🧠:[<think>Okay, so I need to find the maximum value of 2x - 3y given that the real numbers x and y satisfy the inequality 3|x + 1| + 2|y - 1| ≤ 6. Hmm, let me think about how to approach this. First, the inequality given is involving absolute values, which usually defines a region in the coordinate plane. Since absolute values can represent distances, this inequality is probably describing some sort of bounded region, maybe a polygon or a diamond shape. The expression we need to maximize, 2x - 3y, is a linear function, so this sounds like a linear optimization problem with constraints defined by absolute values. I remember that for linear optimization problems, the maximum or minimum of a linear function over a convex region occurs at one of the vertices (corner points) of the region. So, maybe the first step is to figure out the shape of the region defined by 3|x + 1| + 2|y - 1| ≤ 6 and then find its vertices. After that, I can evaluate the objective function 2x - 3y at each vertex to find the maximum value.Let me start by analyzing the inequality 3|x + 1| + 2|y - 1| ≤ 6. Absolute value inequalities can be broken down into different cases depending on the signs of the expressions inside the absolute values. For |x + 1|, the expression inside is x + 1, so the critical point is at x = -1. Similarly, for |y - 1|, the critical point is at y = 1. These critical points divide the coordinate plane into four regions:1. x ≥ -1 and y ≥ 12. x ≥ -1 and y < 13. x < -1 and y ≥ 14. x < -1 and y < 1In each of these regions, the absolute value expressions can be replaced by linear expressions without the absolute value, and then we can rewrite the original inequality accordingly. Let me do that for each case.Case 1: x ≥ -1 and y ≥ 1In this case, |x + 1| = x + 1 and |y - 1| = y - 1. So the inequality becomes:3(x + 1) + 2(y - 1) ≤ 6 Simplifying: 3x + 3 + 2y - 2 ≤ 6 3x + 2y + 1 ≤ 6 3x + 2y ≤ 5So in this region, the inequality is 3x + 2y ≤ 5.Case 2: x ≥ -1 and y < 1Here, |x + 1| = x + 1 and |y - 1| = -(y - 1) = 1 - y. Substituting into the inequality:3(x + 1) + 2(1 - y) ≤ 6 Simplify: 3x + 3 + 2 - 2y ≤ 6 3x - 2y + 5 ≤ 6 3x - 2y ≤ 1So the inequality here is 3x - 2y ≤ 1.Case 3: x < -1 and y ≥ 1In this case, |x + 1| = -(x + 1) = -x - 1 and |y - 1| = y - 1. The inequality becomes:3(-x - 1) + 2(y - 1) ≤ 6 Simplify: -3x - 3 + 2y - 2 ≤ 6 -3x + 2y - 5 ≤ 6 -3x + 2y ≤ 11So here, the inequality is -3x + 2y ≤ 11.Case 4: x < -1 and y < 1Here, both absolute values are negative, so |x + 1| = -x -1 and |y - 1| = 1 - y. The inequality becomes:3(-x -1) + 2(1 - y) ≤ 6 Simplify: -3x -3 + 2 - 2y ≤ 6 -3x - 2y -1 ≤ 6 -3x - 2y ≤ 7Which simplifies to 3x + 2y ≥ -7 (if we multiply both sides by -1 and reverse the inequality sign). Wait, but actually, let's check that step again. Starting from:-3x -3 + 2 - 2y ≤ 6 Combine constants: -3x -2y -1 ≤ 6 Add 1 to both sides: -3x -2y ≤ 7 Multiply both sides by -1 (remembering to reverse the inequality): 3x + 2y ≥ -7Yes, correct. So in this case, the inequality is 3x + 2y ≥ -7. But we also have x < -1 and y < 1 here.So now, in each of these four cases, we have a linear inequality which, combined with the conditions on x and y, defines a region. The overall feasible region is the intersection of all these regions. Since each case corresponds to a different quadrant relative to the points x = -1 and y = 1, the total feasible region should be a convex polygon with vertices at the intersection points of the boundaries of these inequalities.To find the vertices, I need to find the intersection points of the lines that form the boundaries of each case. Let's list out the boundary lines for each case:1. For Case 1: 3x + 2y = 5, with x ≥ -1 and y ≥ 1.2. For Case 2: 3x - 2y = 1, with x ≥ -1 and y < 1.3. For Case 3: -3x + 2y = 11, with x < -1 and y ≥ 1.4. For Case 4: 3x + 2y = -7, with x < -1 and y < 1.But these lines are only parts of the boundaries in their respective regions. The entire feasible region is formed by these four lines along with the lines x = -1 and y = 1, which are the boundaries between the cases. So actually, the feasible region is a polygon bounded by the lines:- 3x + 2y = 5 (from Case 1)- 3x - 2y = 1 (from Case 2)- -3x + 2y = 11 (from Case 3)- 3x + 2y = -7 (from Case 4)- x = -1 (vertical line separating Cases 1/2 from 3/4)- y = 1 (horizontal line separating Cases 1/3 from 2/4)However, the feasible region is the intersection of all the inequalities, so we need to see where these lines intersect each other and form the vertices.To find the vertices of the feasible region, we need to find the intersection points of each pair of adjacent boundary lines.Let me list the possible intersection points:1. Intersection of 3x + 2y = 5 and 3x - 2y = 1 (intersection of Case 1 and Case 2 boundaries)2. Intersection of 3x - 2y = 1 and x = -1 (Case 2 boundary meets x = -1)3. Intersection of x = -1 and 3x + 2y = -7 (x = -1 meets Case 4 boundary)4. Intersection of 3x + 2y = -7 and -3x + 2y = 11 (Case 4 and Case 3 boundaries)5. Intersection of -3x + 2y = 11 and y = 1 (Case 3 boundary meets y =1)6. Intersection of y = 1 and 3x + 2y = 5 (Case 1 boundary meets y =1)Wait, but maybe not all these intersections lie within the feasible region. Let's check each one step by step.1. Intersection of 3x + 2y = 5 and 3x - 2y = 1:Solve the two equations:3x + 2y = 5 3x - 2y = 1Adding both equations: 6x = 6 → x = 1Substitute x = 1 into first equation: 3(1) + 2y = 5 → 3 + 2y = 5 → 2y = 2 → y = 1So intersection at (1,1). But we need to check if this point is in the feasible region. For Case 1, x ≥ -1 and y ≥ 1. Since x =1 ≥ -1 and y =1, it's on the boundary of y ≥1. So this point is a vertex of the feasible region.2. Intersection of 3x - 2y = 1 and x = -1:Substitute x = -1 into 3x - 2y =1:3(-1) -2y =1 → -3 -2y =1 → -2y =4 → y = -2So intersection at (-1, -2). Now check if this is in the feasible region. For Case 2, x ≥ -1 and y <1. Here, x = -1 (which is allowed since x ≥ -1) and y = -2 <1. So this point is on the boundary of x = -1 and within the y <1 region. Also, check if this point satisfies the original inequality 3|x +1| +2|y -1| ≤6.Plugging in x=-1, y=-2:3| -1 +1 | +2| -2 -1 | = 3(0) + 2(3) = 0 +6 =6 ≤6. So yes, it's on the boundary. So (-1, -2) is a vertex.3. Intersection of x = -1 and 3x + 2y = -7:Substitute x = -1 into 3x +2y = -7:3(-1) +2y = -7 → -3 +2y =-7 → 2y = -4 → y = -2So intersection at (-1, -2). Wait, that's the same point as the previous one. So maybe this is a duplicate. Hmm, but 3x +2y = -7 is the boundary from Case 4, and x = -1 is a vertical line. So both Case 2 and Case 4 boundaries intersect x = -1 at (-1, -2). But that's the same point. So that point is shared between two boundaries.4. Intersection of 3x + 2y = -7 and -3x + 2y =11:Solve the system:3x + 2y = -7 -3x + 2y = 11Add the two equations: 4y =4 → y=1Substitute y=1 into first equation: 3x +2(1) =-7 → 3x +2 =-7 → 3x =-9 →x =-3So intersection at (-3,1). Check if this is in the feasible region. For Case 4, x < -1 and y <1. But here, y=1, which is not less than 1, so actually, this point is on the boundary y=1. But let's check which case this belongs to. Since x =-3 < -1 and y =1, which is part of Case 3 (x < -1 and y ≥1). Wait, but the line 3x +2y = -7 is part of Case 4 (x < -1 and y <1). However, the intersection with -3x +2y=11 (Case 3's boundary) gives a point at (-3,1). Let's check if this point satisfies the original inequality.Plugging x=-3, y=1 into 3|x+1| +2|y-1|:3| -3 +1 | +2|1 -1| = 3| -2 | +2(0) = 6 +0=6 ≤6. So it's on the boundary. However, the region for Case 3 is x < -1 and y ≥1. So this point is on the boundary of Case 3 and also on the line between Case 3 and Case 4. So (-3,1) is another vertex.5. Intersection of -3x +2y=11 and y=1:Substitute y=1 into -3x +2y=11:-3x +2(1)=11 → -3x +2=11 → -3x=9 →x= -3So intersection at (-3,1), which is the same point as above. So that's already considered.6. Intersection of y=1 and 3x +2y=5:Substitute y=1 into 3x +2(1)=5:3x +2=5 →3x=3→x=1So intersection at (1,1), which is the same as the first point. So that point is also on y=1.Wait, so so far, we have vertices at (1,1), (-1, -2), (-3,1). Are there more vertices?Wait, let me check the other boundaries. For example, between Case 1 and Case 3. But Case 1 is x ≥ -1, y ≥1 and Case 3 is x < -1, y ≥1. The boundary between them is x = -1. So if we intersect x=-1 with y ≥1, then plugging into 3|x+1| +2|y -1| ≤6 gives 3(0) +2|y -1| ≤6 → 2|y -1| ≤6 → |y -1| ≤3 → -3 ≤ y -1 ≤3 → -2 ≤ y ≤4. But since we are in y ≥1, then y can be from 1 to 4. But we need to check if this is part of the feasible region. Wait, but in the original problem, the inequality is 3|x+1| +2|y -1| ≤6. If x=-1 and y=4, then 3(0) +2(3)=6, which is allowed. So the point (-1,4) is on the boundary. Wait, but how come we didn't get that in previous intersections?Wait, perhaps there's another vertex at (-1,4). Let me check.Wait, when x=-1, the original inequality becomes 0 +2|y-1| ≤6 → |y-1| ≤3 → y ∈ [-2,4]. So when x=-1, y can range from -2 to4. But in the feasible region, when x=-1, in the Cases 2 and 3. For y ≥1, x=-1 is part of Case 3 (x < -1? Wait no, x=-1 is the boundary between x ≥-1 and x < -1. Wait, but when x=-1, in Cases 1 and 2 (since x ≥-1 includes x=-1). Wait, perhaps I made a mistake in the earlier breakdown.Actually, when x=-1, it's part of the regions where x ≥ -1 (Cases 1 and 2). Similarly, y=1 is part of regions where y ≥1 (Cases 1 and 3). So perhaps the line x=-1 is part of Cases 1 and 2, and the line y=1 is part of Cases 1 and 3.Therefore, in the feasible region, along x=-1, the boundary is from y=-2 to y=4. But how does that interact with the other constraints?Wait, but when x=-1 and y is between -2 and4, does that automatically satisfy all the other constraints?Wait, no. The original inequality 3|x+1| +2|y-1| ≤6 must be satisfied. So when x=-1, the term 3|x+1| becomes 0, so 2|y-1| ≤6 → |y-1| ≤3 → y ∈ [-2,4]. So along x=-1, y can range from -2 to4. Similarly, when y=1, the term 2|y-1| becomes 0, so 3|x+1| ≤6 → |x+1| ≤2 → x ∈ [-3,1]. So along y=1, x ranges from -3 to1.But in the feasible region, the boundaries are not only x=-1 and y=1 but also the lines from each case. So perhaps the feasible region is a polygon with vertices at (1,1), (-1,4), (-3,1), (-1,-2), and back to (1,1). Wait, let me visualize this.Wait, let's think of the original inequality as the sum of two scaled absolute values. This is similar to the equation of a diamond (like L1 norm), but scaled. So in general, the shape would be a convex polygon with vertices where each absolute value term switches its sign, i.e., at x=-1 and y=1. But due to the coefficients 3 and 2, the vertices might be stretched.Wait, perhaps to better understand the feasible region, I can parametrize it. Let me consider the inequality 3|x +1| +2|y -1| ≤6. Let's set u = x +1 and v = y -1. Then the inequality becomes 3|u| +2|v| ≤6. So in terms of u and v, this is a diamond with vertices along the u and v axes. Specifically, when v=0, |u| ≤2 (since 3|u| ≤6 → |u| ≤2), so u=±2 → x = -1 ±2, so x=1 and x=-3. Similarly, when u=0, |v| ≤3 (since 2|v| ≤6 → |v| ≤3), so v=±3 → y =1 ±3, so y=4 and y=-2. Therefore, the feasible region in the (u,v) coordinates is a diamond with vertices at (2,0), (-2,0), (0,3), (0,-3). Translating back to (x,y) coordinates:u = x +1, so x = u -1v = y -1, so y = v +1Thus, the vertices in (x,y) coordinates are:(2 -1, 0 +1) = (1,1)(-2 -1, 0 +1) = (-3,1)(0 -1, 3 +1) = (-1,4)(0 -1, -3 +1) = (-1,-2)So the feasible region is a convex quadrilateral with vertices at (1,1), (-3,1), (-1,4), (-1,-2). Wait, but connecting these points: (1,1) to (-3,1), then to (-1,4), then to (-1,-2), then back to (1,1)? Wait, no. Wait, in the (u,v) coordinates, the diamond has four vertices: (2,0), (-2,0), (0,3), (0,-3). Translating these to (x,y):(2,0) → (1,1)(-2,0) → (-3,1)(0,3) → (-1,4)(0,-3) → (-1,-2)So the diamond in (x,y) coordinates is a quadrilateral with these four vertices. Therefore, the feasible region is this quadrilateral. So these four points are the vertices. Therefore, the maximum of 2x -3y must occur at one of these four vertices.Therefore, to find the maximum value of 2x -3y, I just need to compute 2x -3y at each of these four vertices and pick the largest one.Let's compute:1. At (1,1):2(1) -3(1) = 2 -3 = -12. At (-3,1):2(-3) -3(1) = -6 -3 = -93. At (-1,4):2(-1) -3(4) = -2 -12 = -144. At (-1,-2):2(-1) -3(-2) = -2 +6 = 4So among these values: -1, -9, -14, and 4. The maximum is 4, which occurs at the vertex (-1, -2).Wait, but before I conclude, let me verify if I have considered all vertices. Since the feasible region is a quadrilateral with four vertices, and we have evaluated all four, and the maximum is indeed 4. Therefore, the maximum value of 2x -3y under the given constraint is 4.But just to be thorough, let me check if there are any other intersection points or if I missed something.Wait, when I considered the intersection points earlier, I found (1,1), (-1,-2), (-3,1), and the point (-1,4) comes from the translation of the diamond's vertex. So that seems correct.Alternatively, maybe I can use Lagrange multipliers or another method, but since the feasible region is a convex polygon, the maximum must be at a vertex. Since we checked all four vertices and found the maximum at (-1, -2) giving 4, that should be the answer.But let me confirm by considering another approach. Let's try to parametrize the problem.Alternatively, we can consider that 3|x +1| +2|y -1| ≤6 is equivalent to |x +1|/2 + |y -1|/3 ≤1. This resembles the equation of a diamond (L1 norm) centered at (-1,1) with axes scaled by 2 and 3. The maximum of 2x -3y over this diamond would be achieved at one of the vertices, which are the points along the axes. These vertices are at (x +1, y -1) = (±2, 0), (0, ±3). Translating back, the vertices are at (-1 ±2, 1), (-1,1 ±3), which gives (1,1), (-3,1), (-1,4), (-1,-2), as before. So same result.Therefore, evaluating 2x -3y at these points gives the same numbers as above, so the maximum is indeed 4.Alternatively, another method: using the Cauchy-Schwarz inequality.But since this is a linear function, Cauchy-Schwarz might not directly apply. However, we can think of 2x -3y as a dot product of the vector (2, -3) with the vector (x, y). The maximum of this dot product over a convex set is achieved at the point where the vector (2, -3) is pointing in the same direction as the furthest vertex.But perhaps overcomplicating. Since we've already found the vertices and calculated the maximum, that should suffice.So, after verifying through multiple approaches, the maximum value is 4, achieved at (-1, -2). Therefore, the answer is 4.Final AnswerThe maximum value of (2x - 3y) is boxed{4}.</think>Given the inequality (3|x + 1| + 2|y - 1| leq 6), we need to find the maximum value of (2x - 3y).First, we analyze the inequality by considering the critical points (x = -1) and (y = 1), which divide the coordinate plane into four regions. We translate the inequality into a more manageable form by substituting (u = x + 1) and (v = y - 1), transforming the inequality into (3|u| + 2|v| leq 6). This represents a diamond (or convex polygon) in the (uv)-plane with vertices at ((pm 2, 0)), ((0, pm 3)).Translating back to the original coordinates ((x, y)), the vertices of the feasible region are:1. ((1, 1)) corresponding to ((2, 0)) in (uv)-coordinates.2. ((-3, 1)) corresponding to ((-2, 0)) in (uv)-coordinates.3. ((-1, 4)) corresponding to ((0, 3)) in (uv)-coordinates.4. ((-1, -2)) corresponding to ((0, -3)) in (uv)-coordinates.We evaluate the objective function (2x - 3y) at each vertex:1. At ((1, 1)): (2(1) - 3(1) = -1)2. At ((-3, 1)): (2(-3) - 3(1) = -9)3. At ((-1, 4)): (2(-1) - 3(4) = -14)4. At ((-1, -2)): (2(-1) - 3(-2) = 4)The maximum value is found at the vertex ((-1, -2)), giving the value (4).Thus, the maximum value of (2x - 3y) is (boxed{4}).,<think>Okay, let's see. I need to find the maximum value of 2x - 3y given that 3|x+1| + 2|y-1| ≤ 6. Hmm, this looks like a linear optimization problem with absolute value constraints. I remember that absolute value inequalities can define regions shaped like diamonds or polygons in the coordinate plane. So maybe I can graph this region and then check the vertices for the maximum value of the objective function 2x - 3y. That sounds like a plan.First, let me rewrite the constraint: 3|x + 1| + 2|y - 1| ≤ 6. To visualize this, it's helpful to consider the definition of absolute value functions. The expression |x + 1| can be split into cases depending on whether x + 1 is positive or negative, and similarly for |y - 1|. However, instead of splitting into all possible cases, maybe I can recognize that this inequality represents a diamond (a convex polygon) centered at (-1, 1) because the absolute values are shifted. The coefficients 3 and 2 will affect the "spread" of the diamond along the x and y axes.Let me recall that the general form of such an inequality is A|x - h| + B|y - k| ≤ C, which forms a diamond with vertices at (h ± C/A, k) and (h, k ± C/B). But wait, in our case, the equation is 3|x + 1| + 2|y - 1| ≤ 6. Comparing to the general form, h is -1, k is 1, A is 3, B is 2, and C is 6. So the vertices should be at (h ± C/A, k) and (h, k ± C/B). Let me compute those:For the x-direction: h ± C/A = -1 ± 6/3 = -1 ± 2. So the x-vertices are at (-1 + 2, 1) = (1, 1) and (-1 - 2, 1) = (-3, 1).For the y-direction: k ± C/B = 1 ± 6/2 = 1 ± 3. So the y-vertices are at (-1, 1 + 3) = (-1, 4) and (-1, 1 - 3) = (-1, -2).Therefore, the diamond-shaped region has vertices at (1, 1), (-3, 1), (-1, 4), and (-1, -2). Wait, but is that right? Let me check.Actually, the diamond should have four vertices. The ones along the x-axis direction (changing x while keeping y constant at k) and the ones along the y-axis direction (changing y while keeping x constant at h). So yes, (1, 1), (-3, 1), (-1, 4), (-1, -2). Let me confirm this.If I set y = 1, then the inequality becomes 3|x + 1| ≤ 6, so |x + 1| ≤ 2, which gives x between -3 and 1. So the two points at y=1, x=1 and x=-3. Similarly, if x = -1, then 2|y - 1| ≤ 6, so |y - 1| ≤ 3, so y between -2 and 4. So the points at x=-1, y=4 and y=-2. Then the diamond is formed by connecting these points. So the four vertices are indeed (1,1), (-3,1), (-1,4), (-1,-2). Then, to complete the diamond, edges connect these points. So between (1,1) and (-1,4), between (-1,4) and (-3,1), between (-3,1) and (-1,-2), and between (-1,-2) and (1,1). So that's the feasible region.Now, to maximize the linear function 2x - 3y over this convex polygon, the maximum must occur at one of the vertices. This is due to the fundamental theorem of linear programming, which states that the extrema of a linear function over a convex polygon occur at the vertices. So I can evaluate 2x - 3y at each of the four vertices and choose the maximum.Let me compute the values:1. At (1,1):2(1) - 3(1) = 2 - 3 = -12. At (-3,1):2(-3) - 3(1) = -6 - 3 = -93. At (-1,4):2(-1) - 3(4) = -2 - 12 = -144. At (-1,-2):2(-1) - 3(-2) = -2 + 6 = 4So the maximum value among these is 4 at the vertex (-1, -2). Therefore, the maximum value of 2x - 3y is 4. But wait, let me make sure I didn't make any mistakes here.Wait, but is that all the vertices? Let me think. Sometimes, depending on the coefficients, there might be other edges or points where the maximum could occur. But in this case, since the feasible region is a convex polygon (a diamond) with four vertices, and the objective function is linear, then the maximum must indeed be at one of these four vertices. Therefore, the maximum is 4. Hmm, but let me check if there could be other points on the edges where the function is higher. For example, suppose we move along the edge from (-1, -2) to (1,1). Let's parametrize this edge.Parametrizing from (-1, -2) to (1,1), we can use a parameter t from 0 to 1:x(t) = -1 + 2ty(t) = -2 + 3tThen the objective function becomes 2(-1 + 2t) - 3(-2 + 3t) = (-2 + 4t) - (-6 + 9t) = -2 + 4t +6 -9t = 4 -5tThis is a linear function in t, decreasing as t increases. So maximum at t=0, which is (-1, -2), value 4. So indeed, along this edge, the maximum is at (-1, -2). Similarly, checking another edge.Take the edge from (-1, -2) to (-3,1). Let's parametrize that:x(t) = -1 - 2ty(t) = -2 + 3t, t from 0 to1Then 2x -3y = 2(-1 -2t) -3(-2 +3t) = (-2 -4t) - (-6 +9t) = -2 -4t +6 -9t = 4 -13tThis is again decreasing with t, so maximum at t=0, which is (-1, -2), value 4.Similarly, check the edge from (1,1) to (-1,4):Parametrize:x(t) =1 -2ty(t)=1 +3t, t from 0 to12x -3y=2(1 -2t) -3(1 +3t)=2 -4t -3 -9t= -1 -13tThis is decreasing in t, so maximum at t=0 (1,1), value -1.Edge from (-1,4) to (-3,1):x(t) = -1 -2ty(t) =4 -3t, t from0 to12x -3y=2(-1 -2t)-3(4 -3t)= -2 -4t -12 +9t= -14 +5tThis is increasing in t, so maximum at t=1: (-3,1), value -14 +5(1)= -9. Which is the same as calculated earlier.So in all edges, the maximum occurs at the vertices, so the conclusion is correct.Alternatively, maybe there's another approach using Lagrange multipliers or something else, but since the region is a convex polygon, checking vertices is straightforward.But just to confirm, maybe I can think of the problem as linear programming with absolute values. The constraint 3|x+1| +2|y-1| ≤6 can be converted into linear inequalities. Let's try that.First, |x +1| can be written as two inequalities: x +1 ≤ t1 and -(x +1) ≤ t1, similarly |y -1| ≤ t2, and the original constraint becomes 3t1 + 2t2 ≤6. Then, the problem becomes maximize 2x -3y subject to:x +1 ≤ t1- (x +1) ≤ t1y -1 ≤ t2- (y -1) ≤ t23t1 + 2t2 ≤6t1, t2 ≥0But this is a linear program with variables x, y, t1, t2. To solve this, we could use the simplex method, but maybe it's more complicated. However, since we already solved it by geometry, perhaps this is redundant. But maybe we can set up the dual variables or something else. Alternatively, since we can express t1 = |x+1| and t2=|y-1|, but maybe this is not helpful.Alternatively, another method: express x and y in terms of u and v, where u = x +1 and v = y -1. Then, the constraint becomes 3|u| +2|v| ≤6, and the objective function becomes 2(u -1) -3(v +1) = 2u -2 -3v -3 = 2u -3v -5. So we need to maximize 2u -3v -5, given that 3|u| +2|v| ≤6.Therefore, the maximum of 2u -3v is (original maximum) +5. So perhaps it's easier to shift coordinates. Let me see.So let me define u = x +1, v = y -1. Then, the constraint is 3|u| +2|v| ≤6. The objective function is 2x -3y = 2(u -1) -3(v +1) = 2u -2 -3v -3 = 2u -3v -5. Therefore, maximizing 2x -3y is equivalent to maximizing 2u -3v -5, which is equivalent to maximizing 2u -3v (since -5 is a constant). Therefore, if I can find the maximum of 2u -3v subject to 3|u| +2|v| ≤6, then add -5 to get the final result.So let's focus on maximizing 2u -3v with 3|u| +2|v| ≤6.This might be easier since the center is now at (0,0) in uv-coordinates. The feasible region is a diamond with vertices at (u, v) where 3|u| +2|v| =6. Let's find the vertices here. If we set v=0, then 3|u|=6, so u=±2. If we set u=0, 2|v|=6, so v=±3. Therefore, the vertices in uv-coordinates are (2,0), (-2,0), (0,3), (0,-3). Connecting these points gives the diamond.So in uv-coordinates, the vertices are (2,0), (-2,0), (0,3), (0,-3). Then, evaluating 2u -3v at these points:1. (2,0): 2*2 -3*0=42. (-2,0): 2*(-2) -3*0=-43. (0,3): 2*0 -3*3=-94. (0,-3): 2*0 -3*(-3)=9Wait, hold on. The maximum here is 9? But in the original coordinates, we had a maximum of 4. But since we shifted variables, this 9 corresponds to which point?Wait, in uv-coordinates, the point (0,-3) corresponds to u=0, v=-3. Translating back to xy-coordinates:u = x +1 =0 ⇒x = -1v = y -1 =-3 ⇒ y = -2So that's the point (-1, -2), which is the same as before. So in uv-coordinates, the maximum of 2u -3v is 9 at (0,-3), but in original coordinates, the objective function is 2u -3v -5. Therefore, 9 -5=4. Which matches the previous result.So this method gives the same answer, 4. So that's good. Therefore, the maximum value is 4.Alternatively, another approach is to use the Cauchy-Schwarz inequality or other inequalities. Let me think. The expression 2u -3v can be seen as the dot product of the vector (2, -3) with (u, v). The constraint is 3|u| +2|v| ≤6. So perhaps we can bound the dot product using Hölder's inequality. Hölder's inequality states that for conjugate exponents p and q (1/p +1/q =1), |a·b| ≤ ||a||_p ||b||_q. However, Hölder's inequality might not directly apply here since the constraint is a mixed norm.Alternatively, since we have 3|u| +2|v| ≤6, we can consider the dual norm. Let me recall that the dual of the L1 norm is the Linfinity norm, but here the coefficients complicate things.Alternatively, we can write 3|u| +2|v| ≤6 as |u|/(2) + |v|/(3) ≤1, which resembles the L1 norm scaled. Then, the dual problem would involve maximizing the linear function over this region.Alternatively, to maximize 2u -3v, we can set up the Lagrangian. Let me try that. Let me consider variables u and v.The Lagrangian is L = 2u -3v - λ(3|u| +2|v| -6). However, because of absolute values, the function is not differentiable at u=0 or v=0, but perhaps we can consider different cases.Alternatively, since the maximum of 2u -3v under the constraint 3|u| +2|v| ≤6 can be found by considering the direction of the gradient. The gradient of the objective function is (2, -3). The maximum will occur at the point where the direction of (2, -3) is tangent to the constraint boundary 3|u| +2|v|=6.But dealing with absolute values complicates this. Let me consider the first quadrant where u ≥0, v ≥0. Then the constraint becomes 3u +2v ≤6. The objective function is 2u -3v. In this quadrant, to maximize 2u -3v, since increasing u increases the function and increasing v decreases it. So the maximum in this quadrant would be at u as large as possible, v as small as possible. But in the first quadrant, v ≥0, so the minimum v is 0. So setting v=0, then u=2. At (2,0), the objective is 4.Similarly, in other quadrants.Second quadrant: u ≤0, v ≥0. Constraint becomes -3u +2v ≤6. Objective 2u -3v. Here, u ≤0, so increasing u (toward 0) increases the function, but v ≥0, so decreasing v (toward 0) also increases the function. The maximum would be at u=0, v=0? Wait, but substituting u=0, v=0, the constraint is 0 +0 ≤6, which is true, but the objective is 0 -0=0. Which is less than 4. Alternatively, maybe at some other point.Wait, in the second quadrant, u ≤0, v ≥0. So to maximize 2u -3v, since u is negative, making u less negative (closer to 0) helps, and v is positive, making v smaller helps. So the maximum in this quadrant would be at u=0, v=0. But (0,0) gives 0. But perhaps on the edge.The constraint in this quadrant is -3u +2v ≤6, with u ≤0, v ≥0. The boundary is -3u +2v=6. Let's parametrize this. Let me solve for v: v=(6 +3u)/2. Since u ≤0, v=(6 +3u)/2 ≥ (6 +0)/2=3. So on the boundary in this quadrant, u ranges from 0 down to some negative value. Wait, if u approaches negative infinity, but with the constraint 3|u| +2|v| ≤6, actually u can't go beyond -2, as when v=0, u=-2. Wait, but in this quadrant, v is positive. Wait, maybe I need to check.Wait, in the second quadrant (u ≤0, v ≥0), the constraint 3|u| +2|v| ≤6 becomes 3(-u) +2v ≤6, which is -3u +2v ≤6. To find the boundary, set -3u +2v=6. Let me express v in terms of u: v=(6 +3u)/2. Since u ≤0, and v ≥0, then (6 +3u)/2 ≥0 ⇒6 +3u ≥0 ⇒3u ≥-6 ⇒u ≥-2. Therefore, u ∈ [-2,0], v ∈ [0, 3].So on this edge, the objective function is 2u -3v =2u -3*(6 +3u)/2=2u - (18 +9u)/2= (4u -18 -9u)/2= (-5u -18)/2. To maximize this expression over u ∈ [-2,0], we can take derivative with respect to u: derivative is (-5)/2, which is negative. Therefore, the maximum occurs at the leftmost point, u=-2. Then, v=(6 +3*(-2))/2=(6 -6)/2=0. But wait, u=-2 is in the second quadrant? Wait, u=-2 is at the boundary with the third quadrant. Wait, but if u=-2 and v=0, that's actually on the boundary between second and third quadrants. But in the second quadrant, v ≥0. So the point (-2,0) is in the second quadrant? Wait, u=-2, v=0 is on the x-axis, between second and third quadrants. Hmm. Anyway, evaluating the objective at u=-2, v=0: 2*(-2) -3*0=-4. Which is less than 4.So in the second quadrant, the maximum is -4 at (-2,0). Not better than 4.Third quadrant: u ≤0, v ≤0. The constraint becomes 3(-u) +2(-v) ≤6 ⇒-3u -2v ≤6. The objective function is 2u -3v. Since u ≤0 and v ≤0, 2u is negative or zero, -3v is positive or zero. So to maximize 2u -3v, we need to minimize u (most negative) and maximize v (least negative). Wait, but u is being minimized (more negative), which makes 2u more negative, and v is being maximized (closer to zero), which makes -3v less positive. So conflicting. Let's see.The boundary in this quadrant is -3u -2v=6. Expressing u in terms of v: u=(-6 -2v)/3. Since u ≤0, v ≤0.Let me parametrize the boundary: let v go from 0 down to some value. Wait, if v=0, u=(-6)/3=-2. If v becomes more negative, say v=-3, u=(-6 -2*(-3))/3=(-6 +6)/3=0. But u must be ≤0, so when v ≤-3, u would be ≥0, which is not allowed in the third quadrant. Therefore, the valid range for v in this boundary is v ∈ [-3,0]. So u=(-6 -2v)/3. Then, substituting into the objective function:2u -3v =2*(-6 -2v)/3 -3v= (-12 -4v)/3 -3v= -4 - (4v)/3 -3v= -4 - (13v)/3To maximize this expression over v ∈ [-3,0], we can take derivative with respect to v: -13/3, which is negative. Therefore, the maximum occurs at the leftmost point, v=-3, which gives u=(-6 -2*(-3))/3=(-6 +6)/3=0. So the point (0, -3). But in the third quadrant, u ≤0, v ≤0. So (0, -3) is on the boundary with the fourth quadrant. Evaluating the objective at (0, -3): 2*0 -3*(-3)=9. Wait, but in uv-coordinates, this is (0, -3). Translating back to xy-coordinates: u=0⇒x=-1; v=-3⇒y=-2. Wait, but this point (0, -3) in uv is (x,y)=(-1, -2). Which is the same as we found earlier. But in the third quadrant, (0, -3) is on the edge. But hold on, in the third quadrant (u ≤0, v ≤0), but u=0 is the boundary with the fourth quadrant. So actually, (0, -3) is not in the interior of the third quadrant. So the maximum in the third quadrant would be approaching that point but perhaps not including it.Wait, but if we are restricted to the third quadrant (u ≤0, v ≤0), but the boundary includes u=0 when v=-3. However, (0, -3) is on the boundary between third and fourth quadrants. So when parametrizing the third quadrant's boundary, we have u from (-6 -2v)/3 where v ∈ [-3,0]. At v=-3, u=0. At v=0, u=-2. Therefore, the point (0, -3) is part of the third quadrant's boundary. Therefore, in the third quadrant, the maximum of the objective function is achieved at (0, -3), which gives 9. However, (0, -3) in uv is (x,y)=(-1, -2). Therefore, in the original problem, the objective function is 9 -5=4. So this is the maximum. However, in the third quadrant, is (0, -3) considered part of the quadrant? Since u=0, v=-3, which is on the axis. Depending on the definition, quadrants might not include the axes, but in optimization over closed regions, boundary points are included. So even if the third quadrant is strictly u <0, v <0, the closure includes the axes. Therefore, in the feasible region, which includes the boundary, (0, -3) is included.So in this case, the maximum in the third quadrant is 9, which corresponds to the point (0, -3) in uv, translating to 2x -3y=4. Therefore, the maximum is indeed 4.Fourth quadrant: u ≥0, v ≤0. The constraint becomes 3u +2(-v) ≤6 ⇒3u -2v ≤6. The objective function is 2u -3v. Here, u ≥0 and v ≤0. So increasing u increases the function, and decreasing v (more negative) increases the function since it's multiplied by -3. Therefore, to maximize 2u -3v, we want to maximize u and minimize v (most negative). But under the constraint 3u -2v ≤6. Let's see.On the boundary 3u -2v=6. Express v=(3u -6)/2. Since v ≤0, (3u -6)/2 ≤0 ⇒3u -6 ≤0 ⇒u ≤2. So u ∈ [0,2], v=(3u -6)/2. Substitute into the objective function:2u -3v =2u -3*(3u -6)/2=2u - (9u -18)/2=(4u -9u +18)/2=(-5u +18)/2.To maximize this expression over u ∈ [0,2], take derivative with respect to u: -5/2, which is negative. Therefore, maximum at u=0: v=(0 -6)/2=-3. Objective value= (-5*0 +18)/2=9. But u=0, v=-3 is again (0, -3) in uv, which is (-1, -2) in xy. Therefore, same point. So in the fourth quadrant, the maximum is 9 at (0, -3). However, in this quadrant, u ≥0, but at u=0, we are on the boundary with the third quadrant. So again, the maximum in this quadrant is 9 at (0, -3), which translates to the original maximum of 4.Therefore, regardless of the quadrant, the maximum occurs at (0, -3) in uv-coordinates, which is (-1, -2) in xy-coordinates, giving the objective value of 4.Alternatively, another method is to parametrize the boundary of the feasible region and use calculus to find maxima. But since the region is convex and the objective is linear, vertices are the way to go.But let me try another approach for confirmation. Let's consider that 3|x +1| + 2|y -1| ≤6. Let me set variables a = x +1 and b = y -1. Then the constraint becomes 3|a| + 2|b| ≤6, and the objective is 2x -3y = 2(a -1) -3(b +1)=2a -2 -3b -3=2a -3b -5. So we need to maximize 2a -3b -5 with 3|a| +2|b| ≤6.So now, as before, define variables a and b. Then, the maximum of 2a -3b will be achieved at some point on the boundary 3|a| +2|b|=6. Let me consider all four quadrants for a and b.Case 1: a ≥0, b ≥0. Then the constraint is 3a +2b=6. The objective is 2a -3b. Express b=(6 -3a)/2. Substitute into the objective: 2a -3*(6 -3a)/2=2a - (18 -9a)/2=(4a -18 +9a)/2=(13a -18)/2. To maximize this, a is in [0,2]. The derivative is 13/2 >0, so maximum at a=2: (13*2 -18)/2=(26 -18)/2=8/2=4. So 4, then subtract 5 gives -1. Wait, no: in this case, 2a -3b=4, so subtracting 5 gives 4 -5=-1. But that's only in this quadrant.Wait, but in this case, the maximum of 2a -3b is 4, achieved at a=2, b=0. Then, translating back to original variables: a=2 ⇒x=2 -1=1, b=0⇒y=0 +1=1. So (1,1), which gives 2x -3y= -1. Which matches the previous result.Case 2: a ≥0, b ≤0. Constraint is 3a -2b=6. Objective is 2a -3b. Express b=(3a -6)/2. Substitute into objective: 2a -3*(3a -6)/2=2a - (9a -18)/2=(4a -9a +18)/2=(-5a +18)/2. To maximize over a. Since a ≥0, and from b=(3a -6)/2 ≤0 ⇒3a -6 ≤0 ⇒a ≤2. So a ∈[0,2]. The expression (-5a +18)/2 is decreasing in a, so maximum at a=0: (-0 +18)/2=9. Then, 9 -5=4. So in this case, maximum at a=0, b=(0 -6)/2=-3. Translating to x=0 -1=-1, y= -3 +1=-2. So (-1, -2), gives 2*(-1) -3*(-2)= -2 +6=4. Correct.Case3: a ≤0, b ≥0. Constraint is -3a +2b=6. Objective is 2a -3b. Express b=(6 +3a)/2. Substitute into objective: 2a -3*(6 +3a)/2=2a - (18 +9a)/2=(4a -18 -9a)/2=(-5a -18)/2. To maximize over a ≤0. The expression (-5a -18)/2. Since a ≤0, -5a ≥0. To maximize, we need to maximize -5a, which occurs at a as small as possible. But from the constraint b=(6 +3a)/2 ≥0 ⇒6 +3a ≥0 ⇒a ≥-2. So a ∈[-2,0]. The expression is (-5a -18)/2, which is linear in a. The coefficient of a is -5/2, which is negative. Therefore, maximum occurs at a=-2: (-5*(-2) -18)/2=(10 -18)/2=(-8)/2=-4. Then, subtract 5 gives -4 -5=-9. Translating to x=-2 -1=-3, y=(6 +3*(-2))/2 +1=0 +1=1. So (-3,1), gives 2*(-3) -3*1=-6 -3=-9. Correct.Case4: a ≤0, b ≤0. Constraint is -3a -2b=6. Objective is 2a -3b. Express a=(-6 -2b)/3. Substitute into objective: 2*(-6 -2b)/3 -3b= (-12 -4b)/3 -3b= -4 - (4b)/3 -3b= -4 - (13b)/3. To maximize over b ≤0. The expression is linear in b, coefficient -13/3, which is negative. Therefore, maximum at b=0: -4 -0=-4. But b=0, then a=(-6 -0)/3=-2. So a=-2, b=0: x=-2 -1=-3, y=0 +1=1. So same as previous case. Wait, but b can go down to b=-3 when a=0. Let me check. If a=0, then from constraint -3*0 -2b=6 ⇒b=-3. So substituting b=-3 into the objective: 2*0 -3*(-3)=9. Then 9 -5=4. So the point a=0, b=-3. Translating to x=0 -1=-1, y=-3 +1=-2. So (-1, -2). Which gives 2x -3y=4.Wait, this is confusing. In case4, a ≤0, b ≤0. Expressing the objective function in terms of b, we have -4 - (13b)/3. To maximize this, since the coefficient of b is negative, we need to minimize b (most negative). The minimum b is when a=0: from -3a -2b=6, a=0 ⇒-2b=6 ⇒b=-3. So substituting b=-3, a=0. But a=0 is not in the interior of the fourth quadrant (a ≤0, b ≤0), but on the boundary. So (0, -3) is allowed here, and gives the maximum value of 9 for 2a -3b, then subtract 5 to get 4. So yes, the maximum is 4 here as well.Therefore, through all four cases, the maximum of 2a -3b is 9, achieved at (0, -3), which translates to (-1, -2) in original coordinates, giving 2x -3y=4.Thus, all methods confirm that the maximum value is 4. Therefore, the answer is boxed{4}.Final Answerboxed{4}</think>Given the constraint ( 3|x+1| + 2|y-1| leq 6 ), we need to find the maximum value of ( 2x - 3y ).First, we recognize that the constraint defines a convex polygon (a diamond) centered at ((-1, 1)) with vertices calculated as follows:- For ( y = 1 ), ( 3|x+1| leq 6 ) gives ( x = 1 ) and ( x = -3 ). Thus, vertices are ((1, 1)) and ((-3, 1)).- For ( x = -1 ), ( 2|y-1| leq 6 ) gives ( y = 4 ) and ( y = -2 ). Thus, vertices are ((-1, 4)) and ((-1, -2)).The vertices of the feasible region are ((1, 1)), ((-3, 1)), ((-1, 4)), and ((-1, -2)).We evaluate the objective function ( 2x - 3y ) at each vertex:1. At ((1, 1)): ( 2(1) - 3(1) = -1 )2. At ((-3, 1)): ( 2(-3) - 3(1) = -9 )3. At ((-1, 4)): ( 2(-1) - 3(4) = -14 )4. At ((-1, -2)): ( 2(-1) - 3(-2) = 4 )The maximum value is found at the vertex ((-1, -2)), giving the value ( 4 ).To confirm, we transform the problem using ( u = x + 1 ) and ( v = y - 1 ), leading to the constraint ( 3|u| + 2|v| leq 6 ) and the objective function ( 2u - 3v - 5 ). The vertices in ( uv )-coordinates are ((2, 0)), ((-2, 0)), ((0, 3)), and ((0, -3)). Evaluating ( 2u - 3v ) at these vertices, the maximum value is ( 9 ) at ((0, -3)), translating back to ((-1, -2)) and giving ( 2x - 3y = 4 ).Thus, the maximum value of ( 2x - 3y ) is (boxed{4}).,<think>Okay, let me try to figure out how to solve this problem. So, we have two real numbers x and y that satisfy the inequality 3|x + 1| + 2|y - 1| ≤ 6. We need to find the maximum value of 2x - 3y. Hmm, okay.First, I remember that when dealing with absolute values in inequalities, it often helps to consider different cases based on the signs of the expressions inside the absolute values. So, the expressions here are |x + 1| and |y - 1|. Let me break this down.The expression |x + 1| can be split into two cases: when x + 1 is non-negative (so x ≥ -1) and when x + 1 is negative (x < -1). Similarly, |y - 1| can be split into y - 1 ≥ 0 (y ≥ 1) and y - 1 < 0 (y < 1). So, there are four possible cases here based on the combinations of these conditions. But maybe there's a better way than checking all four cases? Because that might take a while.Alternatively, I recall that inequalities involving absolute values can represent regions in the coordinate plane. Specifically, 3|x + 1| + 2|y - 1| ≤ 6 defines a diamond-shaped region (a convex polygon) centered at (-1, 1) because the absolute values are linear and the sum is bounded. The coefficients 3 and 2 will affect the "stretch" of the diamond along the x and y axes respectively. So, maybe the maximum of the linear function 2x - 3y occurs at one of the vertices of this region. That's a common occurrence in linear optimization problems with convex constraints.So, if I can find the vertices of the region defined by 3|x + 1| + 2|y - 1| ≤ 6, then I can evaluate 2x - 3y at each vertex and pick the maximum value. That seems like a solid plan.Let me recall how to find the vertices of such a region. The equation 3|x + 1| + 2|y - 1| = 6 will form the boundary, and the vertices occur where the expressions inside the absolute values switch their signs. That is, where x + 1 = 0 and y - 1 = 0. So, the center of the diamond is at (-1, 1). Then, the diamond extends in the four quadrants around this center.To find the vertices, we can set either x + 1 or y - 1 to zero and solve for the other variable. Wait, actually, perhaps more methodically: the diamond will have four vertices, each where one of the terms 3|x + 1| or 2|y - 1| is maximized. Alternatively, when moving along the x or y direction until the other term becomes zero. Let me think.Alternatively, the diamond is a convex polygon with four edges. Each edge corresponds to one side of the absolute value expressions. For example, when x + 1 ≥ 0 and y - 1 ≥ 0, the inequality becomes 3(x + 1) + 2(y - 1) ≤ 6. Similarly, other edges correspond to different combinations of signs.So, let's consider each case:1. x + 1 ≥ 0 (x ≥ -1) and y - 1 ≥ 0 (y ≥ 1): Then, the inequality becomes 3(x + 1) + 2(y - 1) ≤ 6. Simplify: 3x + 3 + 2y - 2 ≤ 6 → 3x + 2y + 1 ≤ 6 → 3x + 2y ≤ 5.2. x + 1 ≥ 0 (x ≥ -1) and y - 1 < 0 (y < 1): The inequality becomes 3(x + 1) + 2(1 - y) ≤ 6. Simplify: 3x + 3 + 2 - 2y ≤ 6 → 3x - 2y + 5 ≤ 6 → 3x - 2y ≤ 1.3. x + 1 < 0 (x < -1) and y - 1 ≥ 0 (y ≥ 1): The inequality becomes 3(-x - 1) + 2(y - 1) ≤ 6. Simplify: -3x - 3 + 2y - 2 ≤ 6 → -3x + 2y - 5 ≤ 6 → -3x + 2y ≤ 11.4. x + 1 < 0 (x < -1) and y - 1 < 0 (y < 1): The inequality becomes 3(-x - 1) + 2(1 - y) ≤ 6. Simplify: -3x - 3 + 2 - 2y ≤ 6 → -3x - 2y - 1 ≤ 6 → -3x - 2y ≤ 7.So, these are four linear inequalities, each defining an edge of the diamond. The vertices occur where these edges intersect each other. To find the vertices, we need to solve the system of equations obtained by taking two adjacent inequalities.Wait, perhaps the vertices are the points where the individual terms are maximized. Alternatively, since it's a diamond, the vertices can be found by setting each absolute value term to their extreme.Alternatively, note that the diamond can be parametrized. Let's consider the general equation 3|x + 1| + 2|y - 1| = 6. The maximum and minimum values for x and y can be found by setting the other variable to its central value.For example, to find the maximum possible x, set y = 1 (since this minimizes the contribution of the y term). Then, 3|x + 1| + 0 ≤ 6 → |x + 1| ≤ 2 → x + 1 ≤ 2 → x ≤ 1 and x + 1 ≥ -2 → x ≥ -3. Wait, but to find the maximum x, set y such that |y - 1| is minimized. Wait, actually, maybe not. Wait, if we want to maximize x, we need to minimize the other term. Since 3|x +1| + 2|y -1| ≤6, to maximize x, we need to maximize |x +1|, which would require minimizing |y -1|. So, set |y -1| =0, so y=1. Then, 3|x +1| ≤6 → |x +1| ≤2 → x ∈ [-3, 1]. So, the maximum x is 1, achieved when y=1.Similarly, the minimum x would be -3 when y=1.Similarly, for y: to maximize y, set |x +1|=0 (x=-1). Then, 2|y -1| ≤6 → |y -1| ≤3 → y ∈ [-2,4]. So maximum y is 4, minimum y is -2.But how does this help us find the vertices? Well, the vertices are the points where both |x +1| and |y -1| are at their extremal values. So, maybe the four vertices are:1. When x is maximized (x=1) and y is set to 1 (since we minimized |y -1|): (1,1)2. When x is minimized (x=-3) and y=1: (-3,1)3. When y is maximized (y=4) and x=-1: (-1,4)4. When y is minimized (y=-2) and x=-1: (-1,-2)Wait, but these four points would form a diamond centered at (-1,1). Let me check if these points lie on the original inequality.Take point (1,1): 3|1 +1| + 2|1 -1| = 3*2 + 2*0 =6, which satisfies the equality. Similarly, (-3,1): 3|-3 +1| + 2|1 -1| =3*2 +0=6. (-1,4): 3| -1 +1| +2|4 -1| =0 +2*3=6. (-1,-2):3|0| +2*3=6. So yes, these four points are on the boundary and form a diamond. So, these are the vertices.Therefore, the feasible region is a convex quadrilateral with vertices at (1,1), (-3,1), (-1,4), (-1,-2). Wait, but hold on: when x=-1, y can be as high as 4 or as low as -2. But when you connect these points, how do they form the diamond? Let's see:From (1,1), moving to (-1,4): when you move left from x=1 to x=-1, and y increases from 1 to 4. Similarly, from (1,1) to (-1,-2): moving left and y decreases. Similarly, from (-3,1) to (-1,4) and (-1,-2). Hmm, maybe the edges connect (1,1) to (-1,4), (-1,4) to (-3,1), (-3,1) to (-1,-2), and (-1,-2) to (1,1). Wait, that seems a bit off.Wait, actually, when you have 3|x +1| + 2|y -1| =6, the shape is a diamond (a convex polygon) with vertices at (1,1), (-3,1), (-1,4), (-1,-2). Let's check if these are indeed the vertices by looking at the edges.In the first quadrant relative to center (-1,1): moving right (x increases) and up (y increases), but since our center is (-1,1), the "quadrants" are shifted.Wait, maybe it's better to parametrize the edges. Let's take each of the four cases we considered earlier:1. 3x + 2y ≤5 (x ≥ -1, y ≥1). The edge here is 3x + 2y =5. To find the endpoints of this edge, we can intersect with y=1 and x=-1. If y=1, then 3x +2(1)=5 →3x=3→x=1. So, one endpoint is (1,1). If x=-1, then 3(-1) +2y=5→-3+2y=5→2y=8→y=4. So, the other endpoint is (-1,4). So, this edge connects (1,1) to (-1,4).2. 3x - 2y ≤1 (x ≥ -1, y ≤1). The edge is 3x -2y=1. To find endpoints, intersect with x=-1 and y=1. If x=-1: 3(-1) -2y=1→-3 -2y=1→-2y=4→y=-2. So, endpoint (-1,-2). If y=1:3x -2(1)=1→3x=3→x=1. So, endpoint (1,1). So, this edge connects (1,1) to (-1,-2).3. -3x + 2y ≤11 (x ≤-1, y ≥1). The edge is -3x +2y=11. Intersect with x=-1 and y=1. If x=-1: -3(-1)+2y=11→3+2y=11→2y=8→y=4. So, endpoint (-1,4). If y=1: -3x +2(1)=11→-3x=9→x=-3. So, endpoint (-3,1). Thus, this edge connects (-1,4) to (-3,1).4. -3x -2y ≤7 (x ≤-1, y ≤1). The edge is -3x -2y=7. Intersect with x=-1 and y=1. If x=-1: -3(-1) -2y=7→3 -2y=7→-2y=4→y=-2. So, endpoint (-1,-2). If y=1: -3x -2(1)=7→-3x=9→x=-3. Endpoint (-3,1). So, this edge connects (-1,-2) to (-3,1).Therefore, the four edges connect the four vertices (1,1), (-1,4), (-3,1), (-1,-2), and back to (1,1). Wait, actually, connecting (1,1) to (-1,4), (-1,4) to (-3,1), (-3,1) to (-1,-2), (-1,-2) to (1,1). So, the diamond is a quadrilateral with those four vertices.Therefore, the feasible region is the convex hull of these four points. Thus, to find the maximum of 2x -3y over this region, it suffices to evaluate the objective function 2x -3y at each of the four vertices and pick the largest value.Let's compute 2x -3y at each vertex:1. At (1,1): 2(1) -3(1) = 2 -3 = -12. At (-1,4): 2(-1) -3(4) = -2 -12 = -143. At (-3,1): 2(-3) -3(1) = -6 -3 = -94. At (-1,-2): 2(-1) -3(-2) = -2 +6 = 4So, the values are -1, -14, -9, and 4. The maximum among these is 4 at the point (-1,-2). Therefore, the maximum value of 2x -3y is 4.Wait, but hold on a second. Is it possible that the maximum occurs not at a vertex but somewhere along an edge? In linear programming, the maximum of a linear function over a convex polygon occurs at a vertex. So, since our objective function 2x -3y is linear, and the feasible region is convex (a convex polygon), then yes, the maximum must occur at one of the vertices. So, evaluating at the vertices suffices. Therefore, 4 is indeed the maximum value.But just to be thorough, maybe I should check along the edges as well. Let me pick a point on each edge and see if the function can be larger than 4.Take the edge between (-1,-2) and (1,1). Parametrize this edge. Let me write the equation for this edge. From the earlier case, when x ≥ -1 and y ≤1, the edge is 3x -2y=1. Wait, actually, the edge between (-1,-2) and (1,1) is part of the line 3x -2y=1. Let's verify:At (1,1): 3(1) -2(1)=1, yes. At (-1,-2): 3(-1) -2(-2)= -3 +4=1, yes. So, the edge is 3x -2y=1, with x between -1 and 1, and y between -2 and1.So, parametrize this edge as x going from -1 to 1. For each x, y = (3x -1)/2. Let me compute 2x -3y along this edge:Substitute y = (3x -1)/2 into 2x -3y:2x -3*(3x -1)/2 = 2x - (9x -3)/2 = (4x -9x +3)/2 = (-5x +3)/2.To find the maximum of this expression over x ∈ [-1,1]. Since this is a linear function in x, decreasing as x increases (coefficient -5/2), so maximum occurs at x = -1:(-5*(-1) +3)/2 = (5 +3)/2 =8/2=4, which is the value at (-1,-2). And at x=1: (-5(1)+3)/2=(-5 +3)/2=-2/2=-1, which is the value at (1,1). So, along this edge, the maximum is indeed 4.Now check another edge, say between (-1,4) and (1,1). The edge is 3x +2y=5. So, parametrize x from -1 to1. Then, y=(5 -3x)/2. Substitute into 2x -3y:2x -3*(5 -3x)/2 =2x - (15 -9x)/2=(4x -15 +9x)/2=(13x -15)/2.This is a linear function in x with coefficient 13/2, which is positive. So, the maximum occurs at x=1: (13*1 -15)/2=(13-15)/2=(-2)/2=-1, and minimum at x=-1: (13*(-1)-15)/2=(-13-15)/2=-28/2=-14. So, the maximum here is -1, which is at (1,1). So, no improvement over the vertex values.Next, the edge between (-3,1) and (-1,4). This edge is part of the line -3x +2y=11. Let's parametrize x from -3 to -1, then y=(11 +3x)/2. Substitute into 2x -3y:2x -3*(11 +3x)/2 =2x - (33 +9x)/2=(4x -33 -9x)/2=(-5x -33)/2.This is linear in x with coefficient -5/2, so decreasing as x increases. Therefore, maximum at x=-3:(-5*(-3) -33)/2=(15 -33)/2=(-18)/2=-9, which is the value at (-3,1). At x=-1: (-5*(-1)-33)/2=(5 -33)/2=(-28)/2=-14, which is at (-1,4). So, no improvement here.Finally, the edge between (-3,1) and (-1,-2). This edge is part of the line -3x -2y=7. Let's parametrize x from -3 to -1, then y=( -7 -3x)/2. Substitute into 2x -3y:2x -3*(-7 -3x)/2=2x + (21 +9x)/2=(4x +21 +9x)/2=(13x +21)/2.This is linear in x with coefficient 13/2, positive. So, maximum at x=-1:(13*(-1)+21)/2=(-13 +21)/2=8/2=4, which is the value at (-1,-2). At x=-3: (13*(-3)+21)/2=(-39 +21)/2=(-18)/2=-9, which is at (-3,1). So, again, maximum here is 4.Therefore, all edges have their maximum at the vertices, confirming that the maximum value of 2x -3y is indeed 4, achieved at the point (-1,-2).So, the answer should be 4. Let me just make sure I didn't make any arithmetic mistakes.Check vertex (-1,-2):2*(-1) -3*(-2) = -2 +6 =4. Correct.Other vertices:(1,1): 2*1 -3*1= -1. Correct.(-1,4): 2*(-1) -3*4= -2 -12= -14. Correct.(-3,1): 2*(-3) -3*1= -6 -3= -9. Correct.So, all the computations check out. Therefore, the maximum value is 4.Final AnswerThe maximum value of (2x - 3y) is boxed{4}.</think>Given the inequality (3|x + 1| + 2|y - 1| leq 6), we need to find the maximum value of (2x - 3y).First, we recognize that the inequality defines a convex region (a diamond) centered at ((-1, 1)). The vertices of this region are found by considering the extreme values of (x) and (y) when the other variable is set to its central value. The vertices are calculated as follows:1. When (y = 1) (minimizing the contribution of the (y) term), we get (3|x + 1| leq 6 implies |x + 1| leq 2). This gives the vertices ((1, 1)) and ((-3, 1)).2. When (x = -1) (minimizing the contribution of the (x) term), we get (2|y - 1| leq 6 implies |y - 1| leq 3). This gives the vertices ((-1, 4)) and ((-1, -2)).The vertices of the region are ((1, 1)), ((-3, 1)), ((-1, 4)), and ((-1, -2)).Next, we evaluate the objective function (2x - 3y) at each vertex:- At ((1, 1)): (2(1) - 3(1) = -1)- At ((-3, 1)): (2(-3) - 3(1) = -9)- At ((-1, 4)): (2(-1) - 3(4) = -14)- At ((-1, -2)): (2(-1) - 3(-2) = 4)The maximum value of (2x - 3y) is found to be 4 at the vertex ((-1, -2)).Thus, the maximum value of (2x - 3y) is (boxed{4}).,<think>Okay, so I have this problem here where I need to find the maximum value of 2x - 3y given that the real numbers x and y satisfy the inequality 3|x + 1| + 2|y - 1| ≤ 6. Hmm, let me try to figure out how to approach this.First, I remember that inequalities involving absolute values often define a region in the coordinate plane, usually some sort of polygon or diamond shape. Since there are two variables, x and y, this should be a two-dimensional region. The expression 3|x + 1| + 2|y - 1| ≤ 6 seems like a linear combination of absolute values, so maybe it's a diamond or a rectangle?Let me recall that the general form |x - a| + |y - b| ≤ c defines a diamond (a rotated square) centered at (a, b). But here, the coefficients of the absolute values are different: 3 and 2. That probably stretches the diamond into a different shape. Maybe it's a convex polygon with vertices at certain points?Alternatively, maybe I can rewrite the inequality to a more standard form. Let's see:3|x + 1| + 2|y - 1| ≤ 6Divide both sides by 6 to normalize:(|x + 1|)/2 + (|y - 1|)/3 ≤ 1Ah, this looks similar to the equation of an ellipse, but with absolute values instead of squares. Wait, actually, when you have |x/a| + |y/b| ≤ 1, that defines a diamond (a convex polygon) with vertices at (±a, 0) and (0, ±b). But in this case, the center isn't at (0,0); it's shifted. Let me adjust for the shifts.The original inequality can be written as:|x + 1| / 2 + |y - 1| / 3 ≤ 1So comparing to |x - h| / a + |y - k| / b ≤ 1, the center is at (h, k) = (-1, 1), and the "axes" lengths are a = 2 and b = 3. However, since this is the sum of absolute values, not squares, the shape is a diamond (a convex quadrilateral) centered at (-1, 1), extending 2 units left and right along the x-axis, and 3 units up and down along the y-axis. But because the coefficients were 3 and 2 in the original inequality, maybe the stretching is different? Wait, after dividing by 6, we have |x +1| /2 + |y -1| /3 ≤1, so the denominators are the distances from the center. So for x, it's |x - (-1)| /2, meaning that along the x-axis, from the center (-1,1), we can go 2 units left and right. Similarly, along the y-axis, from the center, we can go 3 units up and down. So the diamond's vertices should be at (-1 + 2, 1) = (1,1), (-1 - 2, 1) = (-3,1), (-1, 1 + 3) = (-1,4), and (-1,1 -3) = (-1,-2). So the diamond has vertices at (1,1), (-3,1), (-1,4), (-1,-2). So the feasible region is this diamond shape.But wait, actually, when you have |x - h| /a + |y - k| /b ≤1, the vertices are at (h ± a, k) and (h, k ± b). So yeah, exactly as above. So the feasible region is a convex quadrilateral with those four vertices.So if I can visualize this diamond, the maximum of the linear function 2x - 3y will occur at one of the vertices. Because in linear programming, the maximum or minimum of a linear function over a convex polygon occurs at a vertex.Therefore, maybe I can just evaluate the function 2x - 3y at each of the four vertices and see which one is the largest.Let me check that.First vertex: (1,1). Plug into 2x -3y: 2*1 -3*1 = 2 -3 = -1.Second vertex: (-3,1). 2*(-3) -3*1 = -6 -3 = -9.Third vertex: (-1,4). 2*(-1) -3*4 = -2 -12 = -14.Fourth vertex: (-1,-2). 2*(-1) -3*(-2) = -2 +6 = 4.So among these four, the maximum value is 4 at (-1,-2). Hmm, so is that the maximum?Wait, but hold on. The feasible region is a convex polygon, so the maximum of a linear function is indeed at a vertex. Therefore, 4 is the maximum. But let me just confirm if this is correct.Wait, but maybe I made a mistake in the vertices. Let me double-check the original inequality.Original inequality: 3|x +1| + 2|y -1| ≤6.If we set each absolute value term to zero, we can find the intercepts.For example, when y =1, then 3|x +1| ≤6 → |x +1| ≤2 → x is between -3 and 1. So the points on the x-axis (y=1) are from x=-3 to x=1.Similarly, when x = -1, then 2|y -1| ≤6 → |y -1| ≤3 → y is between -2 and 4. So the points on the y-axis (x=-1) are from y=-2 to y=4.So the diamond has vertices at (x=-1 ± 2, y=1) when y is fixed, but wait, no. Wait, if we set one of the absolute terms to zero, but actually, the vertices are when we have maximum in one direction. Wait, perhaps my initial analysis was wrong.Wait, perhaps the diamond is actually not axis-aligned? Wait, no, because the inequality is 3|x +1| + 2|y -1| ≤6. So each term is independent. So the shape is a diamond, but stretched in the x and y directions. So, actually, the intercepts are at x when y=1, which is from -3 to 1, as above, and y when x=-1, which is from -2 to 4, as above. But to find the vertices, which are the points where both absolute values are maximized. Wait, but in the standard diamond |x| + |y| ≤1, the vertices are at (1,0), (-1,0), (0,1), (0,-1). Similarly, here, if we have 3|x +1| + 2|y -1| ≤6, then the "intercepts" are when one of the terms is zero. So, for example, to find the intercepts on the x-axis (where y=1), we get x from -3 to 1, as above. Similarly, intercepts on the y-axis (x=-1) are y from -2 to 4.But the other vertices would be where both terms are non-zero. Wait, no, in the standard diamond |x/a| + |y/b| ≤1, the vertices are at (±a, 0) and (0, ±b). So in this case, substituting variables:Let me make a substitution. Let u = x +1, v = y -1. Then the inequality becomes 3|u| + 2|v| ≤6. Dividing both sides by 6: |u|/2 + |v|/3 ≤1. So in terms of u and v, it's a diamond with vertices at (2,0), (-2,0), (0,3), (0,-3). Translating back to x and y: u = x +1, so x = u -1; v = y -1, so y = v +1. Therefore, the vertices in (x,y) coordinates are:(2 -1, 0 +1) = (1,1),(-2 -1, 0 +1) = (-3,1),(0 -1, 3 +1) = (-1,4),(0 -1, -3 +1) = (-1,-2).So exactly the vertices I found before. So that seems correct.Therefore, the feasible region is a diamond with those four vertices, so the maximum of 2x -3y is at (-1,-2) with value 4. So is that the answer? Wait, but let me check.But wait, the problem says "real numbers x, y satisfy 3|x +1| + 2|y -1| ≤6". So the feasible region is all points (x,y) such that that inequality holds. The maximum of 2x -3y over that region. Since it's a convex polygon, and the objective function is linear, the maximum must be attained at a vertex. So evaluating at all four vertices gives the maximum as 4. So the answer is 4. Hmm.But let me think again. Maybe there's another point on the edges where 2x -3y is larger? For example, suppose we parametrize one of the edges and check.Let's take the edge between (1,1) and (-1,4). Let me parametrize this edge. Since it's a diamond, each edge is a straight line between two vertices. So between (1,1) and (-1,4), the edge can be parametrized as x going from 1 to -1, y going from 1 to 4.But perhaps a better way is to consider the inequality as a set of linear inequalities in different quadrants.Because absolute value functions can be split into different cases based on the signs of the expressions inside. So let's try that approach.The original inequality is 3|x +1| + 2|y -1| ≤6.To handle the absolute values, we can split into cases depending on the sign of x +1 and y -1. There are four cases:1. x +1 ≥0 and y -1 ≥02. x +1 ≥0 and y -1 ≤03. x +1 ≤0 and y -1 ≥04. x +1 ≤0 and y -1 ≤0For each case, we can remove the absolute values and rewrite the inequality, then find the feasible region in that quadrant.Then, for each case, we can express the inequality as a linear constraint and maximize 2x -3y.But since the maximum must occur at a vertex, but maybe this approach is more complicated. However, maybe it can help confirm the result.Case 1: x +1 ≥0 and y -1 ≥0. So x ≥ -1 and y ≥1.Then the inequality becomes 3(x +1) + 2(y -1) ≤6.Simplify: 3x +3 + 2y -2 ≤6 → 3x +2y +1 ≤6 → 3x +2y ≤5.So in this region, we have 3x +2y ≤5, x ≥-1, y ≥1.We can express 2x -3y as the objective function and try to maximize it under these constraints.Case 2: x ≥-1 and y ≤1. Then |y -1| = 1 - y.So the inequality becomes 3(x +1) + 2(1 - y) ≤6.Simplify: 3x +3 +2 -2y ≤6 → 3x -2y +5 ≤6 → 3x -2y ≤1.So constraints: x ≥-1, y ≤1, 3x -2y ≤1.Maximize 2x -3y here.Case 3: x ≤-1 and y ≥1. Then |x +1| = -x -1.Inequality: 3(-x -1) + 2(y -1) ≤6.Simplify: -3x -3 +2y -2 ≤6 → -3x +2y -5 ≤6 → -3x +2y ≤11 → 3x -2y ≥-11.Constraints: x ≤-1, y ≥1, 3x -2y ≥-11.Maximize 2x -3y here.Case 4: x ≤-1 and y ≤1. Then |x +1| = -x -1 and |y -1| =1 - y.Inequality: 3(-x -1) +2(1 - y) ≤6.Simplify: -3x -3 +2 -2y ≤6 → -3x -2y -1 ≤6 → -3x -2y ≤7 → 3x +2y ≥-7.Constraints: x ≤-1, y ≤1, 3x +2y ≥-7.Maximize 2x -3y here.So now, in each of these four cases, we can set up the problem as a linear programming problem with linear constraints and a linear objective function. The maximum will occur at one of the vertices of the feasible region in each case, and then we can compare across all cases.Alternatively, since we already know the feasible region is a convex quadrilateral with four vertices, each corresponding to one of these cases, then maybe the maximum is indeed at one of those four vertices. But just to be thorough, let's check in each case.Starting with Case 1: 3x +2y ≤5, x ≥-1, y ≥1.We need to maximize 2x -3y. To maximize 2x -3y, we want to maximize x and minimize y. But in this region, x can be as large as possible (since we want to maximize x) and y as small as possible (since we want to minimize y). However, x is bounded by the constraint 3x +2y ≤5. If y is minimized, which is y=1, then 3x +2*1 ≤5 → 3x ≤3 → x ≤1. So the point (1,1) is in this region, which is a vertex we already considered. At (1,1), 2x -3y = -1. If we take another point in this region, say x=1, y=1, which is the vertex. If we take x less than 1, then y can be slightly larger, but since we want to minimize y, the minimal y here is 1. So the maximum of 2x -3y in this case is -1 at (1,1).Case 2: 3x -2y ≤1, x ≥-1, y ≤1.We need to maximize 2x -3y. Here, to maximize 2x -3y, we want to maximize x and maximize (-3y), which is equivalent to minimizing y. However, y is already bounded above by 1 and below by... Well, y can be as low as possible, but subject to 3x -2y ≤1. Wait, but y can be any real number, but in this case, since original problem had y ≤1, but y can be as low as possible. However, the original absolute value inequality 3|x+1| +2|y-1| ≤6 would constrain y even if y is less than 1. Wait, but in Case 2, we split the inequality into 3(x+1) + 2(1 - y) ≤6, which simplifies to 3x -2y ≤1. So, given that x ≥-1 and y ≤1, but also, the other constraints from the original inequality. Wait, but perhaps there's a lower bound on y?Wait, when we split into Case 2, we have x ≥-1 and y ≤1, but the original inequality in this case is 3x -2y ≤1. But we need to ensure that 3|x+1| +2|y -1| ≤6. But since in this case, x+1 ≥0 and y-1 ≤0, so |x+1| =x+1 and |y -1|=1 - y. Therefore, the inequality 3(x+1) +2(1 - y) ≤6 is equivalent to 3x -2y ≤1. So in this case, we have the constraints x ≥-1, y ≤1, and 3x -2y ≤1.But we need to check if there are other constraints. For example, could y be unbounded below? Let's see. If y approaches negative infinity, then 3x -2y ≤1 would require x to also approach negative infinity, but x is bounded below by -1. So if x ≥-1 and y ≤1, then substituting x=-1 into 3x -2y ≤1 gives -3 -2y ≤1 → -2y ≤4 → y ≥-2. So actually, y cannot be less than -2 in this case. Because if y were less than -2, then even with x=-1, 3*(-1) -2y ≤1 → -3 -2y ≤1 → -2y ≤4 → y ≥-2. Therefore, in Case 2, y is between -2 and 1.Therefore, in Case 2, the constraints are x ≥-1, y ≥-2, y ≤1, and 3x -2y ≤1.So in this case, to maximize 2x -3y, we want to maximize x and minimize y. So maximum x and minimum y. The minimum y in this case is y=-2, but we need to check if y can be -2 given the constraints. If y=-2, then 3x -2*(-2) ≤1 → 3x +4 ≤1 → 3x ≤-3 → x ≤-1. But x ≥-1 in this case. Therefore, x=-1 and y=-2 is the point. So (x,y)=(-1,-2) is in this case, as x=-1 ≥-1 (just equal) and y=-2 ≤1. So this point is in Case 2. Therefore, in Case 2, the maximum of 2x -3y is at (-1,-2), which gives 2*(-1) -3*(-2) = -2 +6=4.So that's the same vertex as before. So in Case 2, maximum is 4 at (-1,-2).Case 3: 3x -2y ≥-11, x ≤-1, y ≥1.We need to maximize 2x -3y here. To maximize 2x -3y, we want to maximize x and minimize y. But x is constrained to be ≤-1, so maximum x is -1. y is ≥1, so minimum y is 1. Let's check if x=-1, y=1 satisfies 3x -2y ≥-11. Plugging in: 3*(-1) -2*1 =-3 -2 =-5 ≥-11? Yes, -5 is greater than -11. So the point (-1,1) is in this region. But 2x -3y at (-1,1) is 2*(-1) -3*1=-2 -3=-5. However, maybe there's another point where 2x -3y is larger?Wait, perhaps if we take a different point. Let's express the constraint 3x -2y ≥-11. So 3x -2y ≥-11 → 2y ≤3x +11 → y ≤ (3x +11)/2.But in this case, y ≥1. So combining with y ≥1 and x ≤-1.We can try to parametrize the boundary of the feasible region. The boundary is 3x -2y = -11. Let's express y in terms of x: y = (3x +11)/2. Since x ≤-1, and y ≥1.So when x=-1, y=( -3 +11)/2=8/2=4. So the point (-1,4) is on this boundary. As x decreases, y=(3x +11)/2 decreases. Let's see when y=1: 1=(3x +11)/2 → 3x +11=2 → 3x= -9 → x=-3. So when x=-3, y=1. So the line 3x -2y = -11 intersects the line y=1 at x=-3, y=1. But in this case, x ≤-1, so from x=-3 to x=-1.So the feasible region for Case 3 is the line segment from (-3,1) to (-1,4). So the edge of the diamond. So to maximize 2x -3y on this edge.Let me parametrize x from -3 to -1, then y=(3x +11)/2.So let’s set x = t, where t ranges from -3 to -1. Then y=(3t +11)/2.Then 2x -3y = 2t -3*( (3t +11)/2 ) = 2t - (9t +33)/2 = (4t -9t -33)/2 = (-5t -33)/2.We need to maximize (-5t -33)/2 as t goes from -3 to -1.Since t is increasing from -3 to -1, -5t is decreasing (because multiplied by -5), so -5t -33 is decreasing, so the entire expression is decreasing. Therefore, maximum occurs at t=-3:At t=-3: (-5*(-3) -33)/2=(15 -33)/2=(-18)/2=-9.At t=-1: (-5*(-1) -33)/2=(5 -33)/2=(-28)/2=-14.So on this edge, the maximum is -9 at (-3,1), which is a vertex we already considered. So in Case 3, the maximum is -9 at (-3,1).Case 4: 3x +2y ≥-7, x ≤-1, y ≤1.Maximize 2x -3y here. To maximize 2x -3y, we want to maximize x and minimize y. However, x is constrained to be ≤-1, and y is ≤1. The lower bound for y? Let's check.From the original inequality in Case 4: 3x +2y ≥-7. So 2y ≥-7 -3x → y ≥ (-7 -3x)/2.But x ≤-1, so -3x ≥3. Therefore, (-7 -3x)/2 ≥ (-7 +3)/2 = (-4)/2 = -2. So y ≥-2.Thus, in Case 4, the constraints are x ≤-1, y ≤1, y ≥(-7 -3x)/2, and y ≥-2.So to maximize 2x -3y, we need to maximize x (as close to -1 as possible) and minimize y (as close to -2 as possible). Let's check if the point (-1,-2) is in this case. x=-1 ≤-1, y=-2 ≤1. Check 3x +2y=3*(-1) +2*(-2)=-3 -4=-7 ≥-7. So yes, it's on the boundary. So (-1,-2) is in Case 4. 2x -3y at (-1,-2) is 4.Alternatively, let's check if there's another point. Suppose we take x=-1, y as low as possible. The lower bound for y when x=-1 is y ≥ (-7 -3*(-1))/2= (-7 +3)/2= (-4)/2=-2. So y cannot be lower than -2. So (-1,-2) is the point where both x is maximized (at -1) and y is minimized (at -2). Therefore, the maximum in Case 4 is 4 at (-1,-2).Alternatively, if we move x lower than -1, say x=-2. Then the lower bound for y is (-7 -3*(-2))/2= (-7 +6)/2= (-1)/2= -0.5. So y must be ≥-0.5. So if x=-2, y can be as low as -0.5. Then 2x -3y=2*(-2) -3*(-0.5)=-4 +1.5=-2.5, which is less than 4. Similarly, taking x=-3, y ≥ (-7 -3*(-3))/2= (-7 +9)/2=2/2=1. But y ≤1, so y=1. Then 2x -3y=2*(-3) -3*1=-6 -3=-9, which is worse. So indeed, the maximum in Case 4 is at (-1,-2).Therefore, across all four cases, the maximum of 2x -3y is 4 at (-1,-2), which is one of the vertices of the feasible region. Therefore, the answer should be 4.But wait, just to be thorough, let's check if there's any other point on the edges that might give a higher value. Let's take an edge, for example, between (-1,-2) and (1,1). Let's parametrize this edge. Since it's a diamond edge, it should be a straight line.The edge between (-1,-2) and (1,1). Let's parametrize this line. The direction vector is (1 - (-1), 1 - (-2)) = (2,3). So parametric equations:x = -1 + 2ty = -2 + 3twhere t ranges from 0 to1.Then, substituting into 2x -3y:2*(-1 + 2t) -3*(-2 + 3t) = (-2 +4t) - (-6 +9t) = -2 +4t +6 -9t = 4 -5tSo as t goes from 0 to1, the expression 4 -5t decreases from 4 to 4 -5(1)= -1. Therefore, the maximum on this edge is 4 at t=0, which is (-1,-2). Similarly, other edges can be checked, but since the maximum occurs at a vertex, and we already checked all vertices, there is no need. Therefore, the maximum is indeed 4.So I think I've confirmed through multiple approaches: analyzing vertices, splitting into cases, and checking edges, that the maximum value is 4 at (-1,-2). Therefore, the answer should be boxed{4}.Final Answerboxed{4}</think>Given the inequality ( 3|x+1| + 2|y-1| leq 6 ), we need to find the maximum value of ( 2x - 3y ).First, we rewrite the inequality in a normalized form:[frac{|x + 1|}{2} + frac{|y - 1|}{3} leq 1]This represents a convex polygon (diamond) centered at ((-1, 1)) with vertices at ((1, 1)), ((-3, 1)), ((-1, 4)), and ((-1, -2)).To find the maximum value of the linear function ( 2x - 3y ) over this convex polygon, we evaluate the function at each vertex:- At ((1, 1)): ( 2(1) - 3(1) = 2 - 3 = -1 )- At ((-3, 1)): ( 2(-3) - 3(1) = -6 - 3 = -9 )- At ((-1, 4)): ( 2(-1) - 3(4) = -2 - 12 = -14 )- At ((-1, -2)): ( 2(-1) - 3(-2) = -2 + 6 = 4 )Among these values, the maximum value is ( 4 ) at the vertex ((-1, -2)).We also verified by considering the constraints in different quadrants and parametrizing the edges, confirming that the maximum value occurs at ((-1, -2)).Thus, the maximum value of ( 2x - 3y ) is (boxed{4}).]

🔑:10.4.The figure determined by 3|x+1|+2|y-1| leqslant 6 is the quadrilateral ABCD and its interior, where,A(-1,4) 、 B(1,1) 、 C(-1,-2) 、 D(-3,1) text {. }By the knowledge of linear programming, the maximum value of 2x-3y is 4.The maximum value can be achieved when x=-1, y=-2.

🧠:[<think>Okay, let me see. I need to find the range of possible values for the real number ( a ) given that ( a + b + c = 0 ) and ( a^2 + b^2 + c^2 = 1 ). Hmm. Alright, so there are three variables here, but they have some constraints. Let me think about how to approach this. Maybe I can express ( b ) and ( c ) in terms of ( a ), then substitute into the second equation?Since ( a + b + c = 0 ), that means ( b + c = -a ). If I can find another equation involving ( b ) and ( c ), maybe I can relate them. The second equation is ( a^2 + b^2 + c^2 = 1 ). So if I substitute ( b + c = -a ), maybe I can use that to simplify ( b^2 + c^2 ).Wait, recall that ( (b + c)^2 = b^2 + c^2 + 2bc ). So if I have ( b + c = -a ), then squaring both sides gives ( (b + c)^2 = a^2 ), so ( b^2 + c^2 + 2bc = a^2 ). But from the second equation, ( a^2 + b^2 + c^2 = 1 ), which implies ( b^2 + c^2 = 1 - a^2 ). Substituting that into the previous equation: ( (1 - a^2) + 2bc = a^2 ). Therefore, ( 2bc = a^2 - (1 - a^2) = 2a^2 - 1 ), so ( bc = a^2 - frac{1}{2} ).Hmm, so now I have ( b + c = -a ) and ( bc = a^2 - frac{1}{2} ). That looks like a quadratic equation whose roots are ( b ) and ( c ). Let me write that down: ( x^2 + a x + (a^2 - frac{1}{2}) = 0 ). For real numbers ( b ) and ( c ) to exist, the discriminant of this quadratic must be non-negative. The discriminant ( D ) is given by ( D = a^2 - 4 times 1 times (a^2 - frac{1}{2}) ).Calculating the discriminant: ( D = a^2 - 4a^2 + 2 = -3a^2 + 2 ). For the quadratic to have real roots, ( D geq 0 ), so ( -3a^2 + 2 geq 0 ). Solving this inequality: ( -3a^2 geq -2 ), multiplying both sides by -1 (which reverses the inequality): ( 3a^2 leq 2 ), so ( a^2 leq frac{2}{3} ). Taking square roots: ( |a| leq sqrt{frac{2}{3}} ), which simplifies to ( |a| leq frac{sqrt{6}}{3} ). Therefore, ( a ) must lie in the interval ( left[ -frac{sqrt{6}}{3}, frac{sqrt{6}}{3} right] ).Wait, but let me check if this makes sense. Let me verify with some example values. Suppose ( a = frac{sqrt{6}}{3} ). Then ( b + c = -frac{sqrt{6}}{3} ). Then ( b^2 + c^2 = 1 - left( frac{sqrt{6}}{3} right)^2 = 1 - frac{6}{9} = 1 - frac{2}{3} = frac{1}{3} ). Also, ( bc = left( frac{sqrt{6}}{3} right)^2 - frac{1}{2} = frac{6}{9} - frac{1}{2} = frac{2}{3} - frac{1}{2} = frac{4}{6} - frac{3}{6} = frac{1}{6} ).So the quadratic equation would be ( x^2 + frac{sqrt{6}}{3} x + frac{1}{6} = 0 ). Let's check the discriminant here: ( left( frac{sqrt{6}}{3} right)^2 - 4 times 1 times frac{1}{6} = frac{6}{9} - frac{4}{6} = frac{2}{3} - frac{2}{3} = 0 ). So discriminant is zero, which means ( b = c ), and they are real numbers. Therefore, this works.Similarly, if ( a = 0 ), then ( b + c = 0 ), and ( b^2 + c^2 = 1 ). Then ( bc = 0 - frac{1}{2} = -frac{1}{2} ). So quadratic equation is ( x^2 + 0x - frac{1}{2} = x^2 - frac{1}{2} = 0 ), which has solutions ( x = pm frac{sqrt{2}}{2} ). Therefore, ( b = frac{sqrt{2}}{2} ), ( c = -frac{sqrt{2}}{2} ) or vice versa. Then ( a^2 + b^2 + c^2 = 0 + frac{1}{2} + frac{1}{2} = 1 ), which is correct.Another test: If ( a = frac{sqrt{6}}{3} approx 0.816 ), then ( b + c approx -0.816 ), and ( bc = frac{1}{6} approx 0.166 ). Then the quadratic equation ( x^2 + 0.816x + 0.166 = 0 ). The roots would be ( x = frac{ -0.816 pm sqrt{0.816^2 - 4 times 0.166} }{2} ). Let's compute the discriminant: ( 0.816^2 - 0.664 approx 0.666 - 0.664 = 0.002 ). Close to zero, as expected. So roots are approximately ( frac{ -0.816 }{2} approx -0.408 ). So ( b ) and ( c ) are both approximately -0.408, which sum to -0.816, and their product is approximately 0.166. That checks out.What if ( a ) is outside of this range? Let's say ( a = 1 ). Then ( b + c = -1 ), and ( b^2 + c^2 = 1 - 1 = 0 ). Therefore, ( b = c = 0 ), but then ( b + c = 0 neq -1 ). Contradiction. So no solution exists for ( a = 1 ). Similarly, if ( a = sqrt{frac{2}{3}} + epsilon ), then discriminant would be negative. Therefore, the range must indeed be ( left[ -sqrt{frac{2}{3}}, sqrt{frac{2}{3}} right] ).Alternatively, maybe there's another method using vectors or geometry? Let me think. The conditions ( a + b + c = 0 ) and ( a^2 + b^2 + c^2 = 1 ) can be interpreted as the vector ( (a, b, c) ) lying on the intersection of a plane through the origin and the unit sphere. The intersection should be a circle, and we need to find the range of the first coordinate ( a ) on this circle.The range of ( a ) would be from the minimum to the maximum value of ( a ) on that circle. Since the circle is symmetric, the maximum and minimum values of ( a ) should be negatives of each other. The distance from the origin to the plane ( a + b + c = 0 ) is ( frac{|0|}{sqrt{1^2 + 1^2 + 1^2}}} = 0 ), so the plane passes through the origin. The radius of the circle is then ( sqrt{1 - 0} = 1 ). Wait, but the radius of the intersection of a sphere with radius ( r ) and a plane at distance ( d ) from the center is ( sqrt{r^2 - d^2} ). Here, ( d = 0 ), so the radius is 1, meaning the intersection is a great circle. But how does this help with finding the maximum ( a )?Alternatively, using Lagrange multipliers. Let me set up the optimization problem: maximize (or minimize) ( a ) subject to constraints ( a + b + c = 0 ) and ( a^2 + b^2 + c^2 = 1 ).Let’s use Lagrange multipliers. Let me consider the function ( f(a, b, c) = a ) to maximize, with constraints ( g(a, b, c) = a + b + c = 0 ) and ( h(a, b, c) = a^2 + b^2 + c^2 - 1 = 0 ).The method of Lagrange multipliers with two constraints says that the gradient of ( f ) must be a linear combination of the gradients of ( g ) and ( h ). So:[nabla f = lambda nabla g + mu nabla h]Calculating the gradients:( nabla f = (1, 0, 0) )( nabla g = (1, 1, 1) )( nabla h = (2a, 2b, 2c) )Therefore:1. ( 1 = lambda cdot 1 + mu cdot 2a )2. ( 0 = lambda cdot 1 + mu cdot 2b )3. ( 0 = lambda cdot 1 + mu cdot 2c )From equations 2 and 3:( lambda + 2mu b = 0 )( lambda + 2mu c = 0 )Subtracting these two equations: ( 2mu (b - c) = 0 ). So either ( mu = 0 ) or ( b = c ).If ( mu = 0 ), then from equation 2: ( lambda = 0 ), but from equation 1: ( 1 = 0 + 0 ), which is impossible. So ( mu neq 0 ), hence ( b = c ).Similarly, from equation 2 and 3, since ( b = c ), then.Now, from the constraints:1. ( a + 2b = 0 ) (since ( b = c ))2. ( a^2 + 2b^2 = 1 )From the first equation, ( b = -a/2 ). Substitute into the second equation:( a^2 + 2(-a/2)^2 = 1 )( a^2 + 2(a^2/4) = 1 )( a^2 + a^2/2 = 1 )( (3/2)a^2 = 1 )( a^2 = 2/3 )Therefore, ( a = pm sqrt{2/3} ). So the maximum value of ( a ) is ( sqrt{2/3} ) and the minimum is ( -sqrt{2/3} ). Therefore, the range of ( a ) is ( [ -sqrt{6}/3, sqrt{6}/3 ] ), since ( sqrt{2/3} = sqrt{6}/3 ).This matches the previous result. Therefore, the answer should be ( boxed{left[ -dfrac{sqrt{6}}{3}, dfrac{sqrt{6}}{3} right]} ).Wait, just to confirm, sqrt(2/3) is equal to sqrt(6)/3. Let me compute:sqrt(2/3) = sqrt(2)/sqrt(3) = (sqrt(2)*sqrt(3))/3 = sqrt(6)/3. Yes, correct.Therefore, the range is from -sqrt(6)/3 to sqrt(6)/3. So the answer is boxed{left[ -dfrac{sqrt{6}}{3}, dfrac{sqrt{6}}{3} right]}.Final AnswerThe range of values for ( a ) is boxed{left[ -dfrac{sqrt{6}}{3}, dfrac{sqrt{6}}{3} right]}.</think>Given real numbers (a), (b), and (c) satisfying (a + b + c = 0) and (a^2 + b^2 + c^2 = 1), we need to find the range of values for (a).First, we use the condition (a + b + c = 0) to express (b + c) in terms of (a):[b + c = -a]Next, we substitute this into the second condition (a^2 + b^2 + c^2 = 1). We know that:[(b + c)^2 = b^2 + c^2 + 2bc]Substituting (b + c = -a) into the equation above, we get:[a^2 = b^2 + c^2 + 2bc]From the second condition (a^2 + b^2 + c^2 = 1), we can express (b^2 + c^2) as:[b^2 + c^2 = 1 - a^2]Substituting this into the previous equation, we get:[a^2 = (1 - a^2) + 2bc]Solving for (bc), we find:[2bc = a^2 - (1 - a^2) implies bc = a^2 - frac{1}{2}]We now consider the quadratic equation with roots (b) and (c):[x^2 + ax + left(a^2 - frac{1}{2}right) = 0]For (b) and (c) to be real, the discriminant of this quadratic must be non-negative:[a^2 - 4left(a^2 - frac{1}{2}right) geq 0]Simplifying the discriminant, we get:[a^2 - 4a^2 + 2 geq 0 implies -3a^2 + 2 geq 0 implies 3a^2 leq 2 implies a^2 leq frac{2}{3}]Taking the square root of both sides, we find:[|a| leq sqrt{frac{2}{3}} = frac{sqrt{6}}{3}]Thus, the range of values for (a) is:[boxed{left[ -dfrac{sqrt{6}}{3}, dfrac{sqrt{6}}{3} right]}],<think>Okay, so I need to find the range of possible values for the real number ( a ) given that ( a + b + c = 0 ) and ( a^2 + b^2 + c^2 = 1 ). Hmm, let me think. First, since ( a ), ( b ), and ( c ) are real numbers and they add up to zero, maybe I can express ( b ) and ( c ) in terms of ( a ). That way, I can substitute them into the second equation and maybe find some inequalities or constraints on ( a ). Let me try that.Starting with the first equation: ( a + b + c = 0 ). If I solve for ( c ), I get ( c = -a - b ). Now, substitute this expression for ( c ) into the second equation ( a^2 + b^2 + c^2 = 1 ). Substituting ( c ):( a^2 + b^2 + (-a - b)^2 = 1 )Let me expand that:( a^2 + b^2 + (a^2 + 2ab + b^2) = 1 )Combine like terms:( a^2 + b^2 + a^2 + 2ab + b^2 = 1 )So that's:( 2a^2 + 2b^2 + 2ab = 1 )I can divide both sides by 2 to simplify:( a^2 + b^2 + ab = frac{1}{2} )Hmm, okay. Now, I need to find the range of ( a ). Maybe I can treat this as a quadratic equation in terms of ( b ), and then ensure that the discriminant is non-negative because ( b ) has to be a real number. Let's try that.Let me rearrange the equation:( b^2 + ab + (a^2 - frac{1}{2}) = 0 )Yes, this is a quadratic equation in ( b ). The standard form is ( b^2 + (a)b + (a^2 - 1/2) = 0 ). For this equation to have real solutions for ( b ), the discriminant must be non-negative. The discriminant ( D ) of a quadratic equation ( x^2 + px + q = 0 ) is ( D = p^2 - 4q ).So here, ( p = a ) and ( q = a^2 - 1/2 ). Therefore:( D = a^2 - 4(a^2 - 1/2) = a^2 - 4a^2 + 2 = -3a^2 + 2 )For real solutions, ( D geq 0 ):( -3a^2 + 2 geq 0 )Multiply both sides by -1 (which reverses the inequality):( 3a^2 - 2 leq 0 )( 3a^2 leq 2 )( a^2 leq frac{2}{3} )Taking square roots:( |a| leq sqrt{frac{2}{3}} )Simplify ( sqrt{frac{2}{3}} ):( sqrt{frac{2}{3}} = frac{sqrt{6}}{3} )So, ( |a| leq frac{sqrt{6}}{3} ), which means ( a ) is in the interval ( left[ -frac{sqrt{6}}{3}, frac{sqrt{6}}{3} right] ).Wait, is that correct? Let me check my steps again.1. Started with ( a + b + c = 0 ) so ( c = -a - b ).2. Substituted into ( a^2 + b^2 + c^2 = 1 ) and simplified to ( 2a^2 + 2b^2 + 2ab = 1 ), then divided by 2 to get ( a^2 + b^2 + ab = 1/2 ).3. Rewrote as a quadratic in ( b ): ( b^2 + ab + (a^2 - 1/2) = 0 ).4. Calculated discriminant: ( D = a^2 - 4(a^2 - 1/2) = a^2 - 4a^2 + 2 = -3a^2 + 2 ).5. Set discriminant >=0: ( -3a^2 + 2 >=0 implies 3a^2 <=2 implies |a| <= sqrt(2/3) = sqrt6/3 ).Seems right. But maybe there's another way to verify this. Let me try using vectors or another method.Alternatively, since ( a + b + c = 0 ), the vector (a, b, c) is orthogonal to the vector (1, 1, 1). The condition ( a^2 + b^2 + c^2 =1 ) is the unit sphere. The range of ( a ) would be the projection of this intersection onto the a-axis. Hmm, maybe using Lagrange multipliers?But perhaps that's overcomplicating. Alternatively, since ( a + b + c =0 ), we can use the formula for the variance. Let me see.If ( a + b + c =0 ), then the average of a, b, c is 0. Then the variance is ( frac{a^2 + b^2 + c^2}{3} = frac{1}{3} ). So each variable has variance 1/3. Therefore, by the Cauchy-Schwarz inequality, the maximum value of |a| would be related to the standard deviation?Wait, but variance is 1/3, so standard deviation is sqrt(1/3). But that would suggest |a| <= sqrt(1/3), which is different from sqrt(2/3). Hmm, so that contradicts. Which is right?Wait, maybe the variance approach isn't directly applicable here. Let me think again.If variables have a variance of 1/3, that's the average of the squares. But individual variables can have higher values. For example, if two variables are equal and the third is adjusted to sum to zero, maybe you can get a higher individual value.Wait, but the maximum possible |a| would occur when b = c, perhaps? Let's test that.Suppose b = c. Then since a + 2b =0, so b = -a/2. Then substitute into the second equation:( a^2 + 2b^2 =1 )Substitute b:( a^2 + 2*(a^2/4) =1 )Which is ( a^2 + (a^2/2) =1 implies (3/2)a^2 =1 implies a^2 = 2/3 implies a = ±sqrt(6)/3 ≈ ±0.816 )So that gives the same result as before. So when b and c are equal, we get the maximum |a|. So that confirms the maximum value of |a| is sqrt(6)/3.Therefore, the range of a is from -sqrt(6)/3 to sqrt(6)/3.Alternatively, maybe to check, let's take specific examples.Suppose a = sqrt(6)/3. Then, as above, b = c = -a/2 = -sqrt(6)/6. Then check the sum of squares:a^2 + 2b^2 = (6/9) + 2*(6/36) = (2/3) + 2*(1/6) = 2/3 + 1/3 =1. Correct.Similarly, if a =0, then b +c =0, and b^2 + c^2 =1. Then, since c = -b, we have 2b^2 =1 => b = ±sqrt(1/2). So that works.If a = sqrt(6)/3, maximum, then as above, works. What if a is larger than sqrt(6)/3? Let's say a=1. Then from a +b +c =0, b +c = -1. From a^2 + b^2 +c^2 =1, 1 + b^2 +c^2 =1 => b^2 +c^2=0 => b=c=0. But then b +c =0 ≠ -1. Contradiction. So a cannot be 1. Similarly, if a= sqrt(2)/2 ≈0.707, which is less than sqrt(6)/3≈0.816. Let's check. If a= sqrt(2)/2, then we have to find b and c such that b +c = -sqrt(2)/2 and b^2 +c^2 =1 - (sqrt(2)/2)^2 =1 -1/2=1/2. So, similar to earlier, let b and c be variables such that their sum is -sqrt(2)/2 and sum of squares is 1/2. Let me compute (b +c)^2 = (sqrt(2)/2)^2 = 2/4=1/2. Then, (b +c)^2 = b^2 +c^2 +2bc =1/2 +2bc =1/2. So 2bc=0. Then bc=0. So either b=0 or c=0. Suppose b=0, then c= -sqrt(2)/2. Check sum of squares: 0 + (sqrt(2)/2)^2 =0.5, which matches. So in this case, possible. So a= sqrt(2)/2 is possible, but according to our previous result, maximum is sqrt(6)/3≈0.816, which is larger. So sqrt(2)/2≈0.707 is within the range. So that works.Alternatively, if we try a =1, as before, it's impossible. Therefore, the maximum is indeed sqrt(6)/3.Therefore, the range is from -sqrt(6)/3 to sqrt(6)/3.Hence, the answer is boxed{left[ -dfrac{sqrt{6}}{3}, dfrac{sqrt{6}}{3} right]}.Final AnswerThe range of values for ( a ) is boxed{left[ -dfrac{sqrt{6}}{3}, dfrac{sqrt{6}}{3} right]}.</think>Given real numbers (a), (b), and (c) satisfying (a + b + c = 0) and (a^2 + b^2 + c^2 = 1), we need to determine the range of values for (a).Starting with the equation (a + b + c = 0), we express (c) as (c = -a - b). Substituting this into the second equation (a^2 + b^2 + c^2 = 1), we get:[a^2 + b^2 + (-a - b)^2 = 1]Expanding and simplifying:[a^2 + b^2 + a^2 + 2ab + b^2 = 1 implies 2a^2 + 2b^2 + 2ab = 1 implies a^2 + b^2 + ab = frac{1}{2}]This is a quadratic equation in (b):[b^2 + ab + left(a^2 - frac{1}{2}right) = 0]For (b) to be a real number, the discriminant of this quadratic equation must be non-negative:[a^2 - 4left(a^2 - frac{1}{2}right) geq 0 implies a^2 - 4a^2 + 2 geq 0 implies -3a^2 + 2 geq 0 implies 3a^2 leq 2 implies a^2 leq frac{2}{3}]Taking the square root of both sides, we find:[|a| leq sqrt{frac{2}{3}} = frac{sqrt{6}}{3}]Thus, the range of values for (a) is:[boxed{left[ -dfrac{sqrt{6}}{3}, dfrac{sqrt{6}}{3} right]}]]

🔑:From a+b+c=0, we have c=-a-b. Substituting into a^2+b^2+c^2=1, we get a^2+b^2+(a+b)^2=1, which simplifies to 2b^2+2ab+2a^2=1. Since b is a real number, the discriminant Delta=4a^2-8(2a^2-1) geq 0, solving this gives -frac{sqrt{6}}{3} leq a leq frac{sqrt{6}}{3}. Therefore, the answer is -frac{sqrt{6}}{3} leq a leq frac{sqrt{6}}{3}. By substituting c=-a-b into a^2+b^2+c^2=1, we obtain a^2+b^2+(a+b)^2=1, which simplifies to 2b^2+2ab+2a^2=1. This equation has real roots, which implies Delta geq 0. This problem tests the relationship between the discriminant of a quadratic equation having real roots and reasoning ability, and is considered a medium-level question.Thus, the range of values for a is boxed{-frac{sqrt{6}}{3} leq a leq frac{sqrt{6}}{3}}.

🧠:[<think>Okay, so I need to find the standard deviation of a new dataset where each value is twice the original. The original data has a variance of 2. Hmm, let me think. I remember that variance and standard deviation are related. The standard deviation is the square root of the variance. But how does scaling the data affect these measures?Let me recall the properties of variance. If you multiply each data point by a constant, the variance gets multiplied by the square of that constant. So, if the original variance is σ², then multiplying each x by 2 would make the new variance (2²)σ², which is 4σ². Then, the standard deviation would be the square root of the new variance. Wait, let me verify this. Suppose we have data points x₁, x₂, ..., xₙ with mean μ. The variance is calculated as the average of the squared differences from the mean: Var(X) = (1/n)Σ(xᵢ - μ)². If we create a new dataset where each value is 2xᵢ, then the new mean would be 2μ. The new variance would be (1/n)Σ(2xᵢ - 2μ)². Let's compute that: (1/n)Σ[2(xᵢ - μ)]² = (1/n)Σ4(xᵢ - μ)² = 4*(1/n)Σ(xᵢ - μ)² = 4*Var(X). So, yes, the variance becomes 4 times the original variance.Therefore, if the original variance is 2, the new variance is 4*2 = 8. Then, the standard deviation is the square root of the variance. So sqrt(8). Hmm, sqrt(8) can be simplified as 2*sqrt(2). Let me check that: sqrt(8) = sqrt(4*2) = sqrt(4)*sqrt(2) = 2*sqrt(2). Yep, that's correct.But just to make sure I didn't mix up standard deviations. If instead, the question had asked about adding a constant, that wouldn't affect the variance or standard deviation, right? Because adding a constant shifts all the data points but doesn't change their spread. But multiplying by a constant scales the spread, hence affects variance quadratically and standard deviation linearly. So, if you multiply by 2, the standard deviation becomes twice as much. Wait a second, variance scales by 4, so standard deviation scales by 2, since it's the square root. So original standard deviation is sqrt(2), then after scaling, standard deviation is 2*sqrt(2). But the problem didn't give the original standard deviation, it gave the original variance. So the original standard deviation would be sqrt(2), and the new standard deviation would be 2*sqrt(2). Alternatively, since variance becomes 8, standard deviation is sqrt(8) which is 2*sqrt(2). So either way, the answer should be 2√2. Wait, let's just recap:Original variance (σ²) = 2.After scaling by 2, new variance = 2² * σ² = 4*2 = 8.Therefore, new standard deviation = sqrt(new variance) = sqrt(8) = 2√2.Yes, that seems right. So the key steps are knowing that scaling the data by a factor 'a' scales the variance by a², hence standard deviation scales by |a|. Since 2 is positive, standard deviation scales by 2. So original standard deviation was sqrt(2), new one is 2*sqrt(2). Therefore, the answer is 2√2.I think that's solid. Let me just check with an example. Suppose the original data was simple, like 0, 0, 0, 0, 0. Then variance is 0. If I multiply by 2, variance remains 0. That edge case works. Another example: suppose data is 1, -1, 1, -1, 1. The mean is (1 -1 +1 -1 +1)/5 = (1)/5 = 0.2. Wait, that's not a good example because the variance calculation would be messy. Let me pick data points that have a known variance.Alternatively, take data points 1, 2, 3, 4, 5. Compute the variance. Then multiply each by 2 and compute the variance to check.Original data: 1, 2, 3, 4, 5.Mean = (1+2+3+4+5)/5 = 15/5 = 3.Variance = [(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2]/5 = [4 + 1 + 0 + 1 + 4]/5 = 10/5 = 2.Oh, that's convenient. So original variance is 2. Then, scaling by 2: data becomes 2, 4, 6, 8, 10.New mean: (2+4+6+8+10)/5 = 30/5 = 6.New variance: [(2-6)^2 + (4-6)^2 + (6-6)^2 + (8-6)^2 + (10-6)^2]/5 = [16 + 4 + 0 + 4 + 16]/5 = 40/5 = 8.So variance is 8, standard deviation is sqrt(8) = 2√2. Exactly matches the calculation. So this example confirms that when the original variance is 2, after scaling by 2, the variance becomes 8 and standard deviation becomes 2√2. Therefore, the answer should be 2√2.Final AnswerThe standard deviation of the new data is boxed{2sqrt{2}}.</think>Given a set of data ( x_1, x_2, x_3, x_4, x_5 ) with a variance of 2, we need to find the standard deviation of the data ( 2x_1, 2x_2, 2x_3, 2x_4, 2x_5 ).1. Understanding Variance Scaling: When each data point is multiplied by a constant ( a ), the variance is multiplied by ( a^2 ). For the given problem, the constant is 2. Therefore, the new variance is ( 2^2 times text{original variance} ).2. Calculating New Variance: The original variance is 2. Multiplying by ( 2^2 ) gives the new variance as ( 4 times 2 = 8 ).3. Finding Standard Deviation: The standard deviation is the square root of the variance. Thus, the standard deviation of the new data is ( sqrt{8} = 2sqrt{2} ). Final AnswerThe standard deviation of the new data is boxed{2sqrt{2}}.,<think>Okay, so I need to find the standard deviation of a new data set where each value is twice the original. The original data has a variance of 2. Hmm, let me recall. Variance and standard deviation are related, right? The standard deviation is the square root of the variance. But here, they're asking about the effect of multiplying each data point by 2. I think when you multiply each data point by a constant, the variance gets multiplied by the square of that constant. Wait, let me think. If you have a dataset and you scale it by a factor, say 'a', then the variance becomes a² times the original variance. So in this case, since we're multiplying each x by 2, the variance should be 2² * original variance. The original variance is 2, so the new variance would be 4 * 2 = 8. Then, the standard deviation is the square root of the variance, so sqrt(8). But sqrt(8) simplifies to 2*sqrt(2), right? So the standard deviation is 2√2.Wait, let me check this step by step. Let's start from the definitions. Variance is the average of the squared deviations from the mean. So if each data point is doubled, what happens to the mean? The mean would also double, because if the original mean was μ, then the new mean is (2x₁ + 2x₂ + ... + 2x₅)/5 = 2*(x₁ + x₂ + ... + x₅)/5 = 2μ. Now, the deviations from the mean. Each original deviation was (x_i - μ). The new deviation is (2x_i - 2μ) = 2(x_i - μ). So each deviation is multiplied by 2. Then, when we square those deviations, we get [2(x_i - μ)]² = 4(x_i - μ)². Therefore, the squared deviations are 4 times the original squared deviations. Taking the average of those squared deviations (which is the variance), the new variance should be 4 times the original variance. So 4*2=8. Then the standard deviation is sqrt(8) = 2*sqrt(2). Yes, that makes sense. So scaling the data by a factor scales the standard deviation by the absolute value of that factor. Since we scaled by 2, the standard deviation also scales by 2. The original standard deviation would have been sqrt(2), so after scaling, it's 2*sqrt(2). Therefore, the answer should be 2√2.But just to be thorough, let me consider an example. Suppose we have a simple data set: 1, 2, 3, 4, 5. Let's compute its variance and then double each term and compute the variance again.Original data: 1, 2, 3, 4, 5. The mean is (1+2+3+4+5)/5 = 15/5 = 3. The deviations from the mean are -2, -1, 0, 1, 2. The squared deviations are 4, 1, 0, 1, 4. The average of those squared deviations is (4+1+0+1+4)/5 = 10/5 = 2. So the variance is 2, matching the given problem. The standard deviation is sqrt(2).Now, double each term: 2, 4, 6, 8, 10. The new mean is (2+4+6+8+10)/5 = 30/5 = 6. The deviations from the mean are -4, -2, 0, 2, 4. The squared deviations are 16, 4, 0, 4, 16. The average is (16+4+0+4+16)/5 = 40/5 = 8. So the variance is 8, which is 4 times the original variance. The standard deviation is sqrt(8) = 2*sqrt(2), which is 2 times the original standard deviation. This concrete example works out exactly as my reasoning before. So that confirms that multiplying each data point by 2 multiplies the variance by 4 (2 squared) and the standard deviation by 2. Therefore, in the problem given, since the original variance is 2, the new variance is 8, and the standard deviation is sqrt(8) which is 2√2. Therefore, the answer should be 2 times the square root of 2. Final AnswerThe standard deviation of the new data set is boxed{2sqrt{2}}.</think>Given a set of data ( x_1, x_2, x_3, x_4, x_5 ) with a variance of 2, we need to find the standard deviation of the data ( 2x_1, 2x_2, 2x_3, 2x_4, 2x_5 ).1. Variance Scaling: When each data point is multiplied by a constant ( a ), the variance of the new data set is ( a^2 ) times the original variance. Here, the constant is 2. [ text{New Variance} = 2^2 times text{Original Variance} = 4 times 2 = 8 ]2. Standard Deviation Calculation: The standard deviation is the square root of the variance. [ text{New Standard Deviation} = sqrt{text{New Variance}} = sqrt{8} = 2sqrt{2} ]Thus, the standard deviation of the new data set is (boxed{2sqrt{2}}).]

🔑:Since the variance of the data x_{1}, x_{2}, x_{3}, x_{4}, x_{5} is 2, the variance of the data 2x_{1}, 2x_{2}, 2x_{3}, 2x_{4}, 2x_{5} is 2^{2} cdot 2 = 8, and its standard deviation is: 2 sqrt{2}, thus, the answer is boxed{2 sqrt{2}}. By applying the formula for variance to the data 2x_{1}, 2x_{2}, 2x_{3}, 2x_{4}, 2x_{5}, we can find the variance and thus calculate the standard deviation. This question tests the definition of variance and the method of calculating variance and standard deviation, making it a basic question.

🧠:[<think>Okay, so I have this triangle ABC, and the problem is giving me a relationship between the tangents of angles A and B and the sides opposite to them. Specifically, it says that (tan A)/(tan B) = (2c - b)/b. Then part (1) asks for the measure of angle A, and part (2) gives another equation involving sin(B + C) and wants the value of b/c. Let me tackle part (1) first.First, I remember that in any triangle, the Law of Sines relates the sides and angles: a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. Maybe that could be helpful here. Also, the Law of Cosines might come into play, but let's see.Given the equation (tan A)/(tan B) = (2c - b)/b. Hmm. Let's recall that tan A = sin A / cos A and similarly for tan B. So maybe we can express tan A / tan B as (sin A / cos A) / (sin B / cos B) = (sin A / sin B) * (cos B / cos A). From the Law of Sines, sin A / sin B = a / b. So substituting that in, we get (a / b) * (cos B / cos A) = (2c - b)/b. Then the b in the denominator cancels out on both sides: (a) * (cos B / cos A) = 2c - b. So that gives us a * (cos B / cos A) = 2c - b. Hmm. Let's write that down:a * (cos B / cos A) = 2c - b. Now, maybe we can use the Law of Cosines here. The Law of Cosines says that a² = b² + c² - 2bc cos A. Similarly, cos B = (a² + c² - b²)/(2ac). Let me substitute that expression for cos B into the equation.So cos B = (a² + c² - b²)/(2ac), so substituting into the left side:a * [(a² + c² - b²)/(2ac)] / cos A = 2c - b.Simplify numerator: a * (a² + c² - b²) / (2ac cos A) = 2c - b.The a cancels with the denominator's a: (a² + c² - b²)/(2c cos A) = 2c - b.Multiply both sides by 2c cos A:a² + c² - b² = 2c(2c - b) cos A.Hmm, okay. Let's note that. Now, maybe we can relate this to the Law of Cosines. From the Law of Cosines, we know that:cos A = (b² + c² - a²)/(2bc). Let's substitute this into the equation above.So replacing cos A with (b² + c² - a²)/(2bc):a² + c² - b² = 2c(2c - b) * [ (b² + c² - a²)/(2bc) ]Simplify the right-hand side:First, the 2c cancels with the denominator's 2c, so we get:(2c - b) * (b² + c² - a²)/bSo the equation becomes:a² + c² - b² = (2c - b)(b² + c² - a²)/bLet me multiply both sides by b to eliminate the denominator:b(a² + c² - b²) = (2c - b)(b² + c² - a²)Now, let's expand both sides.Left side: b(a² + c² - b²) = ba² + bc² - b³Right side: (2c - b)(b² + c² - a²) = 2c(b² + c² - a²) - b(b² + c² - a²)= 2c b² + 2c³ - 2c a² - b³ - b c² + b a²Now, let's collect like terms:Left side: ba² + bc² - b³Right side: 2c b² + 2c³ - 2c a² - b³ - b c² + b a²= (2c b² - b c²) + (2c³) + (-2c a² + b a²) - b³= (2b² c - b² c) + 2c³ + a²(-2c + b) - b³Wait, maybe better to just list all terms:2c b², 2c³, -2c a², -b³, -b c², +b a²So, group the a² terms: (-2c a² + b a²) = a²(b - 2c)Group the b² terms: 2c b²Group the c² terms: 2c³ - b c² = c²(2c - b)Then the -b³ term.So altogether:Right side = a²(b - 2c) + 2c b² + c²(2c - b) - b³Now, let's write the equation again:Left side: ba² + bc² - b³Right side: a²(b - 2c) + 2c b² + c²(2c - b) - b³So set left side = right side:ba² + bc² - b³ = a²(b - 2c) + 2c b² + c²(2c - b) - b³First, notice that -b³ appears on both sides, so they cancel out. So we can remove them from both sides.So:ba² + bc² = a²(b - 2c) + 2c b² + c²(2c - b)Now, expand the right side:a²(b - 2c) = b a² - 2c a²2c b² = 2b² cc²(2c - b) = 2c³ - b c²So putting all together:Right side = b a² - 2c a² + 2b² c + 2c³ - b c²Therefore, the equation becomes:Left side: ba² + bc²Right side: b a² - 2c a² + 2b² c + 2c³ - b c²Subtract left side from both sides:0 = -2c a² + 2b² c + 2c³ - b c² - bc²Wait, let's do term by term:Right side - left side:(b a² - 2c a² + 2b² c + 2c³ - b c²) - (ba² + bc²) =b a² - 2c a² + 2b² c + 2c³ - b c² - ba² - bc²Simplify:b a² - ba² = 0-2c a² remains2b² c remains2c³ remains- b c² - bc² = -2b c²So overall:-2c a² + 2b² c + 2c³ - 2b c² = 0Factor out 2c:2c(-a² + b² + c² - b c) = 0Since 2c is not zero (c is a side length of a triangle, so c > 0), we have:- a² + b² + c² - b c = 0Thus:b² + c² - a² - b c = 0Therefore:b² + c² - bc = a²So from here, we can recall that in the Law of Cosines, a² = b² + c² - 2bc cos A.But here, we have a² = b² + c² - bc.Comparing the two expressions:b² + c² - 2bc cos A = b² + c² - bcSubtract b² + c² from both sides:-2bc cos A = -bcDivide both sides by -bc (since bc ≠ 0):2 cos A = 1Therefore:cos A = 1/2So angle A is arccos(1/2), which is 60 degrees, or π/3 radians.Hence, the measure of angle A is 60 degrees.Wait, let me verify this step again. Because if a² = b² + c² - bc, and Law of Cosines says a² = b² + c² - 2bc cos A. Therefore:b² + c² - 2bc cos A = b² + c² - bcSubtract b² + c²: -2bc cos A = -bcDivide by -bc: 2 cos A = 1 → cos A = 1/2. Yes, that's correct. So angle A is 60 degrees. Okay, that seems good.So part (1) answer is 60 degrees. Let's just check if there's any alternative approach that might confirm this. Maybe using area or some other trigonometric identities?Alternatively, starting from tan A / tan B = (2c - b)/b.We could express tan A / tan B as (sin A / cos A) / (sin B / cos B) = (sin A cos B)/(sin B cos A).From the Law of Sines, sin A / sin B = a / b. So tan A / tan B = (a/b) * (cos B / cos A).Set this equal to (2c - b)/b. Then, as before, cancels b, get a * cos B / cos A = 2c - b.Then using the Law of Cosines for cos A and cos B.But seems like the same path. So seems consistent. So angle A is 60 degrees.Okay, so part (1) is 60 degrees. Now part (2): If sin(B + C) = 6 cos B sin C, find the value of b/c.First, note that in triangle ABC, angles A + B + C = 180 degrees. Since part (1) gives angle A as 60 degrees, then B + C = 120 degrees. So sin(B + C) is sin(120°) = √3/2. Wait, but in the equation given, sin(B + C) is equal to 6 cos B sin C. But sin(B + C) is a known value here. Wait, but hold on.Wait, actually, the problem says: "If sin(B + C) = 6 cos B sin C, find the value of b/c."But in any triangle, B + C = 180° - A. Since part (1) is independent, is part (2) dependent on part (1)? The problem is structured as (1) and (2), so maybe part (2) is a separate problem under the same triangle ABC with the given condition. Wait, no, actually, the problem statement is: "In triangle ABC, the sides opposite to angles A, B, C are a, b, c respectively, and they satisfy (tan A)/(tan B) = (2c - b)/b. (1) Find the measure of angle A; (2) If sin(B + C) = 6 cos B sin C, find the value of b/c."So both parts are about the same triangle, so angle A is 60 degrees from part (1), so in part (2), B + C = 120 degrees, so sin(B + C) = sin 120° = √3 / 2. Then the equation given is √3 / 2 = 6 cos B sin C. Then we need to find b/c.But maybe we need to connect this with the previous result. Let me check.But the problem says "If sin(B + C) = 6 cos B sin C", which in our case is sin(120°) = 6 cos B sin C. So √3 / 2 = 6 cos B sin C. So we have √3 / 12 = cos B sin C.Alternatively, maybe there's a different approach. Let's note that in triangle ABC, angles B and C are related since they sum to 120°, so C = 120° - B.So let's set C = 120° - B.Then sin C = sin(120° - B) = sin 120° cos B - cos 120° sin B = (√3/2) cos B - (-1/2) sin B = (√3/2) cos B + (1/2) sin B.Then cos B sin C = cos B [ (√3/2) cos B + (1/2) sin B ] = (√3/2) cos² B + (1/2) cos B sin B.So according to the given equation, sin(B + C) = 6 cos B sin C.But sin(B + C) = sin(120°) = √3/2. So:√3/2 = 6 [ (√3/2) cos² B + (1/2) cos B sin B ]Multiply both sides by 2:√3 = 6 [ √3 cos² B + cos B sin B ]Divide both sides by 6:√3 / 6 = √3 cos² B + cos B sin BLet me write this as:√3 cos² B + cos B sin B = √3 / 6Hmm. Maybe we can factor out cos B:cos B (√3 cos B + sin B) = √3 / 6Alternatively, let's divide both sides by √3:cos² B + (1/√3) cos B sin B = 1/6But not sure if this helps. Alternatively, let's use substitution. Let me let θ = B. Then since C = 120° - θ.Our equation is √3 / 2 = 6 cos θ sin(120° - θ)Let me compute sin(120° - θ):sin(120° - θ) = sin 120° cos θ - cos 120° sin θ = (√3/2) cos θ - (-1/2) sin θ = (√3/2) cos θ + (1/2) sin θTherefore, the equation becomes:√3 / 2 = 6 cos θ [ (√3/2) cos θ + (1/2) sin θ ]Multiply the terms inside:6 cos θ [ (√3/2) cos θ + (1/2) sin θ ] = 6*(√3/2) cos² θ + 6*(1/2) cos θ sin θ = 3√3 cos² θ + 3 cos θ sin θSo:√3 / 2 = 3√3 cos² θ + 3 cos θ sin θMultiply both sides by 2:√3 = 6√3 cos² θ + 6 cos θ sin θDivide both sides by 6:√3 / 6 = √3 cos² θ + cos θ sin θLet me rearrange:√3 cos² θ + cos θ sin θ - √3 / 6 = 0Hmm, maybe divide through by cos θ? But not sure. Alternatively, use substitution. Let me let x = cos θ, y = sin θ. Then x² + y² = 1. Also, θ is between 0 and 120°, so x and y are positive?Wait, θ is angle B, which is between 0 and 120°, but since angle A is 60°, angles B and C must each be less than 120°, but could be in different ranges. However, sin θ and cos θ are positive if θ is between 0 and 90°, but θ could be up to 120°, where cos θ would be negative. But if θ is 120°, then C would be 0°, which is impossible, so θ must be less than 120°, but could be greater than 90°.But maybe we can proceed.The equation is:√3 x² + x y - √3 / 6 = 0But since y = sqrt(1 - x²), but that might complicate things. Alternatively, express in terms of tan θ.Let me divide both sides by cos² θ:√3 + (y / x) - √3 / (6 x²) = 0But y / x = tan θ, and 1 / x² = 1 + tan² θ. So:√3 + tan θ - √3 / 6 (1 + tan² θ) = 0Let me denote t = tan θ. Then:√3 + t - (√3 / 6)(1 + t²) = 0Multiply through by 6 to eliminate denominators:6√3 + 6t - √3(1 + t²) = 0Expand:6√3 + 6t - √3 - √3 t² = 0Combine like terms:(6√3 - √3) + 6t - √3 t² = 05√3 + 6t - √3 t² = 0Rearranged:-√3 t² + 6t + 5√3 = 0Multiply both sides by -1:√3 t² - 6t - 5√3 = 0This is a quadratic equation in t. Let's write it as:√3 t² - 6t - 5√3 = 0Let me multiply all terms by √3 to rationalize:(√3 * √3) t² - 6√3 t - 5√3 * √3 = 0Which simplifies to:3 t² - 6√3 t - 15 = 0Divide all terms by 3:t² - 2√3 t - 5 = 0Now, solve for t using quadratic formula:t = [2√3 ± sqrt( (2√3)^2 + 20 ) ] / 2Compute discriminant:(2√3)^2 + 20 = 12 + 20 = 32So sqrt(32) = 4√2Thus:t = [2√3 ± 4√2]/2 = √3 ± 2√2So tan θ = √3 + 2√2 or tan θ = √3 - 2√2But θ is an angle in the triangle, so between 0 and 120°. Let's check if these values are possible.Compute √3 ≈ 1.732, 2√2 ≈ 2.828. So √3 + 2√2 ≈ 4.56, which is a steep tangent, so angle θ would be arctan(4.56) ≈ 77.5 degrees. The other solution: √3 - 2√2 ≈ 1.732 - 2.828 ≈ -1.096. That's negative, so tan θ negative. So θ would be in the second quadrant (between 90° and 180°), but since θ + C = 120°, and C must be positive, θ must be less than 120°, so θ would be between 90° and 120°, with tan θ negative. Let's check if that works.But let's check validity. If tan θ is negative, θ is between 90° and 120°, but in that case, sin θ is positive, cos θ is negative. Let's see if substituting back into the equation works.But perhaps only the positive tan θ is valid because if θ is between 0 and 90°, tan θ is positive, but if θ is between 90° and 120°, tan θ is negative. Let's check both solutions.First, tan θ = √3 + 2√2 ≈ 4.56. So θ ≈ arctan(4.56) ≈ 77.5 degrees. Then C = 120° - 77.5° ≈ 42.5°. That seems valid.Second, tan θ = √3 - 2√2 ≈ -1.096. Then θ ≈ arctan(-1.096) ≈ -47.5°, but since θ is between 0° and 120°, we add 180° to get 132.5°, which is more than 120°, which is invalid because C = 120° - θ would be negative. So this solution is invalid. Therefore, the only valid solution is tan θ = √3 + 2√2, θ ≈77.5°, C≈42.5°.But perhaps we need an exact value. Let's see. So tan θ = √3 + 2√2. Hmm. Let's see if this corresponds to a specific angle.Alternatively, maybe we can find sin θ and cos θ.Alternatively, since we need to find the value of b/c, which from the Law of Sines is sin B / sin C. Because in Law of Sines, b / c = sin B / sin C. Since angles B and C are θ and 120° - θ, so sin B / sin C = sin θ / sin(120° - θ). Let's compute that.We have:sin θ / sin(120° - θ) = [sin θ] / [sin 120° cos θ - cos 120° sin θ]= sin θ / [ (√3/2) cos θ - (-1/2) sin θ ]= sin θ / [ (√3/2) cos θ + (1/2) sin θ ]Multiply numerator and denominator by 2:[2 sin θ] / [ √3 cos θ + sin θ ]Let me denote this ratio as K = [2 sin θ] / [ √3 cos θ + sin θ ]But from the given equation in part (2), sin(B + C) = 6 cos B sin C. Since B + C = 120°, sin 120° = 6 cos B sin C. So we have √3 / 2 = 6 cos B sin C. Therefore, cos B sin C = √3 / 12.But also, from above, in part (2), we derived that tan θ = √3 + 2√2. Let's see if we can express sin θ and cos θ in terms of tan θ.Given tan θ = √3 + 2√2, let's compute sin θ and cos θ.Let’s denote t = tan θ = √3 + 2√2. Then:sin θ = t / sqrt(1 + t²)cos θ = 1 / sqrt(1 + t²)Compute t²:(√3 + 2√2)² = 3 + 4*2 + 2*√3*2√2 = 3 + 8 + 4√6 = 11 + 4√6So 1 + t² = 12 + 4√6Therefore, sqrt(1 + t²) = sqrt(12 + 4√6) = sqrt(4*(3 + √6)) = 2*sqrt(3 + √6)Therefore, sin θ = (√3 + 2√2) / [2*sqrt(3 + √6)]Similarly, cos θ = 1 / [2*sqrt(3 + √6)]But this seems complicated. Maybe rationalize the denominator or find a better approach.Alternatively, since we need to compute b/c = sin B / sin C = sin θ / sin(120° - θ). Let's compute sin θ and sin(120° - θ).But θ is angle B with tan θ = √3 + 2√2. Let's see:Alternatively, perhaps use the given equation √3 / 2 = 6 cos B sin C, so cos B sin C = √3 / 12.But in part (1), we have angle A = 60°, so sides are related by a² = b² + c² - bc. Maybe we can use that in combination with Law of Sines.From part (1), a² = b² + c² - bc. Also, from Law of Sines, a / sin A = b / sin B = c / sin C. Since angle A is 60°, sin A = √3 / 2. Therefore, a = (2R) * sin A = 2R*(√3 / 2) = R√3. Similarly, b = 2R sin B, c = 2R sin C. So the ratio b/c = sin B / sin C.But we need to find b/c = sin B / sin C. Let’s denote this ratio as k. Let’s let k = sin B / sin C. Then sin B = k sin C.Also, since B + C = 120°, C = 120° - B. So sin C = sin(120° - B) = sin 120° cos B - cos 120° sin B = (√3/2) cos B + (1/2) sin B.Therefore, sin C = (√3/2) cos B + (1/2) sin B.But sin B = k sin C, so substituting:sin C = (√3/2) cos B + (1/2)(k sin C)Rearranged:sin C - (k/2) sin C = (√3 / 2) cos BFactor sin C:sin C (1 - k/2) = (√3 / 2) cos BBut we also have from the given equation in part (2): sin(B + C) = 6 cos B sin C. Which is sin 120° = 6 cos B sin C. So √3/2 = 6 cos B sin C. Therefore, cos B sin C = √3 / 12.So from the previous equation:sin C (1 - k/2) = (√3 / 2) cos BBut cos B sin C = √3 / 12. So substituting:sin C (1 - k/2) = (√3 / 2) * (√3 / 12) / sin C ? Wait, no. Wait, cos B = (√3 / 12) / sin C. Wait, let me think.From cos B sin C = √3 / 12, so cos B = (√3 / 12) / sin C.Therefore, substitute into the equation:sin C (1 - k/2) = (√3 / 2) * (√3 / 12) / sin CSimplify the right side:(√3 / 2) * (√3 / 12) = (3 / 2) * (1 / 12) = 3 / 24 = 1 / 8So:sin C (1 - k/2) = (1 / 8) / sin CMultiply both sides by sin C:sin² C (1 - k/2) = 1/8But sin² C = 1 - cos² C. Hmm, not sure. Alternatively, let's recall that k = sin B / sin C, and B = 120° - C.Wait, since B + C = 120°, then B = 120° - C, so sin B = sin(120° - C) = sin 120° cos C - cos 120° sin C = (√3/2) cos C + (1/2) sin C.Therefore, k = sin B / sin C = [ (√3/2) cos C + (1/2) sin C ] / sin C = (√3/2) cot C + 1/2.So k = (√3/2) cot C + 1/2.But maybe this is getting too complicated. Let's think differently.From part (1), we have a² = b² + c² - bc. Also, from Law of Sines, a = 2R sin A = 2R sin 60° = 2R*(√3/2) = R√3.Similarly, b = 2R sin B, c = 2R sin C.Therefore, substituting into a² = b² + c² - bc:(R√3)² = (2R sin B)^2 + (2R sin C)^2 - (2R sin B)(2R sin C)Simplify:3R² = 4R² sin² B + 4R² sin² C - 4R² sin B sin CDivide both sides by R²:3 = 4 sin² B + 4 sin² C - 4 sin B sin CDivide both sides by 4:3/4 = sin² B + sin² C - sin B sin CBut since B + C = 120°, C = 120° - B. So let's write everything in terms of angle B.sin² B + sin²(120° - B) - sin B sin(120° - B) = 3/4Let me compute each term:sin²(120° - B) = [sin 120° cos B - cos 120° sin B]^2 = [ (√3/2 cos B + 1/2 sin B) ]^2 = (3/4 cos² B + (√3/2) cos B sin B + 1/4 sin² B )Similarly, sin B sin(120° - B) = sin B [ (√3/2 cos B + 1/2 sin B ) ] = (√3/2 sin B cos B + 1/2 sin² B )Therefore, sin² B + sin²(120° - B) - sin B sin(120° - B) becomes:sin² B + [3/4 cos² B + (√3/2) cos B sin B + 1/4 sin² B ] - [ (√3/2 sin B cos B + 1/2 sin² B ) ]Combine terms:sin² B + 3/4 cos² B + √3/2 cos B sin B + 1/4 sin² B - √3/2 sin B cos B - 1/2 sin² BSimplify:sin² B + 1/4 sin² B - 1/2 sin² B = (1 + 1/4 - 1/2) sin² B = (3/4) sin² BSimilarly, 3/4 cos² B remains.The √3/2 cos B sin B terms cancel with -√3/2 sin B cos B.So overall:3/4 sin² B + 3/4 cos² B = 3/4 (sin² B + cos² B) = 3/4 * 1 = 3/4Which matches the left side. So this equation is an identity? Wait, so substituting back, we get 3/4 = 3/4. So this doesn't give us new information. Therefore, the equation from part (1) is consistent but doesn't help us directly for part (2). Hmm.Therefore, perhaps need to use another relation. Let's recall that from part (2), we have:√3 / 2 = 6 cos B sin C => cos B sin C = √3 / 12But also, from the Law of Sines, b / c = sin B / sin C = k. Let's denote k = b / c. So sin B = k sin C.Also, since B + C = 120°, we can write B = 120° - C. Then:cos B = cos(120° - C) = cos 120° cos C + sin 120° sin C = (-1/2) cos C + (√3 / 2) sin CSo cos B = - (1/2) cos C + (√3 / 2) sin CTherefore, substituting into cos B sin C:[ - (1/2) cos C + (√3 / 2) sin C ] * sin C = √3 / 12Simplify:- (1/2) cos C sin C + (√3 / 2) sin² C = √3 / 12Multiply both sides by 2:- cos C sin C + √3 sin² C = √3 / 6Now, let's use substitution. Let’s set x = C. Then, since B = 120° - x, and angles are in degrees.So:- cos x sin x + √3 sin² x = √3 / 6Let’s divide through by sin x (assuming sin x ≠ 0, which it isn't since x is an angle in a triangle):- cos x + √3 sin x = √3 / (6 sin x )But this might not help. Alternatively, express in terms of sin(2x) or other identities.Note that -cos x sin x = - (1/2) sin 2x. And sin² x = (1 - cos 2x)/2.So substituting:- (1/2) sin 2x + √3 * (1 - cos 2x)/2 = √3 / 6Multiply both sides by 2:- sin 2x + √3 (1 - cos 2x) = √3 / 3Expand:- sin 2x + √3 - √3 cos 2x = √3 / 3Bring √3 to the right side:- sin 2x - √3 cos 2x = √3 / 3 - √3 = √3 / 3 - 3√3 / 3 = (-2√3) / 3Multiply both sides by -1:sin 2x + √3 cos 2x = 2√3 / 3Now, this looks like a linear combination of sin and cos. We can write this as R sin(2x + φ) = 2√3 / 3, where R is the amplitude and φ is the phase shift.Compute R:R = sqrt(1² + (√3)²) = sqrt(1 + 3) = 2Then, sin(2x + φ) = (2√3 / 3) / R = (2√3 / 3) / 2 = √3 / 3Also, tan φ = √3 / 1 = √3 => φ = 60°, or π/3 radians.So the equation becomes:2 sin(2x + 60°) = 2√3 / 3Divide both sides by 2:sin(2x + 60°) = √3 / 3Thus:2x + 60° = arcsin(√3 / 3) or 180° - arcsin(√3 / 3) + 360°nBut since x is an angle in a triangle, x is between 0° and 120°, so 2x is between 0° and 240°, so 2x + 60° is between 60° and 300°. arcsin(√3 / 3) is approximately 30°, but let's compute it:√3 / 3 ≈ 0.577, which is sin(35.264°) approximately. So:2x + 60° ≈ 35.264° or 180° - 35.264° = 144.736°Therefore:Case 1: 2x + 60° ≈ 35.264° => 2x ≈ -24.736°, which is invalid since x > 0.Case 2: 2x + 60° ≈ 144.736° => 2x ≈ 84.736° => x ≈ 42.368°So angle C ≈ 42.368°, angle B ≈ 120° - 42.368° ≈ 77.632°, which matches our previous approximate result.Therefore, angle C ≈ 42.37°, angle B ≈77.63°, so sin B / sin C = sin 77.63° / sin 42.37°Let's compute this ratio:sin 77.63° ≈ sin(60° + 17.63°) ≈ sin 60° cos 17.63° + cos 60° sin 17.63° ≈ 0.96 * 0.953 + 0.5 * 0.304 ≈ 0.914 + 0.152 ≈ 1.066Wait, actually, sin 77.63° is approximately 0.975, and sin 42.37° is approximately 0.673. Therefore, ratio ≈ 0.975 / 0.673 ≈ 1.45. But exact value is needed.But maybe we can find the exact value using the relations we have.We know from part (1) that a² = b² + c² - bc. From Law of Sines, a = 2R sin 60°, b = 2R sin B, c = 2R sin C. Therefore:(2R sin 60°)^2 = (2R sin B)^2 + (2R sin C)^2 - (2R sin B)(2R sin C)Divide both sides by (2R)^2:sin² 60° = sin² B + sin² C - sin B sin CBut sin² 60° = 3/4, so 3/4 = sin² B + sin² C - sin B sin C.But we also have angles B and C such that B + C = 120°, and from part (2), we have sin(B + C) = 6 cos B sin C. Which led us to find that angle B is approximately 77.63°, and C ≈ 42.37°, but we need an exact ratio b/c.Alternatively, since we found tan B = √3 + 2√2, and k = b/c = sin B / sin C. Let's express sin B and sin C in terms of tan B.We have tan B = √3 + 2√2. Let's construct a right triangle where the opposite side is √3 + 2√2 and the adjacent side is 1. Then the hypotenuse is sqrt(1 + (√3 + 2√2)^2 ) = sqrt(1 + 3 + 4*2 + 4√6 ) = sqrt(1 + 3 + 8 + 4√6) = sqrt(12 + 4√6) = 2*sqrt(3 + √6)Therefore, sin B = (√3 + 2√2) / (2*sqrt(3 + √6))Similarly, since C = 120° - B, sin C = sin(120° - B). Using sine subtraction formula:sin(120° - B) = sin 120° cos B - cos 120° sin B = (√3/2) cos B + (1/2) sin BBut we can express cos B and sin B from the same triangle:cos B = 1 / sqrt(1 + tan² B) = 1 / (2*sqrt(3 + √6)) as we found before.Therefore, sin C = (√3/2) * [1 / (2*sqrt(3 + √6))] + (1/2) * [ (√3 + 2√2) / (2*sqrt(3 + √6)) ]Simplify:= [ √3 / (4*sqrt(3 + √6)) ) + (√3 + 2√2) / (4*sqrt(3 + √6)) ]Combine terms:= [ √3 + √3 + 2√2 ] / (4*sqrt(3 + √6)) )= [ 2√3 + 2√2 ] / (4*sqrt(3 + √6)) )Factor out 2:= 2(√3 + √2) / (4*sqrt(3 + √6)) )= (√3 + √2) / (2*sqrt(3 + √6)) )So now, the ratio k = sin B / sin C = [ (√3 + 2√2) / (2*sqrt(3 + √6)) ] / [ (√3 + √2) / (2*sqrt(3 + √6)) ] = (√3 + 2√2) / (√3 + √2)Simplify this fraction:Multiply numerator and denominator by (√3 - √2) to rationalize the denominator:[ (√3 + 2√2)(√3 - √2) ] / [ (√3 + √2)(√3 - √2) ]Denominator: (√3)^2 - (√2)^2 = 3 - 2 = 1Numerator: √3*√3 + √3*(-√2) + 2√2*√3 - 2√2*√2 = 3 - √6 + 2√6 - 4 = (3 - 4) + (-√6 + 2√6) = -1 + √6Wait, compute numerator:First term: √3 * √3 = 3Second term: √3 * (-√2) = -√6Third term: 2√2 * √3 = 2√6Fourth term: 2√2 * (-√2) = -2*(2) = -4So overall: 3 - √6 + 2√6 - 4 = (3 - 4) + (-√6 + 2√6) = -1 + √6Therefore, k = (-1 + √6)/1 = √6 - 1Therefore, the ratio b/c = √6 - 1Wait, that's exact. Let me check the calculation again.Numerator after multiplying: (√3 + 2√2)(√3 - √2) = 3 - √6 + 2√6 - 2*2 = 3 - √6 + 2√6 - 4 = (3 - 4) + ( -√6 + 2√6 ) = -1 + √6.Denominator is 1. So yes, k = √6 - 1.Therefore, the value of b/c is √6 - 1.Let me verify this with approximate values:√6 ≈ 2.449, so √6 -1 ≈1.449. Earlier, we had approximately 1.45, which matches. So this seems correct.Hence, the value of b/c is √6 -1.Final Answer(1) The measure of angle ( A ) is (boxed{60^circ}).(2) The value of (frac{b}{c}) is (boxed{sqrt{6} - 1}).</think>Given a triangle ( triangle ABC ) with sides opposite to angles ( A ), ( B ), and ( C ) being ( a ), ( b ), and ( c ) respectively, and the relationship ( frac{tan A}{tan B} = frac{2c - b}{b} ):1. Finding the measure of angle ( A ): - Using the tangent ratio and the Law of Sines: [ frac{tan A}{tan B} = frac{sin A / cos A}{sin B / cos B} = frac{sin A cos B}{sin B cos A} = frac{a cos B}{b cos A} ] Given ( frac{a cos B}{b cos A} = frac{2c - b}{b} ), we simplify to: [ a cos B = (2c - b) cos A ] - Using the Law of Cosines for ( cos A ) and ( cos B ): [ cos A = frac{b^2 + c^2 - a^2}{2bc}, quad cos B = frac{a^2 + c^2 - b^2}{2ac} ] Substituting these into the equation and simplifying, we find: [ a^2 = b^2 + c^2 - bc ] Comparing with the Law of Cosines ( a^2 = b^2 + c^2 - 2bc cos A ), we get: [ 2 cos A = 1 implies cos A = frac{1}{2} implies A = 60^circ ]2. Finding the value of ( frac{b}{c} ): - Given ( sin(B + C) = 6 cos B sin C ) and knowing ( B + C = 120^circ ): [ sin 120^circ = 6 cos B sin C implies frac{sqrt{3}}{2} = 6 cos B sin C implies cos B sin C = frac{sqrt{3}}{12} ] - Using ( B = 120^circ - C ) and trigonometric identities: [ cos B = -frac{1}{2} cos C + frac{sqrt{3}}{2} sin C ] Substituting and simplifying, we find: [ - cos C sin C + sqrt{3} sin^2 C = frac{sqrt{3}}{6} ] - Solving using substitution and quadratic equations, we find the ratio ( frac{b}{c} = sqrt{6} - 1 ). Final Answers1. The measure of angle ( A ) is (boxed{60^circ}).2. The value of (frac{b}{c}) is (boxed{sqrt{6} - 1}).,<think>Alright, let's tackle this geometry problem step by step. It's about triangle ABC with sides a, b, c opposite angles A, B, C respectively. There are two parts: first, find angle A given that (tan A)/(tan B) = (2c - b)/b, and second, given that sin(B + C) = 6 cos B sin C, find the value of b/c.Starting with part (1). The problem gives a ratio of tangents of angles A and B equal to a ratio of sides involving c and b. Hmm, tangents... I remember that in triangles, the tangent of an angle can relate to the sides through the Law of Sines or maybe the Law of Cosines. Let me think. The Law of Sines says that a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius. So perhaps I can express tan A and tan B in terms of sine and cosine, then use the Law of Sines to substitute for the sides.Given that tan A / tan B = (2c - b)/b. Let's write tan A as sin A / cos A and tan B as sin B / cos B. Therefore:(sin A / cos A) / (sin B / cos B) = (2c - b)/bSimplifying the left side: (sin A cos B) / (sin B cos A) = (2c - b)/bLet me rearrange that:(sin A / sin B) * (cos B / cos A) = (2c - b)/bFrom the Law of Sines, sin A / sin B = a / b. So substituting that in:(a / b) * (cos B / cos A) = (2c - b)/bMultiply both sides by b:a * (cos B / cos A) = 2c - bHmm, interesting. Let's write that as:a cos B / cos A = 2c - bNow, maybe use the Law of Cosines here. The Law of Cosines relates sides and angles. For angle B, it says:cos B = (a² + c² - b²) / (2ac)Similarly, for angle A:cos A = (b² + c² - a²) / (2bc)So substituting these into the equation:a * [ (a² + c² - b²)/(2ac) ] / [ (b² + c² - a²)/(2bc) ) ] = 2c - bSimplify numerator and denominator:Numerator: a * (a² + c² - b²)/(2ac) = [a(a² + c² - b²)] / (2ac) = (a² + c² - b²) / (2c)Denominator: (b² + c² - a²)/(2bc)Therefore, dividing them:[ (a² + c² - b²)/(2c) ] / [ (b² + c² - a²)/(2bc) ) ] = [ (a² + c² - b²) / (2c) ] * [ 2bc / (b² + c² - a²) ) ] = [ (a² + c² - b²) * 2bc ] / [ 2c (b² + c² - a²) ) ]Simplify: The 2c cancels out, so:[ (a² + c² - b²) * b ] / [ (b² + c² - a²) )So the entire left side of the equation becomes:[ b(a² + c² - b²) ] / (b² + c² - a²) ) = 2c - bSo set up the equation:[ b(a² + c² - b²) ] / (b² + c² - a²) = 2c - bLet me denote D = a² + c² - b² and E = b² + c² - a². Then, the equation becomes (bD)/E = 2c - b. Let's cross multiply:bD = (2c - b)EExpanding both sides:b(a² + c² - b²) = (2c - b)(b² + c² - a²)This seems messy, but perhaps expanding both sides will lead to some cancellation. Let's try.Left side: b(a² + c² - b²) = a²b + bc² - b³Right side: (2c - b)(b² + c² - a²) = 2c(b² + c² - a²) - b(b² + c² - a²)Compute each part:First term: 2c*b² + 2c*c² - 2c*a² = 2b²c + 2c³ - 2a²cSecond term: -b*b² - b*c² + b*a² = -b³ - bc² + a²bCombine these:Right side = 2b²c + 2c³ - 2a²c - b³ - bc² + a²bNow, write left side and right side:Left: a²b + bc² - b³Right: 2b²c + 2c³ - 2a²c - b³ - bc² + a²bBring all terms to the left side:Left - Right = [a²b + bc² - b³] - [2b²c + 2c³ - 2a²c - b³ - bc² + a²b] = 0Compute term by term:a²b - a²b = 0bc² - (-bc²) = bc² + bc² = 2bc²-b³ - (-b³) = -b³ + b³ = 0Then, subtract the remaining terms:-2b²c - 2c³ + 2a²cSo overall:2bc² - 2b²c - 2c³ + 2a²c = 0Factor out 2c:2c [ b² - b² - c² + a² ]? Wait, let's check:Wait, 2bc² - 2b²c - 2c³ + 2a²c = 2c(b² - b² - c² + a²) Hmm, no. Wait:Wait, 2bc² - 2b²c - 2c³ + 2a²c = 2c(c² - b² - c² + a²) ?Wait, let me factor 2c from each term:2c [ bc - b² - c² + a² ] ?Wait, perhaps better to group terms:2c [ (bc² / c) - (b²c / c) - (c³ / c) + (a²c / c) ) ]? Not sure.Alternatively, factor:2bc² - 2b²c = 2bc(c - b)-2c³ + 2a²c = -2c(c² - a²) = -2c(c - a)(c + a)So altogether:2bc(c - b) - 2c(c - a)(c + a) = 0Factor out 2c:2c [ b(c - b) - (c - a)(c + a) ] = 0Since c is a side length, c ≠ 0, so:b(c - b) - (c² - a²) = 0Expand:bc - b² - c² + a² = 0Rearranged:a² = b² + c² - bcWait, that's interesting. So from all that algebra, we end up with a² = b² + c² - bcNow, compare this to the Law of Cosines. The Law of Cosines says:a² = b² + c² - 2bc cos ASo if we have a² = b² + c² - bc, then setting equal to Law of Cosines:b² + c² - 2bc cos A = b² + c² - bcSubtract b² + c² from both sides:-2bc cos A = -bcDivide both sides by -bc (assuming b, c ≠ 0):2 cos A = 1Thus, cos A = 1/2Therefore, angle A is arccos(1/2), which is 60 degrees or π/3 radians.So part (1) answer is 60 degrees, or π/3.Wait, let me check that again. Because from the equation a² = b² + c² - bc, which comes from all that substitution. Then Law of Cosines gives a² = b² + c² - 2bc cos A. Equate them:b² + c² - 2bc cos A = b² + c² - bcSubtract the common terms: -2bc cos A = -bcDivide by -bc: 2 cos A = 1 → cos A = 1/2. So angle A is 60 degrees. Correct.Okay, so part (1) is 60 degrees.Now part (2): If sin(B + C) = 6 cos B sin C, find the value of b/c.First, note that in triangle ABC, angles A + B + C = π. So B + C = π - A. Since we found in part (1) that angle A is 60 degrees, then B + C = 120 degrees. So sin(B + C) = sin(120°) = √3/2.But wait, the equation is sin(B + C) = 6 cos B sin C. Wait, but if we already know B + C = 120°, then sin(120°) = 6 cos B sin C. So:√3/2 = 6 cos B sin CBut maybe we need to relate this to the sides. Alternatively, perhaps there is a trigonometric identity we can use here.Alternatively, since we need to find b/c, perhaps use the Law of Sines. Let's recall that by Law of Sines, b/sin B = c/sin C, so b/c = sin B / sin C. Therefore, if we can find sin B / sin C, that would be the value of b/c.So the problem reduces to finding sin B / sin C given that sin(B + C) = 6 cos B sin C.But B + C = 120°, so sin(B + C) = sin 120° = √3/2. Therefore:√3/2 = 6 cos B sin CLet me write that equation:6 cos B sin C = √3/2Divide both sides by 6:cos B sin C = √3 / 12But also, since B + C = 120°, we can write C = 120° - B. So sin C = sin(120° - B). Let's use that.So sin C = sin(120° - B) = sin 120° cos B - cos 120° sin BWe know sin 120° = √3/2, cos 120° = -1/2. Therefore:sin C = (√3/2) cos B - (-1/2) sin B = (√3/2) cos B + (1/2) sin BTherefore, plug this into the equation cos B sin C = √3 /12:cos B [ (√3/2) cos B + (1/2) sin B ] = √3 /12Multiply through:(√3/2) cos² B + (1/2) cos B sin B = √3 /12Multiply both sides by 2 to eliminate denominators:√3 cos² B + cos B sin B = √3 /6Now, let's recall that we need to relate this to sin B / sin C. Since C = 120° - B, perhaps express sin C in terms of sin B. Alternatively, we can let x = B, then C = 120° - x, and try to solve for x.Let me let x = B, so C = 120° - x. Then, the equation becomes:√3 cos² x + cos x sin x = √3 /6Hmm, this seems a bit complicated, but maybe we can use some trigonometric identities here.Alternatively, perhaps express everything in terms of sin x and cos x, then see if we can find a relationship.Alternatively, express in terms of sin x.Note that cos² x = 1 - sin² x. Let me substitute that:√3 (1 - sin² x) + cos x sin x = √3 /6So:√3 - √3 sin² x + cos x sin x = √3 /6Bring √3 to the right:-√3 sin² x + cos x sin x = √3 /6 - √3Simplify the right side:√3 /6 - √3 = √3 (1/6 - 1) = √3 (-5/6 )Thus:-√3 sin² x + cos x sin x = -5√3 /6Multiply both sides by -1:√3 sin² x - cos x sin x = 5√3 /6Hmm, not sure if this is helpful. Maybe factor sin x:sin x (√3 sin x - cos x) = 5√3 /6Alternatively, maybe use substitution variables. Let me think.Let me denote t = sin x / cos x = tan x. Then, sin x = t cos x, and since sin² x + cos² x = 1, t² cos² x + cos² x = 1 ⇒ cos² x (1 + t²) = 1 ⇒ cos x = 1 / √(1 + t²). But this might complicate things. Alternatively, maybe express as a single trigonometric function.Looking back at the equation:√3 cos² x + cos x sin x = √3 /6Let me divide both sides by cos² x (assuming cos x ≠ 0):√3 + (sin x / cos x) = √3 / (6 cos² x)But sin x / cos x = tan x, and 1 / cos² x = 1 + tan² x. Therefore:√3 + tan x = √3 (1 + tan² x) /6Multiply both sides by 6:6√3 + 6 tan x = √3 (1 + tan² x)Let me rearrange:√3 tan² x -6 tan x + (√3 - 6√3) = 0. Wait, let's do it step by step.Starting from:6√3 + 6 tan x = √3 + √3 tan² xBring all terms to the left side:6√3 + 6 tan x - √3 - √3 tan² x = 0Combine like terms:(6√3 - √3) + 6 tan x - √3 tan² x = 0Which is:5√3 + 6 tan x - √3 tan² x = 0Multiply both sides by -1 to make the quadratic term positive:√3 tan² x -6 tan x -5√3 =0Now, let me set t = tan x. Then the equation becomes:√3 t² -6 t -5√3 =0Multiply both sides by √3 to eliminate the radical:3 t² -6√3 t -15 =0Divide by 3:t² - 2√3 t -5 =0Now, solve for t using quadratic formula:t = [2√3 ± √( (2√3)^2 + 20 ) ] / 2Compute discriminant:(2√3)^2 +20 = 12 +20=32Thus:t= [2√3 ± √32 ] /2 = [2√3 ± 4√2 ] /2 = √3 ± 2√2So tan x = √3 + 2√2 or tan x = √3 - 2√2But x is angle B in the triangle, which is between 0 and π - A - C, but since A is 60°, x is between 0 and 120°, so tan x can be positive or negative. However, in a triangle, angles are between 0 and π, so x is between 0 and 120°, but since A is 60°, the other angles must be less than 120°, so x is between 0 and 120°, but tan x can be positive or negative. Wait, actually in a triangle, all angles are between 0 and π, so x is between 0 and π, but since A is 60°, B + C = 120°, so x and C are between 0 and 120°, so tan x can be positive or negative? Wait, no. In a triangle, all angles are positive and less than π, so between 0 and π. For angles between 0 and π/2, tan is positive; between π/2 and π, tan is negative. But in a triangle, if angle B is greater than 90°, then angles C and A must sum to less than 90°, but A is already 60°, so C would have to be less than 30°, which might be possible. However, let's check the possible solutions.Compute tan x = √3 + 2√2 ≈ 1.732 + 2.828 ≈ 4.56, which is positive. So angle x would be in the first quadrant. The other solution is tan x = √3 - 2√2 ≈ 1.732 - 2.828 ≈ -1.096, which is negative, implying angle x is in the second quadrant (between 90° and 180°). However, if angle B is in the second quadrant, then angle C = 120° - B would be less than 30°, which is possible. So both solutions might be possible? Wait, but in a triangle, can angle B be greater than 90°? Let's check if possible.If angle B is greater than 90°, then angle C = 120° - B would be less than 30°, so sides would satisfy that. However, we need to check if such a triangle is possible. Since the Law of Sines holds, but we also need to ensure that sides are positive. Let's see.But let's check both possibilities. So tan x can be √3 + 2√2 or √3 - 2√2. Let's compute both possibilities.First, let's take tan x = √3 + 2√2 ≈4.56, so angle x ≈ arctan(4.56) ≈77.5 degrees. Then angle C = 120° - 77.5 ≈42.5 degrees. Then sides ratio b/c = sin B / sin C ≈ sin77.5 / sin42.5 ≈0.97 / 0.676 ≈1.435.Alternatively, if tan x = √3 - 2√2 ≈-1.096, then angle x ≈180° - 47.7°≈132.3°, which would make angle C =120° -132.3°≈-12.3°, which is impossible. Wait, that's negative. Therefore, this solution is invalid. So only the first solution is valid.Therefore, tan x = √3 + 2√2, angle B ≈77.5°, angle C≈42.5°, and the ratio b/c≈1.435. But we need an exact value.Wait, but let's do this algebraically.We have tan B = √3 + 2√2. Then, angle B is arctan(√3 + 2√2). Then angle C = 120° - B. Then, sin B / sin C = [sin B] / [sin(120° - B)].Alternatively, express sin(120° - B) as sin(120°)cos B - cos(120°)sin B.So sin C = sin(120° - B) = (√3/2)cos B - (-1/2)sin B = (√3/2)cos B + (1/2)sin BSo sin B / sin C = sin B / [ (√3/2)cos B + (1/2)sin B ]Multiply numerator and denominator by 2:[2 sin B] / [√3 cos B + sin B]Let me divide numerator and denominator by cos B:[2 tan B] / [√3 + tan B]Since tan B = √3 + 2√2, substitute:[2(√3 + 2√2)] / [√3 + (√3 + 2√2)] = [2√3 + 4√2] / [2√3 + 2√2]Factor numerator and denominator:Numerator: 2(√3 + 2√2)Denominator: 2(√3 + √2)Cancel the 2:(√3 + 2√2) / (√3 + √2)Multiply numerator and denominator by (√3 - √2) to rationalize the denominator:[ (√3 + 2√2)(√3 - √2) ] / [ (√3 + √2)(√3 - √2) ]Denominator: (√3)^2 - (√2)^2 = 3 - 2 =1Numerator:√3 * √3 + √3*(-√2) + 2√2*√3 + 2√2*(-√2)= 3 - √6 + 2√6 - 2*2= 3 - √6 + 2√6 -4= (3 -4) + (-√6 + 2√6)= -1 + √6Therefore, the ratio is (-1 + √6)/1 = √6 -1Thus, sin B / sin C = √6 -1, so b/c = √6 -1.Wait, let me check that calculation again.Wait, numerator:(√3 + 2√2)(√3 - √2) = √3*√3 + √3*(-√2) + 2√2*√3 + 2√2*(-√2)= 3 - √6 + 2√6 - 2*(2)= 3 - √6 + 2√6 -4= (3 -4) + ( -√6 + 2√6 )= -1 + √6Yes, correct. And denominator is 1. Therefore, ratio is √6 -1.Therefore, b/c = √6 -1 ≈2.449 -1≈1.449, which aligns with the approximate value earlier.Hence, the exact value is √6 -1.Alternatively, let me confirm this result using another approach.Given sin(B + C) =6 cos B sin C. Since B + C =120°, sin(120°)=√3/2=6 cos B sin C.Thus, √3 /2 =6 cos B sin C → cos B sin C=√3 /12.But we also know that angle C=120° - B. So sin C=sin(120° - B)=sin120° cos B - cos120° sin B= (√3/2)cos B + (1/2)sin B.So, substituting into cos B sin C:cos B [ (√3/2)cos B + (1/2) sin B ] = √3 /12.Which simplifies to:(√3/2) cos² B + (1/2) cos B sin B = √3 /12.Multiply both sides by 2:√3 cos² B + cos B sin B = √3 /6.Then, perhaps use the identity involving sin(2B). But maybe express in terms of sin B and cos B.Alternatively, note that we need to relate this to the ratio b/c = sin B / sin C. Let’s denote k = sin B / sin C. Then, since sin C = sin(120° - B), we can write k = sin B / sin(120° - B).Let’s set k = sin B / sin(120° - B). Let’s express sin B = k sin(120° - B).But sin(120° - B) = sin120° cos B - cos120° sin B = (√3/2)cos B + (1/2) sin B.Thus:sin B = k [ (√3/2)cos B + (1/2) sin B ]Bring all terms to one side:sin B - (k/2) sin B - (k√3/2) cos B =0Factor:sin B (1 - k/2) - (k√3 /2 ) cos B =0Divide both sides by cos B:tan B (1 - k/2 ) - (k√3 /2 ) =0Solve for tan B:tan B = (k√3 /2 ) / (1 - k/2 ) = [k√3 /2 ] / [ (2 -k)/2 ] = (k√3 ) / (2 -k )But we also have from earlier:From equation √3 cos² B + cos B sin B = √3 /6.Alternatively, maybe express tan B in terms of k and solve for k.But perhaps we can use the expression for tan B from the previous approach. Wait, earlier we found that tan B = √3 + 2√2, leading to k=√6 -1. Let's check if substituting k=√6 -1 into tan B expression gives the same.Compute tan B = (k√3 ) / (2 -k )If k=√6 -1,tan B = ( (√6 -1 )√3 ) / (2 - (√6 -1 )) = ( √18 - √3 ) / (3 -√6 )Simplify √18 = 3√2,= (3√2 - √3 ) / (3 -√6 )Multiply numerator and denominator by (3 +√6 ):Numerator: (3√2 - √3 )(3 +√6 )= 3√2 *3 +3√2 *√6 -√3 *3 -√3 *√6=9√2 + 3√12 -3√3 -√18=9√2 + 3*2√3 -3√3 -3√2=9√2 -3√2 +6√3 -3√3=6√2 +3√3Denominator: (3 -√6 )(3 +√6 )=9 -6=3Thus, tan B=(6√2 +3√3 )/3=2√2 +√3Which matches our previous result of tan B=√3 +2√2.Yes, same thing. Therefore, k=√6 -1.Hence, the value of b/c is √6 -1.Therefore, the answer to part (2) is √6 -1.Final Answer(1) The measure of angle ( A ) is (boxed{60^circ}).(2) The value of (frac{b}{c}) is (boxed{sqrt{6} - 1}).</think>Given a triangle ( triangle ABC ) with sides ( a, b, c ) opposite angles ( A, B, C ) respectively, we need to solve two parts:1. Find the measure of angle ( A ): We start with the given equation: [ frac{tan A}{tan B} = frac{2c - b}{b} ] Using the tangent identity and the Law of Sines, we rewrite the equation as: [ left( frac{sin A}{cos A} right) left( frac{cos B}{sin B} right) = frac{2c - b}{b} ] Substituting ( frac{sin A}{sin B} = frac{a}{b} ) from the Law of Sines, we get: [ frac{a}{b} cdot frac{cos B}{cos A} = frac{2c - b}{b} ] Simplifying, we find: [ a cdot frac{cos B}{cos A} = 2c - b ] Using the Law of Cosines for ( cos A ) and ( cos B ), substituting and simplifying, we eventually derive: [ a^2 = b^2 + c^2 - bc ] Comparing this with the Law of Cosines ( a^2 = b^2 + c^2 - 2bc cos A ), we find: [ 2 cos A = 1 implies cos A = frac{1}{2} ] Thus, ( A = 60^circ ).2. Find the value of ( frac{b}{c} ) given ( sin(B + C) = 6 cos B sin C ): Since ( B + C = 120^circ ) (as ( A = 60^circ )), we have: [ sin(120^circ) = 6 cos B sin C ] [ frac{sqrt{3}}{2} = 6 cos B sin C ] Simplifying, we get: [ cos B sin C = frac{sqrt{3}}{12} ] Expressing ( sin C ) in terms of ( B ) (since ( C = 120^circ - B )): [ sin C = sin(120^circ - B) = frac{sqrt{3}}{2} cos B + frac{1}{2} sin B ] Substituting back, we solve the trigonometric equation and find that: [ frac{b}{c} = sqrt{6} - 1 ] Final Answers1. The measure of angle ( A ) is (boxed{60^circ}).2. The value of (frac{b}{c}) is (boxed{sqrt{6} - 1}).]

🔑:(1) In triangle ABC, we have frac{tan A}{tan B} = frac{2c-b}{b}.By the sine law, we get frac{sin A cos B}{cos A sin B} = frac{2sin C - sin B}{sin B}, ... (3 points)therefore sin A cos B sin B = 2sin C cos A sin B - sin^2B cos A,Since sin B neq 0, we have sin A cos B = 2sin C cos A - sin B cos A, ... (5 points)therefore sin A cos B + sin B cos A = 2sin C cos A,This implies that sin(A+B) = 2sin C cos A, ... (6 points)therefore cos A = frac{1}{2}; ... (5 points)Since 0 < A < pi, we have A = frac{pi}{3}. ... (6 points)(2) From (1), we know that cos A = frac{1}{2},By the cosine law, b^2+c^2-a^2 = bc; ... (8 points)therefore a^2 = b^2+c^2-bc quad quad quad quad quad ;, : , :, :, :, :, :, :, quad quad quad quad quad quad quad quad quad ;, , (1)From sin(B+C)=6cos Bsin C, we get sin A = 6cos Bsin C,therefore frac{a}{c} = 6 times frac{a^{2}+c^{2}-b^{2}}{2ac},therefore a^2 = 3a^2+3c^2-3b^2;therefore 2a^2 = 3b^2-3c^2 quad quad quad quad quad quad quad quad ;, : , :, :, :, :, :, quad quad quad quad quad quad quad quad quad ;, , (2)From (1) and (2), we get b^2+2bc-5c^2=0,This implies that (frac{b}{c})^2 + 2(frac{b}{c}) - 5 = 0,Solving, we get frac{b}{c} = boxed{sqrt{6}-1}. ... (12 points)

🧠:[<think>Okay, so I need to find all pairs of positive integers (p, q) such that both quadratic equations x² - px + q = 0 and x² - qx + p = 0 have integer solutions. Hmm, let's start by recalling that for a quadratic equation x² - sx + t = 0 to have integer solutions, the discriminant must be a perfect square. The discriminant here would be s² - 4t. But also, the solutions are given by [s ± √(s² - 4t)] / 2. Since the solutions need to be integers, both the numerator terms must be even numbers. That means s² - 4t must be a perfect square, and s must have the same parity as the square root of the discriminant. So, applying this to both equations. Let's first consider the equation x² - px + q = 0. The discriminant here is p² - 4q, which must be a perfect square. Let's call this discriminant D1 = p² - 4q = k² for some integer k. Similarly, for the second equation x² - qx + p = 0, the discriminant is q² - 4p, which must also be a perfect square. Let's denote this discriminant D2 = q² - 4p = m² for some integer m.Therefore, we have the system of equations:1. p² - q = k²2. q² - p = m²Wait, actually, no. The discriminants are p² - 4q and q² - 4p. So, the equations should be:1. p² - 4q = k²2. q² - 4p = m²Right. So, we need both p² - 4q and q² - 4p to be perfect squares. And since p and q are positive integers, k and m must also be non-negative integers. Also, since p² - 4q must be non-negative, p² ≥ 4q. Similarly, q² ≥ 4p.So, the key here is to find all positive integers p, q such that both p² - 4q and q² - 4p are perfect squares. Let's see if we can find some relationships between p and q here.Let me consider that both equations have integer roots. So, another approach is to suppose that the roots are integers. Let's think about the first equation, x² - px + q = 0. Suppose its roots are a and b, which are integers. Then by Vieta's formula:a + b = pab = qSimilarly, the second equation x² - qx + p = 0 has roots c and d, integers, so:c + d = qcd = pSo, substituting p and q from Vieta's formulas, we have:From the first equation, p = a + b and q = ab.From the second equation, q = c + d and p = cd.Therefore, substituting q from the second equation into the first, we have:p = a + b = cdq = ab = c + dSo, we need to find integers a, b, c, d such that:cd = a + bab = c + dSo, this seems like a system of equations. Let me think. So, we need to find positive integers a, b, c, d such that:cd = a + b ...(1)ab = c + d ...(2)So, this is a system where we can try to find solutions. Let's see if we can find possible values here.First, note that a, b, c, d are positive integers because p and q are positive integers, and the roots must be integers (since the problem states integral solutions). So, all variables here are positive integers.Let me try small integers for a and b, then compute ab = c + d, and then check if cd = a + b.Alternatively, start with small c and d, compute cd and check if ab = c + d where a and b are such that ab = c + d. Hmm, this is a bit recursive, but maybe manageable.Alternatively, note that since ab = c + d and cd = a + b, perhaps we can express variables in terms of each other.Let me try substituting. Let's say from equation (2), ab = c + d, so c = ab - d. Then substitute into equation (1):cd = a + b(ab - d)d = a + babd - d² = a + bLet's rearrange terms:abd - a - b = d²Hmm, maybe factor a and b:a(bd - 1) - b = d²Not sure. Alternatively, let's bring all terms to one side:abd - d² - a - b = 0Hmm, maybe factor d from the first two terms:d(ab - d) - a - b = 0But ab - d is equal to c from equation (2). So, d*c - a - b = 0. Wait, but from equation (1), cd = a + b. So, d*c - a - b = 0, which is exactly equation (1). Hmm, that brings us back. Maybe this substitution isn't helpful.Alternatively, perhaps we can use symmetry. Let me suppose that a ≤ b and c ≤ d. Then, since a, b, c, d are positive integers, maybe there are only small values possible. Let's try small values.Let's start with a = 1.Case 1: a = 1Then, from equation (2): ab = c + d => 1*b = c + d => b = c + d.From equation (1): cd = a + b = 1 + b.But since b = c + d, substitute into equation (1):cd = 1 + c + dRearranged:cd - c - d = 1Add 1 to both sides:cd - c - d + 1 = 2Factor:(c - 1)(d - 1) = 2Since c and d are positive integers, and c ≤ d, (c - 1) and (d - 1) must be positive integers or zero. But since (c - 1)(d - 1) = 2, possible factors of 2:1*2 or 2*1. But since c ≤ d, (c -1) ≤ (d -1). So possible:(c -1, d -1) = (1, 2) => c = 2, d = 3or (c -1, d -1) = (2,1) which would give c =3, d=2 but since c ≤ d, this is same as the first case. So, only c=2, d=3.Therefore, c=2, d=3. Then, b = c + d = 5. So, we have a=1, b=5, c=2, d=3.Therefore, p = a + b = 1 + 5 = 6q = ab = 1*5 = 5Check the second equation: x² - qx + p = x² -5x +6. Its roots are 2 and 3, which are integers. So this works.So (p, q) = (6, 5) is a solution. Let me check if there are other solutions in this case.But (c -1)(d -1) = 2. Are there other factors? If we consider c-1 and d-1 to be positive integers, then 2 can only be factored as 1*2. But if we allowed c-1=0, then d-1=2/0 which is undefined. So no. So only one solution here.Case 2: a = 2From equation (2): ab = c + d => 2b = c + dFrom equation (1): cd = a + b = 2 + bSo, cd = 2 + b. But since 2b = c + d, we can write b = (c + d)/2. Substitute into cd = 2 + b:cd = 2 + (c + d)/2Multiply both sides by 2:2cd = 4 + c + dRearrange:2cd - c - d = 4Hmm, let's see. Let me rearrange:c(2d -1) - d = 4Alternatively, perhaps factor:Not sure. Let's try to express this as:2cd - c - d = 4Add 1 to both sides:2cd - c - d + 1 = 5Factor:(2c -1)(2d -1) = 5*2 -1? Wait, maybe not. Let me try another approach.Alternatively, multiply both sides by 2 and add 1:4cd - 2c - 2d = 8Then add 1 to both sides:4cd - 2c - 2d + 1 = 9This can be written as (2c -1)(2d -1) = 9Yes, because (2c -1)(2d -1) = 4cd -2c -2d +1 = 9So, (2c -1)(2d -1) =9Since c and d are positive integers, (2c -1) and (2d -1) are positive odd integers. Factor pairs of 9 are (1,9), (3,3), (9,1). Since c ≤ d, (2c -1) ≤ (2d -1). So possible pairs:(1,9): 2c -1 =1 => c=1, 2d -1=9 => d=5(3,3): 2c -1=3 => c=2, 2d -1=3 => d=2So, possible solutions:Case 1: c=1, d=5Check if 2b = c + d =1 +5=6 => b=3Check cd =1*5=5, which should equal 2 +b =5. Yes, 5=5. So this works.Thus, a=2, b=3, c=1, d=5.Therefore, p =a + b=5, q=ab=6.Check second equation: x² -6x +5=0. Its roots are 1 and5, which are integers. So this works. So (p, q)=(5,6).Case 2: c=2, d=2. Then check if 2b =c +d=4 => b=2Check cd=4, which should equal 2 +b=4. 4=4. So this works.Thus, a=2, b=2, c=2, d=2.Therefore, p =a + b=4, q=ab=4.Check second equation: x² -4x +4=0. Its root is 2 (double root). So integers. So (p, q)=(4,4).Wait, but does this satisfy both equations? Let me check.First equation: x² -4x +4=0. Root 2 (twice). Second equation: x² -4x +4=0. Same equation, so roots are 2. So yes. So (4,4) is a solution.So for a=2, we get two solutions: (5,6) and (4,4).Case 3: a=3From equation (2): ab =c + d =>3b =c + dFrom equation (1): cd =a +b =3 +bSo, 3b =c + d => d =3b -cSubstitute into cd =3 +b:c(3b -c)=3 +b=> 3b c -c² =3 +bRearrange:3b c - b -c² -3 =0Hmm, this is getting complicated. Let me see if I can find integer solutions here.Alternatively, express in terms of c:Let me rearrange:3bc - c² - b =3=> c² -3bc +b +3=0This is a quadratic in c:c² -3b c + (b +3)=0For c to be a positive integer, discriminant must be a perfect square:Discriminant D = (3b)^2 -4*1*(b +3) =9b² -4b -12This must be a perfect square. Let me check for small values of b.Since a=3, and ab =c + d, with c and d positive integers, so ab ≥2 (since c and d at least 1 each). But a=3, so 3b ≥2, so b ≥1.Start with b=1:D=9 -4 -12= -7. Not a square. Disregard.b=2:D=36 -8 -12=16. 16 is a square. So c=(3*2 ±4)/2=(6±4)/2. So c=(10)/2=5 or c=2/2=1.So c=5 or c=1.Check c=5:Then d=3*2 -5=6 -5=1Check cd=5*1=5. Should be equal to 3 +b=5. 5=5. So yes. So a=3, b=2, c=5, d=1. But since c ≤d, we need to swap c and d: c=1, d=5.But a=3, b=2, so p =3 +2=5, q=3*2=6. Wait, this is the same as the previous solution (5,6). So maybe this is a duplicate.Similarly, c=1:d=6 -1=5. Same as above.Next, b=3:D=9*9 -4*3 -12=81 -12 -12=57. Not a square.b=4:D=9*16 -4*4 -12=144 -16 -12=116. Not a square.b=5:D=9*25 -4*5 -12=225 -20 -12=193. Not a square.b=6:D=9*36 -4*6 -12=324 -24 -12=288. Not a square.b=7:D=9*49 -4*7 -12=441 -28 -12=401. Not a square.b=8:D=9*64 -4*8 -12=576 -32 -12=532. Not a square.Hmm, seems like only b=2 gives a valid solution here, which we already have.So for a=3, we only get the same solution (5,6) again, which is already covered.Case 4: a=4From equation (2): ab =c + d =>4b =c + dFrom equation (1): cd =a +b =4 +bSo, d=4b -cSubstitute into cd=4 +b:c(4b -c) =4 +b4b c -c² =4 +bRearranged:4b c -c² -b -4=0Again, quadratic in c:c² -4b c +b +4=0Discriminant D=(4b)^2 -4*1*(b +4)=16b² -4b -16D must be a perfect square.Check for small b:Since a=4, ab =c +d ≥2, so 4b ≥2 => b ≥1.b=1:D=16 -4 -16= -4. Not a square.b=2:D=64 -8 -16=40. Not a square.b=3:D=144 -12 -16=116. Not a square.b=4:D=256 -16 -16=224. Not a square.b=5:D=400 -20 -16=364. Not a square.Not promising. Maybe no solution here.Case 5: a=4. No solution. Let's try a=4 with different approach?Alternatively, maybe there's a symmetry here. Note that if (p, q) is a solution, then (q, p) might also be a solution. Let's check the solutions we have so far:We have (6,5), (5,6), (4,4). Indeed, (6,5) and (5,6) are swapped, and (4,4) is symmetric. So maybe these are the only solutions? Let's check.But let's check for a=1,2,3,4... Maybe higher a?Alternatively, let's consider that the problem is symmetric in p and q if we swap the equations. So any solution (p,q) would imply (q,p) is also a solution. So, except for symmetric solutions like (4,4). So, given that we found (5,6) and (6,5), which are distinct, and (4,4), perhaps these are the only solutions.Wait, but let's check for a=1, we had (6,5). For a=2, we had (5,6) and (4,4). For a=3, the same as a=2. For a=4, nothing. Let's check a=5.Case 5: a=5From equation (2): ab =c + d =>5b =c + dFrom equation (1): cd =a +b =5 +bSo, d=5b -cSubstitute into cd=5 +b:c(5b -c)=5 +b5b c -c² =5 +bRearranged:5b c -c² -b -5=0Quadratic in c:c² -5b c +b +5=0Discriminant D=(5b)^2 -4*(b +5)=25b² -4b -20Check for b=1:D=25 -4 -20=1. Which is 1². So c=(5*1 ±1)/2=(5 ±1)/2. So c=3 or 2.c=3:d=5*1 -3=2. Check cd=3*2=6, which should be 5 +1=6. Yes. So c=3, d=2. Since c ≤d, swap: c=2, d=3.Thus, a=5, b=1, c=2, d=3. Therefore, p =a +b=6, q=ab=5. Which is the same as the original solution (6,5). So same solution.c=2:d=5 -2=3. Same as above.So, same solution again. So, no new solutions here.Similarly, for higher b:b=2:D=25*4 -4*2 -20=100 -8 -20=72. Not a square.b=3:25*9 -4*3 -20=225 -12 -20=193. Not a square.So, no new solutions here.Therefore, for a=5, we get the existing solution again.Similarly, if I check a=6, probably same pattern.Alternatively, since we are getting cyclic solutions, perhaps only (4,4), (5,6), and (6,5) are the solutions.But let's check if there are other possible solutions where a and b are different.Wait, but maybe we need to check for a=0? But the problem states positive integers, so a, b, c, d must be positive. So a=0 is invalid.Alternatively, perhaps there are other cases where the quadratic equations have a double root? For example, discriminant zero. Let's check.Suppose discriminant p² -4q=0. Then p²=4q. Similarly, q²=4p. Then, substituting q=p²/4 into the second equation:(q)² -4p = (p²/4)² -4p = p⁴/16 -4p = m²But p is a positive integer. Let's check for small p.p=4: q=4²/4=4. Then q² -4p=16 -16=0=0². So (4,4) which we already have.p=2: q=4/2=2. Then q² -4p=4 -8=-4, which is not a square. Disregard.p=1: q=1/4. Not integer. Disregard.p=8: q=64/4=16. Then q² -4p=256 -32=224. Not a square.p=16: q=256/4=64. Then q² -4p=4096 -64=4032. Not a square.So, the only solution here is (4,4). So double roots only give us (4,4).Therefore, combining all the cases, the solutions are (4,4), (5,6), and (6,5). Let me check if there are other possible solutions.Wait, let's check for a=2, b=2. Then p=4, q=4. Then the equations are x² -4x +4=0, which has root x=2, and x² -4x +4=0 same equation. So yes, integral solutions. So that works.Another case: when a=1, b=5. Then p=6, q=5. First equation x² -6x +5=0, roots 1 and5. Second equation x² -5x +6=0, roots 2 and3. Both integral. Similarly, swapping p and q, (5,6) gives the reverse.Are there other solutions? Let me see.Suppose a=1, b=4. Then p=5, q=4. Then first equation: x² -5x +4=0, roots 1 and4. Second equation: x² -4x +5=0. Discriminant 16 -20=-4, which is not a square. So no.Similarly, if a=1, b=6. Then p=7, q=6. First equation: roots 1 and6. Second equation: x² -6x +7=0. Discriminant 36 -28=8. Not a square. So no.a=1, b=3: p=4, q=3. Second equation: x² -3x +4=0. Discriminant 9 -16=-7. Not a square.Similarly, a=1, b=2: p=3, q=2. Second equation: x² -2x +3=0. Discriminant 4 -12=-8. Not a square.So, only when a=1, b=5 do we get a valid solution.Similarly, for a=2, we have (5,6) and (4,4). Let's see if a=3 can give anything else. When a=3, we tried and only got existing solutions.Wait, what if we consider a=2, b=3. Then p=5, q=6. Then second equation x² -6x +5=0, roots 1 and5. Which is the same as the first solution.Alternatively, maybe check if there are any other solutions where the quadratics have different roots.Wait, suppose we have a=2, b=3. Then p=5, q=6. The first equation is x² -5x +6=0, roots 2 and3. The second equation is x² -6x +5=0, roots 1 and5. So different roots. Similarly, the solution (4,4) has both equations having the same double root.Alternatively, let's check for possible other solutions where roots are different but still integers.Suppose we have a different set of roots. Let's say first equation has roots 2 and4. Then p=6, q=8. Then the second equation would be x² -8x +6=0. Discriminant 64 -24=40, not a square. So no.Or roots 1 and6. Then p=7, q=6. Second equation x² -6x +7=0. Discriminant 36 -28=8. Not a square.Roots 3 and3. Then p=6, q=9. Second equation x² -9x +6=0. Discriminant 81 -24=57. Not a square.Roots 1 and7. Then p=8, q=7. Second equation x² -7x +8=0. Discriminant 49 -32=17. Not a square.Roots 2 and2. Then p=4, q=4. Which we already have.Roots 3 and4. Then p=7, q=12. Second equation x² -12x +7=0. Discriminant 144 -28=116. Not a square.Hmm, seems challenging. Maybe there are no other solutions beyond the ones we found.Alternatively, let's consider the equations p² -4q =k² and q² -4p =m². Let's suppose k and m are integers. Let's check for small p and q.We already have (4,4): 16 -16=0 and 16 -16=0. So k=0, m=0. Valid.(5,6): p=5, q=6. Check p² -4q=25 -24=1=1². q² -4p=36 -20=16=4². So k=1, m=4. Valid.(6,5): p=6, q=5. p² -4q=36 -20=16=4². q² -4p=25 -24=1=1². So k=4, m=1. Valid.Let me check p=5, q=5. Then p² -4q=25 -20=5. Not a square. q² -4p=25 -20=5. Not a square. Not valid.p=5, q=7. p² -4q=25 -28=-3. Not valid.p=6, q=6. p² -4q=36 -24=12. Not a square. q² -4p=36 -24=12. Not a square.p=7, q=5. p² -4q=49 -20=29. Not a square. q² -4p=25 -28=-3. Not valid.p=3, q=3. p² -4q=9 -12=-3. Not valid.p=2, q=2. 4 -8=-4. Not valid.p=5, q=4. 25 -16=9=3². q² -4p=16 -20=-4. Not valid.So, no other small p and q work. So likely only (4,4), (5,6), (6,5).Another approach: Let's suppose p and q are positive integers such that both p² -4q and q² -4p are squares. Let me set p² -4q =k² and q² -4p =m².Assuming p ≥ q. Then, since p and q are positive integers, maybe p and q are bounded.From p² -4q ≥0 => q ≤ p²/4From q² -4p ≥0 => p ≤ q²/4Therefore, combining these: p ≤ q²/4 and q ≤p²/4. So substitute q ≤p²/4 into p ≤ q²/4:p ≤ (p²/4)² /4 = p^4 /64Thus, p ≤ p^4 /64 =>64 ≤p^3. Therefore, p≥4. But if p=4, then q ≤4²/4=4. So q ≤4. Then p=4, q=4. Which works. For p=5, q ≤25/4=6.25. So q≤6. Then q=6. Check if q=6 satisfies q² -4p=36 -20=16=4². Yes. So (5,6). For p=6, q≤36/4=9. But if p=6, then q must satisfy q² -24 is a square. So q² -24=m². Let's check q=5:25 -24=1=1². So (6,5). q=6:36 -24=12. Not a square. q=7:49 -24=25=5². But then p=6 must be equal to cd where c + d=7 and cd=6. But c and d are roots of x² -7x +6=0, which are 1 and6. So cd=6, which is p. But p=6. So yes. Wait, but q=7, p=6. Then in the first equation, p=6, q=7: x² -6x +7=0. Discriminant 36 -28=8. Not a square. Wait, contradiction. Wait, hold on. Wait, if q=7 and p=6, then first equation discriminant is 36 -28=8, not a square. So this is invalid. So even though q² -4p=25=5², the other discriminant is not a square. Therefore, this is not a valid solution. So even though we found q=7 gives q² -4p=25, which is square, the first equation does not have a square discriminant. Therefore, it's invalid. Therefore, q=5 is the only possible for p=6. Therefore, (6,5). Similarly, p=5, q=6. So seems like (4,4), (5,6), (6,5) are the only solutions.Check p=7: q ≤49/4≈12.25. So q up to 12. Let's check q=12: q² -4p=144 -28=116. Not a square. q=10:100 -28=72. Not a square. q=8:64 -28=36=6². So m=6. Then check first equation p² -4q=49 -32=17. Not a square. So invalid.Similarly, q=9:81 -28=53. Not a square.Thus, p=7, no solutions.p=8: q≤64/4=16. Let's try q=16: q² -4p=256 -32=224. Not a square. q=10:100 -32=68. Not a square. q=9:81 -32=49=7². Then check p² -4q=64 -36=28. Not a square. So invalid. q=8:64 -32=32. Not a square. q=7:49 -32=17. Not a square. q=6:36 -32=4=2². Then check first equation p² -4q=64 -24=40. Not a square. So no.p=8, q=6: second equation discriminant is 4, but first equation discriminant is 40. Not a square. Disregard.Therefore, p=8 no solution.Similarly, p=9: q≤81/4≈20.25. Check q=20:400 -36=364. Not a square. q=16:256 -36=220. Not a square. q=12:144 -36=108. Not a square. q=10:100 -36=64=8². Then check first equation p² -4q=81 -40=41. Not a square. So invalid.Similarly, p=9, q=10. First equation discriminant 41. Not a square.So no solution here.Therefore, seems like beyond p=6, no solutions. So the only solutions are (4,4), (5,6), and (6,5).Let me check if there's a mathematical way to bound p and q.Assume p ≥ q. Then p² -4q ≥ q² -4p (since p≥q). Wait, not sure. Alternatively, if p ≥ q, then from p² -4q =k² and q² -4p =m². Since p ≥ q, then k² = p² -4q ≥ q² -4q. Similarly, m² = q² -4p ≤ q² -4q. So k² ≥ m². Thus, k ≥m.But not sure if helpful.Alternatively, add the two equations:p² + q² -4p -4q =k² +m²But I don't see how this helps.Alternatively, subtract them:p² - q² -4q +4p =k² -m²(p - q)(p + q) +4(p - q) = (k -m)(k +m)(p - q)(p + q +4) = (k -m)(k +m)But unless we can factor further, not helpful.Alternatively, think of p and q as variables and try to solve the system:p² - q² =k² -m² +4(p - q)But not sure.Alternatively, consider if p = q. Then both equations become p² -4p =k². So p² -4p must be a square. Let’s solve p² -4p =k².p² -4p -k²=0.Solutions for p: [4 ±√(16 +4k²)]/2=2 ±√(4 +k²). For p to be positive integer, √(4 +k²) must be integer. Let’s set √(4 +k²)=m, so m² -k²=4. Which factors as (m -k)(m +k)=4. Since m and k are positive integers, m >k. The factor pairs of 4 are (1,4) and (2,2). So:Case 1: m -k=1, m +k=4. Solving: m=2.5, k=1.5. Not integers.Case 2: m -k=2, m +k=2. Then m=2, k=0. But k must be non-negative integer. So k=0, m=2. Then p=2 ±√(4 +0)=2 ±2. So p=4 or p=0. p=4 is valid. Thus, when p=q=4, which we already have. So only solution when p=q is (4,4).Thus, all solutions are (4,4), (5,6), (6,5). Therefore, these are the only pairs.Final AnswerThe pairs are boxed{(4, 4)}, boxed{(5, 6)}, and boxed{(6, 5)}.</think>To find all pairs of positive integers ((p, q)) such that both the quadratic equations (x^2 - px + q = 0) and (x^2 - qx + p = 0) have integral solutions, we start by considering the conditions for the discriminants of these quadratic equations to be perfect squares. For the equation (x^2 - px + q = 0) to have integer solutions, the discriminant (p^2 - 4q) must be a perfect square. Similarly, for the equation (x^2 - qx + p = 0) to have integer solutions, the discriminant (q^2 - 4p) must also be a perfect square. We use Vieta's formulas to express the roots of these equations. Suppose the roots of the first equation are (a) and (b), and the roots of the second equation are (c) and (d). Then:- For the first equation: (a + b = p) and (ab = q)- For the second equation: (c + d = q) and (cd = p)This leads to the system of equations:1. (cd = a + b)2. (ab = c + d)By examining small values and using substitution, we find the following pairs ((p, q)):1. Case (a = 1): - (ab = c + d) gives (b = c + d) - Solving (cd = 1 + b) leads to (c = 2), (d = 3), (b = 5), resulting in (p = 6) and (q = 5).2. Case (a = 2): - (ab = c + d) gives (2b = c + d) - Solving (cd = 2 + b) leads to two solutions: - (c = 1), (d = 5), (b = 3), resulting in (p = 5) and (q = 6). - (c = 2), (d = 2), (b = 2), resulting in (p = 4) and (q = 4).3. Case (a = 3) and higher values do not yield new solutions beyond those already found.Additionally, considering the symmetry of the problem, we check the pairs ((5, 6)) and ((6, 5)), which are valid. The pair ((4, 4)) is symmetric and valid as well.Thus, the pairs of positive integers ((p, q)) that satisfy the given conditions are:[boxed{(4, 4)}, boxed{(5, 6)}, boxed{(6, 5)}],<think>Okay, so I need to find all pairs of positive integers (p; q) such that both quadratic equations x² - px + q = 0 and x² - qx + p = 0 have integer solutions. Hmm, let me break this down step by step.First, I remember that if a quadratic equation x² + bx + c = 0 has integer roots, then those roots must be integers, and by Vieta's formulas, the sum of the roots is -b and the product is c. But in our case, the equations are x² - px + q = 0 and x² - qx + p = 0. So, for the first equation, the sum of the roots is p and the product is q. Similarly, for the second equation, the sum of the roots is q and the product is p. Let me write that down.For the first equation x² - px + q = 0:Let the roots be a and b, both integers. Then:a + b = pab = qFor the second equation x² - qx + p = 0:Let the roots be c and d, both integers. Then:c + d = qcd = pSo, p and q are each sums and products of integers. So, we need to find p and q such that these relationships hold for some integers a, b, c, d. Also, since p and q are positive integers, the roots a, b must be positive integers as well? Wait, not necessarily. Because if the roots are negative, their sum would be negative, but p is positive. So, if a and b are integers such that a + b = p (positive), then either both are positive, or one is positive and the other is negative, but the positive one is larger in magnitude. But since their product q is also positive, that means both roots must be of the same sign. Since their sum p is positive, both roots must be positive. So, a and b are positive integers. Similarly for c and d.Therefore, all roots are positive integers. So, we have four positive integers a, b, c, d such that:a + b = pab = qc + d = qcd = pSo, substituting, we can think of p as a sum of a and b, and q as their product. Then q is the sum of c and d, and p is their product. So, p and q are each other's product and sum in terms of different pairs of integers. That seems a bit intertwined. Let me try to write equations connecting these.From the first equation, p = a + b and q = ab.From the second equation, q = c + d and p = cd.Therefore, we can write:ab = c + dcd = a + bSo, we need to find positive integers a, b, c, d such that ab = c + d and cd = a + b. Then p = a + b and q = ab (or equivalently p = cd and q = c + d). So, we need to solve these two equations for positive integers a, b, c, d.Hmm. This seems a bit abstract. Maybe I can assume that a ≤ b and c ≤ d to reduce the number of variables. Let me try substituting one equation into the other.From ab = c + d and cd = a + b, maybe we can substitute c + d with ab in the second equation. Wait, but cd is a + b, which is p, which is also equal to a + b. So, cd = a + b. But since ab = c + d, perhaps we can express c and d in terms of a and b, or vice versa.Alternatively, let's consider that both p and q are positive integers. Let me think if there are small integers that satisfy these conditions. Maybe start with small values of p and q and see if they fit.Let me try p = q. If p = q, then both equations would be x² - px + p = 0. Then the equation x² - px + p = 0 would need to have integer solutions. Let's check discriminant: p² - 4p must be a perfect square. So, p² - 4p = k² for some integer k. Then p² -4p -k² = 0. Hmm, this might not be straightforward. Let me check specific values.If p = q = 4, then the equation x² -4x +4 = 0 has a double root at x=2, which is integer. Similarly, the other equation would also have x=2 as a root. So, (4,4) is a possible pair. Wait, but p = q = 4, both equations have double root at 2. So, that works. But maybe there are other pairs where p ≠ q.Alternatively, let's take the equations:From first equation, p = a + b, q = ab.From the second equation, q = c + d, p = cd.Therefore, substituting p = cd into the first equation's p: cd = a + b.Similarly, substituting q = ab into the second equation's q: ab = c + d.Therefore, we have:cd = a + bab = c + dSo, these are the two equations. Let me try to link them.From the first equation: a + b = cdFrom the second equation: c + d = abSo, substituting a + b = cd into the second equation, c + d = ab = a*b.But a and b are positive integers such that a + b = cd. So, perhaps this is a system:a + b = cdc + d = abThis is a system of equations with variables a, b, c, d. All positive integers. Let me try to solve this system.Let me assume that a ≤ b and c ≤ d. Since they are positive integers, maybe start with small values for a and b and see if they can satisfy the equations.Let's try a = 1. Then, since a =1, from a + b = cd, we have 1 + b = cd. Also, from c + d = ab = 1*b = b. So, c + d = b. But we also have 1 + b = cd. So, substituting b = c + d into 1 + b = cd gives 1 + c + d = cd.So, equation: cd - c - d =1. Which can be rewritten as (c -1)(d -1) = 2.Since c and d are positive integers and c ≤ d, then (c -1)(d -1) = 2. The possible factorizations of 2 are 1*2 or 2*1. But since c ≤ d, then c -1 =1 and d -1=2, so c=2, d=3. Therefore, c=2, d=3. Then b = c + d = 5. Therefore, a=1, b=5. So, p = a + b =6, q = ab=5. Let's check if these satisfy the original equations.First equation: x² -6x +5 =0. Solutions: x=(6 ±√(36-20))/2=(6 ±√16)/2=(6 ±4)/2=5 or 1. Which are integers. Second equation: x² -5x +6=0. Solutions: x=(5 ±√(25-24))/2=(5 ±1)/2=3 or 2. Which are integers. So, (p; q) = (6;5) is a solution. Also, since a and b could be swapped, but since we ordered them, we don't need to consider duplicates here.Similarly, maybe there are other solutions when a=1. Wait, let's check if there are other factorizations. (c-1)(d-1)=2. If we considered c-1=2 and d-1=1, then c=3 and d=2, but since we assumed c ≤ d, this would reverse, so we can ignore that. So, only one solution when a=1.Now, let's try a=2. Then, a=2, so a + b = cd => 2 + b = cd. Also, c + d = ab =2b. So, c + d = 2b. So, substituting cd = 2 + b into c + d =2b. Let me express d =2b -c, then substituting into cd=2 +b:c*(2b -c) =2 +b=> 2b c -c² =2 +bLet's rearrange:2b c - b -c² -2=0Hmm, this seems a bit complicated. Maybe we can think in terms of possible values. Since c and d are positive integers with c ≤ d, and c + d =2b. Also, cd =2 +b. Let's see if possible.Given that c and d are positive integers, c ≤ d, c + d =2b. So, c ≤ b, d ≥b. Since c + d=2b, so average is b.Also, cd=2 +b. Let's try small values for b.Let me take b=1. Then cd=3, c + d=2. But c + d=2 and cd=3. The only positive integers c and d would have to be 1 and1, but 1*1=1≠3. Not possible.b=2: cd=4, c + d=4. So, possible pairs (1,3) product 3≠4, (2,2) product 4. So, c=2, d=2. Then cd=4=2 +2=4. Wait, but cd=2 +b=2 +2=4, which works. So, c=2, d=2. Then p=cd=2*2=4, q=ab=2*2=4. So, p=4, q=4. Let's check the original equations.First equation: x² -4x +4=0, which factors as (x-2)²=0, root x=2 (double root). Second equation: same, x² -4x +4=0. So, both equations have integer solutions. Therefore, (4,4) is a solution.Wait, but when a=2 and b=2, then p=2+2=4, q=2*2=4. But in this case, the roots for the first equation would be 2 and2, sum 4, product 4. For the second equation, roots would be 2 and2, sum 4, product4. So that's consistent. So (4,4) is another solution.Wait, but how did we get here? When a=2, b=2, but we started with a=2, and found that when b=2, c=2 and d=2. So that works. Let's check b=3. Then cd=2 +3=5, and c + d=2*3=6. So, find c and d such that c + d=6 and c*d=5. The solutions are c=1, d=5 (product 5) or c=5, d=1. But since c ≤ d, c=1, d=5. Then check if cd=5=2 +3=5, yes. Then p=cd=1*5=5, q=ab=2*3=6. Wait, p=cd=5? Wait, no. Wait, in this case, p= a + b=2 +3=5, q=ab=6. Then the second equation is x² -6x +5=0. Solutions: x=(6 ±√(36-20))/2=(6 ±√16)/2=(6 ±4)/2=5 or1, which are integers. So, that works. So, (p;q)=(5;6). Wait, but earlier when a=1, we had (6;5). So, perhaps (5,6) is another solution.Wait, but in this case, a=2, b=3, leading to p=5, q=6. Then the second equation's roots c=1, d=5, so sum=6, product=5, which matches. So, that's another solution. So (5,6) and (6,5) would both be solutions? Wait, but in the problem statement, the pair is (p; q). So, both equations must have integer solutions. So, if we have (5,6), then first equation is x² -5x +6=0, which factors as (x-2)(x-3)=0, roots 2 and3, integers. The second equation is x² -6x +5=0, factors as (x-1)(x-5)=0, roots1 and5, integers. Similarly, if we swap p and q, (6,5), first equation x² -6x +5=0, roots1 and5; second equation x² -5x +6=0, roots2 and3. So both (5,6) and (6,5) are solutions. But in our earlier case when a=1, we got (6,5), and when a=2, b=3, we got (5,6). So both are valid. So these are distinct pairs.Wait, so (4,4) is a solution, (5,6) and (6,5) are solutions. Are there more? Let's check.Continuing with a=2, b=4. Then cd=2 +4=6, and c + d=2*4=8. So, c + d=8, c*d=6. Solving, the possible pairs (c,d) would be (2,6), but 2*6=12≠6, or (1,7) product7≠6. So, no solution here.b=4: cd=2 +4=6, c + d=8. No integer solutions. So, no.b=5: cd=2 +5=7, c + d=10. Then c and d must sum to10, product7. But 7 is prime, so only 1 and7, but sum is8≠10. No solution.So, only when b=2 and b=3 in a=2 case, but b=2 gives (4,4), and b=3 gives (5,6). Wait, but when a=2, b=3, p=5, q=6. Then (5,6) is a solution.Now, moving to a=3. Then, a + b =cd, and c + d =3b.So, similar approach. Let's see:From a=3, so 3 + b = cd, and c + d =3b.Express d =3b -c, substitute into cd=3 +b:c*(3b -c) =3 +b3b c -c² =3 +bRearranged:3b c -c² -b -3=0Hmm, maybe try specific values for b. Let's take b=1. Then cd=4, c + d=3*1=3. So c + d=3 and c*d=4. Possible pairs:1 and2, product2≠4. Not possible.b=2: cd=5, c + d=6. Possible pairs:1*5=5, sum6. So c=1, d=5. Then cd=5=3 +2=5. So, p=cd=5, q=ab=3*2=6. Wait, but p should be a + b=3 +2=5, q=ab=6. Then, check second equation: x² -6x +5=0, roots1 and5, which are integers. So, (5,6) again. Wait, but this is the same solution as before.Wait, perhaps this is a different path leading to the same solution. So, when a=3, b=2, gives p=5, q=6. But since a and b are ordered as a ≤ b, perhaps a=3 and b=2 is not allowed, as 3 >2. So, actually, we should have a ≤ b, so a=3 would require b≥3. Wait, but in the previous case when a=2, b=3, but since a ≤ b, 2 ≤3. So, in the case when a=3, b must be at least3.So, for a=3, b ≥3. Let's try b=3. Then cd=3 +3=6, c + d=3*3=9. So, c + d=9, c*d=6. Looking for integers c,d where c + d=9 and c*d=6. Factors of6:1,6;2,3. Neither pair sums to9. So, no solution.b=4: cd=3 +4=7, c + d=12. c + d=12, c*d=7. Again, factors are1,7. Sum8≠12. No solution.b=5: cd=3 +5=8, c + d=15. No solutions.So, no solutions for a=3 except when b=2, which violates a ≤ b. Therefore, no new solutions here.Moving on to a=4. Then, a=4, so 4 +b=cd, c + d=4b.Express d=4b -c, substitute into cd=4 +b:c*(4b -c)=4 +b4b c -c² =4 +bRearranged:4b c -c² -b -4=0Again, try small values for b.b=1: cd=5, c + d=4. So, c + d=4, c*d=5. No integer solutions.b=2: cd=6, c + d=8. Possible pairs: (2,6) product12≠6, (3,5) product15≠6. Not possible.b=3: cd=7, c + d=12. No solutions.b=4: cd=8, c + d=16. Factors of8:1,8;2,4. Neither sum to16. No.So, no solutions here.a=5: Similarly, 5 +b=cd, c +d=5b.Trying b=1: cd=6, c +d=5. Possible pairs: (2,3) sum5, product6. So c=2, d=3. Then p=cd=6, q=ab=5*1=5. So, (p; q)=(6,5). Which is the reverse of the previous solution. So, this is again (6,5), same as before. But here, a=5 and b=1, but since we assume a ≤b, 5 ≤1 is false. So, invalid. So, no new solutions here.Proceeding to a=4, but tried that already. Maybe there's a pattern here that for a ≥3, the solutions either repeat previous ones or don't exist. Let's try to think of another approach.Let me consider that we have two equations:1. a + b = cd2. c + d = abWe can think of this as a system where (a,b) and (c,d) are pairs of positive integers satisfying these equations. Let me try to express variables in terms of each other.From equation1: cd = a + b. From equation2: ab = c + d.Let me substitute c + d = ab into equation1. So, cd = a + b, and ab = c + d. Then, replacing c + d with ab in equation1, we have cd = a + b. But c and d are positive integers such that their sum is ab and their product is a + b. So, c and d are roots of the quadratic equation x² - abx + (a + b) =0. For this equation to have integer solutions, the discriminant must be a perfect square.So discriminant D = (ab)² -4(a + b). This must be a perfect square. Let me write that:(ab)^2 -4(a + b) = k², for some integer k.Similarly, from equation2, ab = c + d. Then, c and d are roots of x² - (ab)x + (a + b)=0. So, the same logic applies here. So, for c and d to be integers, discriminant must be a perfect square.But since we already derived that, perhaps we can use this to find possible a and b. However, this seems complicated.Alternatively, maybe consider symmetry. Suppose that (p, q) is a solution, then (q, p) is also a solution. Because swapping p and q swaps the two equations, so if one pair works, the reverse might also work. Except when p=q, which is its own reverse. So, (4,4), (5,6), and (6,5) are solutions. Are there others?Wait, let's check if there's another possible solution. Let's try a=2, b=4. Then p=6, q=8. Then check if the second equation x² -8x +6=0 has integer roots. The discriminant is 64 -24=40, which is not a perfect square. So, no.Alternatively, trying a=1, b=4. Then p=5, q=4. Then second equation x² -4x +5=0. Discriminant 16 -20=-4, no real roots. So, invalid.Wait, but earlier when a=1, b=5, we had p=6, q=5. Then the second equation x² -5x +6=0, which factors as (x-2)(x-3)=0. So, that works.Alternatively, let's check for a=1, b=2. Then p=3, q=2. Second equation x² -2x +3=0. Discriminant 4 -12=-8, no real roots. Doesn't work.a=1, b=3: p=4, q=3. Second equation x² -3x +4=0. Discriminant9 -16=-7. Nope.a=1, b=4: p=5, q=4. Second equation x² -4x +5=0. Discriminant16 -20=-4. Nope.a=1, b=5: p=6, q=5. Which works. Then a=1, b=6: p=7, q=6. Second equation x² -6x +7=0. Discriminant36 -28=8, not a square. No.So, only when a=1 and b=5 do we get a valid solution. Similarly, a=2 and b=2 gives (4,4), a=2 and b=3 gives (5,6). Let me check a=3, b=3: Then p=6, q=9. Second equation x² -9x +6=0. Discriminant81 -24=57, not a square. No.a=3, b=4: p=7, q=12. Second equation x² -12x +7=0. Discriminant144 -28=116, not a square. No.So, up to here, we found (4,4), (5,6), (6,5). Are there any others?Wait, perhaps there's another approach. Let's consider that p and q must satisfy both equations having integer roots. So, for both quadratics x² -px + q and x² -qx + p to have integer roots. Let me think in terms of possible roots.Suppose the first equation x² - px + q =0 has roots m and n (integers). Then p = m + n, q = mn.The second equation x² - qx + p =0 has roots k and l (integers). Then q =k + l, p = kl.Therefore, we have:p = m + n = klq = mn = k + lSo, similar to before. Let me try to set variables here. Let me call m and n the roots of the first equation, and k and l the roots of the second. So:m + n = klmn = k + lSo, similar equations as before, just with different variables. So, again, need to find integers m, n, k, l such that these hold. Let me think if there's symmetry or some substitution.From mn =k + l and kl =m + n. So, substituting k + l = mn into kl =m + n. So, kl = m + n, but k + l = mn. So, k and l are roots of x² - mnx + (m + n)=0. Then discriminant must be a perfect square.Discriminant D = (mn)^2 -4(m + n). So, again, similar to before. Therefore, (mn)^2 -4(m +n) must be a perfect square.Alternatively, perhaps if we set m =k and n = l. Then, m + n = kl = m*n. So, m +n = mn. So, mn -m -n =0 => (m-1)(n-1)=1. Therefore, m-1=1 and n-1=1, so m=2, n=2. Then, p =2 +2=4, q=2*2=4. Which is the (4,4) solution. So, this is the case where both quadratics have the same roots. But in this case, both quadratics would have roots 2 and2. Wait, but x² -4x +4=0 and x² -4x +4=0, same equation. So, that's why p=q=4.Alternatively, if m and n are different from k and l. For example, in the case (5,6), the first quadratic x² -5x +6=0 has roots2 and3, sum5, product6. The second quadratic x² -6x +5=0 has roots1 and5, sum6, product5. So, here, m=2, n=3, k=1, l=5. Then, p=5=1*5, q=6=2*3. So, p = kl =1*5=5, q= mn=2*3=6. Which satisfies the equations.Similarly, the reverse case (6,5) would have p=6=2*3, q=5=1*5.So, these are solutions where the pairs (m,n) and (k,l) are different but interconnected.So, perhaps the only solutions are (4,4), (5,6), and (6,5). Let me check if there are any more.Let me consider m and n as 1 and5. Then p =1 +5=6, q=1*5=5. Then the second equation x² -5x +6=0 has roots2 and3. So, sum5, product6. Which is consistent. So that gives (6,5).Similarly, if m=1 and n=6, p=7, q=6. Then second equation x² -6x +7=0, which has discriminant 36 -28=8, not a square. So, invalid.Alternatively, m=1, n=4. Then p=5, q=4. Second equation x² -4x +5=0, discriminant negative. No.m=2, n=3. p=5, q=6. Second equation roots1 and5. So, kl=1*5=5, k + l=6. So, kl=5 and k + l=6. Then k and l are roots of x² -6x +5=0, which are1 and5. Correct.m=2, n=4. p=6, q=8. Second equation x² -8x +6=0. Discriminant64 -24=40, not square. No.m=2, n=5. p=7, q=10. Second equation x² -10x +7=0. Discriminant100 -28=72, not square.m=3, n=3. p=6, q=9. Second equation x² -9x +6=0. Discriminant81 -24=57, not square.m=3, n=4. p=7, q=12. Second equation x² -12x +7=0. Discriminant144 -28=116, not square.So, seems like no other solutions.Alternatively, maybe we can consider larger numbers, but likely not. Let me think if there's a mathematical way to bound the solutions.Suppose that p and q are positive integers. Let's assume that p ≤ q. Then, we can look for solutions where p ≤ q and then also consider the reverse. Let's suppose p ≤ q.From the equations:First equation: x² -px + q =0 has integer roots m and n. So, m +n =p, mn =q.Second equation: x² -qx +p =0 has integer roots k and l. So, k + l =q, kl =p.Since p ≤ q, then kl =p ≤ q =k + l. So, kl ≤k + l. Therefore, kl -k -l ≤0. Adding1 to both sides: kl -k -l +1 ≤1. Which is (k -1)(l -1) ≤1.Since k and l are positive integers, (k -1)(l -1) ≤1. Therefore, possible cases:Either (k -1)(l -1)=0 or (k -1)(l -1)=1.Case1: (k -1)(l -1)=0. Then either k=1 or l=1.Case2: (k -1)(l -1)=1. Then (k -1)=1 and (l -1)=1, so k=2, l=2.So, these are the possibilities.Let me analyze both cases.Case1: Either k=1 or l=1.Assume k=1. Then, kl =1*l =p. And k + l=1 +l =q. So, p=l and q=1 +l. Since p ≤q, l ≤1 +l, which is always true. So, possible l is any positive integer. But also, from the first equation, m +n =p =l, and mn =q =1 +l. Therefore, m +n =l and mn =1 + l. So, we need to find positive integers m and n such that m +n = l and mn =1 + l. Let me express n = l -m. Then:m(l -m) =1 + lml -m² =1 + lRearranged:m² -ml + l +1=0This is a quadratic in m:m² - l m + (l +1)=0Discriminant D= l² -4(l +1)= l² -4l -4. For m to be integer, D must be a perfect square. So:l² -4l -4 =k² for some integer k.So, l² -4l -4 -k²=0. Let me complete the square for l:l² -4l +4 -8 -k²=0 => (l -2)² - (k² +8)=0 => (l -2)² =k² +8So, (l -2)^2 -k^2=8 => (l -2 -k)(l -2 +k)=8Since l and k are positive integers, and l ≥1, k ≥1. Let me set a = l -2 -k and b= l -2 +k, so that a*b=8. Since a and b are integers and a ≤b, and both positive or both negative. But since l -2 +k > l -2 -k, and since their product is8, which is positive, both a and b must be positive. So possible pairs (a,b) where a ≤b and a*b=8:(1,8), (2,4)Then:For (1,8):a=1, b=8:l -2 -k=1l -2 +k=8Adding equations: 2(l -2)=9 => l -2=4.5 => l=6.5. Not integer. Disregard.For (2,4):a=2, b=4:l -2 -k=2l -2 +k=4Adding equations: 2(l -2)=6 => l -2=3 => l=5Subtracting equations: 2k=2 =>k=1Therefore, l=5, k=1.Check if this works:From above, l=5, so p=kl=1*5=5, q=k + l=1 +5=6. So, p=5, q=6. Which is a solution we found earlier. Then check m andn:From first equation, m +n =5, mn=6. So, roots2 and3. Valid.Similarly, if we consider l=1, then k=1. Wait, but earlier we considered k=1. Wait, perhaps I need to check another case where l=1. Wait, no. Wait, in this case, if l=5, k=1, which gives p=5, q=6.Alternatively, if we take the other factor pair (a,b)=(2,4), we get l=5. There are no other factor pairs for8. Hence, this is the only solution in this case.Similarly, if l=1, but then from kl =p, k=1, l=1, p=1, q=2. Let's check if this works. Then, first equation x² -1x +2=0. Discriminant1 -8=-7, no real roots. So, invalid.Alternatively, other cases where l=1. Maybe not.Case1: when k=1, leads to only one solution (5,6). Similarly, if we set l=1, then k=1, but that leads to invalid solutions. So, only (5,6) from this case.Case1 also includes when l=1. But if l=1, kl =k*1 =p, and q=k +1. Then p=k, q=k +1. From the first equation, m +n =p=k, mn=q=k +1. Then, similar to before, m andn are roots of x² -kx + (k +1)=0. The discriminant is k² -4(k +1)=k² -4k -4. For this to be a perfect square, same as previous. So, only when k=5, but then l=1. Wait, but l=1, k=5 would mean p=5*1=5, q=5 +1=6. Wait, but l=1, k=5, so p=kl=5*1=5, q=k +l=5 +1=6. Which is the same as previous. So, this gives (5,6). So, only this solution.Case2: (k -1)(l -1)=1. Then, (k -1)=1 and (l -1)=1. So, k=2, l=2. Therefore, kl=4, and k + l=4. Therefore, p=kl=4, q=k +l=4. So, (4,4) is the solution. Check first equation: x² -4x +4=0, roots2,2. Second equation: same. Valid.Therefore, the only possible solutions under the assumption p ≤q are (4,4) and (5,6). Then, considering p ≤q and q ≤p, the full set of solutions would be (4,4), (5,6), and (6,5).Hence, the pairs are (4,4), (5,6), and (6,5).Final AnswerThe pairs are boxed{(4, 4)}, boxed{(5, 6)}, and boxed{(6, 5)}.</think>To find all pairs of positive integers ((p, q)) such that both quadratic equations (x^2 - px + q = 0) and (x^2 - qx + p = 0) have integer solutions, we use Vieta's formulas for the roots of the quadratic equations.For the first equation (x^2 - px + q = 0) with roots (a) and (b) (both integers):- Sum of roots: (a + b = p)- Product of roots: (ab = q)For the second equation (x^2 - qx + p = 0) with roots (c) and (d) (both integers):- Sum of roots: (c + d = q)- Product of roots: (cd = p)We need to find pairs ((p, q)) such that:1. (a + b = p) and (ab = q)2. (c + d = q) and (cd = p)By substituting and solving these equations, we consider the following cases:1. Case (p = q): - Both equations become (x^2 - px + p = 0). The discriminant must be a perfect square: (p^2 - 4p). For (p = 4), the equation (x^2 - 4x + 4 = 0) has a double root (2), leading to the pair ((4, 4)).2. Case (p neq q): - For (p = 5) and (q = 6), the first equation (x^2 - 5x + 6 = 0) has roots (2) and (3), and the second equation (x^2 - 6x + 5 = 0) has roots (1) and (5). This gives the pairs ((5, 6)) and ((6, 5)).By checking all possible pairs and ensuring the discriminants are perfect squares, we find the valid solutions. Final AnswerThe pairs are (boxed{(4, 4)}), (boxed{(5, 6)}), and (boxed{(6, 5)}).]

🔑:To find all pairs of positive integers ((p, q)) such that both the equations (x^2 - px + q = 0) and (x^2 - qx + p = 0) have integral solutions, we proceed as follows:1. Roots of the first equation: The roots of the equation (x^2 - px + q = 0) are given by Vieta's formulas: [ a + b = p quad text{and} quad ab = q ] where (a) and (b) are integers.2. Roots of the second equation: Similarly, the roots of the equation (x^2 - qx + p = 0) are: [ c + d = q quad text{and} quad cd = p ] where (c) and (d) are integers.3. Inequalities derived from positivity: Since (a, b, c, d > 0), we have: [ (a-1)(b-1) geq 0 quad Rightarrow quad p - q + 1 geq 0 ] Similarly, from the second equation: [ (c-1)(d-1) geq 0 quad Rightarrow quad q - p + 1 geq 0 ] Combining these inequalities, we get: [ -1 leq p - q leq 1 ] Thus, (p - q in {-1, 0, 1}).4. Case 1: (p = q): If (p = q), the equation (x^2 - px + p = 0) must have integer solutions. The discriminant of this quadratic equation must be a perfect square: [ D = p^2 - 4p = k^2 quad text{for some integer } k ] This implies: [ p^2 - 4p - k^2 = 0 quad Rightarrow quad (p-2)^2 - k^2 = 4 ] Factoring, we get: [ (p-2-k)(p-2+k) = 4 ] The pairs ((p-2-k, p-2+k)) that multiply to 4 and have the same parity are: [ (2, 2) quad text{or} quad (-2, -2) ] Solving these, we get: [ p-2-k = 2 quad text{and} quad p-2+k = 2 quad Rightarrow quad p = 4 ] For (p = 4), the equation (x^2 - 4x + 4 = 0) has the double root (x = 2), which is an integer.5. Case 2: (p = q + 1): If (p = q + 1), the equation (x^2 - px + q = 0) must have integer solutions. Let the roots be (1) and (q): [ 1 + q = p quad Rightarrow quad p = q + 1 ] The second equation becomes: [ x^2 - qx + (q + 1) = 0 ] The discriminant must be a perfect square: [ D = q^2 - 4(q + 1) = k^2 quad Rightarrow quad q^2 - 4q - 4 = k^2 ] Factoring, we get: [ (q-2-k)(q-2+k) = 8 ] The pairs ((q-2-k, q-2+k)) that multiply to 8 and have the same parity are: [ (2, 4) quad text{or} quad (-2, -4) ] Solving these, we get: [ q-2-k = 2 quad text{and} quad q-2+k = 4 quad Rightarrow quad q = 5 ] Thus, (p = q + 1 = 6). The pair ((p, q) = (6, 5)) satisfies the conditions.6. Case 3: (p = q - 1): If (p = q - 1), the equation (x^2 - qx + p = 0) must have integer solutions. Let the roots be (1) and (p): [ 1 + p = q quad Rightarrow quad q = p + 1 ] The first equation becomes: [ x^2 - px + (p + 1) = 0 ] The discriminant must be a perfect square: [ D = p^2 - 4(p + 1) = k^2 quad Rightarrow quad p^2 - 4p - 4 = k^2 ] Factoring, we get: [ (p-2-k)(p-2+k) = 8 ] The pairs ((p-2-k, p-2+k)) that multiply to 8 and have the same parity are: [ (2, 4) quad text{or} quad (-2, -4) ] Solving these, we get: [ p-2-k = 2 quad text{and} quad p-2+k = 4 quad Rightarrow quad p = 5 ] Thus, (q = p + 1 = 6). The pair ((p, q) = (5, 6)) satisfies the conditions.Conclusion:The pairs ((p, q)) that satisfy the conditions are:[(p, q) = (4, 4), (6, 5), (5, 6)]The final answer is ( boxed{ (p, q) = (4, 4), (6, 5), (5, 6) } )

🧠:[<think>Okay, so I have this problem here: given real numbers x₁, x₂, ..., x₅, I need to find the least positive integer n such that if n distinct sums of the form x_p + x_q + x_r (with 1 ≤ p < q < r ≤ 5) are equal to zero, then all the x's must be zero. The question is asking for the smallest n with this property.First, let me understand the problem. We have five numbers, and we are looking at all possible triplets (since p < q < r). Each triplet sum is a sum of three distinct numbers from the five. There are C(5,3) = 10 such sums. So there are 10 possible sums. The problem is asking: what's the smallest number n such that if n of these sums are zero, then all the x's are forced to be zero. So, we need to find the minimal n where having n zeros among the 10 sums implies all x_i = 0.So the question is about the minimal n where the system of equations (each being a sum x_p + x_q + x_r = 0) has only the trivial solution when at least n such equations are imposed. And we need the least such n.First, let me think: How many equations (sums equal to zero) can we have without forcing all variables to be zero? If we have too few equations, there might be non-trivial solutions. So we need to determine the threshold n where beyond that, only the trivial solution exists.So the problem is similar to determining the minimal number of equations (each corresponding to a triplet sum) needed to force all variables to zero. The key is to find how many such equations can be satisfied with non-zero variables, and then the minimal n would be one more than that maximum.Alternatively, perhaps we need to find the maximum number of such sums that can be zero without all x_i being zero. Then the minimal n would be that maximum plus one.So, first step: What is the maximum number of triplet sums that can be zero without all variables being zero? Then the answer is that maximum plus one.Therefore, my task is to determine that maximum. Let's denote it as m. Then the minimal n is m + 1.So how can I find m? Let me try to construct a non-trivial solution (i.e., not all x_i zero) with as many triplet sums zero as possible.Alternatively, I can model this as a system of equations. Let x₁, x₂, x₃, x₄, x₅ be real numbers. The possible triplet sums are 10 in total. Suppose some of them are zero. How many can be zero without forcing all x_i to be zero?Let me consider variables x₁, x₂, x₃, x₄, x₅. Let me suppose that some of the triplet sums are zero. Each equation is of the form x_i + x_j + x_k = 0 for distinct i, j, k.So, if we have several such equations, they form a linear system. The question becomes: What is the minimal number of equations in such a system that forces the only solution to be the trivial one. Then, the answer would be the minimal n such that any system with n equations has only the trivial solution. But the problem is phrased differently: it's saying "if some n distinct sums are equal to zero", so it's not any n equations, but some n equations. Therefore, perhaps there exists a set of n equations (triplet sums) which, when set to zero, force all x_i to be zero, and n is the minimal such number. But we need to find the least n such that if any n equations (i.e., if there exists a set of n equations) that are all zero, then the variables must all be zero.Wait, the problem states: "if some n distinct sums of the form x_p + x_q + x_r are equal to 0, then x₁=x₂=...=x₅=0". So it's not that every set of n sums being zero implies all x_i zero, but rather that if some set of n sums are zero, then all x_i must be zero. Therefore, the minimal n such that the existence of n zero sums implies all variables zero.Therefore, we need to find the smallest n where having n zero triplet sums (for some choice of n triplets) necessarily leads to all variables being zero. So to find the minimal n where any non-trivial solution cannot have n zero sums.Therefore, the approach is: find the maximum number m such that there exists a non-trivial solution (not all x_i zero) with m zero triplet sums. Then, the minimal n is m + 1.Thus, my task reduces to finding the maximum m.To find m, I need to construct a non-trivial solution with as many zero triplet sums as possible.Let me think of possible configurations. Let me start with small cases.Suppose all variables are zero. Then all 10 sums are zero, but that's trivial.Suppose one variable is non-zero. Let's say x₁ = a ≠ 0, and others are zero. Then any triplet sum that includes x₁ will be a, and others will be zero. How many triplet sums include x₁? The number is C(4,2) = 6. So in this case, 6 sums will be equal to a, and the remaining 4 sums (those not involving x₁) will be zero. Wait, but in this case, if x₁ is non-zero, and the others are zero, then the sums involving x₁ are x₁ + 0 + 0 = x₁, so those 6 sums are equal to x₁, and the other 4 sums (without x₁) are 0. Therefore, if x₁ is non-zero, we have 4 sums equal to zero. So m is at least 4. So here, with this configuration, we can have 4 zero sums. Therefore, m ≥ 4.But maybe we can do better? Let me see.Suppose two variables are non-zero. Let's say x₁ = a, x₂ = b, and others are zero. Then the triplet sums:Triplets with both x₁ and x₂: x₁ + x₂ + x_k for k = 3,4,5. Each such sum is a + b + 0 = a + b. There are 3 such sums.Triplets with x₁ but not x₂: x₁ + x_j + x_k where j,k ≥3. There are C(3,2) = 3 such triplets. Each sum is a + 0 + 0 = a.Triplets with x₂ but not x₁: Similarly, x₂ + x_j + x_k, j,k ≥3. Also 3 triplets, each sum is b.Triplets with neither x₁ nor x₂: These are x₃ + x₄ + x₅, which is 0.Therefore, in this case, how many zero sums? The triplet with neither x₁ nor x₂: 1 zero sum. The triplet with x₁ + x₂ + x_k: if a + b = 0, then these 3 sums are zero. Then total zero sums would be 1 + 3 = 4. If a + b ≠ 0, then only 1 zero sum. So to maximize the number of zero sums, set a + b = 0, i.e., x₂ = -x₁. Then, triplet sums with both x₁ and x₂ will be zero (3 sums), and the triplet without x₁ and x₂ is also zero. So total 4 zero sums. Additionally, the triplets with only x₁ will be a, and triplets with only x₂ will be -a. So if a ≠ 0, then those sums are non-zero. Therefore, in this configuration, we can get 4 zero sums. So same as the previous case. So m is at least 4.Wait, but maybe we can have more.Suppose three variables are non-zero. Let's say x₁ = a, x₂ = b, x₃ = c, and x₄ = x₅ = 0.Then triplet sums:Triplets involving x₁, x₂, x₃: For example, x₁ + x₂ + x₃ = a + b + c. Then, if a + b + c = 0, that sum is zero. But other triplets:Triplets with two of x₁, x₂, x₃ and one zero: For example, x₁ + x₂ + x₄ = a + b + 0. If we set a + b = 0, then this sum is zero. Similarly, x₁ + x₃ + x₄ = a + c + 0, if a + c = 0. Similarly, x₂ + x₃ + x₄ = b + c + 0. Similarly for x₅.Wait, this is getting complex. Let's see:Suppose we set x₁ = a, x₂ = -a, x₃ = b, x₄ = -b, x₅ = 0.Then, let's compute the triplet sums.Triplet sums:1,2,3: a - a + b = b1,2,4: a - a - b = -b1,2,5: a - a + 0 = 01,3,4: a + b - b = a1,3,5: a + b + 0 = a + b1,4,5: a - b + 0 = a - b2,3,4: -a + b - b = -a2,3,5: -a + b + 0 = -a + b2,4,5: -a - b + 0 = -a - b3,4,5: b - b + 0 = 0So, in this case, how many zero sums? Let's check each:1,2,5: 03,4,5: 0So two zero sums. The others depend on a and b. If we set some relationships, perhaps more zeros.Alternatively, suppose x₁ = 1, x₂ = 1, x₃ = -2, x₄ = 1, x₅ = -1. Wait, maybe this is arbitrary. Let me think of a symmetric case.Alternatively, suppose x₁ = x₂ = x₃ = a, x₄ = x₅ = -a. Then:Triplet sums with three a's: C(3,3) = 1 sum: a + a + a = 3a. If we set 3a = 0, then a = 0, trivial.Triplet sums with two a's and one -a: C(3,2)*C(2,1) = 3*2 = 6 sums. Each sum is 2a - a = a.Triplet sums with one a and two -a's: C(3,1)*C(2,2) = 3*1 = 3 sums. Each sum is a - 2a = -a.Triplet sums with three -a's: Not possible, since there are only two -a's.Therefore, if a ≠ 0, then all sums are either 3a, a, -a, none of which can be zero unless a = 0. So in this case, the only zero sum is if 3a = 0, which implies a = 0. So this configuration does not help.Alternatively, suppose two variables are a, two variables are -a, and one is 0. Let's say x₁ = x₂ = a, x₃ = x₄ = -a, x₅ = 0.Compute the triplet sums:1,2,3: a + a - a = a1,2,4: a + a - a = a1,2,5: a + a + 0 = 2a1,3,4: a - a - a = -a1,3,5: a - a + 0 = 01,4,5: a - a + 0 = 02,3,4: a - a - a = -a2,3,5: a - a + 0 = 02,4,5: a - a + 0 = 03,4,5: -a - a + 0 = -2aSo, the zero sums are:1,3,5; 1,4,5; 2,3,5; 2,4,5. That's four zero sums. Also, 3,4,5 is -2a. If we set a = 0, trivial. Otherwise, a ≠ 0, then we have four zero sums. So in this configuration, we have four zero sums. So again, m ≥4.Wait, but in this case, with two a's, two -a's, and one 0, we get four zero sums. So m is at least 4 here as well. So, same as the previous cases.Is it possible to get more than four?Let me try another configuration. Suppose three variables are a, and two variables are -a. Let x₁ = x₂ = x₃ = a, x₄ = x₅ = -a.Then compute the triplet sums:Triplet with three a's: 1,2,3: 3aTriplets with two a's and one -a: 1,2,4;1,2,5;1,3,4;1,3,5;2,3,4;2,3,5: Each is 2a - a = aTriplets with one a and two -a's: 1,4,5;2,4,5;3,4,5: Each is a - 2a = -aSo, zero sums would require 3a =0, a =0, which is trivial. So no non-trivial zeros here. So not helpful.Alternatively, maybe another configuration. Let me try setting some variables in a geometric progression or something. Suppose x₁ = 1, x₂ = ω, x₃ = ω², x₄ = ω³, x₅ = ω⁴, where ω is a fifth root of unity. But maybe over real numbers, this is not possible. Since we're dealing with real numbers, complex roots won't work.Alternatively, suppose arranging the variables such that several triplet sums cancel out. For example, maybe a system where each variable is included in the same number of positive and negative terms. But this is vague.Alternatively, think of the problem as a linear algebra question. Each triplet sum is a linear equation: x_p + x_q + x_r = 0. We have 10 possible equations, and we need to find the minimal n such that any n equations from these 10 force the variables to be zero. But actually, the problem says "if some n distinct sums are equal to zero", so it's not any n equations, but there exists a set of n equations (triplet sums) that are zero, which implies all variables are zero. So we need to find the smallest n such that there does not exist a non-trivial solution with n zeros.Therefore, equivalently, we need the minimal n where the only solution common to any n equations (triplet sums) is the trivial solution. Wait, no. Because the problem states that if there exists a set of n equations (triplet sums) that are zero, then the variables must all be zero. So the contrapositive is: if the variables are not all zero, then no set of n triplet sums are all zero. Therefore, the maximal number of zeros you can have with non-trivial solutions is n-1. Therefore, the minimal n is one more than the maximal number of zeros possible in a non-trivial solution.Therefore, to find n, we need to find the maximum number m such that there exists a non-trivial solution with m triplet sums equal to zero, then n = m +1.So the key is to find the maximal m.In previous examples, we saw that m is at least 4.Is it possible to have m = 5? Let's try to find a non-trivial solution with five zero sums.Suppose we have five equations (triplet sums) equal to zero. Let me try to set up such a system.Let me consider variables x₁, x₂, x₃, x₄, x₅. Let me pick five triplet sums and set them to zero. Then, check if the system has a non-trivial solution.Alternatively, perhaps arranging the variables so that several dependencies occur. For example, if we have variables such that x₁ + x₂ + x₃ = 0, x₁ + x₄ + x₅ = 0, x₂ + x₄ + x₅ = 0, x₃ + x₄ + x₅ = 0, and another equation. Wait, but this might over-constrain.Alternatively, let me attempt to construct such a system.Suppose we have variables a, b, c, d, e.Let me set the following equations:1. a + b + c = 02. a + b + d = 03. a + c + d = 04. b + c + d = 05. a + b + e = 0Let me see if this system has a non-trivial solution.Subtract equation 1 from equation 2: (a + b + d) - (a + b + c) = d - c = 0 ⇒ d = c.From equation 1: a + b + c = 0From equation 3: a + c + d = a + c + c = a + 2c = 0 ⇒ a = -2c.From equation 4: b + c + d = b + c + c = b + 2c = 0 ⇒ b = -2c.From equation 1: a + b + c = (-2c) + (-2c) + c = -3c = 0 ⇒ c = 0.Then, a = 0, b = 0, d = 0.From equation 5: a + b + e = 0 + 0 + e = e = 0 ⇒ e = 0.Thus, only trivial solution. So this system forces all variables to zero.But maybe another system?Let me try another set.Set equations:1. a + b + c = 02. a + d + e = 03. b + d + e = 04. c + d + e = 05. a + b + d = 0Let's solve this system.From equations 2, 3, 4: Each involves d + e. Let me subtract equation 2 from equation 3: (b + d + e) - (a + d + e) = b - a = 0 ⇒ b = a.Similarly, subtract equation 2 from equation 4: (c + d + e) - (a + d + e) = c - a = 0 ⇒ c = a.So, b = a, c = a.From equation 1: a + b + c = a + a + a = 3a = 0 ⇒ a = 0 ⇒ b = 0, c = 0.From equation 2: a + d + e = 0 + d + e = 0 ⇒ d + e = 0.From equation 5: a + b + d = 0 + 0 + d = d = 0 ⇒ d = 0 ⇒ e = 0.Thus, only trivial solution. Hmm.Another attempt. Let's try using equations that overlap more.Let equations be:1. a + b + c = 02. a + b + d = 03. a + c + e = 04. b + d + e = 05. c + d + e = 0Let me solve these equations.From 1 and 2: Subtract 1 from 2: (a + b + d) - (a + b + c) = d - c = 0 ⇒ d = c.From 3: a + c + e = 0 ⇒ e = -a - c.From 4: b + d + e = b + c + (-a - c) = b - a = 0 ⇒ b = a.From 5: c + d + e = c + c + (-a - c) = c - a = 0 ⇒ c = a.Thus, c = a, d = c = a, b = a, e = -a - c = -a - a = -2a.Now, equation 1: a + b + c = a + a + a = 3a = 0 ⇒ a = 0. Then all variables are zero. So again, only trivial solution.Hmm. It's challenging to get a non-trivial solution with five equations. Maybe five is too much. What if I take four equations?Wait, in the previous examples with four equations, like the case with two a's, two -a's, and one zero, we had four zero sums. Let's see if such a system is possible.Suppose variables are a, a, -a, -a, 0. Let me denote them as x₁ = a, x₂ = a, x₃ = -a, x₄ = -a, x₅ = 0.Then, as computed earlier, the zero sums are:1. x₁ + x₃ + x₅ = a + (-a) + 0 = 02. x₁ + x₄ + x₅ = a + (-a) + 0 = 03. x₂ + x₃ + x₅ = a + (-a) + 0 = 04. x₂ + x₄ + x₅ = a + (-a) + 0 = 0So, four zero sums. The other sums are non-zero if a ≠ 0.Therefore, in this configuration, we can have four zero sums with non-trivial solutions. So m = 4. Hence, the minimal n is 5. Therefore, the answer is 5. Wait, but let me check if m can be higher than 4.Suppose we can find a non-trivial solution with five zero sums. Then m would be 5, hence n =6. But in my previous attempts, systems with five equations forced the variables to zero. Therefore, maybe m=4, so n=5.But let me try to see if there exists a non-trivial solution with five zero sums.Suppose we have variables x₁, x₂, x₃, x₄, x₅. Let me try to set five triplet sums to zero.Let me pick the following five triplet sums:1. x₁ + x₂ + x₃ = 02. x₁ + x₄ + x₅ = 03. x₂ + x₄ + x₅ = 04. x₃ + x₄ + x₅ = 05. x₁ + x₂ + x₄ = 0Let me attempt to solve this system.From equations 2 and 3:Equation 2: x₁ + x₄ + x₅ = 0Equation 3: x₂ + x₄ + x₅ = 0Subtract equation 2 from equation 3: x₂ - x₁ = 0 ⇒ x₂ = x₁From equation 5: x₁ + x₂ + x₄ = 0. Since x₂ = x₁, this becomes 2x₁ + x₄ = 0 ⇒ x₄ = -2x₁From equation 2: x₁ + x₄ + x₅ = x₁ -2x₁ + x₅ = -x₁ + x₅ = 0 ⇒ x₅ = x₁From equation 3: x₂ + x₄ + x₅ = x₁ -2x₁ + x₁ = 0 ⇒ 0 = 0, which is redundant.From equation 4: x₃ + x₄ + x₅ = x₃ -2x₁ + x₁ = x₃ - x₁ = 0 ⇒ x₃ = x₁From equation 1: x₁ + x₂ + x₃ = x₁ + x₁ + x₁ = 3x₁ = 0 ⇒ x₁ = 0 ⇒ x₂ = x₃ = x₄ = x₅ = 0.Thus, only trivial solution. So this system also forces all variables to zero.Another attempt with different equations.Let equations be:1. x₁ + x₂ + x₃ = 02. x₁ + x₂ + x₄ = 03. x₁ + x₃ + x₄ = 04. x₂ + x₃ + x₄ = 05. x₁ + x₂ + x₅ = 0Solve:From 1 and 2: Subtract 1 from 2: x₄ - x₃ = 0 ⇒ x₄ = x₃From 3: x₁ + x₃ + x₄ = x₁ + 2x₃ = 0 ⇒ x₁ = -2x₃From 4: x₂ + x₃ + x₄ = x₂ + 2x₃ = 0 ⇒ x₂ = -2x₃From 1: x₁ + x₂ + x₃ = (-2x₃) + (-2x₃) + x₃ = -3x₃ = 0 ⇒ x₃ = 0 ⇒ x₁ = 0, x₂ = 0, x₄ = 0From 5: x₁ + x₂ + x₅ = 0 + 0 + x₅ = x₅ = 0 ⇒ x₅ = 0All variables zero. Again, trivial solution.Alternatively, pick five equations with some overlap.Suppose equations:1. x₁ + x₂ + x₃ = 02. x₁ + x₄ + x₅ = 03. x₂ + x₄ + x₅ = 04. x₃ + x₄ + x₅ = 05. x₁ + x₂ + x₄ = 0Wait, this is similar to a previous system. Let me check:From equations 2 and 3: x₂ = x₁From equation 5: 2x₁ + x₄ = 0 ⇒ x₄ = -2x₁From equation 2: x₁ + x₄ + x₅ = x₁ -2x₁ + x₅ = -x₁ + x₅ = 0 ⇒ x₅ = x₁From equation 4: x₃ + x₄ + x₅ = x₃ -2x₁ + x₁ = x₃ -x₁ =0 ⇒ x₃ = x₁From equation 1: x₁ + x₂ + x₃ = 3x₁ =0 ⇒ x₁=0 ⇒ all zero. Again trivial.It seems challenging to find a non-trivial solution with five zeros. Therefore, perhaps m=4, so n=5. Therefore, the answer is 5.But wait, let's confirm this by another approach. Suppose that the maximum number of zero triplet sums in a non-trivial solution is 4. Then, if we have 5 zero triplet sums, it must be trivial. Therefore, the minimal n is 5.But how can we be sure that m=4 is indeed the maximum?Earlier, we constructed a case with four zeros: two variables as a, two as -a, and one as zero. Then, four triplet sums are zero. Let me count them again:Variables: x₁ = a, x₂ = a, x₃ = -a, x₄ = -a, x₅ = 0.Triplet sums:1. x₁ + x₂ + x₃ = a + a -a = a2. x₁ + x₂ + x₄ = a + a -a = a3. x₁ + x₂ + x₅ = a + a +0 = 2a4. x₁ + x₃ + x₄ = a -a -a = -a5. x₁ + x₃ + x₅ = a -a +0 = 06. x₁ + x₄ + x₅ = a -a +0 = 07. x₂ + x₃ + x₄ = a -a -a = -a8. x₂ + x₃ + x₅ = a -a +0 = 09. x₂ + x₄ + x₅ = a -a +0 = 010. x₃ + x₄ + x₅ = -a -a +0 = -2aSo the zero sums are positions 5,6,8,9. That's four zeros. So indeed, in this case, there are four zero sums. If a ≠0, then this is a non-trivial solution with four zero sums. Therefore, m=4. Therefore, the minimal n is 5. Thus, the answer is 5.But to make sure, we need to confirm that no non-trivial solution can have five zero sums. Suppose there exists a non-trivial solution with five zero sums. Then, we need to derive a contradiction.Assume that there exists a non-trivial solution (not all x_i zero) with five triplet sums equal to zero. Let's try to see if this is possible.Suppose we have five equations (sums) equal to zero. Each equation is x_p + x_q + x_r =0.These five equations form a linear system. If the system has a non-trivial solution, then the rank of the system must be less than 5 (since there are 5 variables). However, each equation is a linear combination of the variables. Let me attempt to analyze the rank.Suppose we have five equations. The rank of the coefficient matrix would determine the solution space. If the rank is 5, then the only solution is trivial. If the rank is less than 5, there are non-trivial solutions.Therefore, to have a non-trivial solution with five zero sums, the coefficient matrix of the five equations must have rank less than 5.But in the previous attempts, when setting up five equations, the system ended up with rank 5, forcing the trivial solution. So perhaps, in any five equations, the system is full rank? Not necessarily. Maybe there exists a set of five equations that are linearly dependent, allowing a non-trivial solution.Wait, for example, take four equations that correspond to the four zeros in the previous case (positions 5,6,8,9), and add another equation. Let's see:In the case where x₁ = a, x₂ = a, x₃ = -a, x₄ = -a, x₅ = 0, the four zero sums are:5. x₁ + x₃ + x₅ = 06. x₁ + x₄ + x₅ = 08. x₂ + x₃ + x₅ = 09. x₂ + x₄ + x₅ = 0If we add another equation, say equation 10: x₃ + x₄ + x₅ = -2a. If a ≠0, this is non-zero. But if we take another equation from the non-zero ones, like equation 1: x₁ + x₂ + x₃ = a. If we set this to zero, we would require a =0, leading to trivial solution. So adding any of the other equations (which are non-zero in this configuration) would require a =0, hence trivial.Alternatively, maybe there's another configuration where five equations can be zero.Alternatively, consider variables arranged in such a way that multiple dependencies exist. For example, if variables satisfy x₁ + x₂ + x₃ =0, x₁ + x₄ + x₅=0, x₂ + x₄ + x₅=0, x₃ + x₄ + x₅=0, and another equation. But as seen before, this leads to trivial solutions.Alternatively, maybe three variables are zero, and two are inverses. Wait, but if three variables are zero, then any triplet sum involving two zeros would be equal to the third variable. So if x₁ = a, x₂ = -a, and x₃ = x₄ = x₅ =0. Then, the triplet sums would be:Triplets with x₁ and x₂: 1,2,3: a -a +0=0; 1,2,4: same; 1,2,5: same; 1,2,3 etc. Wait, actually, there are C(3,1)=3 triplet sums with x₁, x₂, and one zero: each is a -a +0=0. Then, triplet sums with x₁ and two zeros: 1,3,4: a +0 +0 = a; similar for others. Triplet sums with x₂ and two zeros: -a. And triplet sums with three zeros:0.So in this case, the zero sums are the three triplets with x₁, x₂, and one zero, plus the triplet with three zeros. But in this case, the variables x₃, x₄, x₅ are zero, so triplet 3,4,5 is zero. So total zero sums: 3 +1 =4. So again, four zeros. If we set x₁ = a, x₂ = -a, and x₃ = x₄ = x₅ =0, then we have four zero sums. Adding another equation to this would require another sum to be zero, which would force a =0.Therefore, in this case, m=4, same as before.Alternatively, perhaps if we have variables in an arithmetic progression? Let me try.Let x₁ = 1, x₂ = 2, x₃ = 3, x₄ = 4, x₅ = -10. Then the sum x₁ +x₂ +x₃ +x₄ +x₅ =0. Wait, but the problem is about triplet sums. But maybe this is not helpful.Alternatively, maybe consider variables such that each variable is the sum of two others. But I need to think of a different approach.Alternatively, think in terms of vector spaces. Each triplet sum is a vector in the space of linear combinations. The question is about the minimal number of such vectors that span the entire space, such that the only solution is trivial.But the problem is not exactly that. It's about the existence of a set of n vectors (triplet sums) such that if their sum is zero, then all variables are zero. The minimal n for which this is true.Another angle: if we have n triplet sums equal to zero, how much information do we get about the variables? Each triplet sum equation reduces the degrees of freedom by 1, assuming linear independence. If the equations are independent, then n equations would allow solving for n variables. Since there are 5 variables, n=5 equations could potentially determine all variables, but depending on the equations, they might not be independent.But if the equations are independent, then 5 equations would uniquely determine the variables (if the system is consistent), which would force the variables to be zero. However, if the equations are dependent, then even with 5 equations, there could be non-trivial solutions.But in the previous examples, when I set up five equations, they turned out to be independent, forcing the variables to zero. But is there a way to have five dependent equations that allow non-trivial solutions?Suppose we take four equations from the previous four-zero configuration and add a fifth equation that is a linear combination of the previous four. Then, the system would have a non-trivial solution. But since the equations are over real numbers and each equation is a sum of variables, linear combinations would involve coefficients, but we are only considering equations that are exactly the sums. So adding a fifth equation that is a linear combination might not correspond to an actual triplet sum.Alternatively, maybe there is a set of five triplet sums that are linearly dependent as equations. For example, suppose we have equations:1. x₁ + x₂ + x₃ =02. x₁ + x₂ + x₄ =03. x₁ + x₃ + x₄ =04. x₂ + x₃ + x₄ =05. x₁ + x₂ + x₅ =0This is the system we tried earlier, which forces all variables to zero. But what if we take another equation instead of equation 5? Let's replace equation 5 with another equation.Suppose equations:1. x₁ + x₂ + x₃ =02. x₁ + x₂ + x₄ =03. x₁ + x₃ + x₄ =04. x₂ + x₃ + x₄ =05. x₁ + x₃ + x₅ =0Let's solve this system.From equations 1 and 2: subtract 1 from 2: x₄ - x₃ =0 ⇒ x₄ =x₃.From equation 3: x₁ + x₃ + x₄ =x₁ +2x₃=0 ⇒ x₁=-2x₃From equation 4: x₂ + x₃ + x₄ =x₂ +2x₃=0 ⇒x₂=-2x₃From equation 1: x₁ +x₂ +x₃= -2x₃ -2x₃ +x₃= -3x₃=0 ⇒x₃=0 ⇒x₁=0, x₂=0, x₄=0.From equation 5: x₁ +x₃ +x₅=0 +0 +x₅=0⇒x₅=0.Thus, all variables zero. Again, trivial solution.Alternatively, take a different set of equations.Suppose:1. x₁ + x₂ + x₃ =02. x₁ + x₄ + x₅ =03. x₂ + x₄ + x₅ =04. x₃ + x₄ + x₅ =05. x₁ + x₂ + x₄ =0This system, as solved earlier, also forces all variables to zero.It seems that regardless of the combination of five equations, the system forces the variables to zero, implying that five equations are enough to force trivial solution. But wait, in our previous non-trivial example with four zeros, the fifth equation would have to be non-zero. So if we only take four equations that correspond to the zeros in the non-trivial case, and then consider adding a fifth equation, that fifth equation would have to be one of the non-zero sums, which in the non-trivial case is non-zero. Therefore, in such a scenario, the fifth equation cannot be zero unless the variables are trivial. Therefore, any five equations would include at least one non-zero sum in the non-trivial case, hence forcing a contradiction.Therefore, in any non-trivial solution, there can be at most four zero sums. Therefore, m=4, so the minimal n is 5.Hence, the answer is boxed{5}.Final Answerboxed{5}</think>Given real numbers (x_1, x_2, ldots, x_5), we need to find the least positive integer (n) such that if (n) distinct sums of the form (x_p + x_q + x_r) (with (1 le p < q < r le 5)) are equal to zero, then all (x_i) must be zero.First, we note that there are (binom{5}{3} = 10) possible triplet sums. The goal is to determine the smallest (n) such that the existence of (n) zero sums implies all (x_i = 0).We explored various configurations of non-trivial solutions (where not all (x_i) are zero) to find the maximum number of zero sums possible. Key configurations considered include:1. Two variables (a) and (-a) with others zero, resulting in 4 zero sums.2. Two pairs of variables (a, a) and (-a, -a) with one zero, resulting in 4 zero sums.We found that configurations with non-trivial solutions can have at most 4 zero sums. For example, setting (x_1 = a), (x_2 = a), (x_3 = -a), (x_4 = -a), and (x_5 = 0) results in 4 zero sums. Attempting to construct a system with 5 zero sums consistently led to trivial solutions, indicating that 5 equations (triplet sums) force all variables to zero. This suggests that the maximum number of zero sums in a non-trivial solution is 4, and thus the minimal (n) is 5.Therefore, the least positive integer (n) such that if (n) distinct triplet sums are zero, then all (x_i) must be zero is (boxed{5}).,<think>Okay, let's see. The problem is about finding the least positive integer n such that if n distinct sums of the form x_p + x_q + x_r (with 1 ≤ p < q < r ≤ 5) are equal to zero, then all the variables x_1, x_2, ..., x_5 must be zero. So, we need to figure out the smallest n where having that many zero three-term sums forces all the variables to be zero.First, let me understand the setup. We have five variables x_1 to x_5. The sums in question are combinations of three distinct variables. Since we're choosing 3 out of 5, the total number of such sums is C(5,3) = 10. So there are 10 possible sums. The question is: what's the minimal n so that if n of these sums are zero, then all variables must be zero.Our task is to find the least n where this implication holds. So, if we have fewer than n sums equal to zero, it's possible that the variables aren't all zero, but once we hit n zeros, they must all be zero. So, n is the threshold here.To approach this, maybe I can think about constructing non-trivial solutions (i.e., not all variables zero) that have as many three-term sums zero as possible. Then, the maximum number of such sums that can be zero without forcing all variables to be zero would be one less than n. So, if the maximum number of zero sums possible without all variables being zero is m, then n would be m + 1.Therefore, we need to find the maximum m such that there exists a non-trivial solution (not all variables zero) with m zero three-term sums. Then, the answer is m + 1.So, the problem reduces to figuring out the maximum number of zero sums possible in a non-trivial solution, and then adding 1 to get the minimal n that forces all variables to be zero.Alright, let's start by considering the possible configurations of the variables that can lead to multiple zero sums.Since we have five variables, perhaps there's some symmetry or linear dependencies we can exploit. Let's think in terms of linear algebra. Each three-term sum is a linear equation: x_p + x_q + x_r = 0. If we have several such equations, they form a system of linear equations. If the system has full rank (i.e., the equations are independent enough), then the only solution would be the trivial one. So, the question is how many equations (zero sums) are needed for the system to have only the trivial solution.But linear algebra might not directly apply here because the problem is not about linear independence of the equations but rather about the minimal number of equations (zero sums) such that any such set of equations forces the variables to zero. But perhaps linear algebra can still give some insight.Alternatively, maybe we can think of specific cases where variables are set in such a way to maximize the number of zero sums without all being zero. For example, if some variables are equal, or negatives of each other, etc.Let me try to construct such a case.Suppose two variables are equal to a, two variables are equal to b, and one variable is equal to c. Then, perhaps choosing a, b, c such that certain combinations sum to zero.Alternatively, maybe assign variables such that x_1 = x_2 = x_3 = a, and x_4 = x_5 = b. Then, the three-term sums would be combinations of three variables from the first three or mixing with the last two.But this might not be the right approach. Let me try a different angle.Suppose we have variables arranged such that x_1 + x_2 + x_3 = 0, x_1 + x_2 + x_4 = 0, x_1 + x_2 + x_5 = 0. Then, subtracting these equations, we get x_3 = x_4 = x_5. Let's call this common value a. Then, from x_1 + x_2 + a = 0, so x_1 + x_2 = -a. Now, other three-term sums would involve different combinations. For example, x_1 + x_3 + x_4 = x_1 + a + a = x_1 + 2a. If this is zero, then x_1 = -2a. Then x_2 = -a - x_1 = -a - (-2a) = a. So, variables would be x_1 = -2a, x_2 = a, x_3 = x_4 = x_5 = a. Let's check how many three-term sums are zero.First, the three sums mentioned initially: x_1 + x_2 + x_3 = (-2a) + a + a = 0, similarly x_1 + x_2 + x_4 = same, and x_1 + x_2 + x_5 = same. So three zeros.Then, let's check others:x_1 + x_3 + x_4 = (-2a) + a + a = 0. Wait, that's another zero. So that's four.x_1 + x_3 + x_5 = same as above: (-2a) + a + a = 0. That's five.x_1 + x_4 + x_5 = (-2a) + a + a = 0. That's six.x_2 + x_3 + x_4 = a + a + a = 3a. Not zero unless a=0.x_2 + x_3 + x_5 = same as above: 3a.x_2 + x_4 + x_5 = same.x_3 + x_4 + x_5 = a + a + a = 3a.x_1 + x_2 + x_3, x_1 + x_2 + x4, x_1 + x_2 + x5: all zero.x_1 + x3 + x4, x_1 + x3 + x5, x_1 + x4 + x5: all zero.So total of six zeros. Wait, that's six three-term sums equal to zero. So in this case, variables are not all zero (as long as a ≠ 0), and we have six zero sums. So this suggests that m is at least six. Therefore, the minimal n would be seven? Because if six zeros don't force all variables to zero, but seven might.But wait, maybe there's a configuration with more zeros. Let's check.In the example above, six sums are zero, and the other four sums are 3a. So if a is non-zero, then those four sums are non-zero. So total zeros: six.Is it possible to have more than six zeros?Alternatively, maybe arranging variables in such a way that more sums cancel.Suppose we set variables such that x_1 + x_2 + x_3 = 0, x_1 + x_4 + x_5 = 0, x_2 + x_4 + x_5 = 0, x_3 + x_4 + x_5 = 0. Let's see.From the first equation: x_1 + x_2 + x_3 = 0.From the second: x_1 + x_4 + x_5 = 0.Third: x_2 + x_4 + x_5 = 0.Fourth: x_3 + x_4 + x_5 = 0.Subtracting the second and third equations: (x_1 + x_4 + x_5) - (x_2 + x_4 + x_5) = x_1 - x_2 = 0 ⇒ x_1 = x_2.Similarly, subtracting third and fourth: (x_2 + x_4 + x_5) - (x_3 + x_4 + x_5) = x_2 - x_3 = 0 ⇒ x_2 = x_3.From the first equation: x_1 + x_2 + x_3 = 3x_1 = 0 ⇒ x_1 = x_2 = x_3 = 0.Then, from the second equation: 0 + x_4 + x_5 = 0 ⇒ x_4 = -x_5.But then, variables x_4 and x_5 are negatives. So, if x_4 = a, x_5 = -a. Then, the three-term sums involving these two variables would be:For example, x_1 + x_4 + x_5 = 0 + a - a = 0.Similarly, x_2 + x_4 + x_5 = 0 + a - a = 0.Same with x_3 + x_4 + x_5.But the other sums, like x_4 + x_5 + x_1, etc., but since x_1, x_2, x_3 are zero, any sum involving two of them and one of x_4 or x_5 would be x_4 or x_5. For example, x_1 + x_2 + x_4 = 0 + 0 + a = a ≠ 0. Similarly, other sums would be a or -a. So, in this case, the number of zero sums would be the three sums from x_1, x_2, x_3 with x_4, x_5, which are three, plus the sums of x_4, x_5 with two zeros. Wait, but three-term sums with two zeros and one non-zero would just be the non-zero variable. So, if x_1, x_2, x_3 are zero, then any sum involving two of them and another variable would just be that variable. So, for example, x_1 + x_2 + x_4 = x_4. If x_4 is non-zero, then that sum is non-zero.So in this case, the total number of zero sums would be the three sums: x_1 + x_2 + x_3 = 0, x_1 + x_4 + x_5 = 0, x_2 + x_4 + x_5 = 0, x_3 + x_4 + x_5 = 0. Wait, but actually, in this setup, we have four equations:1. x_1 + x_2 + x_3 = 02. x_1 + x_4 + x_5 = 03. x_2 + x_4 + x_5 = 04. x_3 + x_4 + x_5 = 0But as we saw, solving these leads to x_1 = x_2 = x_3 = 0 and x_4 = -x_5. So, then, the three-term sums that are zero would be:- All sums involving x_1, x_2, x_3 (only one sum: x_1 + x_2 + x_3)Wait, no. Wait, the three-term sums are all combinations of three variables. So, in this case:If x_1, x_2, x_3 are zero, and x_4 = a, x_5 = -a, then:- Any sum with all three variables from x_1, x_2, x_3: only one sum, which is x_1 + x_2 + x_3 = 0.- Any sum with two variables from x_1, x_2, x_3 and one from x_4 or x_5: these would be x_1 + x_2 + x_4 = 0 + 0 + a = a, x_1 + x_2 + x5 = -a, and similarly others. Each of these would be either a or -a. Since a ≠ 0, these sums are non-zero. There are C(3,2)*2 = 3*2=6 such sums.- Any sum with one variable from x_1, x_2, x_3 and two from x_4, x_5: but since we need three variables, so one from the first three and two from the last two. But there are only two variables in the last two, so to pick two from them, we need to pick both x4 and x5. So, the sums would be x1 + x4 + x5 = 0, x2 + x4 + x5 = 0, x3 + x4 + x5 = 0. Each of these is 0 + a - a = 0. So that's three more zero sums.- The sum x4 + x5 + x_i for i=1,2,3, but that's the same as the above.- The sum x4 + x5 + x_ something else, but we only have five variables. Wait, perhaps I'm overcomplicating.Wait, total three-term sums:1. All three from x1, x2, x3: 1 sum, which is 0.2. Two from x1, x2, x3 and one from x4, x5: there are C(3,2)*C(2,1) = 3*2=6 sums, each equal to a or -a.3. One from x1, x2, x3 and two from x4, x5: C(3,1)*C(2,2)= 3*1=3 sums, each equal to 0 (since x4 + x5 = 0, and adding a zero from x1, x2, x3).4. All three from x4, x5 and one more: but we can't have three from x4, x5 since there's only two variables. So that's not possible.So total zero sums: 1 (from case 1) + 3 (from case 3) = 4. So, in this case, we have four zero sums. But earlier, in the first example, we had six zero sums. So that first example gives more zeros. Therefore, that's a better example for maximizing m.So, in that first case, we had six zeros. Let me check again.In the first example:x1 = -2a, x2 = a, x3 = a, x4 = a, x5 = a.Then, the three-term sums:1. x1 + x2 + x3 = (-2a) + a + a = 0.2. x1 + x2 + x4 = same as above: 0.3. x1 + x2 + x5 = 0.4. x1 + x3 + x4 = (-2a) + a + a = 0.5. x1 + x3 + x5 = 0.6. x1 + x4 + x5 = 0.So that's six zeros.Then, the remaining four sums:7. x2 + x3 + x4 = a + a + a = 3a.8. x2 + x3 + x5 = 3a.9. x2 + x4 + x5 = 3a.10. x3 + x4 + x5 = 3a.So, six zeros, four non-zeros.Therefore, in this case, m=6. So, the minimal n would be m+1=7.But is this the maximum possible m? Let's see if we can get more than six zeros.Suppose we try a different assignment. Let me think. Maybe assign variables such that three of them are a, and two of them are -a. Let's see:Let x1 = x2 = x3 = a, and x4 = x5 = -a.Then, the three-term sums:1. x1 + x2 + x3 = 3a.2. x1 + x2 + x4 = a + a - a = a.3. x1 + x2 + x5 = a + a - a = a.4. x1 + x3 + x4 = same as above: a.5. x1 + x3 + x5 = a.6. x1 + x4 + x5 = a -a -a = -a.7. x2 + x3 + x4 = a + a -a = a.8. x2 + x3 + x5 = a.9. x2 + x4 + x5 = a -a -a = -a.10. x3 + x4 + x5 = -a.So, in this case, all sums are either 3a, a, -a. If a ≠ 0, then none of these are zero. So, in this case, the number of zero sums is zero, which is worse than the previous example.Hmm. So that's not helpful.Another idea: Maybe set two variables to a, two variables to -a, and one variable to 0. Let's see.x1 = a, x2 = a, x3 = -a, x4 = -a, x5 = 0.Now, compute the three-term sums:1. x1 + x2 + x3 = a + a -a = a.2. x1 + x2 + x4 = a + a -a = a.3. x1 + x2 + x5 = a + a + 0 = 2a.4. x1 + x3 + x4 = a -a -a = -a.5. x1 + x3 + x5 = a -a + 0 = 0.6. x1 + x4 + x5 = a -a + 0 = 0.7. x2 + x3 + x4 = a -a -a = -a.8. x2 + x3 + x5 = a -a + 0 = 0.9. x2 + x4 + x5 = a -a + 0 = 0.10. x3 + x4 + x5 = -a -a + 0 = -2a.So, in this case, we have three zero sums: sums 5, 6, 8, 9. Wait, no:Wait sum 5: x1 + x3 + x5 = 0.Sum 6: x1 + x4 + x5 = 0.Sum 8: x2 + x3 + x5 = 0.Sum 9: x2 + x4 + x5 = 0.So four zero sums. The rest are non-zero. So that's four zeros. Not as good as the previous example with six zeros.Alternatively, maybe another configuration where three variables are a, one is -a, and one is 0.x1 = x2 = x3 = a, x4 = -a, x5 = 0.Compute the three-term sums:1. x1 + x2 + x3 = 3a.2. x1 + x2 + x4 = a + a -a = a.3. x1 + x2 + x5 = a + a + 0 = 2a.4. x1 + x3 + x4 = a + a -a = a.5. x1 + x3 + x5 = a + a + 0 = 2a.6. x1 + x4 + x5 = a -a + 0 = 0.7. x2 + x3 + x4 = a + a -a = a.8. x2 + x3 + x5 = a + a + 0 = 2a.9. x2 + x4 + x5 = a -a + 0 = 0.10. x3 + x4 + x5 = a -a + 0 = 0.So here, sums 6, 9, 10 are zero: three zeros. Not better.So, the first example with six zeros seems better. Let me check another possibility.Suppose three variables are a, and two variables are b, with a and b related in some way.Let x1 = x2 = x3 = a, x4 = x5 = b.Then, the three-term sums:1. x1 + x2 + x3 = 3a.2. x1 + x2 + x4 = 2a + b.3. x1 + x2 + x5 = 2a + b.4. x1 + x3 + x4 = 2a + b.5. x1 + x3 + x5 = 2a + b.6. x1 + x4 + x5 = a + 2b.7. x2 + x3 + x4 = 2a + b.8. x2 + x3 + x5 = 2a + b.9. x2 + x4 + x5 = a + 2b.10. x3 + x4 + x5 = a + 2b.So, in order to have some zeros, we need 3a = 0, 2a + b = 0, or a + 2b = 0.If 3a = 0, then a = 0. Then 2a + b = b, so if we set a = 0, and then set b = 0, all variables are zero. But if a = 0 and b ≠ 0, then the sums 6,9,10 would be 2b, but the other sums (2a + b) would be b, which is non-zero. So, zeros would be only sum 1. Not helpful.If we set 2a + b = 0 and a + 2b = 0. Let's see: solving these equations.From 2a + b = 0 ⇒ b = -2a.From a + 2b = 0 ⇒ a + 2*(-2a) = a -4a = -3a = 0 ⇒ a = 0 ⇒ b = 0. So again, trivial solution.Thus, non-trivial solutions require inconsistent equations here, so it's not possible to have multiple zeros in this configuration.Alternatively, perhaps set variables as x1 = a, x2 = a, x3 = -a, x4 = -a, x5 = 0. Wait, I think we tried this earlier and got four zeros.Alternatively, maybe set two variables to a, two variables to b, and one to c, and find a relationship between them to have multiple zeros.Let x1 = x2 = a, x3 = x4 = b, x5 = c.Then, the three-term sums:1. x1 + x2 + x3 = 2a + b.2. x1 + x2 + x4 = 2a + b.3. x1 + x2 + x5 = 2a + c.4. x1 + x3 + x4 = a + 2b.5. x1 + x3 + x5 = a + b + c.6. x1 + x4 + x5 = a + b + c.7. x2 + x3 + x4 = a + 2b.8. x2 + x3 + x5 = a + b + c.9. x2 + x4 + x5 = a + b + c.10. x3 + x4 + x5 = 2b + c.So, how can we get multiple zeros here?Suppose we set 2a + b = 0, a + 2b = 0, and 2a + c = 0, and 2b + c = 0.From 2a + b = 0 and a + 2b = 0:Multiply the first equation by 2: 4a + 2b = 0.Subtract the second equation: 4a + 2b - (a + 2b) = 3a = 0 ⇒ a = 0.Then, from 2a + b = 0 ⇒ b = 0.From 2a + c = 0 ⇒ c = 0.From 2b + c = 0 ⇒ c = 0.So, only trivial solution. Not helpful.Alternatively, set some of the other equations to zero. For example, set 2a + b = 0 and a + b + c = 0.From 2a + b = 0 ⇒ b = -2a.Then, a + b + c = a - 2a + c = -a + c = 0 ⇒ c = a.Then, check other sums:1. 2a + b = 0.2. same as 1.3. 2a + c = 2a + a = 3a.4. a + 2b = a + 2*(-2a) = a -4a = -3a.5. a + b + c = 0.6. same as 5.7. same as 4.8. same as 5.9. same as 5.10. 2b + c = 2*(-2a) + a = -4a + a = -3a.So, zeros are sums 1,2,5,6,8,9: total of six zeros. The other sums (3,4,7,10) are 3a, -3a, -3a, -3a. If a ≠ 0, these are non-zero. So, here, we have six zero sums again. So, similar to the first example. So this is another configuration with six zeros.So, in this case, variables are x1 = x2 = a, x3 = x4 = -2a, x5 = a. Let's check:x1 = a, x2 = a, x3 = -2a, x4 = -2a, x5 = a.Three-term sums:1. x1 + x2 + x3 = a + a -2a = 0.2. x1 + x2 + x4 = same as above: 0.3. x1 + x2 + x5 = a + a + a = 3a.4. x1 + x3 + x4 = a -2a -2a = -3a.5. x1 + x3 + x5 = a -2a + a = 0.6. x1 + x4 + x5 = a -2a + a = 0.7. x2 + x3 + x4 = a -2a -2a = -3a.8. x2 + x3 + x5 = a -2a + a = 0.9. x2 + x4 + x5 = a -2a + a = 0.10. x3 + x4 + x5 = -2a -2a + a = -3a.So, zeros are sums 1,2,5,6,8,9: six zeros. So, same as before. So, this configuration is similar but different variables. So, same number of zeros.Therefore, seems like six is achievable. So, m=6. Therefore, the minimal n is 7.But is there a way to get more than six zeros? Let's see.Suppose we try seven zeros. If we can construct a non-trivial solution with seven three-term sums zero, then m would be seven, and n would be eight. But if it's impossible, then six is the maximum, so n=7.So, let's try to see if seven zeros are possible.Suppose we have seven three-term sums equal to zero. Let's attempt to construct such a case.Suppose variables x1, x2, x3, x4, x5.Assume that seven of the ten sums are zero. Let's pick seven sums and set them to zero, then solve for variables.But this would involve solving seven equations, which is a lot. Maybe the system is overdetermined, leading to only the trivial solution.Alternatively, maybe there's a clever assignment.Wait, let's think about linear algebra. Each three-term sum is an equation: x_p + x_q + x_r = 0. So, if we have seven such equations, the system could potentially have only the trivial solution, but maybe not. If the seven equations are dependent, then perhaps a non-trivial solution exists.But this is getting complicated. Alternatively, maybe use the previous examples to see if we can add another zero.In the first example, we had six zeros. Let's see if we can modify that to get a seventh.In the first example, variables were x1 = -2a, x2 = a, x3 = a, x4 = a, x5 = a. The zeros were all sums that included x1 and x2, or x1 and two others. The non-zeros were the sums of three variables from x2, x3, x4, x5. Since x2 = a and x3 = x4 = x5 = a, those sums were 3a.Suppose we set x2 = a, x3 = a, x4 = a, x5 = a, and x1 = -3a. Then, let's see:Sum x1 + x2 + x3 = -3a + a + a = -a ≠ 0. So, not helpful.Alternatively, set x1 = -ka, and adjust k to get more zeros.Wait, in the first example, x1 was set to -2a so that x1 + x2 + x3 = 0, and since x2 = a, x3 = a, then x1 = -2a. Then, other sums that included x1 and two of the a's would be zero.But if x1 is -3a, then x1 + x_i + x_j = -3a + a + a = -a ≠ 0. So that doesn't help.Alternatively, maybe set some variables to different values. Let's think of variables such that more combinations sum to zero.Suppose we have two variables equal to a, two variables equal to b, and one variable equal to c, with a, b, c related such that several combinations sum to zero.Suppose x1 = a, x2 = a, x3 = b, x4 = b, x5 = c.Then, the three-term sums:1. x1 + x2 + x3 = 2a + b.2. x1 + x2 + x4 = 2a + b.3. x1 + x2 + x5 = 2a + c.4. x1 + x3 + x4 = a + 2b.5. x1 + x3 + x5 = a + b + c.6. x1 + x4 + x5 = a + b + c.7. x2 + x3 + x4 = a + 2b.8. x2 + x3 + x5 = a + b + c.9. x2 + x4 + x5 = a + b + c.10. x3 + x4 + x5 = 2b + c.To get multiple zeros, we can set 2a + b = 0, 2a + c = 0, a + 2b = 0, and 2b + c = 0.From 2a + b = 0 ⇒ b = -2a.From 2a + c = 0 ⇒ c = -2a.From a + 2b = a + 2*(-2a) = a -4a = -3a = 0 ⇒ a = 0 ⇒ b = 0, c = 0. So trivial solution.Alternatively, set 2a + b = 0 and a + b + c = 0.From 2a + b = 0 ⇒ b = -2a.Then a + (-2a) + c = -a + c = 0 ⇒ c = a.Then, 2b + c = 2*(-2a) + a = -4a + a = -3a.So, unless a = 0, which leads to trivial solution, 2b + c ≠ 0.So, zeros would be sums 1,2,5,6,8,9 (assuming a + b + c = 0). So six zeros again. So, same as previous examples.Therefore, seems like six is the maximum number of zero sums possible with non-trivial solutions.Therefore, the minimal n is seven. So, if seven three-term sums are zero, then all variables must be zero. But we need to confirm that seven zeros indeed force all variables to zero.Alternatively, suppose we assume that seven three-term sums are zero. Let's see if we can derive that all variables must be zero.Alternatively, think of the variables as vectors in some space. Wait, but perhaps using linear algebra. Each three-term sum is an equation. If we have seven equations, how many variables do we have? Five variables. So, a system of seven equations with five variables. If these equations are independent, then the only solution is trivial.But linear dependence is possible. However, with seven equations, even if they are dependent, the system might still have only the trivial solution.But to check whether seven equations can force all variables to zero, we need to see if the rank of the coefficient matrix is five. If the rank is five, then the only solution is trivial.But this requires knowing the specific equations. Since the problem states "if some n distinct sums... are equal to zero", so we need to choose n equations such that any such choice of n equations with zeros forces the variables to zero. But the problem says "if some n distinct sums... are equal to zero", i.e., regardless of which n sums are zero? Or does it mean that if there exists a set of n sums equal to zero?Wait, the problem says: "if some n distinct sums of the form... are equal to zero", then all variables are zero. So, the least n such that any occurrence of n zero sums (regardless of which ones) implies all variables are zero. Wait, no. Actually, the wording is: "if some n distinct sums... are equal to zero", so it's the existence of n zero sums. So, the least n such that if there exists a set of n zero sums, then variables must be all zero.Therefore, the problem is to find the minimal n where there does not exist a non-trivial solution with n zero sums. So, the maximal m (number of zero sums in a non-trivial solution) is six, hence n = m + 1 = 7.But need to confirm that indeed seven zeros cannot exist in a non-trivial solution.Alternatively, suppose there is a non-trivial solution with seven zero sums. Then, we have seven equations of the form x_p + x_q + x_r = 0. Let's attempt to construct such a system.Assuming variables x1, x2, x3, x4, x5.Suppose seven of the ten possible three-term sums are zero. Let's choose these seven sums:1. x1 + x2 + x3 = 02. x1 + x2 + x4 = 03. x1 + x2 + x5 = 04. x1 + x3 + x4 = 05. x1 + x3 + x5 = 06. x1 + x4 + x5 = 07. x2 + x3 + x4 = 0Now, let's try to solve this system.From equations 1, 2, and 3:From 1: x1 + x2 + x3 = 0From 2: x1 + x2 + x4 = 0Subtract 1 - 2: x3 - x4 = 0 ⇒ x3 = x4Similarly, from 1: x1 + x2 + x3 = 0From 3: x1 + x2 + x5 = 0Subtract 1 - 3: x3 - x5 = 0 ⇒ x3 = x5Thus, x3 = x4 = x5 = a (let's say).From equation 1: x1 + x2 + a = 0 ⇒ x1 + x2 = -a.From equation 4: x1 + x3 + x4 = x1 + a + a = x1 + 2a = 0 ⇒ x1 = -2a.Then, from x1 + x2 = -a ⇒ -2a + x2 = -a ⇒ x2 = a.So, x1 = -2a, x2 = a, x3 = x4 = x5 = a.Now, check equation 7: x2 + x3 + x4 = a + a + a = 3a.But equation 7 is supposed to be zero ⇒ 3a = 0 ⇒ a = 0.Therefore, all variables are zero. So, in this case, the system of seven equations forces all variables to zero. But this is a specific set of seven equations. However, in the previous example with six equations, we had a non-trivial solution.But in this case, adding a seventh equation (equation 7) forced a = 0, leading to trivial solution. Therefore, this particular set of seven equations cannot be satisfied by a non-trivial solution. However, the question is whether there exists a different set of seven equations that can be satisfied by a non-trivial solution.Alternatively, try another set of seven equations. Let's pick seven sums that exclude the ones that were problematic in the previous example.Suppose we pick the seven sums:1. x1 + x2 + x3 = 02. x1 + x2 + x4 = 03. x1 + x3 + x4 = 04. x1 + x3 + x5 = 05. x1 + x4 + x5 = 06. x2 + x3 + x4 = 07. x2 + x3 + x5 = 0Let's try solving these.From equations 1 and 2:1: x1 + x2 + x3 = 02: x1 + x2 + x4 = 0Subtracting: x3 - x4 = 0 ⇒ x3 = x4.From equations 1 and 3:1: x1 + x2 + x3 = 03: x1 + x3 + x4 = 0But x3 = x4, so equation 3 becomes x1 + x3 + x3 = x1 + 2x3 = 0.From equation 1: x1 + x2 + x3 = 0 ⇒ x2 = -x1 - x3.From equation 3: x1 + 2x3 = 0 ⇒ x1 = -2x3.Substitute x1 = -2x3 into equation 1: -2x3 + x2 + x3 = 0 ⇒ -x3 + x2 = 0 ⇒ x2 = x3.So, x2 = x3, x1 = -2x3, x4 = x3.Now, from equation 4: x1 + x3 + x5 = 0 ⇒ (-2x3) + x3 + x5 = -x3 + x5 = 0 ⇒ x5 = x3.From equation 5: x1 + x4 + x5 = (-2x3) + x3 + x3 = 0 ⇒ 0 = 0, which is always true.From equation 6: x2 + x3 + x4 = x3 + x3 + x3 = 3x3 = 0 ⇒ x3 = 0 ⇒ x1 = 0, x2 = 0, x4 = 0, x5 = 0.Thus, trivial solution again.Hmm. So, again, the seventh equation forces x3 = 0.Alternatively, try a different set of seven equations.Suppose we exclude equations that involve x5 and focus on equations with variables x1, x2, x3, x4.But we have to include x5 in some equations since it's one of the variables. Let's try another approach.Pick seven equations such that each variable is involved in several equations. For example:1. x1 + x2 + x3 = 02. x1 + x2 + x4 = 03. x1 + x3 + x4 = 04. x1 + x4 + x5 = 05. x2 + x3 + x5 = 06. x2 + x4 + x5 = 07. x3 + x4 + x5 = 0Let's solve this system.From 1: x1 + x2 + x3 = 0From 2: x1 + x2 + x4 = 0Subtract 1 - 2: x3 - x4 = 0 ⇒ x3 = x4From 3: x1 + x3 + x4 = x1 + 2x3 = 0 ⇒ x1 = -2x3From 4: x1 + x4 + x5 = -2x3 + x3 + x5 = -x3 + x5 = 0 ⇒ x5 = x3From 5: x2 + x3 + x5 = x2 + x3 + x3 = x2 + 2x3 = 0 ⇒ x2 = -2x3From 6: x2 + x4 + x5 = -2x3 + x3 + x3 = (-2x3) + x3 + x3 = 0 ⇒ 0 = 0 (redundant)From 7: x3 + x4 + x5 = x3 + x3 + x3 = 3x3 = 0 ⇒ x3 = 0Thus, x3 = 0 ⇒ x4 = 0, x1 = 0, x5 = 0, x2 = 0. Trivial solution again.So, this system also forces all variables to zero.It seems that in any system of seven equations, the solution is forced to be trivial. Therefore, if seven three-term sums are zero, variables must be all zero. Hence, n=7.But wait, we need to confirm that there's no non-trivial solution with seven zeros. From the above attempts, whenever we set up seven equations, it leads to the trivial solution. So, this suggests that seven zeros cannot coexist in a non-trivial solution. Therefore, the minimal n is seven.Therefore, the answer is boxed{7}.Final Answerboxed{7}</think>To find the least positive integer ( n ) such that if ( n ) distinct sums of the form ( x_p + x_q + x_r ) (with ( 1 le p < q < r le 5 )) are equal to 0, then all ( x_1, x_2, ldots, x_5 ) must be zero, we need to determine the maximum number of zero sums possible without all variables being zero, and then add 1.1. Example with Six Zero Sums: - Consider ( x_1 = -2a ), ( x_2 = a ), ( x_3 = a ), ( x_4 = a ), ( x_5 = a ). - The following sums are zero: - ( x_1 + x_2 + x_3 ) - ( x_1 + x_2 + x_4 ) - ( x_1 + x_2 + x_5 ) - ( x_1 + x_3 + x_4 ) - ( x_1 + x_3 + x_5 ) - ( x_1 + x_4 + x_5 ) - This configuration results in six zero sums, showing that six zero sums are possible without all variables being zero.2. Attempting Seven Zero Sums: - Constructing a system of seven equations with non-trivial solutions leads to contradictions, forcing all variables to zero. - For example, solving a system of seven equations such as: - ( x_1 + x_2 + x_3 = 0 ) - ( x_1 + x_2 + x_4 = 0 ) - ( x_1 + x_2 + x_5 = 0 ) - ( x_1 + x_3 + x_4 = 0 ) - ( x_1 + x_3 + x_5 = 0 ) - ( x_1 + x_4 + x_5 = 0 ) - ( x_2 + x_3 + x_4 = 0 ) - Solving these equations leads to all variables being zero, showing that seven zero sums force a trivial solution.Thus, the maximum number of zero sums possible without all variables being zero is six. Therefore, the least positive integer ( n ) such that any ( n ) zero sums force all variables to zero is ( 6 + 1 = 7 ).The answer is (boxed{7}).]

🔑:1. Claim: The least positive integer ( n ) such that if some ( n ) distinct sums of the form ( x_p + x_q + x_r ) (with ( 1 le p < q < r le 5 )) are equal to ( 0 ), then ( x_1 = x_2 = cdots = x_5 = 0 ) is ( 7 ).2. Verification that ( n = 6 ) is not sufficient: - Consider the set of numbers ( {1, 1, 1, 1, -2} ). - Any sum of the form ( x_p + x_q + x_r ) where ( x_p, x_q ) are ( 1 ) and ( x_r ) is (-2) will be ( 0 ). - There are ( binom{4}{2} = 6 ) such sums, which are all equal to ( 0 ). - However, ( x_1, x_2, x_3, x_4 neq 0 ) and ( x_5 = -2 ), so ( x_1 = x_2 = x_3 = x_4 = 1 neq 0 ). - Thus, ( n = 6 ) is not sufficient.3. Proof that ( n = 7 ) is sufficient: - There are ( binom{5}{3} = 10 ) possible sums of the form ( x_p + x_q + x_r ). - If 7 of these sums are zero, we need to show that ( x_1 = x_2 = x_3 = x_4 = x_5 = 0 ).4. Application of the Pigeonhole Principle: - Each sum ( x_p + x_q + x_r ) involves 3 terms. - With 7 sums, there are ( 7 times 3 = 21 ) terms in total. - Since there are only 5 numbers, by the Pigeonhole Principle, at least one number must appear in at least ( lceil frac{21}{5} rceil = 5 ) of these sums. - Without loss of generality, assume ( x_1 ) appears in at least 5 of these sums.5. Analysis of the sums involving ( x_1 ): - Suppose ( x_1 ) appears in the sums ( x_1 + x_2 + x_3 = 0 ), ( x_1 + x_2 + x_4 = 0 ), ( x_1 + x_2 + x_5 = 0 ), ( x_1 + x_3 + x_4 = 0 ), and ( x_1 + x_3 + x_5 = 0 ). - From ( x_1 + x_2 + x_3 = 0 ) and ( x_1 + x_2 + x_4 = 0 ), we get: [ x_1 + x_2 + x_3 = 0 quad text{and} quad x_1 + x_2 + x_4 = 0 ] Subtracting these equations, we get: [ x_3 - x_4 = 0 implies x_3 = x_4 ] - Similarly, from ( x_1 + x_2 + x_3 = 0 ) and ( x_1 + x_2 + x_5 = 0 ), we get: [ x_3 - x_5 = 0 implies x_3 = x_5 ] - Thus, ( x_3 = x_4 = x_5 ).6. Further simplification: - From ( x_1 + x_3 + x_4 = 0 ) and ( x_1 + x_3 + x_5 = 0 ), we get: [ x_1 + x_3 + x_4 = 0 quad text{and} quad x_1 + x_3 + x_5 = 0 ] Since ( x_3 = x_4 = x_5 ), these equations are consistent.7. Conclusion: - Since ( x_1 ) appears in at least 5 sums, and we have 7 sums in total, there must be at least one sum that does not include ( x_1 ). - Suppose ( x_2 + x_3 + x_4 = 0 ). Given ( x_3 = x_4 ), we have: [ x_2 + x_3 + x_3 = 0 implies x_2 + 2x_3 = 0 implies x_2 = -2x_3 ] - Since ( x_2 = x_3 ), we get: [ x_3 = 0 implies x_2 = 0 ] - Therefore, ( x_2 = x_3 = x_4 = x_5 = 0 ). - Finally, from any of the initial sums involving ( x_1 ), we get ( x_1 = 0 ).Thus, ( x_1 = x_2 = x_3 = x_4 = x_5 = 0 ).The final answer is ( boxed{ 7 } ).