.Wait, no. Wait, the gradient of g is (2a, 2b, 2c). So, the equations are:2a v w - c u w - b u v = 2λ a,2b u w - a u w - c u v = 2λ b,2c u v - b u v - a u w = 2λ c.This seems complicated. Let me see if we can find a solution that matches our previous case where a = b and c = -2a.Let’s assume a = b, then we can set a = b, and see if the Lagrange equations are satisfied.Let’s denote a = b, and c = k a as before. Then, u = a² - bc = a² -a *c = a² - a*(k a) = a²(1 -k).Similarly, v = b² - ca = a² - c a = a²(1 -k).w = c² -ab = k² a² -a² = a²(k² -1).So f = uvw = a^6 (1 -k)^2 (k² -1).The partial derivatives:df/da = 2a v w - c u w - b u v.Substituting:2a*(v w) -c*(u w) -b*(u v).But since a = b and c = k a,Compute v w: v = a²(1 -k), w = a²(k² -1). So v w = a^4 (1 -k)(k² -1).Similarly, u w = u =a²(1 -k), w =a²(k² -1). So u w = a^4 (1 -k)(k² -1).Similarly, u v = a²(1 -k)*a²(1 -k)=a^4 (1 -k)^2.Therefore:df/da = 2a*(a^4 (1 -k)(k² -1)) - k a*(a^4 (1 -k)(k² -1)) -a*(a^4 (1 -k)^2 )= 2a^5 (1 -k)(k² -1) - k a^5 (1 -k)(k² -1) -a^5 (1 -k)^2.Factor out a^5 (1 -k):= a^5 (1 -k)[ 2(k² -1) -k(k² -1) - (1 -k) ].= a^5 (1 -k)[ (k² -1)(2 -k) - (1 -k) ].Similarly, df/da = 2λ a.But a ≠0 (unless trivial case), so:a^4 (1 -k)[ (k² -1)(2 -k) - (1 -k) ] = 2λ.Similarly, compute df/db. But since a = b, df/db should be equal to df/da.Similarly, df/dc would need to be computed. But this is getting too involved.Alternatively, in the case where we found a solution with k=-2, let's verify if it satisfies the Lagrange conditions.From previous case: a = b=1/√6, c=-2/√6, then u = a² - bc =1/6 - (1/√6)(-2/√6)=1/6 +2/6=1/2.Similarly, v=1/2, w= c² -ab=4/6 -1/6=3/6=1/2. So f=1/8.Now compute partial derivatives:df/da = 2a v w -c u w -b u v.Plugging in the values:2*(1/√6)*(1/2)*(1/2) - (-2/√6)*(1/2)*(1/2) - (1/√6)*(1/2)*(1/2)= 2*(1/√6)*(1/4) + (2/√6)*(1/4) - (1/√6)*(1/4)= (2/4√6) + (2/4√6) - (1/4√6) = (2 +2 -1)/4√6 =3/(4√6).On the other hand, 2λ a=2λ*(1/√6). So equate:3/(4√6) = 2λ*(1/√6) => 3/4 = 2λ => λ=3/8.Similarly, compute df/db. Since a =b, df/db should be the same as df/da, which is 3/(4√6). Which would also equal 2λ b=2λ*(1/√6)= same as above, which checks out.Now compute df/dc:df/dc=2c u v -b u v -a u w.Wait, original expression:df/dc = 2c u v -b u v -a u w.Wait, no, let's re-express:From the earlier general expression:df/dc = u v (2c) - b u v -a u w.Plug in values:u=1/2, v=1/2, w=1/2, a=b=1/√6, c=-2/√6.So df/dc = (1/2)(1/2)(2*(-2/√6)) - (1/√6)(1/2)(1/2) - (1/√6)(1/2)(1/2).Compute term by term:First term: (1/4)(-4/√6) = -1/√6.Second term: -(1/√6)(1/4) = -1/(4√6).Third term: -(1/√6)(1/4) = -1/(4√6).Total df/dc= -1/√6 -1/(4√6) -1/(4√6) = -1/√6 -1/(2√6)= - (2/2√6 +1/2√6)= -3/(2√6).On the other hand, the Lagrange condition is df/dc=2λ c.Compute 2λ c=2*(3/8)*(-2/√6)= (3/4)*(-2/√6)= -6/(4√6)= -3/(2√6), which matches df/dc= -3/(2√6). Thus, the solution satisfies the Lagrange multiplier conditions. Therefore, this critical point is valid.Therefore, this critical point is a valid extremum. Since we found the value 1/8 here, and other tested cases give lower or zero, it is likely that this is the global maximum.To confirm, let's check the second derivative or the Hessian, but that's quite involved. Alternatively, since we found a valid critical point with a positive value higher than other tested points, and given the symmetry, it's reasonable to conclude that this is the maximum.Therefore, the maximum value of the given expression under the constraint is (boxed{dfrac{1}{8}}).

✅:Given the constraint (a^2 + b^2 + c^2 = 1), we aim to find the maximum value of the expression ((a^2 - bc)(b^2 - ca)(c^2 - ab)).Key Steps and Reasoning:1. Symmetry Consideration: - We first explored symmetric cases where variables are equal or zero. These cases resulted in the product being zero, indicating the maximum might occur elsewhere.2. Case with Two Equal Variables: - Assuming (a = b) and (c = -2a), we derived the constraint (2a^2 + c^2 = 1) leading to (a = pm frac{1}{sqrt{6}}) and (c = mp frac{2}{sqrt{6}}).3. Substitution and Simplification: - Substituting (a = b = frac{1}{sqrt{6}}) and (c = -frac{2}{sqrt{6}}) into the expression: - Each term ((a^2 - bc)), ((b^2 - ca)), and ((c^2 - ab)) evaluates to (frac{1}{2}). - The product is (left(frac{1}{2}right)^3 = frac{1}{8}).4. Verification via Lagrange Multipliers: - Confirmed the critical point satisfies the necessary conditions for extrema, validating (frac{1}{8}) as a maximum.5. Comparison with Other Configurations: - Tested other configurations (e.g., one variable zero, different ratios) but found lower or zero values, reinforcing (frac{1}{8}) as the highest value.Final Answer:The maximum value of (left(a^{2} - b cright)left(b^{2} - c aright)left(c^{2} - a bright)) under the constraint (a^2 + b^2 + c^2 = 1) is (boxed{dfrac{1}{8}}).

🔑:: To find the maximum value of ( left(a^{2} - bcright)left(b^{2} - caright)left(c^{2} - abright) ) given (a^2 + b^2 + c^2 = 1), we consider the two cases: 1. When all terms ( a^2 - bc ), ( b^2 - ca ), and ( c^2 - ab ) are positive.2. When one term is positive and the other two are negative.Let's investigate these cases in detail.1. Case 1: All terms are positive If ( a^2 - bc > 0 ), ( b^2 - ca > 0 ), and ( c^2 - ab > 0 ), then applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we get: [ left(a^2 - bcright)left(b^2 - caright)left(c^2 - abright) leq left( frac{a^2 + b^2 + c^2 - ab - bc - ca}{3} right)^3 ] Utilizing the given condition ( a^2 + b^2 + c^2 = 1 ), we can substitute: [ frac{a^2 + b^2 + c^2 - ab - bc - ca}{3} = frac{1 - (ab + bc + ca)}{3} ] To find the maximum value, the expression within must be minimized. The minimum for ( ab + bc + ca ) occurs when ( a = 0, b = -frac{1}{sqrt{2}}, c = frac{1}{sqrt{2}} ). Thus, [ boxed{frac{1}{2}} ]2. Case 2: One term is positive and the other two are negative Without loss of generality, assume ( a^2 - bc > 0 ), ( b^2 - ca < 0 ), and ( c^2 - ab < 0 ). For such cases: [ left(a^2 - bcright)left(b^2 - caright)left(c^2 - abright) = left(a^2 - bcright)left(ca - b^2right)left(ab - c^2right) ] Again, applying the AM-GM inequality for the positive and negatives: [ left(a^2 - bcright)left(b^2 - caright)left(c^2 - abright) leq left(frac{a^2 - b^2 - c^2 + ab + ac - bc}{3}right)^3 ] Here, [ left(a^2 - b^2 - c^2 + ab + ac - bcright) leq frac{1 - (ab + bc + ca)}{3} ] Again, the minimum for ( ab + bc + ca ) equates, where ( a = 0, b = -frac{1}{sqrt{2}}, c = frac{1}{sqrt{2}} ).Therefore for all cases, the maximum value is given by[boxed{frac{1}{8}}]

🧠:Okay, so I need to find two points on a line that's parallel to the first picture plane, and these two points should form an equilateral triangle with a given point on the projection axis. Hmm, let me start by visualizing the problem. First, let me recall some concepts. The first picture plane in projection geometry is typically the plane onto which objects are projected. A line parallel to this plane would maintain its direction and not converge towards any vanishing points. The projection axis is probably the line where the two picture planes (like the horizontal and vertical planes in orthographic projection) intersect. So the given point is somewhere along this intersection line.Wait, maybe I should clarify the coordinate system here. Let me assume a 3D coordinate system where the first picture plane is the XY-plane, and the projection axis is the Z-axis? Or maybe it's the X-axis? Hmm, different sources might use different setups. Let me think. In standard multiview projections, the projection axes are the lines where the principal planes intersect. For example, the front view (elevation) is on the XY-plane, the top view (plan) is on the XZ-plane, so their intersection is the X-axis. Similarly, the side view might be on the YZ-plane, intersecting on the Y-axis. But maybe in this case, since it's the first picture plane, perhaps the projection axis is the X-axis?Alternatively, maybe the projection axis is the line of intersection between the first and second picture planes. If the first picture plane is, say, the XY-plane, and there's another picture plane (maybe a profile plane) which is the YZ-plane, their intersection is the Y-axis. So the projection axis could be the Y-axis? Hmm, this is getting a bit confusing. Let me try to define the coordinate system properly.Let's assume the first picture plane is the XY-plane. A line parallel to this plane would have a direction that doesn't change in the X or Y directions. Wait, no, a line parallel to the plane doesn't necessarily have to be in the plane. If the line is parallel to the XY-plane, it would have a constant Z-coordinate. So, for example, a line in 3D space that's parallel to the XY-plane could be something like z = c, where c is a constant. But if the line is parallel to the first picture plane (XY-plane), then it lies in a plane parallel to XY-plane, so yes, it's of the form z = c. Now, the projection axis. If the projection axis is the line where two picture planes intersect, like the X-axis (intersection of XY and XZ planes) or Y-axis (intersection of XY and YZ planes). The given point is on the projection axis. Let me suppose that the projection axis is the X-axis. So the given point is (a, 0, 0) for some a. Alternatively, if the projection axis is the Z-axis, then the point is (0, 0, a). Wait, but in standard projections, the line of intersection between the horizontal and profile planes is the Z-axis, but the front view and top view intersect on the X-axis. Hmm. Maybe the problem is using a specific setup, but since it's not specified, I need to make assumptions.Let me try to formalize the problem. Let's set up a coordinate system where the first picture plane is the XY-plane, and the projection axis is the X-axis (common in multiview projections). Therefore, the given point is on the X-axis, say point P (p, 0, 0). The line parallel to the first picture plane (XY-plane) must be a line that is parallel to the XY-plane, so all points on this line have the same Z-coordinate. Let's say the line is in the plane z = c, and it's parallel to some direction in the XY-plane. Wait, but the problem states it's a line parallel to the first picture plane. A line can be parallel to a plane if it lies within a plane that's parallel to the given plane. Wait, no. A line being parallel to a plane actually means that the line is either lying on a parallel plane or is skew but directionally parallel. Wait, actually, in geometry, a line is parallel to a plane if it lies on a plane parallel to the given plane or if it doesn't intersect the given plane and is not skew. Hmm, maybe this is confusing. Alternatively, maybe the line is parallel to the picture plane in the sense that its direction vector is parallel to the picture plane. So, for the XY-plane, a line with direction vector (a, b, 0), i.e., no component in the Z-direction. So such a line would be parallel to the XY-plane. Yes, that makes sense. So the given line has a direction vector that's parallel to the XY-plane, meaning it can be represented parametrically as (x0 + ta, y0 + tb, z0), where z0 is constant. But the problem states "a line parallel to the first picture plane", which is the XY-plane. So such a line would have a constant Z-coordinate. Therefore, the line can be written as (x(t), y(t), z0), where x(t) and y(t) are linear functions of t. So, for example, a line in the plane z = z0, which is parallel to the XY-plane.But the problem says "a line parallel to the first picture plane and a point on the projection axis". Wait, no, correction: the problem says "There is a line parallel to the first picture plane and a point on the projection axis. Find two points on the line such that they form an equilateral triangle with the given point."Wait, so the line is parallel to the first picture plane (so, as above, in a plane z = c if the first picture plane is XY). The given point is on the projection axis. So, projection axis is likely the line of intersection between the first picture plane and another picture plane. If the first picture plane is XY, then the projection axis could be the X-axis (intersection with XZ-plane) or Y-axis (intersection with YZ-plane) or Z-axis? Wait, in standard multiview projection, the projection axes are the lines where the planes intersect. So front view (XY) and top view (XZ) intersect on the X-axis; front and side view (YZ) intersect on the Y-axis. So, depending on the setup, the projection axis could be X or Y. But the problem mentions "the projection axis", so perhaps it's a specific one.Alternatively, maybe in the context of the problem, the projection axis is the Z-axis. Wait, but if the first picture plane is XY, the projection axis would be orthogonal to it, but that might not be the case. Hmm. Maybe I need to check.Alternatively, perhaps the projection axis is the line that is the intersection of the picture plane and the horizontal plane in a perspective projection setup. But this is getting too vague. Let me make an assumption.Assume the first picture plane is the XY-plane, and the projection axis is the Z-axis. So the given point is on the Z-axis, say (0, 0, p). The line parallel to the first picture plane (XY-plane) would be a line in some plane z = c, with direction vector in the XY-plane. So parametric equations: x = x0 + at, y = y0 + bt, z = c. The problem is to find two points on this line such that they form an equilateral triangle with the point (0, 0, p). So, points A and B on the line, such that triangle PAB is equilateral, where P is (0, 0, p).Alternatively, if the projection axis is the X-axis, then the given point is (p, 0, 0). The line is parallel to the first picture plane (XY-plane), so it's in some z = c plane. Then the line has parametric equations (x0 + at, y0 + bt, c). Then, we need two points on this line such that the triangle with vertices (p, 0, 0), A, and B is equilateral.Wait, but depending on the coordinate system, the calculations will differ. Maybe the problem is in 2D? Wait, but projection geometry is inherently 3D. However, maybe the line is in the first picture plane? Wait, the line is parallel to the first picture plane. If the first picture plane is considered as a 2D plane, then a line parallel to it would be in a different plane but parallel. Hmm, this is confusing.Alternatively, maybe the problem is in 2D. Wait, if it's in 2D, then a line parallel to the picture plane would just be a line in the plane, and the projection axis is a point? That doesn't make sense. Hmm. Maybe I need to think in 3D.Alternatively, let's consider the first picture plane as the plane z = 0 (the XY-plane). The projection axis is the line where this picture plane intersects another picture plane, say the XZ-plane (which would be the X-axis). So the projection axis is the X-axis, lying in both the XY and XZ planes. Therefore, the given point is on the X-axis, say (a, 0, 0). The line parallel to the first picture plane (z=0) must be a line in some plane z = c (parallel to z=0) with direction vector in the XY-plane. So the line can be parametrized as (x0 + dt, y0 + et, c), where d and e are constants determining the direction, and t is a parameter.We need two points on this line, say A and B, such that the triangle PAB is equilateral, where P is (a, 0, 0). So the distances PA, PB, and AB must all be equal.So let's formalize this. Let’s denote point P as (p, 0, 0). The line L is given by parametric equations:x = x0 + dty = y0 + etz = cWe need to find two points A = (x0 + d t1, y0 + e t1, c) and B = (x0 + d t2, y0 + e t2, c) on line L such that PA = PB = AB.So, PA = distance from P to A, PB = distance from P to B, AB = distance from A to B.Given that PA = PB = AB.First, compute PA^2:PA^2 = (x0 + d t1 - p)^2 + (y0 + e t1 - 0)^2 + (c - 0)^2Similarly, PB^2 = (x0 + d t2 - p)^2 + (y0 + e t2)^2 + c^2AB^2 = (d(t2 - t1))^2 + (e(t2 - t1))^2 + 0 = (d^2 + e^2)(t2 - t1)^2Since PA = PB = AB, their squares must be equal:1. (x0 + d t1 - p)^2 + (y0 + e t1)^2 + c^2 = (x0 + d t2 - p)^2 + (y0 + e t2)^2 + c^22. (x0 + d t1 - p)^2 + (y0 + e t1)^2 + c^2 = (d^2 + e^2)(t2 - t1)^2From equation 1, since c^2 is on both sides, they cancel out. So:(x0 + d t1 - p)^2 + (y0 + e t1)^2 = (x0 + d t2 - p)^2 + (y0 + e t2)^2Let’s expand both sides:Left side: (x0 - p + d t1)^2 + (y0 + e t1)^2= (x0 - p)^2 + 2 d (x0 - p) t1 + d^2 t1^2 + y0^2 + 2 e y0 t1 + e^2 t1^2Right side: (x0 - p + d t2)^2 + (y0 + e t2)^2= (x0 - p)^2 + 2 d (x0 - p) t2 + d^2 t2^2 + y0^2 + 2 e y0 t2 + e^2 t2^2Subtract left side from right side:0 = 2 d (x0 - p)(t2 - t1) + d^2(t2^2 - t1^2) + 2 e y0 (t2 - t1) + e^2(t2^2 - t1^2)Factor terms:0 = (t2 - t1)[2 d (x0 - p) + d^2(t2 + t1) + 2 e y0 + e^2(t2 + t1)]Let’s denote s = t2 + t1 and r = t2 - t1. Then the equation becomes:0 = r [2 d (x0 - p) + d^2 s + 2 e y0 + e^2 s]Assuming r ≠ 0 (since t2 ≠ t1, otherwise points A and B coincide), we have:2 d (x0 - p) + d^2 s + 2 e y0 + e^2 s = 0Factor s:s (d^2 + e^2) + 2 d (x0 - p) + 2 e y0 = 0So:s = [ -2 d (x0 - p) - 2 e y0 ] / (d^2 + e^2 )But s = t1 + t2, so:t1 + t2 = [ -2 d (x0 - p) - 2 e y0 ] / (d^2 + e^2 )Now, moving to equation 2:PA^2 = AB^2From PA^2: (x0 + d t1 - p)^2 + (y0 + e t1)^2 + c^2 = (d^2 + e^2)(t2 - t1)^2Let’s expand PA^2:= (x0 - p + d t1)^2 + (y0 + e t1)^2 + c^2= (x0 - p)^2 + 2 d (x0 - p) t1 + d^2 t1^2 + y0^2 + 2 e y0 t1 + e^2 t1^2 + c^2Set equal to AB^2:(d^2 + e^2)(t2 - t1)^2So:(x0 - p)^2 + y0^2 + c^2 + 2 d (x0 - p) t1 + 2 e y0 t1 + (d^2 + e^2) t1^2 = (d^2 + e^2)(t2^2 - 2 t1 t2 + t1^2)Bring all terms to the left side:(x0 - p)^2 + y0^2 + c^2 + 2 d (x0 - p) t1 + 2 e y0 t1 + (d^2 + e^2) t1^2 - (d^2 + e^2) t2^2 + 2 (d^2 + e^2) t1 t2 - (d^2 + e^2) t1^2 = 0Simplify:(x0 - p)^2 + y0^2 + c^2 + 2 d (x0 - p) t1 + 2 e y0 t1 - (d^2 + e^2) t2^2 + 2 (d^2 + e^2) t1 t2 = 0This looks complicated. Maybe substitute t2 from the earlier relation. Since s = t1 + t2 = [ -2 d (x0 - p) - 2 e y0 ] / (d^2 + e^2 ), we can express t2 = s - t1. Let’s substitute t2 = s - t1 into the equation.So:(x0 - p)^2 + y0^2 + c^2 + 2 d (x0 - p) t1 + 2 e y0 t1 - (d^2 + e^2)(s - t1)^2 + 2 (d^2 + e^2) t1 (s - t1) = 0Expand (s - t1)^2:s^2 - 2 s t1 + t1^2So:(x0 - p)^2 + y0^2 + c^2 + 2 d (x0 - p) t1 + 2 e y0 t1 - (d^2 + e^2)(s^2 - 2 s t1 + t1^2) + 2 (d^2 + e^2) t1 (s - t1) = 0Expand the other term:+ 2 (d^2 + e^2) t1 s - 2 (d^2 + e^2) t1^2Combine terms:= (x0 - p)^2 + y0^2 + c^2 + 2 d (x0 - p) t1 + 2 e y0 t1 - (d^2 + e^2)s^2 + 2 (d^2 + e^2) s t1 - (d^2 + e^2) t1^2 + 2 (d^2 + e^2) t1 s - 2 (d^2 + e^2) t1^2Simplify:= (x0 - p)^2 + y0^2 + c^2 + 2 d (x0 - p) t1 + 2 e y0 t1 - (d^2 + e^2)s^2 + 4 (d^2 + e^2) s t1 - 3 (d^2 + e^2) t1^2This is getting too messy. Maybe there's a better approach. Let's recall that we have two equations:1. t1 + t2 = s = [ -2 d (x0 - p) - 2 e y0 ] / (d^2 + e^2 )2. PA^2 = AB^2But maybe instead of substituting, think geometrically. The two points A and B on the line must be such that their midpoint is a certain point, and the distance between them is equal to PA.Alternatively, in 3D space, forming an equilateral triangle with a point not on the line. The set of points forming an equilateral triangle with P and a line L would be the intersection of two spheres: one centered at P with radius PA, and another centered at A with radius PA, but this seems recursive.Wait, actually, given point P and line L, we need two points A and B on L such that PA = PB = AB. So, the points A and B must lie on the intersection of two spheres: one centered at P with radius AB, and another centered at A with radius AB. But this is not straightforward.Alternatively, consider that for any point A on L, the locus of points B such that triangle PAB is equilateral is a circle perpendicular to PA at a certain angle. But this might be complex in 3D.Alternatively, maybe project the problem onto a plane. Since the line L is parallel to the first picture plane (XY-plane), and the projection axis is X (assuming), so point P is on the X-axis. Then, perhaps the problem can be simplified by projecting onto the XY-plane?Wait, but in 3D, the line L is in a plane z = c. The point P is (p, 0, 0). If we project orthographically onto the XY-plane, the line L would project to a line in XY-plane (since z is constant), and point P would project to (p, 0). Then, in the projection, the problem reduces to finding two points on the projected line such that they form an equilateral triangle with (p, 0). However, since the original line is in 3D, the distances would involve the z-coordinate as well. So this complicates things.Alternatively, if the line L is horizontal (in the XY-plane), but the problem states it's parallel to the first picture plane, which is the XY-plane, so it's in a plane z = c. The distance from P (p, 0, 0) to any point on L would involve the z-coordinate. Therefore, even if the projection on the XY-plane forms an equilateral triangle, the actual 3D distances might not.Therefore, perhaps the problem is inherently 3D, and we need to solve it in 3D space.Let me consider specific coordinates to simplify. Let’s assume some values to make the math easier. Suppose the line L is along the X-axis at z = c, so parametric equations: x = t, y = 0, z = c. Then point P is on the projection axis, say the X-axis at (p, 0, 0). So we need two points on L: A(t1, 0, c) and B(t2, 0, c) such that PA = PB = AB.Compute PA: distance from (p,0,0) to (t1,0,c): sqrt( (t1 - p)^2 + 0 + c^2 )Similarly, PB: sqrt( (t2 - p)^2 + c^2 )AB: distance between A and B: sqrt( (t2 - t1)^2 + 0 + 0 ) = |t2 - t1|Set PA = PB:sqrt( (t1 - p)^2 + c^2 ) = sqrt( (t2 - p)^2 + c^2 )Squaring both sides: (t1 - p)^2 = (t2 - p)^2Thus, t1 - p = ±(t2 - p)Case 1: t1 - p = t2 - p ⇒ t1 = t2, which can't be since A and B must be distinct.Case 2: t1 - p = - (t2 - p ) ⇒ t1 - p = -t2 + p ⇒ t1 + t2 = 2pSo t1 + t2 = 2p.Now, set PA = AB:sqrt( (t1 - p)^2 + c^2 ) = |t2 - t1|Square both sides:(t1 - p)^2 + c^2 = (t2 - t1)^2But from t1 + t2 = 2p, we have t2 = 2p - t1.Substitute t2 = 2p - t1 into the equation:(t1 - p)^2 + c^2 = (2p - t1 - t1)^2 = (2p - 2t1)^2 = 4(p - t1)^2Expand left side:(t1 - p)^2 + c^2 = (p - t1)^2 + c^2 = (p - t1)^2 + c^2Right side:4(p - t1)^2Thus:(p - t1)^2 + c^2 = 4(p - t1)^2Rearranged:c^2 = 3(p - t1)^2Therefore:(p - t1)^2 = c^2 / 3 ⇒ p - t1 = ± c / sqrt(3)Thus, t1 = p ± c / sqrt(3)Then, since t2 = 2p - t1:If t1 = p + c / sqrt(3), then t2 = 2p - (p + c / sqrt(3)) = p - c / sqrt(3)If t1 = p - c / sqrt(3), then t2 = 2p - (p - c / sqrt(3)) = p + c / sqrt(3)Therefore, the two points are:A: (p + c / sqrt(3), 0, c) and B: (p - c / sqrt(3), 0, c)Or vice versa.So in this specific case where the line is the X-axis in the plane z = c, the points are located at p ± c / sqrt(3) along the X-axis.But this is a specific case. Let's check if this makes sense. The distance PA would be sqrt( (c / sqrt(3))^2 + c^2 ) = sqrt( c^2 / 3 + c^2 ) = sqrt(4c^2 / 3) = (2c)/sqrt(3). The distance AB is |t2 - t1| = | (p - c / sqrt(3)) - (p + c / sqrt(3)) | = | -2c / sqrt(3) | = 2c / sqrt(3). So indeed PA = AB, which satisfies the equilateral triangle condition.Therefore, in this specific case, the solution exists as long as c ≠ 0. If c = 0, the line L would be on the projection axis (X-axis), and the points would collapse.But the problem is more general. The line is an arbitrary line parallel to the first picture plane, not necessarily aligned with the X-axis. So let's try to generalize.Suppose the line L is given by parametric equations:x = x0 + dty = y0 + etz = cWhere d and e are not both zero (since it's a line). The point P is (p, 0, 0) on the projection axis (X-axis). We need to find t1 and t2 such that points A(x0 + d t1, y0 + e t1, c) and B(x0 + d t2, y0 + e t2, c) form an equilateral triangle with P.So, the conditions are:1. PA = PB2. PA = ABStarting with PA = PB:Distance PA = sqrt( (x0 + d t1 - p)^2 + (y0 + e t1)^2 + c^2 )Distance PB = sqrt( (x0 + d t2 - p)^2 + (y0 + e t2)^2 + c^2 )Setting them equal:( x0 + d t1 - p )^2 + ( y0 + e t1 )^2 = ( x0 + d t2 - p )^2 + ( y0 + e t2 )^2Expanding both sides:Left: (x0 - p)^2 + 2 d (x0 - p) t1 + d² t1² + y0² + 2 e y0 t1 + e² t1²Right: (x0 - p)^2 + 2 d (x0 - p) t2 + d² t2² + y0² + 2 e y0 t2 + e² t2²Subtract left from right:0 = 2 d (x0 - p)(t2 - t1) + d²(t2² - t1²) + 2 e y0 (t2 - t1) + e²(t2² - t1²)Factor:0 = (t2 - t1)[2 d (x0 - p) + d²(t2 + t1) + 2 e y0 + e²(t2 + t1)]Let’s let s = t1 + t2 and r = t2 - t1 (as before). Then:0 = r [2 d (x0 - p) + (d² + e²) s + 2 e y0 ]Assuming r ≠ 0, then:2 d (x0 - p) + (d² + e²) s + 2 e y0 = 0Therefore:s = [ -2 d (x0 - p) - 2 e y0 ] / (d² + e² )Now, PA = AB.Compute PA² = (x0 + d t1 - p)^2 + (y0 + e t1)^2 + c²AB² = (d(t2 - t1))² + (e(t2 - t1))² = (d² + e²)(t2 - t1)^2Setting PA² = AB²:(x0 + d t1 - p)^2 + (y0 + e t1)^2 + c² = (d² + e²)(t2 - t1)^2Let’s express t2 = s - t1 = [ -2 d (x0 - p) - 2 e y0 ] / (d² + e² ) - t1But this might not be helpful directly. Alternatively, note that from PA = AB, we have:PA² = AB² ⇒ (x0 + d t1 - p)^2 + (y0 + e t1)^2 + c² = (d² + e²)(t2 - t1)^2But from PA = PB, we have t1 + t2 = s = [ -2 d (x0 - p) - 2 e y0 ] / (d² + e² )Let’s denote t2 = s - t1So substitute into AB²:AB² = (d² + e²)(s - 2 t1)^2Wait, no, t2 - t1 = (s - t1) - t1 = s - 2 t1. Wait, no:t2 = s - t1 ⇒ t2 - t1 = s - 2 t1But then AB² = (d² + e²)(s - 2 t1)^2But PA² is also equal to this. So:(x0 + d t1 - p)^2 + (y0 + e t1)^2 + c² = (d² + e²)(s - 2 t1)^2This is a quadratic equation in t1. Let’s expand both sides.Left side:(x0 - p + d t1)^2 + (y0 + e t1)^2 + c²= (x0 - p)^2 + 2 d (x0 - p) t1 + d² t1² + y0² + 2 e y0 t1 + e² t1² + c²= (x0 - p)^2 + y0² + c² + [2 d (x0 - p) + 2 e y0 ] t1 + (d² + e²) t1²Right side:(d² + e²)(s - 2 t1)^2= (d² + e²)(s² - 4 s t1 + 4 t1² )So equate left and right:(x0 - p)^2 + y0² + c² + [2 d (x0 - p) + 2 e y0 ] t1 + (d² + e²) t1² = (d² + e²)s² - 4 (d² + e²) s t1 + 4 (d² + e²) t1²Bring all terms to left side:(x0 - p)^2 + y0² + c² + [2 d (x0 - p) + 2 e y0 ] t1 + (d² + e²) t1² - (d² + e²)s² + 4 (d² + e²) s t1 - 4 (d² + e²) t1² = 0Simplify:(x0 - p)^2 + y0² + c² - (d² + e²)s² + [2 d (x0 - p) + 2 e y0 + 4 (d² + e²)s ] t1 + [ (d² + e²) - 4 (d² + e²) ] t1² = 0Simplify coefficients:For t1²: (1 - 4)(d² + e²) = -3(d² + e²)For t1 term: 2 d (x0 - p) + 2 e y0 + 4 (d² + e²)sSo:(x0 - p)^2 + y0² + c² - (d² + e²)s² + [2 d (x0 - p) + 2 e y0 + 4 (d² + e²)s ] t1 - 3 (d² + e²) t1² = 0Now, substitute s = [ -2 d (x0 - p) - 2 e y0 ] / (d² + e² )First, compute (d² + e²)s:= (d² + e²) * [ -2 d (x0 - p) - 2 e y0 ] / (d² + e² )= -2 d (x0 - p) - 2 e y0Similarly, compute (d² + e²)s²:= (d² + e²) * [ (-2 d (x0 - p) - 2 e y0 ) / (d² + e² ) ]^2= [ (-2 d (x0 - p) - 2 e y0 )^2 ] / (d² + e² )Also, compute 4 (d² + e²)s:= 4 * [ -2 d (x0 - p) - 2 e y0 ] Now, substitute into the equation:Left side becomes:(x0 - p)^2 + y0² + c² - [ (-2 d (x0 - p) - 2 e y0 )^2 / (d² + e² ) ] + [2 d (x0 - p) + 2 e y0 + 4 (-2 d (x0 - p) - 2 e y0 ) ] t1 - 3 (d² + e²) t1² = 0Simplify the coefficients:First term: (x0 - p)^2 + y0² + c² - [4 d² (x0 - p)^2 + 8 d e (x0 - p) y0 + 4 e² y0² ] / (d² + e² )Second term: [2 d (x0 - p) + 2 e y0 - 8 d (x0 - p) - 8 e y0 ] t1 = [ -6 d (x0 - p) -6 e y0 ] t1Third term: -3 (d² + e²) t1²Let’s handle the first term:Let’s write it as:[ (x0 - p)^2 + y0² ] + c² - [4 d² (x0 - p)^2 + 8 d e (x0 - p) y0 + 4 e² y0² ] / (d² + e² )Factor 4 in the numerator:= [ (x0 - p)^2 + y0² ] + c² - 4 [ d² (x0 - p)^2 + 2 d e (x0 - p) y0 + e² y0² ] / (d² + e² )Notice that d² (x0 - p)^2 + 2 d e (x0 - p) y0 + e² y0² = [ d(x0 - p) + e y0 ]^2So:= [ (x0 - p)^2 + y0² ] + c² - 4 [ ( d(x0 - p) + e y0 )^2 ] / (d² + e² )Now, this expression can be simplified if we express [ (x0 - p)^2 + y0² ] as something. Let’s see:Let’s denote vector v = (d, e), and vector w = (x0 - p, y0). Then, the term [ d(x0 - p) + e y0 ] is the dot product v · w, and [ (x0 - p)^2 + y0² ] is ||w||². So:= ||w||² + c² - 4 (v · w)^2 / ||v||²Using the Cauchy-Schwarz inequality, (v · w)^2 ≤ ||v||² ||w||², so this term is ||w||² - 4 (v · w)^2 / ||v||² + c².Not sure if that helps, but perhaps proceed numerically.Alternatively, let’s plug in specific values to check. Suppose x0 = 0, y0 = 0, d = 1, e = 0, so the line is x = t, y = 0, z = c. Then point P is (p, 0, 0). This is the case we considered earlier.Then:s = [ -2 * 1 * (0 - p) - 2 * 0 * 0 ] / (1² + 0² ) = [ 2 p ] / 1 = 2pThen, the first term:(x0 - p)^2 + y0² + c² - [4 d² (x0 - p)^2 + ... ] / (d² + e² )But x0 = 0, y0 = 0, d=1, e=0:= ( - p )² + 0 + c² - [4 *1²*(-p)^2 + 0 +0 ] / (1 +0 )= p² + c² - 4p² = -3p² + c²The coefficient for t1:[ -6 *1*(-p) -6*0*0 ] t1 = 6p t1Third term: -3*(1 + 0)*t1² = -3 t1²Thus, the equation becomes:-3p² + c² + 6p t1 -3 t1² = 0Multiply through by -1:3p² - c² -6p t1 +3 t1² =0Divide by 3:p² - c²/3 - 2p t1 + t1² =0This is a quadratic equation in t1:t1² -2p t1 + (p² - c² /3 )=0Solving:t1 = [2p ± sqrt(4p² - 4*(p² - c² /3 ))] /2= [2p ± sqrt(4p² -4p² + 4c² /3 )]/2= [2p ± (2c / sqrt(3))]/2= p ± c / sqrt(3)Which matches our earlier result. So the general approach works in this specific case.But for the general case, this seems quite involved. Let’s see if we can find a geometric interpretation.Given line L and point P, we need two points A and B on L such that triangle PAB is equilateral. This is similar to constructing an equilateral triangle with one vertex at P and the other two on line L.In 2D, this can be done by rotating the line L by 60 degrees around point P and finding the intersection points. Maybe in 3D, a similar approach can be used with rotations.However, in 3D, there are infinitely many planes containing point P and line L. In each such plane, the problem reduces to the 2D case. But since we need the triangle to be equilateral in 3D space, not projected onto a plane, this complicates things.Alternatively, consider that the set of points A such that PA = AB and B is another point on L can be found by solving the equations. But this might not simplify easily.Alternatively, use vectors. Let’s denote vector PA and vector PB. For triangle PAB to be equilateral, we need |PA| = |PB| = |AB|, and the angle between PA and PB should be 60 degrees. However, in 3D, the angle condition might not necessarily hold because the triangle could be in a non-planar orientation? Wait, no. A triangle is always planar, so the three points P, A, B must lie in a plane, and in that plane, the triangle is equilateral. Therefore, the problem reduces to finding a plane containing P and line L such that in this plane, there exists two points A and B on L forming an equilateral triangle with P.But line L is parallel to the first picture plane (say XY-plane), so it's in a plane z = c. The point P is on the projection axis (X-axis). The plane containing P and L must include the line L and the point P. Since L is in z = c and P is at z = 0, the plane would be a non-horizontal plane cutting through z = 0 and z = c.Wait, but any plane containing line L and point P must contain all points of L and P. Since L is in z = c and P is at (p, 0, 0), the plane is defined by line L and point P. However, such a plane would intersect the XY-plane (z=0) at point P and some line. But the line L is parallel to the XY-plane, so the plane containing L and P is not parallel to XY-plane, hence would intersect the XY-plane along a line passing through P.In this plane, the problem reduces to 2D: find two points A and B on line L (which is a line in this plane) such that triangle PAB is equilateral. So, in this plane, we can apply the standard 2D construction.Therefore, perhaps the solution involves finding the intersection of the line L with two circles: one centered at P with radius PA, and another rotated by 60 degrees.But since we are in 3D, the distance PA involves the z-coordinate. However, within the plane containing P and L, the distance from P to A is sqrt( (x_A - p)^2 + y_A^2 + c^2 ). But in the plane itself, the distance would be the Euclidean distance within that plane.Wait, no. In the 3D plane containing P and L, the distance between P and A is the straight-line distance, which includes the z-component. So even within that plane, it's a 3D distance. Therefore, constructing an equilateral triangle in that plane would require the 3D distances to be equal.But in the plane, we can use 2D coordinates by establishing a coordinate system within the plane. Let me define a coordinate system where P is the origin, and the line L is represented appropriately.Let’s denote the plane as π, containing point P(p, 0, 0) and line L. Since line L is parallel to the XY-plane (z = c) and has direction vector (d, e, 0), the plane π can be defined by the point P and the direction vectors of L and the vector from P to a point on L.Take a point A0(x0, y0, c) on line L. The vector PA0 is (x0 - p, y0, c). The direction vector of L is (d, e, 0). Therefore, the plane π is spanned by vectors (d, e, 0) and (x0 - p, y0, c).But this might not simplify things. Alternatively, parametrize the plane.Alternatively, consider that within the plane π, we can perform a coordinate transformation to 2D. Let’s define a coordinate system with P as the origin, one axis along the projection of line L onto the plane, and another axis perpendicular. However, this might be complex.Alternatively, use the earlier algebraic approach. Given that the general solution leads to a quadratic equation in t1, which would have two solutions, corresponding to the two points A and B. Therefore, in general, there are two such points, provided the discriminant is non-negative.Given the complexity of the general case, maybe the answer is that there are two such points, found by solving the quadratic equation derived above, but the exact coordinates depend on the specific parameters of the line and the point.But the problem statement says "Find two points on the line such that they form an equilateral triangle with the given point." It doesn’t provide specific coordinates for the line or the point, so the solution must be general.Alternatively, perhaps the problem expects a geometric construction explanation rather than algebraic. But given the initial problem statement, likely an algebraic solution is expected.Wait, but the problem might be in a specific projection setup where the line is parallel to the picture plane and the projection axis is considered as a line in the same space. Maybe using descriptive geometry methods.In descriptive geometry, to solve such a problem, one might use auxiliary views. For example, since the line is parallel to the picture plane, its projection onto the picture plane is a true-length line. The projection axis is the intersection of two projection planes. If we create an auxiliary view where the line is seen in true length and the point is projected, then the equilateral triangle can be constructed in this view.But I'm not sure. Alternatively, in the first picture plane (say, front view), the line is parallel, so its front view is true length. The projection axis is the X-axis. The given point is on the X-axis. Then, construct two points on the line's front view such that they form an equilateral triangle with the given point. Then, since the line is parallel to the picture plane, the distances in the front view are true distances, so the construction is valid.Wait, but in reality, the line is in 3D space at a certain height (z = c), so the true distance from the point (p,0,0) to a point (x, y, c) is sqrt( (x - p)^2 + y^2 + c^2 ). However, in the front view (projection onto XY-plane), the distance would appear as sqrt( (x - p)^2 + y^2 ), neglecting the z-coordinate. Therefore, constructing an equilateral triangle in the front view would not account for the z-coordinate, leading to incorrect distances.Therefore, the construction must consider the third dimension. So, perhaps an auxiliary view that shows the true distance from P to the line.To find points A and B on line L such that PA = PB = AB, we can consider the following steps:1. Find the true distance from point P to line L. Since L is parallel to the XY-plane, the shortest distance from P to L is the perpendicular distance in 3D, which would involve the z-coordinate.2. For an equilateral triangle, the height h satisfies h = (sqrt(3)/2) * side length. So, if we can find the height from P to line L, we can determine the side length.However, the issue is that the points A and B are not necessarily the foot of the perpendicular from P to L. They can be any two points on L such that the triangle is equilateral.Alternatively, consider that in the plane containing P and L, the locus of points A such that PA = AB is a circle with center at the midpoint of PB and radius PA/2. But this might not help directly.Given the time I've spent on this and the complexity, perhaps it's best to recall that in 3D space, the set of points A such that PA = AB and B lies on L is the intersection of two spheres: one centered at P with radius AB, and another centered at B with radius AB. But since B is variable, this is a varying sphere.Alternatively, consider that for each point B on L, the set of points A such that PA = AB is a sphere centered at B with radius PB. The intersection of this sphere with line L will give possible points A. But this seems recursive.Wait, but we need both PA = PB = AB. So for points A and B on L:PA = PB = AB.From PA = PB, we derived that t1 + t2 = 2 [ -d (x0 - p) - e y0 ] / (d² + e² )From PA = AB, we have a quadratic equation in t1. Solving this quadratic would give the values of t1, and subsequently t2 = s - t1.The solution exists if the discriminant is non-negative. Therefore, in general, there can be 0, 1, or 2 solutions. However, given the geometric nature of the problem, there should be two points in general position.Given that the problem asks to "find two points", we can assume solutions exist.Therefore, the final answer would be expressed in terms of the parameters of the line and the coordinates of P. However, since the problem doesn’t provide specific numbers, the solution must be general.But the user might expect a method or formula to find these points. Alternatively, since the problem is in a projection context, there might be a specific construction.Wait, recalling that the line is parallel to the first picture plane, meaning that in a projection onto that plane, the line appears in true length. Therefore, if we ignore the z-coordinate (height), the problem reduces to finding two points on the line's projection such that they form an equilateral triangle with the projection of P. However, since the actual distances involve the z-coordinate, this approach would not work. But maybe if we adjust the side length to account for the height.Suppose in the XY-plane, the projection of line L is a line L', and the projection of P is P'(p, 0). If we find two points A' and B' on L' such that P'A'B' is equilateral, then the true points A and B would be vertically above A' and B' at height c. However, the true distances PA and PB would be sqrt( (distance in XY)^2 + c^2 ), so unless c is zero, this won't form an equilateral triangle in 3D.Therefore, this approach is invalid.Alternatively, scale the equilateral triangle in the XY-plane by a factor that accounts for the height c. Let’s say in the XY-plane, the side length is s, then in 3D, the side length would be sqrt(s² + c² ). To have an equilateral triangle in 3D, we need sqrt(s² + c² ) = s', which must be equal for all sides. Therefore, set sqrt(s² + c² ) = s, which implies c = 0, but the line is at height c ≠ 0. Contradiction. Hence, this approach doesn't work.Therefore, the solution must genuinely be found in 3D space, leading to the quadratic equation we derived earlier.Given the complexity, I think the answer involves solving the quadratic equation for t1, leading to two solutions, hence two points. The coordinates of these points depend on the specific parameters of the line and the position of P.But since the problem doesn’t give specific numbers, the answer should be expressed in terms of the given parameters. However, in an exam or homework problem, usually specific coordinates or a general geometric construction would be expected.Wait, but the original problem statement might be in a specific context where the line is parallel to the picture plane and the projection axis is the X-axis, and the line is horizontal. For instance, in the earlier example where the line is along the X-axis at z = c, the solution was p ± c /√3 along the X-axis. Maybe in general, if the line is parallel to the X-axis in the XY-plane, then the displacement from the projection of P onto the line is c /√3. Alternatively, if the line is arbitrary parallel to the XY-plane, then the points A and B are located at a distance of c√(4/3) from P, but I need to check.Wait, in the specific case where the line was the X-axis at z = c, the distance PA was 2c /√3, as we saw earlier. The displacement along the X-axis was c /√3 from point P(p,0,0). So if the line is general, perhaps the displacement from the projection of P onto the line should be c /√3 in the direction of the line.Wait, the projection of P onto the line L (which is parallel to the XY-plane) would be the point on L closest to P. Since L is parallel to the XY-plane, the closest point would be the orthogonal projection onto L's direction in XY-plane, plus maintaining the z-coordinate. Wait, but the line L is in 3D space, parallel to the XY-plane. The shortest distance from P(p,0,0) to L is the minimal distance in 3D, which involves both the XY distance and the z-coordinate.But perhaps if we project P onto the plane containing L (z = c), the projection would be (p, 0, c). Then, the distance from this projection to the line L in the plane z = c is the minimal XY-distance. Then, the equilateral triangle points would be offset from this projection by c /√3 in the direction of the line.But this is speculative. Given the time I've spent and the different approaches, I think the precise answer requires solving the quadratic equation we derived earlier, leading to two solutions for t1 and t2, which depend on the line's parameters and the position of P.Therefore, the final answer is that there exist two such points, and their coordinates can be found by solving the system of equations derived from the distances PA = PB = AB. The explicit solution, given the line's parametric equations and the coordinates of P, involves solving a quadratic equation, resulting in two points that are symmetrically placed relative to the projection of P onto the line.However, since the problem likely expects a concrete method or formula, and given the earlier specific case, I think the solution involves displacing the projection of P onto the line by c /√3 along the line's direction. Hence, the two points are located at a distance of c /√3 from the projection of P onto the line, in both directions.But to confirm, let’s consider another example. Suppose the line L is in the plane z = c, has direction vector (0, 1, 0), i.e., vertical line along Y-axis at some x = x0. So parametric equations: x = x0, y = t, z = c. Point P is (p, 0, 0).Then, following the earlier steps:PA = PB = AB.Compute PA: sqrt( (x0 - p)^2 + t1^2 + c^2 )PB: sqrt( (x0 - p)^2 + t2^2 + c^2 )AB: |t2 - t1|Setting PA = PB:sqrt( (x0 - p)^2 + t1^2 + c^2 ) = sqrt( (x0 - p)^2 + t2^2 + c^2 )Squaring: t1^2 = t2^2 ⇒ t1 = ±t2Since points A and B must be distinct, t1 = -t2.Then, t1 + t2 = 0 ⇒ s = 0.From earlier equation:s = [ -2 d (x0 - p) - 2 e y0 ] / (d² + e² )Here, d = 0, e = 1, y0 = 0 (assuming the line is x = x0, y = t, z = c, so x0 is x0, y0 = 0 when t=0).Thus:s = [ -2*0*(x0 - p) - 2*1*0 ] / (0 + 1 ) = 0So t1 + t2 = 0 ⇒ t2 = -t1Now, set PA = AB:sqrt( (x0 - p)^2 + t1^2 + c^2 ) = |t2 - t1| = | -t1 - t1 | = | -2 t1 | = 2 |t1|Square both sides:(x0 - p)^2 + t1^2 + c^2 = 4 t1^2Rearrange:(x0 - p)^2 + c^2 = 3 t1^2Thus:t1 = ± sqrt( [ (x0 - p)^2 + c^2 ] / 3 )Therefore, the points A and B are (x0, sqrt( [ (x0 - p)^2 + c^2 ] / 3 ), c ) and (x0, - sqrt( [ (x0 - p)^2 + c^2 ] / 3 ), c )This shows that the displacement along the Y-axis depends on both the horizontal distance (x0 - p) and the height c. Therefore, the general solution isn't a simple displacement by c /√3 but involves the horizontal distance as well.Hence, the points are located at parameters t1 and t2 = -t1 along the line, with t1 = sqrt( [ (x0 - p)^2 + c^2 ] / 3 ). This complicates the general case.Therefore, the precise answer requires solving the quadratic equation derived earlier, which takes into account the line's direction vector, its position, and the position of point P. The solution involves two points whose parameters t1 and t2 satisfy the derived equations.Since the problem doesn't provide specific numbers, the final answer must be expressed in terms of these parameters. However, the problem might be expecting a geometric construction or a specific formula based on the relative positions.Given the complexity and the lack of specific information, I think the answer is that there are two such points, and their coordinates can be determined by solving the system of equations PA = PB = AB, which results in a quadratic equation with two solutions. The explicit coordinates depend on the specific parameters of the line and the point.But since the user likely expects a boxed final answer, and given the problem's nature, I might have missed a simpler geometric insight. However, based on the earlier specific case where the line is along the X-axis, the solution was to offset along the line by c /√3 from the projection of P. If this pattern holds, perhaps the general solution is to displace along the line's direction by c /√3 from the point where the perpendicular from P to the line meets the line.But in the previous example where the line was vertical, the displacement wasn't along the line's direction but depended on the horizontal offset. So that approach doesn't hold.Alternatively, the equilateral triangle's side length is equal to the distance from P to the line multiplied by 2 / sqrt(3). But the distance from P to the line in 3D is sqrt( (x0 - p)^2 + y0^2 + c^2 ) / sqrt(d^2 + e^2 + 0^2 ) )? Wait, no. The distance from point P(p,0,0) to the line L parametrized as (x0 + dt, y0 + et, c) is given by the formula:Distance = |(AP × v)| / |v|, where AP is the vector from a point A on L to P, and v is the direction vector of L.Let’s take point A0(x0, y0, c) on L. Vector AP = (x0 - p, y0, c). Direction vector v = (d, e, 0).Cross product AP × v = |i   j   k|                     |x0-p y0 c|                     |d   e   0|= i*(y0*0 - c*e) - j*( (x0 - p)*0 - c*d ) + k*( (x0 - p)*e - y0*d )= -c e i + c d j + [ (x0 - p)e - y0 d ] kMagnitude squared: c² e² + c² d² + [ (x0 - p)e - y0 d ]²= c² (d² + e² ) + [ (x0 - p)e - y0 d ]²Divide by |v|² = d² + e²So distance squared = [ c² (d² + e² ) + ( (x0 - p)e - y0 d )² ] / (d² + e² )This complicates the relation. Therefore, the side length of the equilateral triangle would be related to this distance, but not straightforwardly.Given all this, I think the most precise answer is that the two points are found by solving the quadratic equation resulting from setting PA = PB = AB, which generally has two solutions, and their exact coordinates depend on the specific parameters of the line and the point P. Without loss of generality, these points can be expressed in terms of the line’s parametric equations and the coordinates of P, leading to the boxed expressions.However, since the problem is presented in a projection context, there might be an expected geometric construction or specific formula. But without additional information, the solution is the algebraic resolution leading to two points whose parameters satisfy the derived equations.Final AnswerThe two points are located at positions (boxed{A}) and (boxed{B}) on the line, determined by solving the system of equations derived from the equidistant conditions for an equilateral triangle.

✅:To solve the problem of finding two points on a line parallel to the first picture plane (assumed to be the XY-plane) that form an equilateral triangle with a given point on the projection axis (assumed to be the X-axis), we follow these steps:1. Define the Coordinate System and Line Parameters: - The first picture plane is the XY-plane. - The projection axis is the X-axis, with the given point ( P ) at ( (p, 0, 0) ). - The line ( L ) parallel to the XY-plane is parametrized as ( (x_0 + dt, y_0 + et, c) ), where ( c ) is a constant.2. Set Up Distance Equations: - Points ( A ) and ( B ) on line ( L ) are given by ( A(x_0 + dt_1, y_0 + et_1, c) ) and ( B(x_0 + dt_2, y_0 + et_2, c) ). - The distances ( PA ), ( PB ), and ( AB ) must be equal for an equilateral triangle.3. Equidistant Conditions: - ( PA = PB ): This leads to the equation involving the parameters ( t_1 ) and ( t_2 ), resulting in ( t_1 + t_2 = frac{-2d(x_0 - p) - 2ey_0}{d^2 + e^2} ). - ( PA = AB ): This results in a quadratic equation in ( t_1 ) after substituting ( t_2 ) from the previous result.4. Solve the Quadratic Equation: - The quadratic equation derived from ( PA = AB ) is solved to find the values of ( t_1 ) and ( t_2 ).5. General Solution: - The solutions for ( t_1 ) and ( t_2 ) depend on the specific parameters of the line and the coordinates of ( P ). These parameters are substituted back into the parametric equations of the line to find the coordinates of points ( A ) and ( B ). Final AnswerThe two points on the line that form an equilateral triangle with the given point on the projection axis are determined by solving the system of equations derived from the equidistant conditions. These points are:[boxed{A} text{ and } boxed{B}]Their exact coordinates are found by solving the quadratic equation resulting from the conditions ( PA = PB = AB ), which involves the specific parameters of the line and the given point.

🔑:1. Planar Alignment and Rotation: - We begin by placing the plane ( n ) such that it contains both the given point ( P ) and the line ( g ) (which is parallel to the first projection plane).2. Rotation of Line ( g ): - Rotate the line ( g ) onto the first projection plane. Let the projection of ( g ) on the first projection plane be denoted as ( g_0 ).3. Perpendicular from Point ( P ): - From the point ( P ), draw a perpendicular line to ( g_0 ). The intersection of this perpendicular with ( g_0 ) will be a point we denote as ( D ). 4. Calculation of ( PD ): - Here, ( PD ) is essentially the height of our equilateral triangle, where ( PD ) forms the altitude.5. Construct Angles: - Measure ( 30^circ ) angles on both sides of ( PD ). This results from the fact that in an equilateral triangle, splitting an angle through the altitude results in two ( 30^circ ) angles (as each angle of an equilateral triangle is ( 60^circ )).6. Identification of Points ( A ) and ( B ): - The ( 30^circ ) measurements result in two new points, ( A ) and ( B ), on the line ( g ) such that triangles ( triangle PAB ) meet the requirement of being equilateral.Thus, the set of points (A) and (B) form the vertices of the equilateral triangle together with the point (P). [boxed{A text{ and } B}]