🧠:[<think>Okay, so I need to figure out the smallest possible length of the sum of 13 vectors on a cube. The vectors are chosen on one main diagonal and all edges. The cube has edge length 1, and the main diagonal is √3. Let me try to visualize this.First, let me recall that a cube has 12 edges. Each edge is a vector of length 1. The main diagonal goes from one corner of the cube to the opposite corner, and its length is √3. The problem says directions are chosen on one main diagonal and all edges, so that's 12 edges plus 1 main diagonal, totaling 13 vectors. We need to assign a direction to each of these vectors (either positive or negative along their respective axes or diagonals) such that when we add all these vectors together, the resulting vector has the smallest possible length.Wait, actually, the problem says "directions are chosen on a main diagonal and all edges". So does that mean each edge and the main diagonal can be assigned a direction, either forwards or backwards? For edges, each edge is between two vertices, so direction could be from one vertex to the other. Similarly, the main diagonal has a direction from one corner to the opposite. So we can choose the direction (sign) for each edge and the main diagonal, and then sum all these vectors. Our goal is to choose these directions such that the magnitude of the sum is minimized.So, first, let me model the cube in a coordinate system. Let's place the cube with one corner at the origin (0,0,0) and extending to (1,1,1). The edges are along the x, y, and z axes. Each edge can be represented as a vector in one of the coordinate directions. The main diagonal would be the vector from (0,0,0) to (1,1,1), which is (1,1,1). But we can choose its direction either as (1,1,1) or (-1,-1,-1).Similarly, for each edge, we can choose the direction. For example, an edge along the x-axis from (0,0,0) to (1,0,0) can be represented as vector (1,0,0) or (-1,0,0) if we reverse it. But actually, each edge is a separate vector. Wait, but there are 12 edges. Each edge is between two vertices. Let me think: a cube has 12 edges: for each of the three axes (x, y, z), there are 4 edges along that axis. For example, along the x-axis, there are edges from (0,0,0) to (1,0,0), (0,1,0) to (1,1,0), (0,0,1) to (1,0,1), and (0,1,1) to (1,1,1). Similarly for y and z axes.But if we can choose the direction of each edge, then each edge vector can be either in the positive or negative direction along its axis. Wait, but each edge is a physical edge of the cube; their directions are fixed in space. Wait, no, maybe the problem is allowing us to choose the direction (orientation) of each edge vector. That is, for each edge, instead of having a fixed vector, we can choose whether it points from, say, (0,0,0) to (1,0,0) or the reverse. So each edge's vector can be +e or -e, where e is the standard edge direction. Similarly, the main diagonal can be either from (0,0,0) to (1,1,1) or the reverse.So, effectively, we have 12 edge vectors, each of which is a unit vector along one of the coordinate axes, but with a sign we can choose, and one main diagonal vector, which is a vector of length √3 in the direction (1,1,1) or (-1,-1,-1). Then, we need to choose signs for all these vectors such that their sum has minimal length.Therefore, the problem reduces to choosing ±1 for each of the 12 edge vectors (each along x, y, or z) and ±1 for the main diagonal vector (along (1,1,1)), such that the vector sum of all these has the smallest possible magnitude.To formalize this, let me denote:- For each coordinate axis (x, y, z), there are 4 edges along that axis. Each of these can be assigned a direction: + or - along the axis. So for each axis, we have 4 vectors that are ±1 in that axis direction.Therefore, for each axis, the total contribution from the edges will be the sum of 4 ±1's in that axis. Let me denote for the x-axis: sum of 4 terms each ±1, so total x-component is S_x = a_1 + a_2 + a_3 + a_4, where each a_i is ±1. Similarly, S_y for y-axis and S_z for z-axis.Then, the main diagonal vector is either (1,1,1) or (-1,-1,-1), so its contribution is either (d, d, d) where d is ±√3. Wait, no. Wait, the main diagonal has length √3, but if we choose its direction, it's a vector of length √3 in the direction (1,1,1) or (-1,-1,-1). So the vector is either (√3/√3, √3/√3, √3/√3) scaled by √3? Wait, no. Wait, the main diagonal vector is from (0,0,0) to (1,1,1), which is (1,1,1), which has length √(1^2 +1^2 +1^2)=√3. So the main diagonal vector is (1,1,1) with magnitude √3, or (-1,-1,-1) with magnitude √3. Therefore, the main diagonal vector is either (1,1,1) or (-1,-1,-1), each with magnitude √3. Wait, but (1,1,1) has magnitude √3, so if we take direction vector, it's (1,1,1) normalized, but here the problem says "directions are chosen", so probably the actual vector is (1,1,1) or (-1,-1,-1), each of length √3. So, yes, the main diagonal is a vector of length √3, so either (1,1,1) or (-1,-1,-1).Therefore, when we sum all vectors, the total vector will be:Total_x = S_x + d_x,Total_y = S_y + d_y,Total_z = S_z + d_z,where d_x, d_y, d_z are each either +1 or -1 (since the main diagonal vector is (1,1,1) or (-1,-1,-1)), multiplied by √3? Wait, wait, hold on. Wait, the main diagonal vector is (1,1,1), which has length √3, but if we are just choosing the direction, maybe we have the vector as (1,1,1) with magnitude √3 or (-1,-1,-1) with magnitude √3. Therefore, when adding to the edge vectors, which have magnitude 1 each, the main diagonal is a single vector with magnitude √3. So, the total sum is:Sum = (sum of edge vectors) + (main diagonal vector).Each edge vector is either + or - unit vector along their respective axes, and the main diagonal vector is either (1,1,1) or (-1,-1,-1), scaled by √3. Wait, but (1,1,1) has length √3, so if we take the vector as (1,1,1), its magnitude is √3, but if we take direction, is the problem considering the main diagonal as a vector of length √3, or just direction with magnitude 1? Wait, the problem says "the edge length is 1, and the length of the main diagonal is √3". So perhaps the main diagonal is considered as a vector of length √3. So when choosing the direction, we can have either (1,1,1) scaled to length √3 or (-1,-1,-1) scaled to length √3. Wait, but (1,1,1) already has length √3, so maybe the main diagonal vector is simply (1,1,1) or (-1,-1,-1). Therefore, its contribution to the sum is either (1,1,1) or (-1,-1,-1), each of length √3.But the edges are each of length 1. So each edge is a unit vector in one of the coordinate directions. Therefore, the 12 edge vectors can be grouped by their axes: 4 along x, 4 along y, 4 along z. Each of these can be assigned + or - direction. So, for each axis, the sum of the 4 edge vectors along that axis will be some integer between -4 and +4, in steps of 2, because each is ±1. For example, if all four are positive, sum is +4. If three positive and one negative, sum is +2. If two and two, sum is 0, etc.Similarly, the main diagonal vector is either (1,1,1) or (-1,-1,-1). Therefore, adding that to the edge sums, the total sum in each component will be:Total_x = S_x + d,Total_y = S_y + d,Total_z = S_z + d,where d is either +1 or -1, multiplied by the main diagonal's contribution. Wait, hold on. Wait, the main diagonal vector is (1,1,1) or (-1,-1,-1), but those vectors have length √3. Wait, but if we are just adding vectors, the edges are length 1, and the main diagonal is length √3. Therefore, the main diagonal vector is either (1,1,1) or (-1,-1,-1), but scaled by √3? Wait, no. Wait, (1,1,1) has length √3, so if we take the vector as (1,1,1), it's already length √3. So the main diagonal vector is either (1,1,1) or (-1,-1,-1), each of length √3. So, when we add the main diagonal vector to the edges, the edges are each of length 1 in their respective directions.Wait, maybe the problem is not about vectors in 3D space, but rather each edge and the main diagonal are represented as vectors, each with their own direction. The edges are along the coordinate axes, and the main diagonal is along (1,1,1). Each can be assigned a direction (positive or negative). Then, the sum is the vector sum of all these directed edges and the directed main diagonal.Therefore, for the edges:Each of the 12 edges can be represented as vectors:- Along the x-axis: from (0,0,0) to (1,0,0), vector (1,0,0). Similarly, three other edges along x-axis, but at different positions. But since we can choose direction, each edge can be (1,0,0) or (-1,0,0), regardless of their position. Wait, but actually, their position in space doesn't matter for the vector sum, because vectors are direction and magnitude, regardless of where they are placed. So even though the edges are at different positions in the cube, when we consider their vector contributions, it's just 12 vectors along the coordinate axes (each can be + or -) and one vector along the main diagonal (can be + or -).But wait, actually, if the cube is considered as a graph with edges, each edge is a vector between two vertices, but if we are allowed to choose the direction of each edge (i.e., the vector from one vertex to the other or vice versa), then each edge vector is either in the positive or negative direction of its respective axis. So for example, the edge from (0,0,0) to (1,0,0) is (1,0,0), but we can choose to assign it as (1,0,0) or (-1,0,0). Similarly for all edges.But since all edges along the x-axis are equivalent in direction (they are all parallel), their positions don't affect the vector sum. So, regardless of where they are in the cube, their vector contributions are just ±1 along their respective axes. So, effectively, for each axis, we have 4 edges, each contributing ±1 along that axis. So the total contribution from x-edges is S_x = sum_{i=1}^4 ±1, similarly for S_y and S_z.Then, the main diagonal is either (1,1,1) or (-1,-1,-1). So the total sum vector is:Total = (S_x + d_x, S_y + d_y, S_z + d_z),where d_x, d_y, d_z are each either +1 or -1, but all the same sign, since the main diagonal is either (1,1,1) or (-1,-1,-1). So d_x = d_y = d_z = d, where d is ±1.Wait, but the main diagonal vector is (1,1,1) or (-1,-1,-1), but since the main diagonal has length √3, actually, the vector is (1,1,1) with magnitude √3. However, if we are allowed to choose the direction, then the main diagonal vector is either (1,1,1) or (-1,-1,-1), each with magnitude √3. However, the edges are each of length 1. So, when adding vectors, the main diagonal vector is (1,1,1) or (-1,-1,-1), each of length √3, while the edges are each of length 1. Therefore, when we sum them, we need to consider the main diagonal as a vector of length √3.But the problem states: "the sum of the resulting 13 vectors can have, if the edge length is 1, and the length of the main diagonal is √3". So the edge vectors have length 1, the main diagonal vector has length √3. So each edge vector is a unit vector, and the main diagonal is a vector of length √3. Therefore, the total sum is the vector sum of 12 unit vectors (each along a coordinate axis, direction chosen) plus one vector of length √3 along the main diagonal (direction chosen).Therefore, the total vector is:Total = (sum of x-edge vectors) + (sum of y-edge vectors) + (sum of z-edge vectors) + (main diagonal vector).Each sum of x-edge vectors is the sum of 4 unit vectors along x-axis, each either + or -. So that sum is (s_x, 0, 0), where s_x is an integer between -4 and 4. Similarly, sum of y-edge vectors is (0, s_y, 0), and sum of z-edge vectors is (0, 0, s_z). The main diagonal vector is either (√3, √3, √3) normalized to length √3? Wait, no. Wait, if the main diagonal vector is from (0,0,0) to (1,1,1), then it's (1,1,1) with magnitude √3. So the main diagonal vector is (1,1,1) scaled by 1, but that vector has length √3. So the main diagonal vector is (1,1,1). But if we choose the direction, it can be (-1,-1,-1). So the main diagonal vector is either (1,1,1) or (-1,-1,-1), each with magnitude √3.Therefore, the total vector is:Total = (s_x + d, s_y + d, s_z + d),where d is either +1 or -1 (for the main diagonal direction), multiplied by √3? Wait, no. Wait, the main diagonal vector is (1,1,1) or (-1,-1,-1), each of length √3. So if we are adding vectors, the edges are unit vectors, but the main diagonal is a vector of length √3. Wait, hold on. There seems to be confusion here.Wait, in the problem statement: "the edge length is 1, and the length of the main diagonal is √3". So each edge is length 1, the main diagonal (from corner to opposite corner) is length √3. But when we talk about vectors, if we represent the main diagonal as a vector, it's (1,1,1) which has magnitude √3, so that's consistent. So the main diagonal vector is (1,1,1) or (-1,-1,-1), with magnitude √3. The edges, when represented as vectors, are unit vectors along the axes.Therefore, the total sum is:Total = (sum of x-edge vectors) + (sum of y-edge vectors) + (sum of z-edge vectors) + (main diagonal vector).Each x-edge vector is ±1 in the x-direction, so the sum of 4 x-edges is (±1 ±1 ±1 ±1, 0, 0). Similarly for y and z. The main diagonal vector is either (1,1,1) or (-1,-1,-1), scaled by √3? Wait, no, (1,1,1) is already of length √3, so if we want to represent the main diagonal as a vector of length √3, it's exactly (1,1,1) or (-1,-1,-1). Therefore, the main diagonal vector is either (1,1,1) or (-1,-1,-1).But then, adding these to the edge sums:Total_x = sum_x + main_x,Total_y = sum_y + main_y,Total_z = sum_z + main_z,where sum_x, sum_y, sum_z are integers between -4 and 4 (since each is the sum of 4 ±1s), and main_x, main_y, main_z are each either +1 or -1 (all same sign), multiplied by √3? Wait, no. Wait, the main diagonal vector is (1,1,1) or (-1,-1,-1). But (1,1,1) has magnitude √3, so if we want to represent it as a vector with magnitude √3, then each component is 1, but scaled by something. Wait, no. Wait, the vector (1,1,1) has magnitude √3, so if we take it as is, its components are (1,1,1), and its magnitude is √3. So when adding to the edge vectors, which have components like (1,0,0) or (-1,0,0), etc., the main diagonal vector is (1,1,1) or (-1,-1,-1), with magnitude √3, and the edges each have magnitude 1.Therefore, the total vector is:Total = (sum_x + d, sum_y + d, sum_z + d),where d is either +1 or -1 (from the main diagonal vector (1,1,1) or (-1,-1,-1)), and sum_x, sum_y, sum_z are the sums of the edge vectors along each axis (each sum_x is between -4 and 4). Wait, but the main diagonal vector has components (1,1,1), not (d, d, d). Wait, if the main diagonal is (1,1,1), then adding that to the edges would mean adding 1 to each component. If it's (-1,-1,-1), then subtracting 1 from each component. But hold on, the edges are summed as vectors. For example, sum_x is the sum of all x-components of the edges (each ±1), sum_y similarly, sum_z similarly. Then the main diagonal vector is (1,1,1) or (-1,-1,-1). Therefore, the total vector components are:Total_x = sum_x + 1 or sum_x -1,Total_y = sum_y + 1 or sum_y -1,Total_z = sum_z + 1 or sum_z -1,But the main diagonal vector is (1,1,1) or (-1,-1,-1), each with magnitude √3. Wait, but the edges are vectors of magnitude 1. So if we have 12 edge vectors, each of magnitude 1, and one main diagonal vector of magnitude √3. Then the total vector sum is:(sum_x_edges, sum_y_edges, sum_z_edges) + (main_x, main_y, main_z),where each component of the main diagonal is either +1 or -1, but all components have the same sign, and the main diagonal vector has magnitude √3. Wait, but if each component is 1 or -1, then the magnitude is √(1^2 +1^2 +1^2)=√3, which matches. So yes, the main diagonal vector is (1,1,1) or (-1,-1,-1). Therefore, when adding, we have:Total_x = sum_x + 1 or sum_x -1,Total_y = sum_y + 1 or sum_y -1,Total_z = sum_z + 1 or sum_z -1,But sum_x, sum_y, sum_z are each the sum of 4 ±1's. So each can be even numbers between -4 and 4. Wait, 4 terms each ±1: the possible sums are -4, -2, 0, 2, 4. Because if you have four ±1's, the number of +1's and -1's must differ by an even number. For example, 4 positive: sum 4, 3 positive and 1 negative: sum 2, 2 and 2: sum 0, 1 and 3: sum -2, 0 and 4: sum -4. So sum_x, sum_y, sum_z ∈ {-4, -2, 0, 2, 4}.Therefore, for each axis, the total component after adding the main diagonal is sum_axis ±1. Therefore, possible total components for each axis are:If sum_axis is even, then sum_axis ±1 is odd. So for example:If sum_x = 4, then Total_x = 4 ±1 → 5 or 3,If sum_x = 2, Total_x = 2 ±1 → 3 or 1,If sum_x = 0, Total_x = 0 ±1 → 1 or -1,If sum_x = -2, Total_x = -2 ±1 → -1 or -3,If sum_x = -4, Total_x = -4 ±1 → -3 or -5.Therefore, the total components for each axis can be odd integers from -5 to 5.But we need to find the configuration where the magnitude of the total vector is minimized. The magnitude is √(Total_x² + Total_y² + Total_z²). To minimize this, we need to minimize the sum of squares of the components.So our variables are:For each axis (x, y, z), choose sum_axis ∈ {-4, -2, 0, 2, 4} (by choosing directions of the edges), and choose d ∈ {+1, -1} for the main diagonal, such that Total_x = sum_x + d, Total_y = sum_y + d, Total_z = sum_z + d, and the magnitude √(Total_x² + Total_y² + Total_z²) is minimized.But note that d must be the same for all components, since the main diagonal is either (1,1,1) or (-1,-1,-1). Therefore, we can't choose d_x, d_y, d_z independently; they all have to be +1 or all -1.So the problem is to choose for each axis the sum_axis (which is the sum of 4 ±1's, so -4, -2, 0, 2, 4), and choose d = +1 or -1, such that the vector (sum_x + d, sum_y + d, sum_z + d) has minimal length.Our task is to find the minimal possible value of √[(sum_x + d)^2 + (sum_y + d)^2 + (sum_z + d)^2], given that sum_x, sum_y, sum_z ∈ {-4, -2, 0, 2, 4} and d ∈ {+1, -1}.To approach this problem, perhaps we can consider that for each axis, we can choose sum_axis and d such that (sum_axis + d) is as close to 0 as possible. Because the closer each component is to 0, the smaller the total magnitude.Therefore, for each axis, we can choose sum_axis and d such that sum_axis + d is minimized in absolute value. Since sum_axis is even (as it's the sum of four ±1's) and d is ±1, then sum_axis + d is odd. Therefore, the minimal |sum_axis + d| is 1, achieved when sum_axis is 0 (then 0 ±1 = ±1) or when sum_axis is ±2 (then ±2 ±1 = ±1 or ±3). If sum_axis is ±4, then ±4 ±1 = ±5 or ±3.Therefore, to minimize each component's absolute value, we need sum_axis to be as close to -d as possible. For example, if we choose d = +1, then to minimize |sum_axis +1|, sum_axis should be -2 (giving |-2 +1| = 1), 0 (|0 +1| =1), or 2 (|2 +1| =3). Wait, no. Wait, sum_axis can be -4, -2, 0, 2, 4.If d = +1:sum_axis +1 can be:-4 +1 = -3,-2 +1 = -1,0 +1 = +1,2 +1 = +3,4 +1 = +5.Similarly, if d = -1:sum_axis + (-1) can be:-4 -1 = -5,-2 -1 = -3,0 -1 = -1,2 -1 = +1,4 -1 = +3.Therefore, for each axis, the minimal |sum_axis + d| is 1, achieved when sum_axis is 0 and d = +1 or -1 (resulting in +1 or -1), or when sum_axis is 2 and d = -1 (resulting in +1), or sum_axis is -2 and d = +1 (resulting in -1). So to get the minimal |sum_axis + d| of 1, we need that sum_axis is 0 and choose d accordingly, or sum_axis is ±2 and d is chosen to flip it to ∓1.But since d has to be the same for all three components, we can't choose d independently for each axis. So we need to choose d once, either +1 or -1, and then for each axis, choose sum_axis such that sum_axis + d is as small as possible.Therefore, if we fix d = +1, then for each axis, the minimal |sum_axis +1| is 1 (achieved when sum_axis = 0 or -2). Similarly, if we fix d = -1, then for each axis, the minimal |sum_axis -1| is 1 (achieved when sum_axis = 0 or +2).Therefore, the strategy would be:- Choose d = +1, then set sum_x, sum_y, sum_z to be either 0 or -2, so that sum_axis +1 = 1 or -1.- Choose d = -1, then set sum_x, sum_y, sum_z to be either 0 or +2, so that sum_axis -1 = -1 or +1.In both cases, the components of the total vector would be ±1, leading to a total magnitude of √(1^2 +1^2 +1^2) = √3, if all components are ±1. However, we need to check if this is possible.But wait, each sum_axis (for x, y, z) is determined by the directions of the edges. For each axis, sum_axis is the sum of 4 ±1's. So sum_axis can be -4, -2, 0, 2, 4. Therefore, if we choose d = +1, to get sum_axis +1 as close to 0 as possible, we need sum_axis to be either -2 (so sum_axis +1 = -1) or 0 (sum_axis +1 = +1). But we need to set sum_axis for each axis to either 0 or -2. Similarly, if we choose d = -1, set sum_axis to 0 or +2.But the problem is that for each axis, the sum_axis is determined by the 4 edges along that axis. For each axis, we can independently choose the directions of the 4 edges to get sum_axis = -4, -2, 0, 2, 4. So for each axis, we can choose sum_axis to be -2 or 0 if d = +1, or 0 or +2 if d = -1.But we need to choose sum_axis for all three axes such that they are either 0 or -2 (if d = +1), or 0 or +2 (if d = -1). Let's check feasibility.First, consider d = +1. For each axis, we need sum_axis to be 0 or -2. So for each axis, we need to set the 4 edges along that axis such that their sum is 0 or -2.Similarly, for d = -1, sum_axis needs to be 0 or +2 for each axis.But is it possible to have sum_axis = -2 or 0 for all three axes when d = +1? Let's see.For each axis, sum_axis = 0 requires that two edges are +1 and two are -1. Because 2*(+1) + 2*(-1) = 0. For sum_axis = -2, we need three edges as -1 and one as +1, since 3*(-1) +1*(+1) = -2. Similarly, for sum_axis = +2, three edges +1 and one -1.So for each axis, we can choose sum_axis = 0 or -2 (if d = +1). Similarly, sum_axis = 0 or +2 (if d = -1). The question is, can we set each axis independently to these sums? For example, for d = +1, can we set x-axis sum to 0, y-axis sum to -2, z-axis sum to 0, etc.? Yes, because each axis is independent.Therefore, the minimal total vector would have components each ±1. So the magnitude would be √3. But wait, that's the same as the main diagonal's magnitude. But we are adding 12 edge vectors and the main diagonal vector. If the total sum is √3, that seems too good. Let me verify.Wait, if we choose d = +1, and for each axis, set sum_axis to 0 or -2 such that sum_axis +1 is ±1. For example, if x-axis sum is 0, then Total_x = 0 +1 =1. If y-axis sum is -2, Total_y = -2 +1 = -1. If z-axis sum is 0, Total_z = 0 +1 =1. Then the total vector is (1, -1, 1), which has magnitude √(1 +1 +1) = √3. Similarly, other combinations would give vectors with components ±1, leading to magnitude √3.But can we achieve this? Let's check.Suppose d = +1.For each axis, we can set sum_axis to 0 or -2.For example, x-axis: sum_axis =0 (requires two +1 and two -1 edges).y-axis: sum_axis = -2 (requires three -1 and one +1 edges).z-axis: sum_axis =0.Then, total vector is (0 +1, -2 +1, 0 +1) = (1, -1,1), magnitude √3.Similarly, if we set all three axes to have sum_axis =0, then total vector is (1,1,1), magnitude √3. But that would require all edges to cancel out (sum_axis=0 for each axis), and then adding the main diagonal (1,1,1). But if all edge vectors cancel, then the sum is just the main diagonal, which has magnitude √3. However, the problem states "the sum of the resulting 13 vectors". If all edges cancel, then the sum is just the main diagonal, so indeed √3. But maybe we can do better.Wait, but the problem is asking for the smallest possible length. If we can get the total vector to be zero, that would be minimal. But is that possible? To have the sum of all edges and the main diagonal equal to zero.But the edges sum to (sum_x, sum_y, sum_z), and the main diagonal is (d,d,d). So (sum_x + d, sum_y + d, sum_z + d) = (0,0,0). Therefore, sum_x = -d, sum_y = -d, sum_z = -d. But sum_x, sum_y, sum_z are even numbers (-4, -2, 0, 2, 4), while -d is either -1 or +1. But -1 or +1 are odd, and sum_x is even. Therefore, sum_x cannot equal -d. Therefore, the total vector cannot be zero. The closest we can get is each component being ±1, leading to magnitude √3. Therefore, the minimal possible magnitude is √3.But wait, the problem says "the smallest length that the sum of the resulting 13 vectors can have". If we can achieve √3, is that indeed the minimum? Let me think.Suppose instead of having all components ±1, maybe we can have some components zero? For example, if Total_x =0, Total_y=0, Total_z= something. But according to our earlier analysis, each Total component has to be odd, since sum_axis is even and d is ±1. Therefore, the components can't be zero. The possible components are odd integers: ±1, ±3, ±5. Therefore, the minimal possible magnitude is √3 when all components are ±1, or if some are ±1 and others are ±3. Wait, but √(1^2 +1^2 +1^2)=√3, √(1^2 +1^2 +3^2)=√11, etc. So √3 is indeed smaller.Therefore, the minimal possible magnitude is √3. But can we actually achieve that?Yes, as follows:Case 1: Choose d = +1. For each axis, set sum_axis to 0 or -2 such that Total_x, Total_y, Total_z are ±1.For example:Set x-axis sum_axis = 0 (Total_x = 1),y-axis sum_axis = -2 (Total_y = -1),z-axis sum_axis = 0 (Total_z =1).This gives Total = (1, -1, 1), magnitude √3.Similarly, other permutations.Alternatively, set all axes sum_axis =0 and choose d=+1, Total=(1,1,1), magnitude √3.Similarly, choose d=-1 and set sum_axis=0 or +2 to get Total components ±1.For example:d=-1,x-axis sum=0 (Total_x= -1),y-axis sum=+2 (Total_y= +1),z-axis sum=0 (Total_z= -1),Total=(-1,1,-1), magnitude √3.Therefore, in both cases, we can achieve a total vector magnitude of √3.But wait, the problem says "the smallest length that the sum of the resulting 13 vectors can have". So is √3 indeed the minimal possible? Let's confirm.Is there a way to get a smaller magnitude? For example, if two components are +1 and one is -1, or some other combination. But the magnitude is still √3. If all components are ±1, regardless of the combination, the magnitude is √3. If some components are ±1 and others are ±3, then the magnitude would be larger.Therefore, √3 is the minimal possible magnitude. Hence, the answer should be √3. But let's check if there's a mistake in reasoning.Wait, but let's consider the actual sum of vectors. When we set the edges such that their sum is (0, -2, 0) and add the main diagonal (1,1,1), the total is (1, -1,1). But the edges sum to (0, -2, 0). So to get sum_edges = (0, -2, 0), we need:For x-axis: 2 edges +1 and 2 edges -1 ⇒ sum_x=0,For y-axis: 3 edges -1 and 1 edge +1 ⇒ sum_y=-2,For z-axis: 2 edges +1 and 2 edges -1 ⇒ sum_z=0.This is possible. Then adding main diagonal (1,1,1) gives (1, -1,1).Similarly, if we set sum_edges=(0,0,0) and add main diagonal (1,1,1), total=(1,1,1). But to get sum_edges=(0,0,0), each axis must have two +1 and two -1 edges. Is that possible? Yes. For each axis, choose two edges in one direction and two in the other. Then sum per axis is 0.Therefore, adding the main diagonal (1,1,1) gives a vector of (1,1,1) with magnitude √3.But wait, in this case, we're adding 12 edge vectors that cancel each other out (sum to zero) and then adding the main diagonal vector. So the total is just the main diagonal vector, which has magnitude √3. Therefore, that's a valid configuration.But in this case, the edges are arranged such that their total sum is zero, so the total sum is just the main diagonal. However, in other configurations, the edges partially cancel the main diagonal. But regardless, the minimal possible total magnitude is √3.Therefore, the answer should be √3. But let me check if there's a way to get a smaller total. Suppose we have the edges sum to (2, 2, 2) and choose the main diagonal as (-1,-1,-1). Then total would be (2 -1, 2 -1, 2 -1)=(1,1,1), magnitude √3. Similarly, edges sum to (-2,-2,-2) and main diagonal (1,1,1), total=(-1,-1,-1), magnitude √3. But edges can only sum to even numbers. So sum_edges=(2,2,2) is possible (each axis has three +1 and one -1 edges). Then adding main diagonal (-1,-1,-1) gives (1,1,1). So again, magnitude √3.Alternatively, can we get cancellation in one or more components? For example, if sum_x + d =0. But sum_x is even and d is ±1, so sum_x + d cannot be zero. The closest is ±1. Therefore, no component can be zero. Thus, the minimal possible magnitude is √3.Hence, the answer is √3.Final AnswerThe smallest possible length is boxed{sqrt{3}}.</think>To determine the smallest possible length of the sum of 13 vectors on a cube (12 edges and one main diagonal), we need to consider the directions of these vectors. The edge vectors are unit vectors along the coordinate axes, and the main diagonal vector is either ((1,1,1)) or ((-1,-1,-1)) with magnitude (sqrt{3}).1. Edge Vectors: Each edge vector can be directed either positively or negatively along its respective axis. For each axis (x, y, z), the sum of the 4 edge vectors along that axis can be (-4, -2, 0, 2, 4) (since each edge vector is (pm 1)).2. Main Diagonal Vector: The main diagonal vector is either ((1,1,1)) or ((-1,-1,-1)), contributing (pm 1) to each component of the total sum.3. Total Vector Calculation: The total vector sum is the combination of the sums of the edge vectors and the main diagonal vector. For each component (x, y, z), the sum is ( text{sum}_{text{axis}} + d ), where ( d ) is (pm 1).4. Minimizing the Magnitude: To minimize the magnitude of the total vector, we need each component ( text{sum}_{text{axis}} + d ) to be as close to zero as possible. Since (text{sum}_{text{axis}}) is even and ( d ) is (pm 1), the closest possible values are (pm 1).5. Feasibility Check: By choosing the appropriate directions for the edge vectors, we can set the sums for each axis such that the resulting components of the total vector are (pm 1). For example: - If the main diagonal is ((1,1,1)), setting the edge sums to (0, -2, 0) results in the total vector ((1, -1, 1)). - If the main diagonal is ((-1,-1,-1)), setting the edge sums to (2, 2, 2) results in the total vector ((1, 1, 1)).In both cases, the magnitude of the total vector is (sqrt{3}).Thus, the smallest possible length of the sum of the 13 vectors is (boxed{sqrt{3}}).,<think>Alright, let me try to figure this problem out. So, the question is about a cube with edge length 1, and the main diagonal is √3. We need to choose directions on one main diagonal and all edges, then find the smallest possible length of the sum of these 13 vectors. Hmm, directions are chosen... so each of these 13 vectors can be either in the positive or negative direction along their respective edges or diagonal. The goal is to minimize the total length of their sum. First, let me visualize the cube. A cube has 12 edges and 4 main diagonals. Wait, the problem mentions "one main diagonal" so maybe they're just considering a single main diagonal here, not all four. So, in total, there are 12 edges and 1 main diagonal, making 13 vectors. Each of these vectors can be directed either way along their respective lines. We need to assign a direction (either + or -) to each of these 13 vectors such that when we add them all up vectorially, the resulting vector has the smallest possible length. Then, report that minimal length.Okay, so the cube edges: each edge is a vector of length 1, and the main diagonal is a vector of length √3. The main diagonal goes from one corner of the cube to the opposite corner. So, if we consider the cube in a coordinate system with one corner at (0,0,0) and the opposite corner at (1,1,1), then the main diagonal vector would be (1,1,1) with length √(1² + 1² + 1²) = √3. But since the problem says "directions are chosen on one main diagonal and all edges," we have 12 edge vectors and 1 main diagonal vector. Each can be pointed in either direction. So, for each edge, we can choose, say, from vertex A to vertex B or from B to A. Similarly, for the main diagonal, we can choose from (0,0,0) to (1,1,1) or the reverse. The challenge is to choose these directions such that when we add all these vectors together, the resultant vector has the smallest possible magnitude.Let me think about how vectors add. If we can arrange the directions so that as many vectors as possible cancel each other out, the total sum will be smaller. Since the cube is symmetrical, maybe there's a symmetrical way to assign the directions to maximize cancellation.First, let's consider the 12 edge vectors. Each edge is part of three axes: x, y, and z. For each axis, there are 4 edges: two in each direction. Wait, no. Let me correct that. In a cube, each axis (x, y, z) has 4 edges parallel to it. For example, along the x-axis, there are 4 edges: each of the top and bottom edges in the front and back faces. Similarly for y and z. Each of these 4 edges for a given axis can be directed either in the positive or negative direction of that axis.If we can direct these edges such that their contributions along each axis cancel out as much as possible. For each axis, suppose we have 4 vectors. If we can set two of them in the positive direction and two in the negative, then their sum would cancel out completely for that axis. That would be ideal. Similarly, for each of the x, y, and z axes, if we can balance the directions of the edges so that each axis component cancels, then the total contribution from the edges would be zero. But since we have 4 edges per axis, which is an even number, this is possible: 2 in each direction. So that's 4 edges per axis × 3 axes = 12 edges. If each axis has 2 vectors in positive and 2 in negative direction, then the sum of the edge vectors would be zero.But then we also have the main diagonal vector. The main diagonal is along (1,1,1) direction. Depending on its direction, it will contribute either (1,1,1) or (-1,-1,-1). So, even if we cancel all the edge vectors, the main diagonal will add a vector of length √3. But maybe we can arrange the edge vectors in such a way that they partially cancel the main diagonal vector. Wait, but if we arrange the edge vectors so that their sum is non-zero, but in the opposite direction of the main diagonal, then the total sum could be smaller.But is that possible? Let's think. Let's suppose that we first cancel all edges, so their sum is zero, and then the main diagonal adds √3. But maybe if instead, we arrange edges so that their sum is a vector that partially cancels the main diagonal, resulting in a smaller total vector.However, to do that, we need the edges to sum up to a vector in the direction opposite to the main diagonal. Let me formalize this.Let me assign coordinates. Let’s model the cube with vertices at (0,0,0) to (1,1,1). The edges are along the x, y, z axes. Each edge vector can be represented as ±(1,0,0), ±(0,1,0), or ±(0,0,1), depending on their direction. The main diagonal is either (1,1,1) or (-1,-1,-1).Suppose we choose directions for the edges such that their sum is S = (S_x, S_y, S_z). Then the main diagonal vector is D = ±(1,1,1). The total sum is S + D. We need to choose the directions of the edges and the direction of D such that the magnitude |S + D| is minimized.But since the edges can be directed either way, we can choose the sign of each edge vector. For each edge vector, which is along a coordinate axis, we can choose its direction to be positive or negative. The challenge is to pick these signs such that S + D is as small as possible.Since there are 12 edges, 4 along each axis. For each axis, say x, we have 4 edges. Each edge along x can be directed either +x or -x. Similarly for y and z.The sum S_x along the x-axis will be equal to (number of edges directed +x minus number directed -x) times 1 (since each edge is length 1). Since there are 4 edges, if we direct k of them in the +x direction and (4 - k) in the -x direction, then S_x = k - (4 - k) = 2k - 4. Similarly for S_y and S_z.Our goal is to choose k, m, n (for x, y, z axes respectively) such that S = (2k - 4, 2m - 4, 2n - 4) and D = ±(1,1,1). Then, |S + D| is minimized.But wait, hold on. Since we can choose the direction of D, which is either (1,1,1) or (-1,-1,-1), we can choose D to be either + or - such that S + D is minimized. So, for a given S, the minimal |S + D| would be the minimum of |S + (1,1,1)| and |S - (1,1,1)|.Therefore, for each possible S, we can compute both possibilities and take the smaller one. But since we can choose the direction of D, the minimal |S + D| would be the minimum between |S + (1,1,1)| and |S - (1,1,1)|.But we also need to choose S by choosing k, m, n for each axis (each from 0 to 4, since there are 4 edges per axis). However, the values of k, m, n must be integers between 0 and 4, inclusive.Wait, but actually, for each axis, the number of edges directed in the positive direction can be 0,1,2,3,4. Then S_x = 2k - 4, so possible values are -4, -2, 0, 2, 4. Similarly for S_y and S_z.Therefore, the possible components of S are even integers from -4 to 4. Then, the vector S can be any triple (even_x, even_y, even_z) where each component is in {-4, -2, 0, 2, 4}.Our goal is to choose S (by choosing directions for the edges) and choose D (either (1,1,1) or (-1,-1, -1)) such that |S + D| is minimized.So, we need to find the minimum of |(S_x ± 1, S_y ± 1, S_z ± 1)|, where S_x, S_y, S_z are even integers between -4 and 4, and the sign in each component is the same (since D is either (1,1,1) or (-1,-1,-1)).Alternatively, we can think of it as |S + D| where D is either (1,1,1) or (-1,-1,-1). So, for each possible S, compute |S + D| for both choices of D and take the minimum, then find the S that gives the smallest such minimum.But since S is determined by k, m, n, which are the number of edges directed positively along each axis, this seems like a problem of choosing k, m, n (each from 0 to 4) to minimize the minimal |S ± (1,1,1)|.Alternatively, perhaps we can frame this as an optimization problem where we want to choose S such that S is as close as possible to (-1, -1, -1) or (1,1,1), depending on the direction of D. Wait, but since D can be either direction, we can choose D to be whichever direction brings S closer to the origin. That is, for each S, we can choose D to be either (1,1,1) or (-1,-1,-1) such that S + D is as small as possible.So, in effect, the problem reduces to finding the S in the lattice of even integers (per component) within [-4, 4]^3 such that the distance from S to either (1,1,1) or (-1,-1,-1) is minimized. But since we can choose D's direction, the minimal distance would be the minimal |S - (-D)|, where D is either (1,1,1) or (-1,-1,-1). Hmm, maybe this is getting a bit convoluted.Alternatively, think of it as S + D. Since D can be either (1,1,1) or (-1,-1,-1), we can write the total sum as S ± (1,1,1). So, we need to choose the sign such that this sum is as small as possible. Therefore, for each S, compute both possibilities and take the smaller one. Then, find the S that gives the smallest result.But since S has even components and (1,1,1) has odd components, S ± (1,1,1) will have odd components. So, the sum S + D will have components that are odd integers. The magnitude of such a vector would be sqrt((2a + 1)^2 + (2b + 1)^2 + (2c + 1)^2) where a, b, c are integers. Wait, but S_x is even, so S_x ± 1 is odd. Similarly for the others. So, each component of S + D is an odd integer.But to minimize the magnitude, we need the components of S + D to be as close to zero as possible. Since they have to be odd integers, the closest they can get to zero is ±1. Therefore, the minimal possible |S + D| would be sqrt(1^2 + 1^2 + 1^2) = √3 ≈ 1.732, but is this achievable?Wait, that would require S + D = (±1, ±1, ±1). But S is a vector with even components, and D is (±1, ±1, ±1). So, if S = (0,0,0) and D is (1,1,1), then S + D = (1,1,1), which has length √3. Alternatively, if S is (2, 2, 2) and D is (-1,-1,-1), then S + D = (1,1,1). But S can't be (2,2,2) because S_x = 2k - 4, so to get S_x = 2, we need 2k - 4 = 2 ⇒ 2k = 6 ⇒ k = 3. So, for each axis, directing 3 edges in the positive direction and 1 in the negative. Similarly for y and z. Then, S would be (2,2,2). Then, if we choose D = (-1,-1,-1), the total sum is (1,1,1), which is length √3. But is this the minimal possible?But wait, can we get S + D to be (0,0,0)? That would require S = -D. Since S has even components and D is (±1, ±1, ±1), but -D would be (±1, ±1, ±1) with signs flipped, but S needs to have even components. Since 1 is odd, S cannot equal -D. Therefore, the closest S can get to -D is differing by 1 in each component. So, the minimal possible |S + D| is √3. But is there a way to get a smaller sum?Wait, suppose we have S = (2, 0, -2) and D = (1,1,1). Then, S + D = (3,1,-1). The magnitude is sqrt(9 + 1 + 1) = sqrt(11) ≈ 3.316, which is larger. Alternatively, if S is (2, 2, 0) and D is (-1,-1,-1), then S + D = (1,1,-1), magnitude sqrt(1 + 1 + 1) = √3 again. Hmm, same as before. What if S is (0,0,0) and D is (1,1,1), then total is √3. If S is (2,0,0) and D is (-1,-1,-1), then total is (1, -1, -1), which is sqrt(1 + 1 + 1) = √3. So regardless of some S, we can get √3. But is there a way to get lower?Wait, what if S is (2, -2, 0) and D is (-1,1,-1). Wait, but D has to be all the same sign. So, D is either (1,1,1) or (-1,-1,-1). So, if S is (2, -2, 0), and D is (-1,-1,-1), then the total is (1, -3, -1). The magnitude is sqrt(1 + 9 + 1) = sqrt(11). Not better. If D is (1,1,1), then total is (3, -1, 1), which is sqrt(9 + 1 + 1) = sqrt(11). So same.Alternatively, if S is (2, 0, 0) and D is (-1,-1,-1), sum is (1, -1, -1), magnitude √3. Alternatively, if S is (0, -2, 0) and D is (1,1,1), sum is (1, -1, 1), magnitude √3. So in all these cases, the minimal total sum seems to be √3. But is there a way to get a smaller sum?Wait, suppose S is (0,0,0). Then, regardless of D's direction, sum is either (1,1,1) or (-1,-1,-1), both with magnitude √3. If S is (2,2,2) and D is (-1,-1,-1), sum is (1,1,1). If S is (-2,-2,-2) and D is (1,1,1), sum is (-1,-1,-1). So in all these cases, the minimal possible sum is √3. But can we get a smaller total?Wait, maybe if we have some components cancel more? For example, suppose S is (2, -2, 0) and D is (-1,1,-1). But D can't be a mixture, it's either all positive or all negative. So, if D is (1,1,1), then S + D is (3, -1, 1). If D is (-1,-1,-1), then S + D is (1, -3, -1). Both have magnitude sqrt(11). Not better.Alternatively, if S is (2, 2, -2) and D is (-1,-1,1), but again D has to be same direction. So, D is either (1,1,1) or (-1,-1,-1). So, S + D would be (3,3,-1) or (1,1,-3). Both have magnitudes sqrt(9 + 9 + 1) = sqrt(19) or sqrt(1 + 1 + 9) = sqrt(11). Still worse.Wait, maybe if S has components that are closer to -D's components. For example, if D is (1,1,1), then S should be close to (-1,-1,-1). But S has even components, so the closest S can get is (0,0,0), (-2,0,0), etc. Similarly, if D is (-1,-1,-1), then S should be close to (1,1,1). But again, S can only have even components. So the closest is (2,2,2) or (0,0,0). Wait, but (0,0,0) + (-1,-1,-1) is (-1,-1,-1). Similarly, (2,2,2) + (-1,-1,-1) is (1,1,1). So in both cases, the distance is √3.Is there a way to get a vector sum with components closer to zero? For example, (1, -1, 0). Its magnitude is sqrt(1 + 1 + 0) = sqrt(2) ≈ 1.414, which is smaller. But can we get such a sum?To get (1, -1, 0), for example, we need S + D = (1, -1, 0). Therefore, S = (1, -1, 0) - D. But D is either (1,1,1) or (-1,-1,-1). Let's see:Case 1: D = (1,1,1). Then S = (1 -1, -1 -1, 0 -1) = (0, -2, -1). But S must have even components. Here, the z-component is -1, which is not even. So invalid.Case 2: D = (-1,-1,-1). Then S = (1 +1, -1 +1, 0 +1) = (2, 0, 1). Again, z-component is 1, which is odd. Not allowed.So, impossible. Similarly, trying to get other combinations with smaller magnitudes but mixed signs would result in components that are not even, hence S cannot be adjusted to get those sums.Therefore, it seems like the minimal possible magnitude is √3. But the problem is asking for the smallest length that the sum of the resulting 13 vectors can have. So, maybe √3 is the minimal. But wait, let me confirm.Wait, is there a way to have S be such that S + D is (0,0,0)? That would require S = -D. But since S has even components and D is (±1,±1,±1), which are all odd, it's impossible. Because even numbers minus odd numbers can't be zero. Hence, the closest possible is when each component of S is either -1 or 1 away from the corresponding component of -D, leading to the vector (1,1,1) or (-1,-1,-1) with magnitude √3. Therefore, the minimal possible total sum is √3.But let me check if there's another possibility. Suppose we can have some components cancel out more. For example, if S + D has two components zero and one component non-zero. Let's see.Suppose S + D = (0, 0, 2). Then magnitude is 2. Is this possible? Let's check.If S + D = (0,0,2), then S = (0,0,2) - D. If D is (1,1,1), then S = (-1,-1,1). But S must have even components. The components here are -1, -1, 1. Not even. If D is (-1,-1,-1), then S = (1,1,3). Again, components 1,1,3, which are odd. Not allowed.Alternatively, S + D = (0,0,1). Then S = (0,0,1) - D. If D is (1,1,1), S = (-1,-1,0). Components: -1, -1, 0. Not all even. If D is (-1,-1,-1), S = (1,1,2). Components 1,1,2. Again, two odd components. Not allowed.Alternatively, S + D = (1, -1, 0). As before, this leads to S having odd components. Therefore, seems impossible.Alternatively, S + D = (2,0,0). Then magnitude is 2. Let's check if possible.If D is (1,1,1), then S = (2 -1, 0 -1, 0 -1) = (1, -1, -1). Not even components. If D is (-1,-1,-1), then S = (2 +1, 0 +1, 0 +1) = (3,1,1). Again, odd components. Not possible.Therefore, getting a sum with magnitude less than √3 seems impossible because all components have to be odd integers, and the minimal Euclidean norm with odd integers is √3 when all components are ±1.Therefore, the minimal total length is √3. But wait, the problem says "the smallest length that the sum of the resulting 13 vectors can have". So, is the answer √3?But let me check if this is actually achievable. For example, if we set all edges such that their sum is (0,0,0), and then the main diagonal adds (1,1,1) or (-1,-1,-1). But is it possible to set the edges so that their sum is (0,0,0)?Yes. For each axis, direct 2 edges in the positive direction and 2 in the negative. Then, the sum for each axis is 0. Therefore, total sum S = (0,0,0). Then, adding the main diagonal vector D = (1,1,1) gives total sum (1,1,1) with length √3. Alternatively, adding D = (-1,-1,-1) gives (-1,-1,-1), same length. So yes, this is achievable.Therefore, the minimal possible total length is √3. However, wait, the problem says "directions are chosen on one main diagonal and all edges of a cube". So, maybe I misunderstood the problem. Wait, the problem says: "On one main diagonal and all edges of a cube, directions are chosen. What is the smallest length that the sum of the resulting 13 vectors can have...".Wait, does this mean that on each of the edges and on one main diagonal, directions are chosen? So, each edge has a direction (vector) chosen, and the main diagonal also has a direction chosen. Then, sum all these 13 vectors. So, the edges are 12 vectors, each of length 1, direction chosen along their edge; and the main diagonal is 1 vector, length √3, direction chosen along the diagonal. Then, sum all 13 vectors. What's the minimal possible length of this sum.So, that is exactly as I was thinking. So, the answer is √3. But before finalizing, let me check again. Because sometimes in these problems, there might be a trick. For example, maybe the main diagonal is considered as two directions, but no, the problem says directions are chosen on each edge and one main diagonal. So, 12 edges, each with two possible directions, and one main diagonal with two possible directions. Total 13 vectors.If we set edges such that their sum is zero, then add the main diagonal vector. Then total sum is the main diagonal vector, with length √3. But is there a way to have the edges and the diagonal vector partially cancel each other?Suppose instead of setting edges to cancel, we set them to have a total sum of (-1,-1,-1), then choose the main diagonal direction as (1,1,1), so that the total sum is zero. But is that possible?But the edges can only sum to vectors with even components. Because each edge is a unit vector along an axis, and for each axis, the number of edges is 4, so sum per axis is (number of positive edges - number of negative edges). Since there are 4 edges, if k are positive, 4 - k are negative. Sum is k - (4 - k) = 2k - 4. So possible sums per axis are even numbers: -4, -2, 0, 2, 4.So, if we need the edge sum S to be (-1,-1,-1), that's impossible because each component must be even. The closest we can get is S = (0,0,0) or (-2,-2,-2). If S = (-2,-2,-2), then adding D = (1,1,1) would give (-1,-1,-1), which has length √3. Alternatively, adding D = (-1,-1,-1) would give (-3,-3,-3), length √27 = 3√3.Alternatively, if S = (0,0,0), adding D = (1,1,1) gives √3. So, regardless, the minimal total is √3. Therefore, I think √3 is indeed the minimal possible length.But let me confirm once more with an example. Let's take the cube with edges along the axes. Let's direct 2 edges in the positive x and 2 in the negative x, same for y and z. Then, sum of edges is (0,0,0). Then, choose the main diagonal direction as (1,1,1). Total sum is (1,1,1), length √3. Yes, that works.Alternatively, if we could arrange edges such that their sum is (-1,-1,-1), then adding D = (1,1,1) would give zero. But since we can't get (-1,-1,-1) from edge sums, that's impossible.Therefore, the minimal total length is √3. So, the answer should be √3. Therefore, boxed answer: boxed{sqrt{3}}.Final Answerboxed{sqrt{3}}</think>To solve the problem of finding the smallest length of the sum of 13 vectors (12 edges and 1 main diagonal) on a cube with edge length 1 and main diagonal length (sqrt{3}), we need to consider the directions of each vector to minimize the resultant vector's length.1. Edge Vectors: Each edge vector can be directed either positively or negatively along the x, y, or z-axis. For each axis, there are 4 edges. By directing 2 edges in the positive direction and 2 in the negative direction, the sum of the vectors along each axis cancels out, resulting in a zero vector. This configuration is possible because each axis has an even number of edges (4), allowing for equal cancellation.2. Main Diagonal Vector: The main diagonal can be directed either from ((0,0,0)) to ((1,1,1)) or the reverse. This vector has a length of (sqrt{3}).3. Sum of Vectors: If the sum of the 12 edge vectors is zero (achieved by balancing directions as described), adding the main diagonal vector (either ((1,1,1)) or ((-1,-1,-1))) results in a vector of length (sqrt{3}). This is because the main diagonal vector, regardless of its direction, has a magnitude of (sqrt{3}) and cannot be canceled by the edge vectors since their sum is zero.Thus, the smallest possible length of the sum of the 13 vectors is (sqrt{3}).[boxed{sqrt{3}}]]

🔑:Answer: sqrt{3}Solution: Let's choose a basis of three vectors along the edges of the cube, such that the vector of the diagonal equals overrightarrow{e_{1}}+overrightarrow{e_{2}}+overrightarrow{e_{3}}. Then, four vectors will be equal to pm overrightarrow{e_{1}}, another four to pm overrightarrow{e_{2}}, and another four to pm overrightarrow{e_{3}}. Thus, the sum vec{S} of the vectors has the form k cdot overrightarrow{e_{1}}+m cdot overrightarrow{e_{2}}+n cdot overrightarrow{e_{3}}, where k, m, and n are odd integers, and |vec{S}|=sqrt{k^{2}+m^{2}+n^{2}} geq sqrt{3}, since k^{2} geq 1, m^{2} geq 1, n^{2} geq 1. Clearly, by pairing the parallel edges of the cube into oppositely directed vectors, we get that k=m=n=1 and |vec{S}|=sqrt{3}.

🧠:[<think>Okay, so I have this problem about an ellipse and points symmetric with respect to a line. Let me try to figure this out step by step. The ellipse equation is x²/4 + y²/3 = 1. The question is about finding the range of m such that there are two distinct points A and B on the ellipse that are symmetric with respect to the line y = 4x + m. The answer choices are given, so maybe after working through the problem, I can check which one matches.First, symmetry with respect to a line means that the line is the perpendicular bisector of the segment AB. So, if two points are symmetric over a line, then the line must be the axis of symmetry for those two points. That implies that the line is perpendicular to the line segment AB and passes through its midpoint.So, given that, for points A and B on the ellipse, the line y = 4x + m must be the perpendicular bisector of AB. Therefore, the midpoint of AB lies on the line y = 4x + m, and the line AB is perpendicular to y = 4x + m.The slope of the line y = 4x + m is 4, so the slope of AB, which is perpendicular, should be -1/4, since the product of the slopes of two perpendicular lines is -1.So, steps I need to take:1. Find the condition that for two points A and B on the ellipse, the line AB has slope -1/4, and their midpoint lies on y = 4x + m.2. Determine for which m such points A and B exist on the ellipse.Alternatively, another approach is that for the line y = 4x + m to be a symmetry axis of the ellipse, but since the ellipse is fixed (not transformed), we need to find lines y = 4x + m such that the ellipse has pairs of points symmetric with respect to this line.But the ellipse's standard form is x²/4 + y²/3 =1, which is centered at the origin, major axis along the x-axis with semi-major axis 2, and semi-minor axis sqrt(3). The given line is y = 4x + m, which is a straight line with slope 4. So, this line is not aligned with any of the ellipse's axes.The problem states that there exist two distinct points A and B on the ellipse symmetric with respect to this line. So, such points must exist for some m. The question is to find the range of m for which this is possible.Hmm. Let's recall that if two points are symmetric with respect to a line, then that line is the perpendicular bisector of the segment connecting them. Therefore, for points A and B on the ellipse, the line y =4x + m must be the perpendicular bisector of AB. So, the midpoint M of AB lies on the line y=4x +m, and the line AB is perpendicular to y=4x +m. Since the slope of y=4x +m is 4, the slope of AB is -1/4.Therefore, given that, we can parametrize the points A and B as follows: Let the midpoint M be (h, k), which lies on y=4x +m, so k =4h +m. The line AB has slope -1/4, so the line connecting A and B has slope -1/4. Let’s denote point A as (x1, y1) and point B as (x2, y2). Then, since M is the midpoint, we have:h = (x1 + x2)/2k = (y1 + y2)/2Also, the slope of AB is (y2 - y1)/(x2 - x1) = -1/4. Therefore, (y2 - y1) = -1/4 (x2 - x1).Moreover, both A and B lie on the ellipse, so:x1²/4 + y1²/3 =1x2²/4 + y2²/3 =1So, the problem reduces to solving these equations with the conditions above. Let’s see if we can express everything in terms of h and k, since M is (h,k) on the line y=4x +m.Alternatively, perhaps parametrize the line AB with slope -1/4 passing through (h, k). The line AB would then be y -k = -1/4 (x - h). This line intersects the ellipse at points A and B. Since AB is a chord of the ellipse with slope -1/4 and midpoint (h,k), we can use the concept of the equation of the chord with a given midpoint.For an ellipse, the equation of the chord with midpoint (h, k) is given by:T = S1Where T is (xx1)/a² + (yy1)/b² =1 and S1 is x1²/a² + y1²/b². Wait, but the equation of the chord with midpoint (h,k) is actually derived by the equation:( h x ) / a² + ( k y ) / b² = ( h² ) / a² + ( k² ) / b²Is that right? Let me recall. For a conic section, the equation of the chord with midpoint (h, k) is T = S1, where T is the tangent at (h,k) if (h,k) is the midpoint. Wait, actually, maybe it's called the equation of the polar line or something.Alternatively, perhaps use parametric equations. Let’s think differently.Given that the line AB has slope -1/4 and midpoint (h, k). The line AB is y = -1/4 x + c, for some c. But since it passes through (h,k), then k = -1/4 h + c => c = k + h/4. So the equation is y = -1/4 x + k + h/4.This line must intersect the ellipse x²/4 + y²/3 =1 at two points A and B. The points A and B must be symmetric with respect to the line y=4x +m, which requires that the midpoint (h,k) lies on y=4x +m, so k =4h +m.Therefore, substituting k =4h +m into the equation of line AB, we get:y = -1/4 x + (4h +m) + h/4 = -1/4 x + (4h + h/4 +m) = -1/4 x + (17h/4 + m)So, the line AB is y = -1/4 x + 17h/4 + m.Now, we need this line to intersect the ellipse x²/4 + y²/3 =1 at two distinct points. For the line to intersect the ellipse, the system of equations must have real solutions. Let’s substitute y from the line equation into the ellipse equation.Substitute y = -1/4 x + 17h/4 + m into the ellipse:x²/4 + [ (-1/4 x + 17h/4 + m )² ] /3 =1Let me compute this:First, expand the y term:(-1/4 x + 17h/4 + m )² = [ (-1/4 x) + (17h/4 + m) ]²= ( (-1/4 x)^2 ) + 2*(-1/4 x)*(17h/4 + m) + (17h/4 + m)^2= (x²)/16 - (2*(17h/4 + m)*x)/4 + (17h/4 + m)^2Simplify:= x²/16 - ( (17h/2 + 2m ) x ) /4 + (17h/4 + m)^2Wait, let's check:First term: (-1/4 x)^2 = x²/16Second term: 2*(-1/4 x)*(17h/4 + m) = 2*(-1/4)*(17h/4 + m)x = (-1/2)*(17h/4 + m)x = - (17h/4 + m)x /2Third term: (17h/4 + m)^2So, the entire y² term becomes [x²/16 - (17h/4 + m)x/2 + (17h/4 + m)^2 ] /3Therefore, the ellipse equation substitution:x²/4 + [x²/16 - (17h/4 + m)x/2 + (17h/4 + m)^2 ] /3 =1Multiply through by 48 to eliminate denominators:12x² + 16[ x²/16 - (17h/4 + m)x/2 + (17h/4 + m)^2 ] = 48Compute each term:12x² + 16*(x²/16) - 16*( (17h/4 + m)x /2 ) + 16*(17h/4 + m)^2 =48Simplify term by term:12x² + x² - 8*(17h/4 + m)x + 16*(17h/4 + m)^2 =48Combine like terms:13x² -8*(17h/4 + m)x +16*(17h/4 + m)^2 -48=0Simplify coefficients:First, compute 8*(17h/4 + m):8*(17h/4) = 34h8*m =8mSo, the second term is -34h x -8m xThird term: 16*(17h/4 + m)^2. Let's expand (17h/4 + m)^2:= (17h/4)^2 + 2*(17h/4)*m + m^2= 289h²/16 + 34hm/4 + m²Multiply by 16:289h² + 136hm +16m²Thus, the third term is 289h² +136hm +16m²Therefore, the entire equation becomes:13x² -34h x -8m x +289h² +136hm +16m² -48=0Combine the x terms:13x² - (34h +8m)x +289h² +136hm +16m² -48=0This is a quadratic in x. For the line to intersect the ellipse at two distinct points, the discriminant of this quadratic must be positive.Compute discriminant D:D = [-(34h +8m)]² -4*13*(289h² +136hm +16m² -48 )Let me compute each part:First, [-(34h +8m)]² = (34h +8m)^2 = (34h)^2 + 2*34h*8m + (8m)^2 = 1156h² + 544hm +64m²Second, compute 4*13*(289h² +136hm +16m² -48 )= 52*(289h² +136hm +16m² -48 )Compute 52*289h²: 289*52. Let's compute 289*50=14450, 289*2=578, so total 14450+578=15028h²52*136hm: 136*52. 100*52=5200, 36*52=1872, so 5200+1872=7072hm52*16m²= 832m²52*(-48)= -2496Therefore, D = [1156h² +544hm +64m²] - [15028h² +7072hm +832m² -2496]= 1156h² +544hm +64m² -15028h² -7072hm -832m² +2496Combine like terms:For h²: 1156h² -15028h² = -13872h²For hm: 544hm -7072hm = -6528hmFor m²:64m² -832m² = -768m²Constants: +2496Therefore, discriminant D = -13872h² -6528hm -768m² +2496For the quadratic to have real roots, D >0:-13872h² -6528hm -768m² +2496 >0Multiply both sides by (-1) (which reverses the inequality):13872h² +6528hm +768m² -2496 <0Let me factor out common terms. Let's see, 13872, 6528, 768, and 2496. Let's divide all terms by 48 to simplify:13872 /48 = 13872 ÷ 48: 48*288=13824, so 13872-13824=48, so 288 +1=2896528 /48= 6528 ÷48= 136 (since 48*136=6528)768 /48=162496 /48=52So, after dividing by 48:289h² +136hm +16m² -52 <0Therefore:289h² +136hm +16m² <52Now, note that 289 is 17², 16 is 4², and 136=4*34=4*2*17=8*17. So, perhaps this quadratic form can be written as (17h +4m)^2?Let’s check: (17h +4m)^2 =289h² +136hm +16m². Yes! Exactly.So, the inequality becomes:(17h +4m)^2 <52Therefore,-√52 <17h +4m <√52Divide all parts by 17:(-√52)/17 < h + (4m)/17 < √52/17But h is related to m via the midpoint lying on y=4x +m, which is k=4h +m. But we also need to relate h and k through the ellipse. Wait, no, h and k are coordinates of the midpoint, which lies on the line y=4x +m, so k=4h +m. However, the midpoint (h,k) must also be such that the chord AB with midpoint (h,k) exists. But we already derived that (17h +4m)^2 <52.But we need to relate h and m. Since k =4h +m, can we substitute k in terms of h and m? Wait, but in our previous steps, we have already considered that.Wait, actually, we arrived at (17h +4m)^2 <52. So, this inequality must hold for some real h. Therefore, the question is: For which m does there exist h such that (17h +4m)^2 <52. But since (17h +4m)^2 is always non-negative, and we need it to be less than 52, the maximum possible value of (17h +4m)^2 is 52. So, for such h to exist, m must satisfy that the minimum value of (17h +4m)^2 is less than 52. Wait, perhaps more precise.Alternatively, treating this as a quadratic in h: 289h² +136hm +16m² <52.This is equivalent to:289h² +136hm +16m² -52 <0.This is a quadratic inequality in h. For real solutions h to exist, the quadratic equation 289h² +136hm +16m² -52 =0 must have real roots, so the discriminant of this quadratic (in h) must be non-negative.Wait, but since the quadratic is 289h² +136hm + (16m² -52) <0, which can be written as:289h² +136hm + (16m² -52) <0To have real solutions h, the quadratic must open upwards (since coefficient of h² is 289>0) and achieve negative values somewhere. Therefore, the minimum of the quadratic must be less than 0. The minimum occurs at h = -b/(2a) = -136m/(2*289) = -68m/289 = -4m/17.Plugging h = -4m/17 into the quadratic:289*(-4m/17)^2 +136m*(-4m/17) +16m² -52Compute each term:289*(16m²/289) =16m²136m*(-4m/17)= - (544m²)/17= -32m²16m² -52 remainsSo total:16m² -32m² +16m² -52= (16-32+16)m² -52=0*m² -52= -52So, the minimum value of the quadratic is -52, which is less than 0. Therefore, the quadratic will always have two real roots, meaning that for any m, there exist h such that (17h +4m)^2 <52. Wait, but that can't be. Because when we set h = -4m/17, the quadratic evaluates to -52, which is negative. Therefore, the quadratic 289h² +136hm +16m² -52 is less than zero for some h near h = -4m/17, regardless of m. But that would imply that for any m, there exists h such that the discriminant D>0, hence the line AB intersects the ellipse at two points, but we need that points A and B are distinct and lie on the ellipse. However, the problem states that there exist two distinct points A and B. So, perhaps the answer is all real m? But the options given are intervals around zero, so I must have messed up somewhere.Wait, maybe I made a mistake here. Let's recap.We found that the discriminant condition reduces to (17h +4m)^2 <52. Therefore, for some h, this inequality must hold. Since (17h +4m)^2 is a square, it's non-negative, and it can take any value from 0 upwards. But we need it to be less than 52. Therefore, for each m, there's a range of h that satisfies this inequality. But since h can be chosen to make 17h +4m as small as needed, but we need to check if the corresponding midpoint (h,k) is such that the chord AB with midpoint (h,k) exists on the ellipse. Wait, perhaps not. Wait, but (h,k) must lie on the line y=4x +m, which is k=4h +m. However, even if h is chosen such that (17h +4m)^2 <52, there might be additional constraints because the midpoint (h,k) must also lie inside the ellipse. Wait, midpoints of chords of an ellipse lie inside the ellipse. So, (h,k) must satisfy the condition that it's inside the ellipse or on it. Wait, but actually, the midpoint of a chord can lie outside the ellipse only if the chord is a diameter, but in general, for an ellipse, midpoints of chords are inside the ellipse. Wait, no. Wait, if you have a chord passing through the ellipse, the midpoint would be inside the ellipse. For example, take two points on the ellipse: the line segment connecting them must lie inside the ellipse, so the midpoint is inside. Therefore, (h,k) must lie inside the ellipse. Hence, (h,k) must satisfy h²/4 +k²/3 <1.But since k=4h +m, we can substitute:h²/4 + (4h +m)^2 /3 <1This is another condition that must be satisfied. Therefore, even if the discriminant condition (17h +4m)^2 <52 is satisfied, we also need (h,k) inside the ellipse. Therefore, combining both conditions:1. (17h +4m)^2 <522. h²/4 + (4h +m)^2 /3 <1Therefore, m must satisfy that there exists h such that both inequalities hold.Therefore, the problem reduces to finding the range of m for which there exists h such that both:1. |17h +4m| < sqrt(52) ≈7.2112. h²/4 + (4h +m)^2 /3 <1So, we need to find m such that the system of inequalities above has a solution h.This seems complex, but perhaps we can parameterize h in terms of m from the first inequality and substitute into the second.Let me solve inequality 1 for h:|17h +4m| < sqrt(52)=> -sqrt(52) <17h +4m <sqrt(52)=> (-sqrt(52) -4m)/17 <h < (sqrt(52) -4m)/17Therefore, h is in the interval [ (-sqrt(52) -4m)/17 , (sqrt(52) -4m)/17 ]Now, substitute h into inequality 2:h²/4 + (4h +m)^2 /3 <1Let’s write h in terms of m. Let’s denote h = t, so we can express the inequality as:t²/4 + (4t +m)^2 /3 <1Which is a quadratic in t:(1/4 + 16/3)t² + (8m/3)t + (m²/3 -1) <0Compute coefficients:1/4 +16/3= (3 +64)/12=67/128m/3 remainsm²/3 -1= (m² -3)/3Therefore, inequality:(67/12)t² + (8m/3)t + (m² -3)/3 <0Multiply through by 12 to eliminate denominators:67t² +32mt +4(m² -3) <0So:67t² +32mt +4m² -12 <0This is a quadratic in t. For there to exist t (i.e., h) satisfying this inequality, the quadratic must attain negative values. Since the coefficient of t² is 67>0, the quadratic opens upwards. Therefore, it will have real solutions if its discriminant is positive.Compute discriminant D:D=(32m)^2 -4*67*(4m² -12)Calculate each term:(32m)^2=1024m²4*67*(4m² -12)=268*(4m² -12)=1072m² -3216Thus, D=1024m² -1072m² +3216= -48m² +3216For the quadratic to have real roots, D≥0:-48m² +3216 ≥0=> -48m² ≥-3216Multiply both sides by -1 (inequality reverses):48m² ≤3216=> m² ≤3216/48=67But 3216/48=3216 ÷48. 48*67=3216. So m² ≤67, so m ∈ [-sqrt(67), sqrt(67)]. But this is a very large interval, and the answer choices are much smaller. So, clearly, this approach is missing something.Wait, perhaps I messed up the substitution. Let's check again.Original inequality 2: h²/4 + (4h +m)^2 /3 <1Expanding:h²/4 + (16h² +8hm +m²)/3 <1Multiply all terms by 12 to eliminate denominators:3h² +64h² +32hm +4m² <12Combine like terms:(3h² +64h²) =67h²So, 67h² +32hm +4m² <12Yes, that's correct. So, 67h² +32hm +4m² -12 <0Which is the same as before.So, discriminant D=(32m)^2 -4*67*(4m² -12)=1024m² -1072m² +3216= -48m² +3216So, D= -48m² +3216 ≥0 → m² ≤67, which is approximately m² ≤8.185, so m ∈ (-8.185,8.185). But this is inconsistent with the answer choices given, which are much smaller intervals. So, where's the mistake?Wait, perhaps the problem is not just that the quadratic in t (h) must have real solutions, but also that the solutions for t must lie within the interval derived from the first inequality. That is, h must satisfy both |17h +4m| < sqrt(52) and 67h² +32hm +4m² -12 <0. Therefore, even though for m² ≤67, the quadratic in h has real roots, the roots must lie within h ∈ [ (-sqrt(52) -4m)/17 , (sqrt(52) -4m)/17 ]Therefore, the problem is more involved. Perhaps I need to find m such that the quadratic 67h² +32hm +4m² -12 <0 has solutions h within h ∈ [ (-sqrt(52) -4m)/17 , (sqrt(52) -4m)/17 ]Alternatively, maybe there's a better approach.Let me think again. The problem says there exist two distinct points on the ellipse symmetric with respect to the line y=4x +m. Which means that the line is a perpendicular bisector of the chord AB. The maximum distance between the center of the ellipse and the line y=4x +m must be such that the line is within the "width" of the ellipse along that direction.Wait, another approach: The set of all midpoints of chords of the ellipse with slope -1/4 (since AB is perpendicular to y=4x +m, so slope -1/4) lies on the line y=4x +m. So, the locus of midpoints of chords with slope -1/4 is a line, called the equation of the chord midpoints. For an ellipse, the equation of the locus of midpoints of chords with a given slope is a line called the diameter of the ellipse corresponding to that slope.The equation of the diameter (locus of midpoints) for chords with slope -1/4 can be found using the formula:For the ellipse x²/a² + y²/b² =1, the diameter conjugate to the slope m is given by y = (-a²/(b² m))xWait, let me check. The conjugate diameter. For an ellipse, if you have a set of parallel chords with slope m, then the midpoints lie on a line called the conjugate diameter, which has slope -a²/(b² m). So in our case, the slope of the chords is -1/4, so the conjugate diameter has slope -a²/(b²*(-1/4)) = (4a²)/b².Given the ellipse x²/4 + y²/3 =1, so a²=4, b²=3.Thus, the slope of the conjugate diameter is (4*4)/3 =16/3. Therefore, the midpoints of chords with slope -1/4 lie on a line with slope 16/3. However, the problem states that these midpoints lie on y=4x +m. So, this line y=4x +m must coincide with the conjugate diameter. But the slope of the conjugate diameter is 16/3 ≈5.333, which is different from 4. Hence, unless the line y=4x +m has slope 16/3, which it doesn't, the lines do not coincide. Therefore, this approach might not be directly applicable.Wait, maybe there's a different method. Let me recall that for each line y=4x +m, we can find the points on the ellipse that are symmetric with respect to this line. The existence of such points implies that the line is at a certain distance from the center, such that reflection across the line maps the ellipse to itself in such a way that at least two points are fixed (i.e., their reflections are also on the ellipse). But this might be more complicated.Alternatively, using the concept that if two points are symmetric with respect to a line, then their midpoint lies on the line, and the line is the perpendicular bisector. So, given the line y=4x +m, then for any pair of symmetric points, their midpoint is on the line, and the line is perpendicular to AB. As before, we have slope of AB is -1/4.Therefore, the problem reduces to finding m such that there exists a chord AB of the ellipse with slope -1/4 whose midpoint lies on y=4x +m. So, the set of all midpoints of such chords is a line (the conjugate diameter), which we found has slope 16/3. But the line y=4x +m has slope 4, which is different. Therefore, these two lines (the conjugate diameter and y=4x +m) must intersect, and for some m, the line y=4x +m will intersect the conjugate diameter. But since their slopes are different (16/3 and 4), they will intersect at exactly one point. Therefore, the only midpoint that lies on both lines is their intersection point. However, the problem states that there exist two distinct points A and B, so their midpoint is on y=4x +m, but if the line y=4x +m is not the conjugate diameter, then there can be at most one such midpoint? But this seems conflicting with the previous analysis.Wait, maybe this is not the case. Wait, the conjugate diameter is the locus of all midpoints of chords with slope -1/4. So, if the line y=4x +m intersects this conjugate diameter, then the intersection point is the midpoint of such a chord. But since the line y=4x +m has a different slope, it can only intersect the conjugate diameter at one point. Therefore, for each m such that y=4x +m intersects the conjugate diameter, there is exactly one such midpoint, and hence exactly one chord AB with slope -1/4 whose midpoint lies on y=4x +m. But the problem states "there exist two distinct points A and B", so such chord AB exists. But if there is exactly one chord for each m where lines intersect, then for m such that the line y=4x +m intersects the conjugate diameter, there exists such a chord. But the conjugate diameter is a line, so y=4x +m can intersect it at exactly one point for each m, but the range of m would be such that the distance between the lines y=4x +m and the conjugate diameter is within certain limits?Wait, I think I need to compute the conjugate diameter.As per earlier, for the ellipse x²/4 + y²/3 =1, the conjugate diameter for chords with slope -1/4 is a line with slope 16/3. Let’s compute its equation.The conjugate diameter passes through the center of the ellipse (since all diameters pass through the center). Wait, no. Wait, conjugate diameters are such that each bisects the chords parallel to the other. However, for central conics like ellipses, conjugate diameters pass through the center.But in this case, the slope of the conjugate diameter is 16/3, as calculated before, so the equation is y = (16/3)x. Because it passes through the origin, as it's a diameter.Therefore, the conjugate diameter for chords with slope -1/4 is y = (16/3)x.Therefore, the midpoints of all chords with slope -1/4 lie on the line y=(16/3)x.Therefore, if we want the midpoint of chord AB to lie on y=4x +m, then the line y=4x +m must intersect the conjugate diameter y=(16/3)x. The intersection point is found by solving:4x +m = (16/3)x=> m = (16/3)x -4x = (16/3 -12/3)x = (4/3)x=> x= (3/4)mSubstitute back into y=4x +m:y=4*(3/4)m +m=3m +m=4mTherefore, the intersection point is ( (3/4)m, 4m )But this point must lie on the conjugate diameter y=(16/3)x, which it does by construction. However, this point is also the midpoint of the chord AB. Since the chord AB lies on the line with slope -1/4 passing through this midpoint, we need to ensure that this chord actually exists on the ellipse, i.e., the line intersects the ellipse at two distinct points.Therefore, the chord AB is the line passing through ( (3/4)m, 4m ) with slope -1/4. Let’s write its equation:y -4m = -1/4 (x - (3/4)m )Simplify:y = -1/4 x + (3/4)m*(1/4) +4mWait, expanding:y = -1/4 x + (3/16)m +4m = -1/4 x + (3/16 +64/16)m = -1/4x + (67/16)mTherefore, the equation of chord AB is y = -1/4 x + (67/16)mNow, this line must intersect the ellipse x²/4 + y²/3 =1 at two distinct points. Let's substitute y from the line equation into the ellipse:x²/4 + [ (-1/4 x +67m/16 )² ] /3 =1Compute y²:(-1/4 x +67m/16 )² = (1/16)x² - (2*(1/4)(67m/16))x + (67m/16)^2= (1/16)x² - (67m/32)x + (4489m²)/256Therefore, substituting into ellipse:x²/4 + [ (1/16)x² - (67m/32)x +4489m²/256 ] /3 =1Multiply through by 256*3=768 to eliminate denominators:768*(x²/4) +768*[ (1/16)x² - (67m/32)x +4489m²/256 ] /3 =768*1Simplify each term:768*(x²/4)=192x²768*[ (1/16)x² - (67m/32)x +4489m²/256 ] /3 =256*[ (1/16)x² - (67m/32)x +4489m²/256 ]Compute each part:256*(1/16)x²=16x²256*(-67m/32)x= -536mx256*(4489m²/256)=4489m²Thus, total:192x² +16x² -536mx +4489m² =768Combine like terms:208x² -536mx +4489m² -768=0Divide entire equation by common factor if possible. Let's check if 208,536,4489,768 have common factors.208=16*13, 536=16*33.5, but 536=16*33.5 isn't integer. Wait, 536 ÷16=33.5. Not integer. 4489 is 67², and 768=256*3=16*16*3. So no common factor besides 1.Therefore, the equation is:208x² -536mx +4489m² -768=0For this quadratic equation in x to have real solutions, the discriminant must be non-negative.Compute discriminant D:D=( -536m )² -4*208*(4489m² -768 )Calculate each term:( -536m )²= (536)^2 m²= 536²= (500+36)^2=500² +2*500*36 +36²=250000 +36000 +1296=287,296. So D=287,296m²Next term:4*208*(4489m² -768 )=832*(4489m² -768 )Compute 832*4489m². Let's compute 4489*800=3,591,200 and 4489*32=143,648. So total 3,591,200 +143,648=3,734,848m²832*768: Compute 800*768=600, 800*700=560,000; 800*68=54,400; total 560,000+54,400=614,400. Then 32*768=24,576. So total 614,400 +24,576=638,976.Thus, D=287,296m² -3,734,848m² +638,976= (287,296 -3,734,848)m² +638,976= (-3,447,552)m² +638,976For real solutions, D≥0:-3,447,552m² +638,976 ≥0=> -3,447,552m² ≥ -638,976Multiply both sides by (-1) (inequality reverses):3,447,552m² ≤638,976=> m² ≤638,976 /3,447,552Simplify:Divide numerator and denominator by 48:638,976 ÷48=13,3123,447,552 ÷48=71,824So, m² ≤13,312 /71,824= Divide numerator and denominator by 16: 832 /4,489Wait, 13,312 ÷16=832, 71,824 ÷16=4,489But 832 and4489: 832=64*13, 4489=67². So no common factors.Thus, m² ≤832/4489Therefore, m² ≤(64*13)/(67²)= (64/67²)*13Thus, m ≤ sqrt(832/4489)=sqrt(832)/sqrt(4489)= (28.6356)/67≈0.427Wait, compute sqrt(832): 28.843. Wait, 28²=784, 29²=841, so sqrt(832)=28.84 approx. sqrt(4489)=67.Therefore, sqrt(832/4489)=sqrt(832)/67≈28.84/67≈0.430Thus, m must satisfy |m| ≤0.430 approximately.But let's compute 832/4489 exactly:832 ÷4489:Divide numerator and denominator by GCD(832,4489). 4489 ÷832=5*832=4160, 4489-4160=329.GCD(832,329). 832 ÷329=2*329=658, 832-658=174.GCD(329,174). 329 ÷174=1*174=174, 329-174=155.GCD(174,155). 174 ÷155=1*155=155, 174-155=19.GCD(155,19). 155 ÷19=8*19=152, 155-152=3.GCD(19,3). 19 ÷3=6*3=18, 19-18=1.GCD(3,1)=1.So GCD is 1. Therefore, 832/4489 cannot be simplified.Compute sqrt(832/4489):sqrt(832)/sqrt(4489)=sqrt(832)/67But 832=64*13, so sqrt(832)=8*sqrt(13)Therefore, sqrt(832)/67=8*sqrt(13)/67= (8/67)sqrt(13)But 8/67 is approximately0.119, sqrt(13)≈3.606, so 0.119*3.606≈0.430Thus, m^2 ≤ (8√13)^2 /67²=64*13 /4489= 832/4489, which we had.Therefore, m ∈ [ -8√13/67, 8√13/67 ]But simplify 8/67: Let's see, 8 and67 have no common factors. Wait, 8√13/67. But the answer options are given in terms of 2√13/13, etc. Let's see.Wait, 8/67. Let me compute 8/67. But 8√13 /67. Let's rationalize or simplify:Note that 8/67 can be written as (8/67)*(√13/√13)= (8√13)/(67√13). Not helpful.Alternatively, perhaps rationalizing the denominator. Wait, but the answer choices have denominators like 13. Let me see:Compute 8√13 /67. Let's see if 67 and13 have common factors. 67 is prime, 13 is prime. So, no. Alternatively, express 8/67 as (8/67)= (8/67)*(13/13)=104/871. Not helpful.Wait, but perhaps there's a mistake here. Because when we derived m² ≤832/4489, which is 64*13/(67²). Therefore, sqrt(m²)=8√13 /67. Which is equal to 8/67 *√13. But 8/67 is not simplifying to 2/13. Since 67 divided by 13 is 5.15, so no relation. Therefore, perhaps my approach is wrong.But the answer options are B: (-2√13/13, 2√13/13), etc.Wait, 2√13 /13 ≈2*3.605/13≈7.21/13≈0.555, which is larger than our computed≈0.430. So, discrepancy here. Therefore, my calculation must have an error.Wait, going back through the steps:The equation of chord AB is y = -1/4 x + (67/16)m. Substituting into ellipse:x²/4 + [ (-1/4 x +67m/16 )² ] /3 =1Expand y²:(-1/4 x +67m/16 )² = (1/16)x² - (2*(1/4)(67m/16))x + (67m/16)^2= (1/16)x² - (67m/32)x + (4489m²)/256Then substituting into the ellipse:x²/4 + [ (1/16)x² - (67m/32)x +4489m²/256 ] /3 =1Multiply through by 768:768*(x²/4) +768*[ (1/16)x² - (67m/32)x +4489m²/256 ] /3 =768*1Which is:192x² +256*( (1/16)x² - (67m/32)x +4489m²/256 )=768Calculate inside the brackets:256*(1/16)x²=16x²256*(-67m/32)x= -8*67mx= -536mx256*(4489m²/256)=4489m²Therefore, total equation:192x² +16x² -536mx +4489m²=768Which gives 208x² -536mx +4489m² -768=0Yes, correct.Discriminant D=( -536m )² -4*208*(4489m² -768 )= 287,296m² -4*208*4489m² +4*208*768Calculate 4*208*4489=832*4489=3,734,8484*208*768=832*768=638,976Thus, D=287,296m² -3,734,848m² +638,976= -3,447,552m² +638,976Set D≥0:-3,447,552m² +638,976 ≥0 →3,447,552m² ≤638,976 →m² ≤638,976 /3,447,552=0.1852 approximately.Which is m²≈0.1852 →m≈±0.430, which matches previous result.However, none of the answer choices have 8√13/67 or similar. But let's compute 8√13 /67 and 2√13 /13:Compute 2√13 /13 ≈2*3.605/13≈7.21/13≈0.555Compute 8√13 /67≈8*3.605/67≈28.84/67≈0.430So, my answer is approximately (-0.430,0.430), while option B is (-2√13/13,2√13/13)≈(-0.555,0.555). Therefore, there is a discrepancy. So, either my approach is wrong, or the answer options are incorrect.Alternatively, maybe there's a miscalculation.Wait, let's check the step where I substituted the chord midpoint into the ellipse. Wait, no, the midpoint ( (3/4)m,4m ) must lie inside the ellipse. Therefore, we have the condition:( (3/4 m )² )/4 + ( (4m )² ) /3 <1Compute:(9/16 m²)/4 + (16m²)/3 = (9/64 m²) + (16/3 m²)Convert to common denominator 192:(9/64)m² =27/192 m²(16/3)m²=1024/192 m²Sum:27/192 +1024/192=1051/192 m² <1Thus, m² <192/1051≈0.1826Therefore, m≈±0.427, which matches the previous result. Therefore, the midpoint must lie inside the ellipse, leading to m²<192/1051≈0.1826, which is approximately (±0.427)^2.But 192/1051 is approximately0.1826, and sqrt(0.1826)=0.427, which is approximately 8√13 /67≈0.430, which is very close. Due to exact calculation:192/1051= (64*3)/1051. However, 1051 is a prime number? Let's check.1051 ÷7=150.142… no. 1051 ÷13=80.846… no. Likely prime.So, exact value is messy, but the approximate is around ±0.427. However, none of the answer choices match this. The closest is option D: (-2√3/13, 2√3/13). Compute 2√3 /13≈2*1.732/13≈3.464/13≈0.266. Not close.Option B is ±2√13 /13≈±2*3.606/13≈±7.212/13≈±0.555. Hmm, larger than our value.Wait, maybe my mistake is in assuming that the midpoint must lie inside the ellipse. But actually, the midpoint of a chord always lies inside the ellipse, right? Because the chord connects two points on the ellipse, so the midpoint is inside the ellipse.Therefore, that condition must hold. But according to our calculation, that gives a tighter bound on m than the discriminant condition. But according to the earlier analysis, the discriminant condition gives a larger interval, but combining with midpoint condition gives a smaller interval.But the answer options do not include this approximate interval. So, perhaps my entire approach is incorrect.Alternative approach:Let me consider the reflection of the ellipse over the line y=4x +m. If there exists a point A on the ellipse such that its reflection B is also on the ellipse, then A and B are symmetric with respect to the line. Therefore, the intersection of the reflected ellipse with the original ellipse must contain at least two points.The reflection of the ellipse over the line y=4x +m will be another ellipse. For these two ellipses to intersect, the distance between their centers must be less than the sum of their major axes, but this might be too vague.Alternatively, the line y=4x +m must be such that the reflection across it maps some point on the ellipse to another point on the ellipse. The set of all lines for which this is true forms a certain locus.Reflection across a line is an affine transformation. Computing the condition that the reflection of a point on the ellipse over the line y=4x +m is also on the ellipse.Let me denote a general point (x,y) on the ellipse. Its reflection over the line y=4x +m is (x', y'). The formula for reflection over a line ax + by +c=0 is known, but let's recall it.The formula for reflecting a point (x,y) over the line ax + by +c =0 is:x' = x - 2a(ax + by +c)/(a² +b²)y' = y - 2b(ax + by +c)/(a² +b²)In our case, the line is y=4x +m, which can be rewritten as 4x - y +m=0. So, a=4, b=-1, c=m.Therefore, the reflection formulas are:x' = x - 2*4*(4x - y +m)/(16 +1)=x - 8*(4x - y +m)/17y' = y - 2*(-1)*(4x - y +m)/17=y + 2*(4x - y +m)/17Simplify:x' = x - (32x -8y +8m)/17= (17x -32x +8y -8m)/17= (-15x +8y -8m)/17y' = y + (8x -2y +2m)/17= (17y +8x -2y +2m)/17= (8x +15y +2m)/17Therefore, the reflection of (x,y) over the line is (x', y')= ( (-15x +8y -8m)/17, (8x +15y +2m)/17 )For the reflected point (x', y') to lie on the ellipse x²/4 + y²/3=1, substitute:[ (-15x +8y -8m )² ] / (4*17² ) + [ (8x +15y +2m )² ] / (3*17² ) =1Multiply both sides by 4*3*17²=12*289=3468 to eliminate denominators:3*(-15x +8y -8m )² +4*(8x +15y +2m )² =3468Expand both terms:First term:3*(225x² -240xy +240xm +64y² -128ym +64m² )Second term:4*(64x² +240xy +32xm +225y² +60ym +4m² )Compute first term:3*225x²=675x²3*(-240xy)=-720xy3*240xm=720xm3*64y²=192y²3*(-128ym)=-384ym3*64m²=192m²Second term:4*64x²=256x²4*240xy=960xy4*32xm=128xm4*225y²=900y²4*60ym=240ym4*4m²=16m²Combine all terms:675x² -720xy +720xm +192y² -384ym +192m² +256x² +960xy +128xm +900y² +240ym +16m²=3468Combine like terms:x²:675 +256=931xy:-720 +960=240xm:720 +128=848y²:192 +900=1092ym:-384 +240= -144m²:192 +16=208Thus, the equation becomes:931x² +240xy +848xm +1092y² -144ym +208m²=3468But the original point (x,y) lies on the ellipse x²/4 + y²/3=1. So, we can use this to replace one of the variables. For instance, solve for y²: y²=3(1 -x²/4 )Substitute y²=3 - (3x²)/4 into the equation:931x² +240xy +848xm +1092*(3 -3x²/4 ) -144ym +208m²=3468Compute each term:1092*3=32761092*(-3x²/4)= -819x²Therefore:931x² +240xy +848xm +3276 -819x² -144ym +208m²=3468Combine like terms:x²:931 -819=112xy:240xyxm:848xmy terms: -144ymconstants:3276m²:208m²Thus, equation becomes:112x² +240xy +848xm -144ym +208m² +3276=3468Subtract 3468-3276=192 from both sides:112x² +240xy +848xm -144ym +208m²=192This equation must hold for some (x,y) on the ellipse. This seems complex, but maybe we can find conditions on m such that this equation has a solution.Alternatively, since this equation is complicated, perhaps assume that the reflected point is distinct from the original point, so that there exist two distinct points. Therefore, the equation has at least one solution (x,y) on the ellipse, with (x,y) ≠(x',y').But this approach seems too involved.Alternative idea: The distance from the center of the ellipse (0,0) to the line y=4x +m must be less than or equal to the maximum distance such that the line is close enough to the ellipse to have points reflecting over it.The distance from (0,0) to the line y=4x +m is |0 -4*0 -m| /sqrt(1 +16)= | -m | /sqrt(17)= |m| /sqrt(17)For there to exist points A and B on the ellipse symmetric over the line, this distance must be less than the maximum possible distance from the center to a point on the ellipse along the direction perpendicular to the line.The maximum distance from the center to the ellipse in any direction is the length of the major axis, but since the line has slope 4, the direction perpendicular has slope -1/4.The maximum distance from the center in the direction perpendicular to the line y=4x +m can be found using the formula for the distance in a given direction for an ellipse.For an ellipse x²/a² + y²/b²=1, the maximum distance in the direction (p,q) is sqrt( (a²p² + b²q²)/(p² + q²) )In our case, the direction perpendicular to y=4x +m is along the vector (4,1) (since the line has slope 4, perpendicular direction has slope -1/4, which corresponds to the direction vector (4,1) after normalization).But actually, the direction vector perpendicular to the line is (4, -1), because the line has slope 4, so perpendicular slope is -1/4, so direction vector is (1, -1/4), or scaled to (4, -1).Therefore, the direction vector is (4, -1). The maximum distance from the center in this direction is sqrt( (a²*4² + b²*(-1)^2 ) / (4² + (-1)^2 ) )Given a²=4, b²=3, so:sqrt( (4*16 +3*1 ) / (16 +1 ) )=sqrt( (64 +3)/17 )=sqrt(67/17)=sqrt(67)/sqrt(17)=sqrt(67)/4.123≈8.185/4.123≈1.985Therefore, the maximum distance from the center in the direction perpendicular to the line is approximately 1.985. The distance from the center to the line must be less than or equal to this for the line to intersect the ellipse or for there to exist points symmetric with respect to the line.Wait, but the distance from the center to the line is |m| /sqrt(17). Therefore, |m| /sqrt(17) ≤ sqrt(67)/sqrt(17)Wait, sqrt(67)/sqrt(17)=sqrt(67/17)=sqrt(3.941)≈1.985. Therefore, |m| /sqrt(17) ≤ sqrt(67)/sqrt(17) => |m| ≤sqrt(67)But sqrt(67)≈8.185, which is much larger than the previous estimates. Therefore, this condition is not sufficient.However, this approach might not directly apply because reflection symmetry is different from intersection.Given the time I've spent and the lack of progress toward the answer choices, perhaps the correct approach is to consider the condition that the line y=4x +m is a valid perpendicular bisector of some chord of the ellipse.Given that the midpoint (h,k) lies on y=4x +m and the line AB has slope -1/4. Using the concept that the equation of the chord with midpoint (h,k) is:(xh)/4 + (yk)/3 = (h²)/4 + (k²)/3But since the chord AB has slope -1/4, we can relate h and k through the slope.The slope of the chord AB is -1/4, which is also given by the derivative at the midpoint (h,k), but wait, no, the slope of the chord is -1/4, but the slope of the tangent at the midpoint is different.Alternatively, the slope of the chord AB is given by - (b² h)/(a² k). For an ellipse x²/a² + y²/b²=1, the slope of the chord bisected at (h,k) is - (b² h)/(a² k). This is a formula from the equation of the chord with a given midpoint.Therefore, in our case, slope of AB is -1/4, so:-1/4 = - (b² h)/(a² k )=> 1/4 = (b² h)/(a² k )Given a²=4, b²=3:1/4 = (3h)/(4k )=> (3h)/(4k )=1/4=> 3h =kBut also, since midpoint (h,k) lies on the line y=4x +m, we have k=4h +mTherefore, substituting k=3h into this:3h=4h +m=> -h =m=> h= -mTherefore, k=3h= -3mTherefore, the midpoint is (-m, -3m). Now, since this midpoint must lie inside the ellipse (as midpoints of chords are inside the ellipse), substituting into the ellipse equation:h²/4 +k²/3 <1=> (-m)²/4 + (-3m)²/3 <1=> m²/4 +9m²/3 <1Simplify:m²/4 +3m² <1=> (1/4 +3)m² <1=> (13/4)m² <1=> m² <4/13=> |m| <2/sqrt(13)= (2√13)/13≈0.555Therefore, the range of m is (-2√13/13, 2√13/13), which is option B.Ah! This is the correct approach. Previously, I was overcomplicating it. The key was using the formula for the slope of the chord with midpoint (h,k): slope = - (b² h)/(a² k ). Setting this equal to -1/4 and combining with the line equation gives h= -m, k= -3m, and then the midpoint must lie inside the ellipse, leading to m² <4/13. Thus, the answer is option B.Final Answerboxed{B}</think>Given the ellipse (frac{x^2}{4} + frac{y^2}{3} = 1), we need to find the range of values for the real number (m) such that there exist two distinct points (A) and (B) on the ellipse that are symmetric with respect to the line (y = 4x + m).1. Slope of the Line and Midpoint Condition: - The line (y = 4x + m) has a slope of 4. Therefore, the slope of the line segment (AB) must be (-frac{1}{4}) (perpendicular to the given line). - The midpoint ((h, k)) of segment (AB) must lie on the line (y = 4x + m), so (k = 4h + m).2. Slope of the Chord: - For an ellipse (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), the slope of the chord with midpoint ((h, k)) is given by (-frac{b^2 h}{a^2 k}). - Here, (a^2 = 4), (b^2 = 3), and the slope of the chord (AB) is (-frac{1}{4}). Setting up the equation: [ -frac{1}{4} = -frac{3h}{4k} ] - Solving this, we get (k = 3h).3. Midpoint on the Line: - Substituting (k = 3h) into the line equation (k = 4h + m), we get: [ 3h = 4h + m implies h = -m ] - Therefore, the midpoint ((h, k)) is ((-m, -3m)).4. Midpoint Inside the Ellipse: - The midpoint ((-m, -3m)) must lie inside the ellipse. Substituting into the ellipse equation: [ frac{(-m)^2}{4} + frac{(-3m)^2}{3} < 1 ] - Simplifying: [ frac{m^2}{4} + frac{9m^2}{3} < 1 implies frac{m^2}{4} + 3m^2 < 1 implies left(frac{1}{4} + 3right)m^2 < 1 implies frac{13}{4}m^2 < 1 implies m^2 < frac{4}{13} ] - Therefore, (|m| < frac{2sqrt{13}}{13}).Thus, the range of values for (m) is (left(-frac{2sqrt{13}}{13}, frac{2sqrt{13}}{13}right)), which corresponds to option (boxed{B}).,<think>Okay, let me try to figure out this problem. So, we have an ellipse given by the equation x²/4 + y²/3 = 1. The question is asking about the range of values for a real number m such that there exist two distinct points A and B on the ellipse which are symmetric with respect to the line y = 4x + m. The options are A, B, C, D, and I need to figure out which one is correct.First, let me recall that if two points are symmetric with respect to a line, then that line is the perpendicular bisector of the segment connecting the two points. So, for points A and B to be symmetric about y = 4x + m, the line y = 4x + m must be the perpendicular bisector of AB. That means two things: the midpoint of AB lies on the line y = 4x + m, and the line AB is perpendicular to y = 4x + m.The slope of the given line y = 4x + m is 4, so the slope of the perpendicular line AB must be -1/4, since perpendicular lines have slopes that are negative reciprocals of each other.So, let's denote points A and B on the ellipse. Let their coordinates be (x₁, y₁) and (x₂, y₂) respectively. Since AB is perpendicular to the line y = 4x + m, the slope of AB is -1/4. Therefore, (y₂ - y₁)/(x₂ - x₁) = -1/4. Also, the midpoint of AB, which is ((x₁ + x₂)/2, (y₁ + y₂)/2), must lie on the line y = 4x + m. Therefore, (y₁ + y₂)/2 = 4*( (x₁ + x₂)/2 ) + m. Simplifying, we get (y₁ + y₂) = 4(x₁ + x₂) + 2m.Moreover, since both A and B lie on the ellipse, they must satisfy the ellipse equation. So, x₁²/4 + y₁²/3 = 1 and x₂²/4 + y₂²/3 = 1.Now, let me think about how to approach this. Since the points A and B are symmetric about the line, and the line is their perpendicular bisector, perhaps we can parametrize the points A and B such that their midpoint lies on the line, and the line AB is perpendicular to the given line.Alternatively, another method might be to consider that for any such line y = 4x + m, the ellipse must intersect the line in such a way that there are two points symmetric about it. But actually, the points A and B don't necessarily lie on the line; they are symmetric with respect to the line. Hmm, so maybe there's a chord of the ellipse that is perpendicular to the line y = 4x + m, and whose midpoint is on the line y = 4x + m. That seems to be the case.So, perhaps the set of all such chords perpendicular to y = 4x + m and with midpoints on y = 4x + m can be found, and then we can determine the values of m for which such chords exist.This seems similar to the concept of the equation of the chord of an ellipse with a given midpoint. The equation of the chord with midpoint (h, k) is given by T = S1, where T is the tangent equation at (h, k) but adjusted. Wait, maybe that's for the chord of contact. Alternatively, the equation of the chord with midpoint (h, k) is (x h)/a² + (y k)/b² = (h²)/a² + (k²)/b². Wait, maybe not exactly. Let me recall.For an ellipse x²/a² + y²/b² = 1, the equation of the chord with midpoint (h, k) is (xx₁)/a² + (yy₁)/b² = (x₁²)/a² + (y₁²)/b², but that's when (x₁, y₁) is a point on the ellipse. Wait, maybe another approach.Alternatively, if we have two points (x₁, y₁) and (x₂, y₂) on the ellipse, then the midpoint (h, k) is ((x₁ + x₂)/2, (y₁ + y₂)/2). The slope of the chord AB is (y₂ - y₁)/(x₂ - x₁). Since AB is perpendicular to y = 4x + m, the slope of AB is -1/4. Therefore, (y₂ - y₁)/(x₂ - x₁) = -1/4.So, given that, maybe we can express y₂ - y₁ = -1/4 (x₂ - x₁). Let me denote the midpoint (h, k). Then x₁ = h - t, x₂ = h + t, similarly y₁ = k - s, y₂ = k + s. Then, since (y₂ - y₁) = 2s and (x₂ - x₁) = 2t, so 2s / 2t = s/t = -1/4. Therefore, s = -t/4.So, substituting back, x₁ = h - t, x₂ = h + t; y₁ = k - (-t/4) = k + t/4, y₂ = k - t/4. Wait, hold on, if s = -t/4, then y₂ - y₁ = 2s = -t/2. But the slope is (y₂ - y₁)/(x₂ - x₁) = (-t/2)/(2t) = -1/4, which checks out.So, we can express the coordinates of points A and B in terms of h, k, and t. Then, since both points lie on the ellipse, we can substitute into the ellipse equation.So, substituting (h - t, k + t/4) into x²/4 + y²/3 = 1:[(h - t)²]/4 + [(k + t/4)²]/3 = 1Similarly, substituting (h + t, k - t/4) into the ellipse equation:[(h + t)²]/4 + [(k - t/4)²]/3 = 1Since both points are on the ellipse, these two equations must hold. Let's subtract the two equations to eliminate the quadratic terms. Let me write them out:First equation: [(h² - 2ht + t²)/4] + [(k² + (kt)/2 + t²/16)/3] = 1Second equation: [(h² + 2ht + t²)/4] + [(k² - (kt)/2 + t²/16)/3] = 1Subtracting the second equation from the first:[ (-4ht)/4 ] + [ ( (kt)/2 - (-kt)/2 ) /3 ] = 0Wait, let me compute each part:First, expand both equations:First equation:(h² - 2ht + t²)/4 + (k² + (kt)/2 + t²/16)/3 = 1Second equation:(h² + 2ht + t²)/4 + (k² - (kt)/2 + t²/16)/3 = 1Subtract the second equation from the first:[ (h² - 2ht + t²)/4 - (h² + 2ht + t²)/4 ] + [ (k² + (kt)/2 + t²/16)/3 - (k² - (kt)/2 + t²/16)/3 ] = 0Simplify each bracket:First bracket:[ (h² - 2ht + t² - h² - 2ht - t²)/4 ] = [ (-4ht)/4 ] = -htSecond bracket:[ ( (k² + (kt)/2 + t²/16) - (k² - (kt)/2 + t²/16) ) /3 ] = [ (kt)/2 + (kt)/2 ) /3 ] = [ kt / 3 ]Therefore, combining both brackets:-ht + (kt)/3 = 0Factor out t:t( -h + k/3 ) = 0So, either t = 0 or -h + k/3 = 0.But t = 0 would mean that the two points A and B coincide, which contradicts the requirement that they are distinct. Therefore, we must have -h + k/3 = 0 => k = 3h.So, the midpoint (h, k) lies on the line k = 3h. But we also know that the midpoint lies on the line y = 4x + m. Therefore, k = 4h + m. Combining these two equations: 3h = 4h + m => -h = m => h = -m.Therefore, the midpoint (h, k) is (-m, k), where k = 3h = -3m. So, the midpoint is (-m, -3m). Therefore, this point must lie on the line y = 4x + m, which it does by construction. Wait, but how?Wait, actually, we derived that the midpoint lies on both the line y = 4x + m and the line y = 3x. Therefore, the intersection of these two lines is the midpoint. So, solving y = 4x + m and y = 3x, we get 3x = 4x + m => -x = m => x = -m, and then y = 3*(-m) = -3m. So, the midpoint is (-m, -3m). Therefore, the midpoint must lie on both lines, so it's the intersection point. Therefore, this is fixed once m is fixed.But we also need that there exists such points A and B on the ellipse. So, we need to ensure that there exists a chord through (-m, -3m) with slope -1/4. Wait, but actually, we parametrized the points as (h ± t, k ∓ t/4), where h = -m, k = -3m. So, points A and B are (-m - t, -3m + t/4) and (-m + t, -3m - t/4). These points must lie on the ellipse.Therefore, substituting these coordinates into the ellipse equation x²/4 + y²/3 = 1.Let me take point A: (-m - t, -3m + t/4). Plugging into the ellipse equation:[ (-m - t )² ] /4 + [ (-3m + t/4 )² ] /3 = 1Similarly for point B: (-m + t, -3m - t/4):[ (-m + t )² ] /4 + [ (-3m - t/4 )² ] /3 = 1Since both points lie on the ellipse, both equations must hold. Let's compute one of them and see.First equation:[ (m + t)² ] /4 + [ (3m - t/4 )² ] /3 = 1Expand the terms:First term: (m² + 2mt + t²)/4Second term: (9m² - (3mt)/2 + t²/16)/3 = [9m² - (3mt)/2 + t²/16]/3 = 3m² - (mt)/2 + t²/(48)Therefore, total equation:(m² + 2mt + t²)/4 + 3m² - (mt)/2 + t²/(48) = 1Multiply all terms by 48 to eliminate denominators:12(m² + 2mt + t²) + 144m² - 24mt + t² = 48Expand:12m² + 24mt + 12t² + 144m² -24mt + t² = 48Combine like terms:(12m² + 144m²) + (24mt -24mt) + (12t² + t²) = 48So:156m² + 13t² = 48Similarly, the second equation for point B would give the same result, since the substitution would lead to the same expression. Therefore, we get 156m² + 13t² = 48.But t is a real number, so 13t² = 48 - 156m²Therefore, t² = (48 - 156m²)/13Since t² must be non-negative, (48 - 156m²)/13 ≥ 0 => 48 - 156m² ≥ 0 => 156m² ≤ 48 => m² ≤ 48/156 => m² ≤ 12/39 => m² ≤ 4/13Therefore, m² ≤ 4/13 => |m| ≤ 2/sqrt(13) => |m| ≤ 2√13/13Therefore, m must lie in the interval (-2√13/13, 2√13/13). So, that's option B.Wait, but let me check if this is correct. So, we derived that for real t to exist (i.e., for the points A and B to exist), the expression under t² must be non-negative, which gives m² ≤ 4/13. Therefore, m must be in the interval from -2√13/13 to 2√13/13. Therefore, the range of m is (-2√13/13, 2√13/13), which is option B.But let me double-check to ensure I didn't make a mistake in the algebra.Starting from plugging point A into the ellipse equation:[ (-m - t )² ] /4 + [ (-3m + t/4 )² ] /3 = 1Expand:(m² + 2mt + t²)/4 + (9m² - (3mt)/2 + t²/16)/3 = 1Breaking it down:First term: (m² + 2mt + t²)/4Second term: [9m² - (3mt)/2 + t²/16]/3 = 9m²/3 - (3mt)/(2*3) + t²/(16*3) = 3m² - mt/2 + t²/48Therefore, combining:First term: (m² + 2mt + t²)/4Second term: 3m² - mt/2 + t²/48So, total equation:(m² + 2mt + t²)/4 + 3m² - mt/2 + t²/48 = 1To combine, multiply all terms by 48:12(m² + 2mt + t²) + 144m² - 24mt + t² = 48Expanding:12m² + 24mt + 12t² + 144m² -24mt + t² = 48Combine terms:(12m² +144m²) + (24mt -24mt) + (12t² + t²) = 48156m² +13t²=48Yes, that seems correct. Therefore, t²=(48-156m²)/13, which must be non-negative, leading to m² ≤48/156= 12/39=4/13. So, m is in (-2√13/13, 2√13/13), which is option B.Therefore, the answer should be B.But let me check the options again:A: (-2√13/13, 2√2/13)B: (-2√13/13, 2√13/13)C: (-√2/13, 2√13/13)D: (-2√3/13, 2√3/13)So, yes, B is the correct interval.Alternatively, maybe there's an error in assuming that the chord must have midpoints lying on both the given line and the line y=3x. Wait, but we derived that the midpoint lies on both y=4x + m and y=3x. Wait, how did we get y=3x?Wait, in the step where we subtracted the two equations, we obtained the condition k = 3h, which is the equation y = 3x. Therefore, the midpoint must lie on the intersection of y = 4x + m and y = 3x, which is the point (-m, -3m). Therefore, that point must lie inside the ellipse? Wait, no. The midpoint doesn't need to lie on the ellipse, just the points A and B.But perhaps the midpoint is inside the ellipse. However, even if the midpoint is outside the ellipse, as long as there are points A and B on the ellipse such that their midpoint is on the line y=4x + m and the line AB is perpendicular to y=4x + m, then such points exist.But in our derivation, we found that for such points to exist, the parameter m must satisfy |m| ≤ 2√13/13, hence the interval in option B.Therefore, unless there's an error in the derivation, the correct answer should be B. Let me check once more.The key steps were:1. The slope of AB is -1/4.2. The midpoint lies on y =4x + m.3. By substituting the coordinates of A and B in terms of midpoint (h,k) and parameter t, we derived that the midpoint must lie on y=3x.4. Therefore, the midpoint is (-m, -3m).5. Then, substituting into the ellipse equations led us to 156m² +13t²=48, requiring that 48 -156m² ≥0, leading to m² ≤4/13, hence m in (-2√13/13,2√13/13).Yes, that seems correct. So the answer is B.Final Answerboxed{B}</think>Given the ellipse (frac{x^2}{4} + frac{y^2}{3} = 1), we need to determine the range of values for the real number (m) such that there exist two distinct points (A) and (B) on the ellipse that are symmetric with respect to the line (y = 4x + m).1. Symmetry and Perpendicular Bisector: The line (y = 4x + m) must be the perpendicular bisector of the segment (AB). This implies the midpoint of (AB) lies on the line (y = 4x + m) and the slope of (AB) is the negative reciprocal of 4, which is (-frac{1}{4}).2. Midpoint and Slope: Let the coordinates of (A) and (B) be ((x_1, y_1)) and ((x_2, y_2)). The midpoint ((frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2})) must lie on (y = 4x + m), giving (frac{y_1 + y_2}{2} = 4 cdot frac{x_1 + x_2}{2} + m). Simplifying, we get (y_1 + y_2 = 4(x_1 + x_2) + 2m).3. Parametrization: Using the midpoint ((h, k)) and parameter (t), we express (A) and (B) as ((h - t, k + frac{t}{4})) and ((h + t, k - frac{t}{4})). The midpoint ((h, k)) must lie on both (y = 4x + m) and (y = 3x) (derived from the slope condition), leading to (k = 3h) and (k = 4h + m). Solving these, we find (h = -m) and (k = -3m).4. Substitution into Ellipse Equation: Substituting the coordinates of (A) and (B) into the ellipse equation, we derive the equation (156m^2 + 13t^2 = 48). For real (t), the term (13t^2) must be non-negative, leading to (48 - 156m^2 geq 0). Solving this, we find (m^2 leq frac{4}{13}), hence (|m| leq frac{2sqrt{13}}{13}).Thus, the range of values for (m) is (left(-frac{2sqrt{13}}{13}, frac{2sqrt{13}}{13}right)), corresponding to option B.(boxed{B})]

🔑:The equation of the ellipse is frac {x^{2}}{4} + frac {y^{2}}{3} = 1, which can be rewritten as 3x^{2} + 4y^{2} - 12 = 0.Let the two points on the ellipse A(x_{1}, y_{1}) and B(x_{2}, y_{2}) be symmetric with respect to the line y = 4x + m, and let the midpoint of AB be M(x_{0}, y_{0}).Then we have 3x_{1}^{2} + 4y_{1}^{2} - 12 = 0 and 3x_{2}^{2} + 4y_{2}^{2} - 12 = 0.Subtracting the second equation from the first gives: 3(x_{1} + x_{2})(x_{1} - x_{2}) + 4(y_{1} + y_{2})(y_{1} - y_{2}) = 0, which simplifies to 3 cdot 2x_{0} cdot (x_{1} - x_{2}) + 4 cdot 2y_{0} cdot (y_{1} - y_{2}) = 0.Therefore, frac {y_{1} - y_{2}}{x_{1} - x_{2}} = -frac {3}{4} cdot frac {x_{0}}{y_{0}} = -frac {1}{4}.Thus, y_{0} = 3x_{0}. Substituting into the line equation y = 4x + m gives x_{0} = -m and y_{0} = -3m.Since (x_{0}, y_{0}) is inside the ellipse, we have 3m^{2} + 4(-3m)^{2} < 12, which simplifies to 3m^{2} + 36m^{2} < 12. Solving this inequality gives -frac {2 sqrt {13}}{13} < m < frac {2 sqrt {13}}{13}.Therefore, the correct choice is: boxed{text{B}}.This problem examines the comprehensive issue of a line and a conic section, focusing on the application of the method of difference of squares and highlighting the examination of the idea of reduction, making it a challenging question.

🧠:[<think>Alright, so I need to solve this triangle problem. Let's see. The question says: In triangle ABC, if 2 sin A cos B = sin C, then the triangle is which type: isosceles, right, scalene, or equilateral. The options are given as (A) to (D). First, I should recall some trigonometric identities related to triangles. Since we're dealing with a triangle, the sum of the angles A, B, and C is 180 degrees. Also, the Law of Sines and Law of Cosines might come in handy here. The given equation is 2 sin A cos B = sin C. Hmm, maybe I can express sin C in terms of angles A and B? Since in a triangle, C = 180° - A - B. So, sin C = sin(180° - A - B). And since sin(180° - x) = sin x, that means sin C = sin(A + B). So the equation becomes 2 sin A cos B = sin(A + B). Wait, that seems like a trigonometric identity. Let me check. The sine of a sum formula is sin(A + B) = sin A cos B + cos A sin B. So substituting that into the equation, we have:2 sin A cos B = sin A cos B + cos A sin BSubtracting sin A cos B from both sides gives:sin A cos B = cos A sin BHmm, that simplifies to sin A cos B - cos A sin B = 0. Wait, that's the sine of (A - B) because sin(A - B) = sin A cos B - cos A sin B. So, sin(A - B) = 0.If sin(A - B) = 0, then A - B must be an angle whose sine is zero. Since A and B are angles in a triangle, they must be between 0° and 180°, so the difference A - B must be between -180° and 180°. The sine of an angle is zero at 0°, 180°, -180°, etc. But since A and B are both positive and less than 180°, the difference A - B can only be 0°, because if it were 180° or -180°, one of the angles would have to be greater than 180°, which isn't possible. Therefore, A - B = 0°, so A = B.If A = B, then the triangle is isosceles with sides opposite angles A and B being equal. So, sides a and b (opposite angles A and B respectively) are equal. That would make the triangle isosceles, which is option (A). But wait, the problem is giving options including right triangle, scalene, etc. Let me check if there's another possibility. The equation led us to A = B, which would make it isosceles. But maybe there are other conditions? Let's verify with an example.Suppose triangle ABC is isosceles with A = B. Then angles A and B are equal. Let’s pick some values. Let’s say angle A = angle B = 50°, then angle C = 80°. Let's check if 2 sin A cos B equals sin C.Calculating left-hand side (LHS): 2 sin 50° cos 50°. Let me compute that. Sin 50° is approximately 0.7660, cos 50° is approximately 0.6428. Multiplying them: 0.7660 * 0.6428 ≈ 0.492. Then multiply by 2: ≈ 0.984.Right-hand side (RHS): sin 80° ≈ 0.9848. The values are approximately equal, considering rounding errors. So that works. Now, let's check if a right triangle could also satisfy this condition. Suppose the triangle is right-angled. Let's say angle C is 90°, then angles A and B sum to 90°. Let's test if 2 sin A cos B = sin C. Sin C is sin 90° = 1. Let's pick angles A = 30°, B = 60°, then LHS: 2 sin 30° cos 60° = 2*(0.5)*(0.5) = 2*(0.25) = 0.5. Which is not equal to 1. So that doesn't work. Wait, but maybe another right triangle? Suppose angle A is 45°, angle B is 45°, angle C is 90°. Then LHS: 2 sin 45° cos 45° = 2*(√2/2)*(√2/2) = 2*(0.5) = 1. RHS: sin 90° = 1. So in this case, it works. But here, angles A and B are both 45°, so the triangle is both isosceles and right-angled. So is that possible?So if a triangle is both isosceles and right-angled, then it's a 45-45-90 triangle, which is indeed an isosceles right triangle. So in that case, the triangle would satisfy both (A) and (B). But the answer options are exclusive. So which one is correct?Wait, the question is "this triangle is" and the options are (A) an isosceles triangle, (B) a right triangle, etc. So if the condition is satisfied by both isosceles triangles and right triangles, but in the case of a right triangle that is not isosceles, the condition doesn't hold. For example, the 30-60-90 triangle doesn't satisfy the condition, but the 45-45-90 triangle does. So in the case of a right triangle, only if it's isosceles does the equation hold. Therefore, the triangle must be isosceles. Because even the right-angled case that works is a subset of isosceles triangles. Therefore, the correct answer is (A). Alternatively, maybe there's a scenario where the triangle is scalene or equilateral? Let's check. An equilateral triangle has all angles 60°. Then LHS: 2 sin 60° cos 60° = 2*(√3/2)*(0.5) = 2*(√3/4) = √3/2 ≈ 0.866. RHS: sin 60° = √3/2 ≈ 0.866. So that works. So an equilateral triangle would also satisfy the equation. So does that mean the answer is both (A) and (D)? Wait, but equilateral triangles are a special case of isosceles triangles, where all sides are equal. However, the problem is from 1990, and in some definitions, isosceles triangles are considered to have exactly two equal sides, while equilateral triangles are a separate category. But in mathematical definitions, usually, an equilateral triangle is a special case of an isosceles triangle. However, the options here are given as separate: (A) isosceles, (D) equilateral. So in that case, if the equation holds for both isosceles and equilateral triangles, but the answer must be one of the options, which is it?Wait, let's re-examine. If we derived that angle A = angle B, then the triangle is isosceles (at least two sides equal). If it's equilateral, then all three angles are equal, so A = B = C. Therefore, equilateral triangles also satisfy A = B, hence are a subset of isosceles triangles. So if the answer is isosceles (A), then technically, equilateral triangles are included in that. But since (D) is an option, perhaps the answer is both (A) and (D). But in the original problem, it's a multiple-choice question with single answer? Let me check.Wait, the original problem says "this triangle is" followed by four options: (A), (B), (C), (D). So if the answer could be both isosceles and equilateral, but (A) and (D) are separate options, which one is correct?Wait, let's test with an equilateral triangle. If all angles are 60°, then as I calculated before, LHS: 2 sin 60° cos 60° = √3/2. RHS: sin 60° = √3/2. So equation holds. Therefore, equilateral triangles do satisfy the condition. But since in the equation, we derived that angle A = angle B. So if angle A = angle B, but angle C can be different, unless angle C is also equal. So in the case of an equilateral triangle, angle C is equal to A and B, so it's a special case. So the general solution is angle A = angle B, which includes the possibility of angle C being equal (i.e., equilateral). So does the problem consider equilateral as a separate case? If the answer options are (A) isosceles, (D) equilateral, then both could technically be correct. But in most cases, equilateral triangles are considered a subset of isosceles. So depending on the competition's answer key, maybe (A) is the correct answer. But in the problem statement, since (D) is an option, perhaps the answer is (D). Wait, but in our derivation, we only get that angle A = angle B. For the triangle to be equilateral, angle C must also equal angle A. Is there a way to show that angle C must equal angle A?Wait, let's go back to the equation 2 sin A cos B = sin C. If A = B, then angle C = 180° - 2A. So substituting back into the equation: 2 sin A cos A = sin(180° - 2A). Since sin(180° - 2A) = sin 2A. Therefore, the equation becomes 2 sin A cos A = sin 2A. But 2 sin A cos A is equal to sin 2A, so this is an identity. Therefore, when A = B, the equation holds regardless of the value of angle C. Therefore, angle C can be any angle such that A = B and A + B + C = 180°, so C = 180° - 2A. Therefore, the triangle is isosceles with A = B, but angle C can vary. For example, if A = B = 50°, then C = 80°, which is an isosceles triangle. If A = B = 45°, then C = 90°, which is an isosceles right triangle. If A = B = 60°, then C = 60°, which is equilateral. Therefore, the equation holds for any isosceles triangle with A = B, not necessarily equilateral. So the triangle must be isosceles, but not necessarily equilateral. Therefore, the correct answer is (A). But wait, the problem is from a competition in 1990. Maybe the answer is (A). Let me check again. So the key step is that 2 sin A cos B = sin C led us to A = B. Therefore, regardless of what angle C is, as long as A = B, the equation holds. So the triangle is isosceles (A). However, if the equation had led us to a necessity for all angles to be equal, then it would be equilateral, but in this case, it's only necessary for A and B to be equal. Therefore, the answer is (A). Therefore, the answer should be option (A), an isosceles triangle.Final Answerboxed{A}</think>Given the problem: In triangle ABC, if 2 sin A cos B = sin C, then this triangle is which type?We start by using the identity for sin C in a triangle. Since C = 180^circ - A - B, we have sin C = sin(A + B). Substituting this into the given equation, we get:[2 sin A cos B = sin(A + B)]Using the sine addition formula, sin(A + B) = sin A cos B + cos A sin B, the equation becomes:[2 sin A cos B = sin A cos B + cos A sin B]Subtracting sin A cos B from both sides, we get:[sin A cos B = cos A sin B]This simplifies to:[sin(A - B) = 0]Since sin(A - B) = 0 implies A - B = 0^circ or A - B = 180^circ (but 180^circ is not possible in a triangle), we have A = B. This means triangle ABC is isosceles with A = B.To verify, we consider different cases:- If A = B, then C = 180^circ - 2A. The equation holds for any A, making the triangle isosceles.- Testing a right triangle (e.g., 45^circ-45^circ-90^circ) also works, but this is a special case of an isosceles triangle.- An equilateral triangle (a special case of isosceles) also satisfies the equation, but the general solution only requires A = B.Thus, the triangle must be isosceles, leading to the answer:[boxed{A}],<think>Okay, so I need to solve this problem: In triangle ABC, if 2 sin A cos B = sin C, then what type of triangle is it? The options are isosceles, right, scalene, or equilateral. Hmm, let me think step by step.First, let me recall some trigonometric identities related to triangles. In any triangle, the sum of the angles is π radians (or 180 degrees). Also, the Law of Sines might be useful here: (a/sin A) = (b/sin B) = (c/sin C) = 2R, where R is the radius of the circumscribed circle. But I'm not sure if that's needed right away. Let me see.The given equation is 2 sin A cos B = sin C. Maybe I can express sin C in terms of A and B since in a triangle, C = π - A - B. So, sin C = sin(π - A - B) = sin(A + B). Because sin(π - x) = sin x. So, sin(A + B) = sin A cos B + cos A sin B. Therefore, the equation becomes:2 sin A cos B = sin A cos B + cos A sin B.Let me subtract sin A cos B from both sides:2 sin A cos B - sin A cos B = cos A sin B.That simplifies to:sin A cos B = cos A sin B.Hmm, so sin A cos B - cos A sin B = 0. Wait, this is the sine of (A - B) because sin(A - B) = sin A cos B - cos A sin B. Therefore, sin(A - B) = 0.If sin(A - B) = 0, then A - B = nπ where n is an integer. But since A and B are angles in a triangle, they must be between 0 and π. The only possibility here is n = 0, so A - B = 0, which means A = B. Therefore, triangle ABC is isosceles with angles A and B equal. So, sides opposite these angles (a and b) are equal as well. Therefore, the triangle is isosceles. So, the answer should be option A.Wait, but let me check if there's another possibility. Suppose that the equation 2 sin A cos B = sin C leads to another condition. Let me verify.Alternatively, maybe I can use the Law of Sines here. Let me denote the sides opposite angles A, B, C as a, b, c respectively.From the Law of Sines, a/sin A = b/sin B = c/sin C = 2R.But maybe it's more straightforward to use the sine addition formula as before. Let me re-examine my steps.Starting from 2 sin A cos B = sin C.But sin C = sin(π - A - B) = sin(A + B) = sin A cos B + cos A sin B.So:2 sin A cos B = sin A cos B + cos A sin B.Subtracting sin A cos B from both sides:sin A cos B = cos A sin B.Which rearranges to:sin A cos B - cos A sin B = 0 => sin(A - B) = 0.So, A - B = 0 (mod π), but since angles in a triangle are between 0 and π, the only solution is A = B. Therefore, angles A and B are equal. Therefore, sides a and b are equal, so it's an isosceles triangle. Therefore, the answer is A.Wait, but maybe there's a case where the triangle is right-angled as well? Let's check.Suppose the triangle is right-angled. Let's assume angle C is 90 degrees (π/2 radians). Then sin C = 1. Then the equation becomes 2 sin A cos B = 1. But in a right-angled triangle, A + B = π/2. So, B = π/2 - A. Therefore, cos B = cos(π/2 - A) = sin A. So, substituting back into the equation:2 sin A * sin A = 1 => 2 sin² A = 1 => sin² A = 1/2 => sin A = √2/2 => A = 45 degrees. So, angles would be A = 45°, B = 45°, C = 90°. Therefore, a right-angled isosceles triangle. So, in this case, the triangle is both right-angled and isosceles. But the options are A, B, C, D. So, if both A and B are correct, but in the options, they are separate. Wait, the answer should be unique.Wait, but according to the problem, the equation given is 2 sin A cos B = sin C. If we have a right-angled isosceles triangle, then that equation holds. But in the general case where A = B, which is isosceles, does the equation hold?Wait, let's check another case where the triangle is isosceles but not right-angled. Suppose angles A and B are both 50 degrees, then angle C is 80 degrees. Then let's compute 2 sin A cos B and sin C.2 sin 50° cos 50° ≈ 2 * 0.7660 * 0.6428 ≈ 2 * 0.492 ≈ 0.984.sin 80° ≈ 0.9848. So, approximately equal. Close enough due to rounding. So, that equation holds. So, in general, if A = B, then 2 sin A cos B = 2 sin A cos A = sin 2A. But angle C = π - 2A. So, sin C = sin(2A). Therefore, 2 sin A cos A = sin 2A. But sin 2A = 2 sin A cos A, so equality holds. Therefore, in that case, 2 sin A cos B = sin C is equivalent to sin 2A = sin C. But since C = π - 2A, sin C = sin(2A). Therefore, identity holds. Therefore, whenever A = B, the equation holds. Therefore, the triangle must be isosceles.Alternatively, if the triangle is right-angled, but not isosceles, does the equation hold? Let's take a right-angled triangle where angle C is 90°, angle A is 30°, angle B is 60°. Then 2 sin A cos B = 2 sin 30° cos 60° = 2 * 0.5 * 0.5 = 2 * 0.25 = 0.5. Sin C = sin 90° = 1. So 0.5 ≠ 1. Therefore, the equation does not hold. So, only if the right-angled triangle is also isosceles (i.e., 45-45-90) does the equation hold. So in that case, the triangle is both right-angled and isosceles. But according to the problem statement, the options are A, B, C, D. So, which one is correct? The answer should be both A and B? But since the options are exclusive, maybe the problem considers that a right-angled isosceles triangle is a special case of isosceles triangle, so answer A. However, in the problem statement, since in the case of a right-angled isosceles triangle, the triangle is both right-angled and isosceles, but if the only necessary condition is that the triangle is isosceles (could be acute or right-angled), then answer A. However, in the problem statement, if the equation holds only for isosceles triangles (including the right-angled isosceles), then the answer is A. But when we derived earlier, the conclusion was that A = B, so the triangle is isosceles. Therefore, regardless of the type, if angles A and B are equal, it's isosceles. Therefore, answer A.But wait, the problem is from a competition, so maybe there is a trick here. Let me check the original equation again. 2 sin A cos B = sin C.If A = B, then C = π - 2A, so sin C = sin(2A) = 2 sin A cos A. Then 2 sin A cos B = 2 sin A cos A = sin C. So that works. So if A = B, the equation holds.But could there be a case where A ≠ B, but the equation still holds?Suppose that sin(A - B) = 0, so A - B = 0, so A = B. So that is the only solution. Therefore, the triangle must be isosceles. So the answer is A.But wait, the problem gives an option B, right triangle. So, is there a case where the triangle is right-angled and satisfies the equation? Yes, when it's a right-angled isosceles triangle. But in that case, it is also isosceles. So, if the answer options are exclusive, then both A and B are correct? But in the options, it's either A, B, C, or D. So, the problem is from 1990, so maybe in the original answer, it's A. Or maybe in Chinese competitions, sometimes they accept multiple answers, but here it's single choice.Wait, let's verify again. If the triangle is right-angled but not isosceles, then the equation does not hold. So the only right-angled triangle that satisfies the equation is the isosceles right-angled triangle. Therefore, all triangles that satisfy the equation are isosceles, some of which may be right-angled. So, the answer is isosceles triangle, option A. Therefore, A is the correct answer. The right-angled triangle is a special case of isosceles here, but since the question is asking for the type, and isosceles is a broader category, the answer should be A.Alternatively, perhaps the problem considers that a right-angled triangle is not necessarily isosceles, so in that case, since there exists a right-angled triangle that satisfies the equation (the isosceles one), but the general solution is isosceles. So the answer must be A.Yes, because when we derived the equation, we found that A must equal B, which is the definition of an isosceles triangle. So regardless of the other angle, as long as A = B, the triangle is isosceles, and the equation holds. Therefore, the correct answer is A.Final Answerboxed{A}</think>Given the problem: In triangle ABC, if 2 sin A cos B = sin C, we need to determine the type of triangle.First, we use the identity for sin C since C = pi - A - B:[sin C = sin(pi - A - B) = sin(A + B)]Using the sine addition formula:[sin(A + B) = sin A cos B + cos A sin B]Substituting this into the given equation:[2 sin A cos B = sin A cos B + cos A sin B]Subtracting sin A cos B from both sides:[sin A cos B = cos A sin B]This simplifies to:[sin A cos B - cos A sin B = 0]Recognizing this as the sine of a difference:[sin(A - B) = 0]Since sin(A - B) = 0, we have A - B = 0 or A = B. Therefore, triangle ABC is isosceles with angles A and B equal, making the sides opposite these angles equal.Checking if the triangle could be right-angled, we consider a right-angled isosceles triangle where A = B = 45^circ and C = 90^circ. This satisfies the equation, but it is still a special case of an isosceles triangle. For a right-angled triangle that is not isosceles, the equation does not hold.Thus, the only necessary condition derived from the equation is that the triangle is isosceles.[boxed{A}]]

🔑:[Solution] Given A+B+C=180^{circ}, then sin C=sin (A+B).Also, from the given condition, we have sin (A+B)=2 sin A cos B, which meansbegin{array}{ll} & sin A cos B=cos A sin B, text { or } tan A=tan B . because quad 0^{circ}<A, B<180^{circ}, therefore quad A=B .end{array}Therefore, the answer is (A).

🧠:[<think>Okay, so I have this problem where I need to find the radii of the inscribed and circumscribed circles of an isosceles triangle. The base is 24 units, and each of the lateral sides is 15 units. Hmm, let me start by visualizing the triangle. It's isosceles, so two sides are equal, which are the lateral sides here. The base is the unequal side, which is longer in this case—24 compared to 15. Wait, actually, hold on, the base is 24, which is longer than the lateral sides of 15. So the triangle is kind of "squat" if I imagine it. That might affect the radii calculations. First, I need to recall the formulas for the radii of the inscribed (inradius) and circumscribed (circumradius) circles. For any triangle, the inradius (r) can be found using the formula r = A/s, where A is the area of the triangle and s is the semi-perimeter. The circumradius (R) can be found using the formula R = (a*b*c)/(4*A), where a, b, c are the sides of the triangle and A is the area again. Alternatively, for the circumradius, there's also the formula involving the sides and the sine of the angles: a/(2*sin(A)) = b/(2*sin(B)) = c/(2*sin(C)) = R. Since this triangle is isosceles, maybe using the Law of Sines might be helpful here. The problem mentions the Law of Sines in brackets, so that might be a hint.But let me start with the inradius. To find r = A/s, I need to compute the area A and the semi-perimeter s. Let's compute the semi-perimeter first. The sides of the triangle are 24, 15, and 15. So the perimeter is 24 + 15 + 15 = 54. Therefore, the semi-perimeter s is 54/2 = 27. Okay, so s = 27.Now, to find the area A. Since it's an isosceles triangle, I can drop an altitude from the apex (the vertex opposite the base) to the base, splitting the base into two equal parts of 12 each. Then, the height h can be found using the Pythagorean theorem. The lateral side is 15, which becomes the hypotenuse of a right triangle with base 12 and height h. So h^2 + 12^2 = 15^2. Let me compute that. 12 squared is 144, 15 squared is 225. Therefore, h^2 = 225 - 144 = 81. Taking the square root, h = 9. So the height is 9. Then, the area A is (base * height)/2 = (24 * 9)/2 = (216)/2 = 108. So the area is 108.Therefore, the inradius r = A/s = 108/27 = 4. So the inradius is 4. That seems straightforward. Let me check if that makes sense. For an isosceles triangle with base 24 and sides 15, getting an inradius of 4. Hmm. Let me just verify the calculations. Yes, semi-perimeter 27, area 108, 108 divided by 27 is 4. That seems correct.Now, the circumradius R. Using the formula R = (a*b*c)/(4*A). Here, the sides are 24, 15, 15. So a = 24, b = 15, c = 15. Then R = (24*15*15)/(4*108). Let me compute the numerator and denominator. Numerator: 24*15 = 360, then 360*15 = 5400. Denominator: 4*108 = 432. So R = 5400/432. Let's simplify that. Dividing numerator and denominator by 12: 5400 ÷12 = 450, 432 ÷12 = 36. So 450/36. Then divide numerator and denominator by 18: 450 ÷18 = 25, 36 ÷18 = 2. So 25/2 = 12.5. So R = 12.5 or 25/2. So circumradius is 25/2. Let me check if that makes sense. Alternatively, using the Law of Sines.Law of Sines states that a/sin(A) = 2R. Let me try that. Let me label the triangle. Let’s call the triangle ABC, where AB = AC = 15, and BC = 24. So angle at A is the vertex angle, angles at B and C are the base angles. Since it's isosceles, angles at B and C are equal. Let me compute angle A first. Using the Law of Cosines. Law of Cosines: cos(A) = (b² + c² - a²)/(2bc). Here, sides opposite angles: angle A is opposite side BC (24), and angles B and C are opposite sides AC and AB, which are 15 each. So cos(A) = (15² + 15² - 24²)/(2*15*15). Let me compute that. 15² is 225, so 225 + 225 = 450. 24² is 576. So numerator is 450 - 576 = -126. Denominator is 2*15*15 = 450. So cos(A) = -126/450. Simplify: divide numerator and denominator by 6: -21/75. Then divide by 3: -7/25. So cos(A) = -7/25. Then angle A is arccos(-7/25). The sine of angle A can be found using sin²(A) + cos²(A) = 1. So sin²(A) = 1 - (49/625) = (625 - 49)/625 = 576/625. So sin(A) = 24/25 (since angle A is between 0 and 180, sin is positive). Then, by Law of Sines: a/sin(A) = 2R. Here, a is 24, sin(A) is 24/25. So 24 / (24/25) = 24 * (25/24) = 25 = 2R. Therefore, R = 25/2 = 12.5. So that matches the previous result. So the circumradius is indeed 25/2. Alternatively, maybe using the formula specific to isosceles triangles. But since both methods gave the same result, that's reassuring. So, putting it all together: inradius is 4, circumradius is 25/2. Let me just check once more. Wait, let me make sure that the Law of Sines approach is applicable here. Yes, in any triangle, Law of Sines holds. So since we had side a opposite angle A, which we calculated, and we found sin(A) = 24/25, then 24/(24/25) = 25 = 2R => R = 12.5. That seems correct. Another way to calculate circumradius for a triangle with sides a, b, c is R = abc/(4K), where K is the area. So plugging in 24, 15, 15, and area 108, as we did earlier, gives (24*15*15)/(4*108) = 5400/432 = 12.5. So that checks out. Is there another method? Maybe using the formula for circumradius in terms of the sides and the inradius? Not that I can recall. But since both methods give the same answer, I think that's solid. Therefore, the radii are 4 and 25/2. Let me write them as R = 25/2 and r = 4. Wait, but the problem says "Find the radii of the inscribed and circumscribed circles." So inscribed is r = 4, circumscribed is R = 25/2. Let me just verify once again the inradius. Inradius is area divided by semi-perimeter. Area is 108, semi-perimeter is 27, so 108/27 = 4. Correct. Alternatively, is there a formula specific to isosceles triangles for inradius? Let me think. If we have an isosceles triangle with base b and equal sides a, then area is (b/4)*sqrt(4a² - b²). Wait, let me check that. The height h = sqrt(a² - (b/2)^2), so area is (b/2)*h = (b/2)*sqrt(a² - (b/2)^2). For our case, a = 15, b = 24. Then area is (24/2)*sqrt(15² - 12²) = 12*sqrt(225 - 144) = 12*sqrt(81) = 12*9 = 108. Which matches. Then semi-perimeter is (24 + 15 + 15)/2 = 27. So inradius is 108/27 = 4. Correct. So that formula gives the same result. Alternatively, for the circumradius, using formula R = (a*b*c)/(4*K) is general for any triangle, so that works here. Therefore, I think the answers are correct. Final AnswerThe radius of the inscribed circle is boxed{4} and the radius of the circumscribed circle is boxed{dfrac{25}{2}}.</think>To find the radii of the inscribed and circumscribed circles of an isosceles triangle with base 24 and lateral sides 15, we start by calculating the semi-perimeter and area of the triangle.1. Semi-perimeter (s): [ s = frac{24 + 15 + 15}{2} = 27 ]2. Area (A): The height (h) of the triangle can be found using the Pythagorean theorem. The height splits the base into two equal parts of 12 each: [ h = sqrt{15^2 - 12^2} = sqrt{225 - 144} = sqrt{81} = 9 ] The area is then: [ A = frac{1}{2} times 24 times 9 = 108 ]3. Inradius (r): Using the formula ( r = frac{A}{s} ): [ r = frac{108}{27} = 4 ]4. Circumradius (R): Using the formula ( R = frac{abc}{4A} ): [ R = frac{24 times 15 times 15}{4 times 108} = frac{5400}{432} = frac{25}{2} ] Verification using the Law of Sines confirms this result. For angle ( A ) opposite the base: [ cos(A) = frac{15^2 + 15^2 - 24^2}{2 times 15 times 15} = -frac{7}{25} ] [ sin(A) = sqrt{1 - left(-frac{7}{25}right)^2} = frac{24}{25} ] Using ( frac{a}{sin(A)} = 2R ): [ 2R = frac{24}{frac{24}{25}} = 25 implies R = frac{25}{2} ] Final AnswerThe radius of the inscribed circle is boxed{4} and the radius of the circumscribed circle is boxed{dfrac{25}{2}}.,<think>Okay, so I need to find the radii of the inscribed and circumscribed circles of an isosceles triangle where the base is 24 and the lateral sides are 15 each. Hmm, let me start by recalling some formulas related to circles in triangles. The radius of the inscribed circle (inradius) can be found using the formula ( r = frac{A}{s} ), where ( A ) is the area of the triangle and ( s ) is the semi-perimeter. The radius of the circumscribed circle (circumradius) can be found using ( R = frac{a b c}{4 A} ), where ( a ), ( b ), and ( c ) are the sides of the triangle. First, let me note down the given sides. The base is 24, so the two equal sides are 15 each. So the sides are 15, 15, and 24. Let me sketch the triangle in my mind: an isosceles triangle with base 24 and equal sides 15. Wait, but hold on, before I proceed, I need to make sure that such a triangle exists. Let me check the triangle inequality. The sum of the two sides must be greater than the third side. For the two equal sides: 15 + 15 = 30, which is greater than 24. For the other combinations: 15 + 24 = 39, which is greater than 15. So, yes, the triangle is valid.Next, I need to find the area of the triangle. Since it's an isosceles triangle, maybe I can drop a height from the apex to the base, which will split the base into two equal parts of 12 each. Then, the height can be calculated using the Pythagorean theorem. Let me do that. Let's call the height h. Then, in one of the right triangles formed, the legs are h and 12, and the hypotenuse is 15. So, by Pythagoras:( h^2 + 12^2 = 15^2 )Calculating:( h^2 + 144 = 225 )Subtract 144 from both sides:( h^2 = 225 - 144 = 81 )Therefore, h = 9. So the height of the triangle is 9. Then, the area A is (base * height)/2 = (24 * 9)/2 = 12 * 9 = 108. Got that. The area is 108.Now, let's compute the semi-perimeter s. The perimeter of the triangle is 15 + 15 + 24 = 54. So the semi-perimeter is 54 / 2 = 27. Therefore, the inradius r is ( r = frac{A}{s} = frac{108}{27} = 4 ). So the inradius is 4. That seems straightforward.Now for the circumradius R. The formula is ( R = frac{a b c}{4 A} ). Here, a, b, c are the sides: 15, 15, 24. So substituting the values:( R = frac{15 * 15 * 24}{4 * 108} )Let me compute numerator and denominator separately:Numerator: 15 * 15 = 225; 225 * 24. Let's compute that. 225 * 20 = 4500; 225 * 4 = 900; so total is 4500 + 900 = 5400.Denominator: 4 * 108 = 432.Therefore, R = 5400 / 432. Let's simplify this. Dividing numerator and denominator by 12: 5400 ÷ 12 = 450; 432 ÷ 12 = 36. So now, 450 / 36. Dividing numerator and denominator by 6: 450 ÷ 6 = 75; 36 ÷ 6 = 6. So 75 / 6 = 12.5. Wait, 75 divided by 6 is 12.5. But 12.5 is 25/2. Hmm, let me check that again. 6 * 12 = 72, so 75 - 72 = 3. So 75/6 = 12.5? Wait, no, that's 12.5? Wait, 6*12=72, 6*12.5=75. Yes, so 75/6=12.5. So R = 12.5, which is 25/2. So the circumradius is 25/2 or 12.5.Wait, but let me verify this another way to make sure. Alternatively, since the triangle is isosceles, maybe using the formula for circumradius in terms of sides and angles. Since we know the sides and the area, perhaps using the Law of Sines. The Law of Sines states that ( frac{a}{sin A} = 2 R ). So maybe I can find one of the angles and use that to compute R.Let me try that. Let's consider the vertex angle opposite the base. Let's call the triangle ABC, with AB = AC = 15, and BC = 24. Then angle at A is the vertex angle. To use the Law of Sines, we can relate angle A with side BC. The formula is ( frac{BC}{sin A} = 2 R ). So ( frac{24}{sin A} = 2 R ). Therefore, ( R = frac{24}{2 sin A} = frac{12}{sin A} ). If I can compute sin A, then I can find R.Alternatively, since we know the area is 108, the area can also be expressed as ( frac{1}{2}ab sin C ), where a and b are two sides and C is the included angle. In triangle ABC, the area is ( frac{1}{2} AB * AC * sin A ). So 108 = ( frac{1}{2} * 15 * 15 * sin A ). Let's compute that:108 = ( frac{1}{2} * 225 * sin A )108 = 112.5 * sin ATherefore, sin A = 108 / 112.5 = 0.96So sin A = 0.96. Then, R = 12 / 0.96 = 12.5. That's the same result as before. So that confirms R is 12.5 or 25/2. So that seems correct.Alternatively, maybe compute angle A using the height. Since we have the height h = 9, then in the right triangle formed by splitting the original triangle, the legs are 12 and 9, hypotenuse 15. So angle at the base (angle B or angle C) can be found. Let's find angle B. The base angles are equal in an isosceles triangle. Then, angle B is adjacent to the base, so in the right triangle, angle B is adjacent to 12 and opposite to 9. So tan B = 9 / 12 = 3/4. Therefore, angle B = arctan(3/4). Then, angle A is 180 - 2 angle B.But maybe using Law of Cosines to find angle A. Let's try that. In triangle ABC, with sides AB = AC = 15, BC = 24. Then, angle at A is between sides AB and AC. The Law of Cosines says:( BC^2 = AB^2 + AC^2 - 2 AB * AC cos A )Plugging in:( 24^2 = 15^2 + 15^2 - 2 * 15 * 15 cos A )Calculating:576 = 225 + 225 - 450 cos A576 = 450 - 450 cos ASubtract 450 from both sides:126 = -450 cos ADivide both sides by -450:cos A = -126 / 450 = -14 / 50 = -7 / 25 ≈ -0.28Therefore, angle A is arccos(-7/25). Then, sin A can be found using sin^2 A + cos^2 A = 1. So sin A = sqrt(1 - (49/625)) = sqrt(576/625) = 24/25. Wait, but angle A is obtuse since cos A is negative, so sin A should be positive. So sin A = 24/25. Then, using Law of Sines: ( frac{24}{sin A} = 2 R ). Therefore, 24 / (24/25) = 2 R → 25 = 2 R → R = 25/2 = 12.5. So that's consistent with the previous results. Therefore, R = 25/2.So, inradius is 4, circumradius is 25/2. Let me check if these make sense. Inradius is 4, which is smaller than the circumradius, which is 12.5. That seems reasonable, as the circumradius is generally larger than the inradius. Also, given the triangle is quite "flat" because the base is longer compared to the sides, the circumradius being significantly larger than the inradius makes sense.Alternatively, maybe use some other formulas to verify. For instance, in any triangle, the relationship between inradius, circumradius, and other elements. For example, inradius formula: ( r = frac{A}{s} ), which we used. Circumradius formula ( R = frac{a}{2 sin A} ), which we also used. So seems consistent.Alternatively, maybe use the formula involving the sides and angles for inradius. But perhaps not necessary here since we already cross-verified. Therefore, I think the inradius is 4 and the circumradius is 25/2. Let me just check if I did all the calculations correctly.First, the height: 12^2 + h^2 = 15^2 → 144 + h^2 = 225 → h^2 = 81 → h = 9. Correct.Area: (24 * 9)/2 = 108. Correct.Semi-perimeter: (15 + 15 + 24)/2 = 54/2 = 27. Correct.Inradius: 108 / 27 = 4. Correct.For circumradius: (15 * 15 * 24) / (4 * 108). 15*15=225, 225*24=5400. 4*108=432. 5400 / 432. Divide numerator and denominator by 12: 5400 ÷ 12 = 450; 432 ÷ 12 = 36. 450 / 36 = 12.5. Correct.Alternatively, using Law of Sines: area = 108 = (1/2)*15*15*sin A → sin A = (108*2)/225 = 216/225 = 24/25. Then, R = (15)/(2*(24/25)) = 15/(48/25) = (15*25)/48 = 375/48 = 12.5. Wait, hold on, another way. Wait, Law of Sines states that a / sin A = 2 R. Here, side BC is 24, opposite angle A. So 24 / sin A = 2 R → R = 24 / (2 sin A) = 12 / sin A. Since sin A = 24/25, R = 12 / (24/25) = 12 * (25/24) = 12.5. So that works. Correct.Alternatively, using Law of Cosines to find angle A: cos A = (15² + 15² - 24²)/(2*15*15) = (225 + 225 - 576)/450 = (450 - 576)/450 = (-126)/450 = -7/25. Then sin A = sqrt(1 - (49/625)) = sqrt(576/625) = 24/25. So again, R = 12 / (24/25) = 12.5. Correct.Therefore, all methods confirm that R = 25/2 and r = 4. So the answer should be inradius 4 and circumradius 25/2.Final AnswerThe radius of the inscribed circle is boxed{4} and the radius of the circumscribed circle is boxed{dfrac{25}{2}}.</think>Given an isosceles triangle with a base of 24 and lateral sides of 15 each, we need to find the radii of the inscribed and circumscribed circles.First, we calculate the height of the triangle using the Pythagorean theorem. The height splits the base into two equal parts of 12 each:[h^2 + 12^2 = 15^2 implies h^2 + 144 = 225 implies h^2 = 81 implies h = 9]The area ( A ) of the triangle is:[A = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 24 times 9 = 108]The semi-perimeter ( s ) is:[s = frac{15 + 15 + 24}{2} = 27]The radius of the inscribed circle (inradius) ( r ) is:[r = frac{A}{s} = frac{108}{27} = 4]For the radius of the circumscribed circle (circumradius) ( R ), we use the formula ( R = frac{abc}{4A} ):[R = frac{15 times 15 times 24}{4 times 108} = frac{5400}{432} = frac{5400 div 12}{432 div 12} = frac{450}{36} = frac{75}{6} = 12.5 = frac{25}{2}]Verification using the Law of Sines confirms the circumradius. The angle ( A ) opposite the base is found using the Law of Cosines:[cos A = frac{15^2 + 15^2 - 24^2}{2 times 15 times 15} = frac{-126}{450} = -frac{7}{25}][sin A = sqrt{1 - left(-frac{7}{25}right)^2} = sqrt{frac{576}{625}} = frac{24}{25}][R = frac{24}{2 sin A} = frac{24}{2 times frac{24}{25}} = frac{24}{frac{48}{25}} = frac{24 times 25}{48} = frac{25}{2}]Thus, the radius of the inscribed circle is boxed{4} and the radius of the circumscribed circle is boxed{dfrac{25}{2}}.]

🔑:To find the radius of the inscribed circle, we can use the property of the angle bisector of a triangle. For the radius of the circumscribed circle, we use the formula a=2 R sin boldsymbol{alpha}.## SolutionLet A M be the height of the isosceles triangle A B C, where A B=A C=15, B C=24, and O is the center of the inscribed circle, with r and R being the radii of the inscribed and circumscribed circles, respectively.Since the triangle is isosceles, the point O lies on the segment A M, and B O is the angle bisector of angle A B C.Therefore, frac{O M}{O A}=frac{B M}{A B}. Consequently,begin{gathered}r=O M=frac{A M cdot M B}{M B+A B}=frac{9 cdot 12}{12+15}=4 R=frac{A C}{2 sin angle A B C}=frac{A C}{2 cdot frac{A M}{A B}}=frac{15}{2 cdot frac{9}{15}}=frac{25}{2}end{gathered}## Answer4 ; 12.5.

🧠:[<think>Alright, let me try to figure this problem out step by step. It's about different types of trains and their speeds. There are three types mentioned: EMU (Electric Multiple Unit), ordinary fast train, high-speed train, and an express train. Wait, actually, four types? Let me check again.Wait, the problem states: EMU, ordinary fast train, high-speed train, and express train. Hmm, but the question is asking for the speeds of the high-speed train and the ordinary fast train. So maybe the express train is another one, but the key ones are EMU, ordinary fast, high-speed, and express. Let me parse the problem again carefully.First sentence: "It is known that the speed of the EMU (Electric Multiple Unit) train is twice that of the ordinary fast train." So, EMU speed = 2 * ordinary fast train speed.Second sentence: "The speed of the EMU train increases by 25% to reach the speed of the high-speed train." So, if EMU's speed is increased by 25%, it becomes equal to the high-speed train's speed. So, high-speed train speed = EMU speed * 1.25.Third sentence: "The average speed of the high-speed train and the ordinary fast train is 15 kilometers/hour faster than the express train, and the average speed of the EMU train and the ordinary fast train is 10 kilometers/hour slower than the express train."So, let's break this down. First part: average of high-speed and ordinary fast is 15 km/h faster than express train. Second part: average of EMU and ordinary fast is 10 km/h slower than express train.Therefore, they want us to find the speeds of the high-speed train and the ordinary fast train. Let me define variables for each.Let me assign variables:Let ordinary fast train speed = x km/hThen EMU speed = 2x km/h (since it's twice the ordinary)High-speed train speed = EMU speed * 1.25 = 2x * 1.25 = 2.5x km/hExpress train speed = let's call that E km/h.Now, the average speed of high-speed (2.5x) and ordinary (x) is (2.5x + x)/2 = 3.5x / 2 = 1.75x. This is 15 km/h faster than the express train. So:1.75x = E + 15Similarly, the average speed of EMU (2x) and ordinary (x) is (2x + x)/2 = 1.5x. This is 10 km/h slower than the express train. So:1.5x = E - 10Now we have two equations:1.75x = E + 151.5x = E - 10We can solve these two equations for x and E. Let's subtract the second equation from the first:1.75x - 1.5x = (E + 15) - (E - 10)0.25x = 25x = 25 / 0.25x = 100So, the ordinary fast train speed is 100 km/h.Then, EMU speed is 2x = 200 km/hHigh-speed train speed is 2.5x = 250 km/hNow, let's check the express train speed. Using the second equation:1.5x = E - 101.5 * 100 = E - 10150 = E - 10E = 160 km/hCheck with the first equation:1.75x = 1.75*100 = 175E + 15 = 160 + 15 = 175. Correct.So, express train is 160 km/h. But the problem asks for the speeds of the high-speed train and the ordinary fast train. So, high-speed is 250 km/h, ordinary is 100 km/h. Let me confirm if all the conditions are satisfied.EMU speed is 200 km/h, which is twice the ordinary (100), correct.EMU increased by 25%: 200 * 1.25 = 250, which is the high-speed train's speed. Correct.Average of high-speed (250) and ordinary (100): (250 + 100)/2 = 175. Express is 160, and 175 - 160 = 15. Correct.Average of EMU (200) and ordinary (100): (200 + 100)/2 = 150. Express is 160, so 160 - 150 = 10. Correct.All conditions are satisfied. Therefore, the high-speed train is 250 km/h and the ordinary fast train is 100 km/h.Final AnswerThe speeds of the high-speed train and the ordinary fast train are boxed{250} kilometers/hour and boxed{100} kilometers/hour.</think>Let the speed of the ordinary fast train be ( x ) km/h. Then, the speed of the EMU train is ( 2x ) km/h. The speed of the high-speed train is ( 2.5x ) km/h (since it is 25% more than the EMU train's speed). Let the speed of the express train be ( E ) km/h.We are given two conditions:1. The average speed of the high-speed train and the ordinary fast train is 15 km/h faster than the express train.2. The average speed of the EMU train and the ordinary fast train is 10 km/h slower than the express train.These conditions translate to the following equations:1. (frac{2.5x + x}{2} = E + 15)2. (frac{2x + x}{2} = E - 10)Simplifying these equations:1. (frac{3.5x}{2} = E + 15 implies 1.75x = E + 15)2. (frac{3x}{2} = E - 10 implies 1.5x = E - 10)Subtracting the second equation from the first:[1.75x - 1.5x = (E + 15) - (E - 10)][0.25x = 25 implies x = frac{25}{0.25} = 100]Thus, the speed of the ordinary fast train is ( x = 100 ) km/h. Substituting ( x = 100 ) into the equations for the high-speed train and the express train:- The speed of the high-speed train is ( 2.5x = 2.5 times 100 = 250 ) km/h.- The speed of the express train ( E ) is found using ( 1.5x = E - 10 ): [ 1.5 times 100 = E - 10 implies 150 = E - 10 implies E = 160 ]Verification:- The average speed of the high-speed train and the ordinary fast train is ( frac{250 + 100}{2} = 175 ) km/h, which is 15 km/h faster than the express train (160 + 15 = 175).- The average speed of the EMU train and the ordinary fast train is ( frac{200 + 100}{2} = 150 ) km/h, which is 10 km/h slower than the express train (160 - 10 = 150).Therefore, the speeds of the high-speed train and the ordinary fast train are (boxed{250}) kilometers/hour and (boxed{100}) kilometers/hour.,<think>Okay, so I have this problem about trains and their speeds. Let me try to parse it step by step. First, the problem says: "It is known that the speed of the EMU (Electric Multiple Unit) train is twice that of the ordinary fast train." Okay, so if I let the speed of the ordinary fast train be something, say, F, then the EMU's speed is 2F. That part seems straightforward.Next: "The speed of the EMU train increases by 25% to reach the speed of the high-speed train." Hmm. So, if the EMU's speed is increased by 25%, then it becomes equal to the high-speed train's speed. Let me note that. If the EMU's original speed is 2F, then increasing that by 25% would be 2F * 1.25, right? So, 2F * 1.25 = 2.5F. Therefore, the high-speed train's speed is 2.5F. Let me check that again. 25% of 2F is 0.5F, so adding that to 2F gives 2.5F. Yep, that seems right. So, high-speed train speed is 2.5F.Then the problem says: "The average speed of the high-speed train and the ordinary fast train is 15 kilometers/hour faster than the express train..." Wait, express train? Wait, the problem mentioned EMU, ordinary fast train, high-speed train. Now express train is introduced. Hmm. So, express train is another category here. Let me make sure I track all the types:- Ordinary fast train: speed F- EMU: speed 2F- High-speed train: speed 2.5F (after a 25% increase from EMU)- Express train: speed E (this is a new variable; let me denote it as E)So, the average speed of the high-speed train (2.5F) and the ordinary fast train (F) is (2.5F + F)/2 = 3.5F/2 = 1.75F. This average speed is 15 km/h faster than the express train. Therefore, 1.75F = E + 15. Wait, no. Wait, the average is 15 km/h faster than the express train. So, 1.75F = E + 15? Wait, actually, the wording is "the average speed of the high-speed train and the ordinary fast train is 15 km/h faster than the express train." So, the average of high-speed and ordinary fast is 15 more than express. So, that would be (2.5F + F)/2 = E + 15. So, 1.75F = E + 15. Got it.Then the next part: "and the average speed of the EMU train and the ordinary fast train is 10 kilometers/hour slower than the express train." So, average of EMU (2F) and ordinary fast (F) is (2F + F)/2 = 1.5F. This is 10 km/h slower than the express train. So, 1.5F = E - 10.So now, we have two equations:1. 1.75F = E + 152. 1.5F = E - 10So, we can solve these two equations to find F and E. Let me write them again:Equation 1: 1.75F - E = 15Equation 2: 1.5F - E = -10Let me subtract equation 2 from equation 1:(1.75F - E) - (1.5F - E) = 15 - (-10)Simplify left side: 1.75F - E -1.5F + E = 0.25FRight side: 25Therefore, 0.25F = 25 => F = 25 / 0.25 = 100. So, F is 100 km/h. Then, we can find E from either equation. Let's take equation 2: 1.5F = E -10. So, 1.5*100 = E -10 => 150 = E -10 => E = 160 km/h. Let me check with equation 1: 1.75*100 = 175. 175 = E +15 => E = 175 -15 = 160. Yep, that matches. So, E is 160 km/h.But the problem asks for the speeds of the high-speed train and the ordinary fast train. So, ordinary fast train is F, which is 100 km/h. High-speed train is 2.5F, which is 2.5*100 = 250 km/h. Therefore, the answers should be 250 and 100.Wait, let me make sure I didn't make a mistake. Let's go back through the problem again.Speed of EMU is twice ordinary fast train: EMU = 2F = 200 km/h. Then, increasing EMU's speed by 25% gives 200 * 1.25 = 250 km/h, which is high-speed train. So, that's correct.Average of high-speed (250) and ordinary (100): (250 + 100)/2 = 175 km/h. This is supposed to be 15 km/h faster than express train. So, express train speed is 175 -15 = 160 km/h. Which matches E = 160.Average of EMU (200) and ordinary (100): (200 + 100)/2 = 150 km/h. This is 10 km/h slower than express train (160). 160 -10 = 150. Correct.So, all the numbers check out.Therefore, the speeds are high-speed train: 250 km/h, ordinary fast train: 100 km/h.Final AnswerThe speeds of the high-speed train and the ordinary fast train are boxed{250} kilometers/hour and boxed{100} kilometers/hour.</think>Let the speed of the ordinary fast train be ( F ) km/h. Then, the speed of the EMU train is ( 2F ) km/h. When the speed of the EMU train is increased by 25%, it becomes the speed of the high-speed train, which is ( 2.5F ) km/h.We are given two conditions:1. The average speed of the high-speed train and the ordinary fast train is 15 km/h faster than the express train.2. The average speed of the EMU train and the ordinary fast train is 10 km/h slower than the express train.Let the speed of the express train be ( E ) km/h.From the first condition:[frac{2.5F + F}{2} = E + 15]Simplifying, we get:[frac{3.5F}{2} = E + 15 implies 1.75F = E + 15 quad text{(Equation 1)}]From the second condition:[frac{2F + F}{2} = E - 10]Simplifying, we get:[frac{3F}{2} = E - 10 implies 1.5F = E - 10 quad text{(Equation 2)}]Subtracting Equation 2 from Equation 1:[1.75F - 1.5F = 15 - (-10)][0.25F = 25 implies F = 100]Substituting ( F = 100 ) back into Equation 2:[1.5 times 100 = E - 10 implies 150 = E - 10 implies E = 160]The speed of the ordinary fast train is ( F = 100 ) km/h. The speed of the high-speed train is ( 2.5F = 2.5 times 100 = 250 ) km/h.Thus, the speeds of the high-speed train and the ordinary fast train are (boxed{250}) kilometers/hour and (boxed{100}) kilometers/hour.]

🔑:【Analysis】Let the speed of the ordinary express train be x km/h, then the speed of the EMU (Electric Multiple Unit) train is 2 x km/h, and the speed of the high-speed train is (1+25 %) times 2 x=2.5 x km/h. According to the given conditions: the average speed of the high-speed train and the ordinary express train is 15 km/h faster than the special express train, and the average speed of the EMU train and the ordinary express train is 10 km/h slower than the special express train, which means the average speed of the high-speed train and the ordinary express train is 25 km/h faster than the average speed of the EMU train and the ordinary express train. We can set up an equation to solve for x.【Solution】Solution: Let the speed of the ordinary express train be x km/h, then the speed of the EMU train is 2 x km/h, and the speed of the high-speed train is (1+25 %) times 2 x=2.5 x km/h, thenbegin{array}{c}frac{mathrm{x}+2.5 mathrm{x}}{2}-frac{mathrm{x}+2 mathrm{x}}{2}=15+10, 1.75 x-1.5 x=25 0.25 x=25 0.25 x div 0.25=25 div 0.25 2.5 x=2.5 times 100=250 quad x=100end{array}Answer: The speeds of the high-speed train and the ordinary express train are 250 km/h and 100 km/h, respectively. Therefore, the answer is: 250, 100.

🧠:[<think>Okay, let's see. I need to compute this integral from 0 to π of (2 sin θ + 3 cos θ - 3) divided by (13 cos θ - 5) dθ. Hmm, that looks a bit complicated. Let me start by writing down the integral properly:∫₀^π [ (2 sin θ + 3 cos θ - 3) / (13 cos θ - 5) ] dθFirst, maybe I can simplify the numerator. Let me check if the numerator can be expressed as a multiple of the denominator plus some remainder. That is, perform polynomial division or something similar. But here both numerator and denominator are trigonometric functions. Alternatively, maybe there's a substitution that can help here.Alternatively, perhaps split the fraction into separate terms:[2 sin θ / (13 cos θ - 5)] + [3 cos θ / (13 cos θ - 5)] - [3 / (13 cos θ - 5)]So split the integral into three parts:∫ [2 sin θ / (13 cos θ - 5)] dθ + ∫ [3 cos θ / (13 cos θ - 5)] dθ - ∫ [3 / (13 cos θ - 5)] dθLet's handle each integral separately. Maybe starting with the first one: ∫ [2 sin θ / (13 cos θ - 5)] dθ.For the first integral, substitution might work. Let u = 13 cos θ - 5. Then du/dθ = -13 sin θ, so du = -13 sin θ dθ. Therefore, sin θ dθ = -du/13. So substituting:2 ∫ [ sin θ dθ / (13 cos θ - 5) ] = 2 ∫ [ (-du/13) / u ] = -2/13 ∫ (1/u) du = -2/13 ln |u| + C = -2/13 ln |13 cos θ - 5| + CSo the first integral is -2/13 ln |13 cos θ - 5| evaluated from 0 to π.But wait, let's check the denominator 13 cos θ -5. When θ is between 0 and π, cos θ ranges from 1 to -1, so 13 cos θ -5 ranges from 13*1 -5 = 8 to 13*(-1) -5 = -18. So the denominator is positive when cos θ > 5/13 and negative otherwise. So ln |13 cos θ -5| is valid except when 13 cos θ -5 =0, but cos θ =5/13. So θ = arccos(5/13). Let's check if 5/13 is within the range of cos θ from 0 to π. Since 5/13 is approximately 0.3846, which is between -1 and 1, so θ = arccos(5/13) is in (0, π/2). So the denominator crosses zero at θ = arccos(5/13), which is within the interval [0, π]. Therefore, the integral has a singularity there. Wait, but is the original integral an improper integral? Let's check if the denominator is zero anywhere between 0 and π. As we saw, yes, at θ = arccos(5/13). Therefore, we need to check if the integral is convergent there.But the original problem just gives the integral from 0 to π. Hmm, perhaps I need to check whether the integrand is defined there or if it's an improper integral. Let me compute 13 cos θ -5 at θ = arccos(5/13). That would be 13*(5/13) -5 = 5 -5 =0. So at that point, the denominator is zero. So the integrand has a vertical asymptote there. Therefore, the integral is improper at θ = arccos(5/13). So we need to evaluate it as a limit approaching that point from the left and right. But before that, perhaps check if the numerator is also zero there. If the numerator is zero at the same point, maybe the limit exists.Compute numerator at θ = arccos(5/13). The numerator is 2 sin θ +3 cos θ -3. Let's compute:cos θ =5/13, so sin θ = sqrt(1 -25/169)=sqrt(144/169)=12/13. Therefore, numerator = 2*(12/13) + 3*(5/13) -3 = (24/13 +15/13 -39/13) = (24 +15 -39)/13 =0/13=0.Ah, so both numerator and denominator are zero at θ = arccos(5/13). Therefore, the integrand has a removable discontinuity there. Therefore, the integral is actually proper. So we can compute it directly. So when integrating, even though the denominator is zero at that point, the numerator is also zero, so the function can be redefined there by the limit, making it continuous. Therefore, the integral is proper. So we can proceed.Therefore, going back to the first integral:First integral evaluated from 0 to π:[ -2/13 ln |13 cos θ -5| ] from 0 to πAt θ=π: 13 cos π -5 =13*(-1)-5=-18, absolute value is 18.At θ=0:13 cos0 -5=13*1 -5=8, absolute value 8.Therefore, first integral part:-2/13 [ ln 18 - ln 8 ] = -2/13 ln(18/8) = -2/13 ln(9/4) = -2/13 (ln9 - ln4) = -2/13 (2 ln3 - 2 ln2)= -4/13 (ln3 - ln2) = (-4/13) ln(3/2)Wait, maybe better to leave it as ln(18/8)=ln(9/4). So first integral is -2/13 ln(9/4).Okay, moving to the second integral: ∫ [3 cos θ / (13 cos θ -5)] dθ.Let me try to simplify this. Let me write numerator as A*(13 cos θ -5) + B, so that:3 cos θ = A*(13 cos θ -5) + BExpanding right side: 13 A cos θ -5A + BEquate coefficients:13 A = 3 => A = 3/13-5A + B =0 => B =5A =5*(3/13)=15/13Therefore, 3 cos θ = (3/13)(13 cos θ -5) +15/13Therefore, 3 cos θ / (13 cos θ -5) = [ (3/13)(13 cos θ -5) +15/13 ] / (13 cos θ -5) = 3/13 + (15/13)/(13 cos θ -5)Therefore, the second integral becomes:∫ [3/13 + 15/(13*(13 cos θ -5))] dθ = 3/13 ∫ dθ + 15/(13*13) ∫ 1/(13 cos θ -5) dθSo that's 3θ/13 + 15/169 ∫ 1/(13 cos θ -5) dθBut then the third integral is -3 ∫ 1/(13 cos θ -5) dθ, so combining the second and third integrals:Second integral: 3θ/13 + 15/169 ∫1/(13 cos θ -5) dθThird integral: -3 ∫1/(13 cos θ -5) dθTherefore, combined second and third integrals:3θ/13 + (15/169 -3) ∫1/(13 cos θ -5) dθConvert 3 to 507/169:15/169 -507/169 = -492/169So combined second and third integrals:3θ/13 - (492/169) ∫1/(13 cos θ -5) dθTherefore, the entire original integral is:First integral + combined second and third integrals:(-2/13 ln(9/4)) + [3θ/13 - (492/169) ∫1/(13 cos θ -5) dθ ] evaluated from 0 to π.Wait, but hold on. The first integral was already evaluated from 0 to π. The second and third integrals are being split into their evaluated parts. Wait, no. Wait, the first integral is an antiderivative evaluated from 0 to π. The second and third integrals need to be evaluated as well.Wait, perhaps I need to combine all steps properly. Let me rephrase.Original integral:I = ∫₀^π [ (2 sin θ +3 cos θ -3)/(13 cos θ -5) ] dθ = I1 + I2 + I3Where:I1 = ∫₀^π [2 sin θ/(13 cos θ -5)] dθI2 = ∫₀^π [3 cos θ/(13 cos θ -5)] dθI3 = ∫₀^π [ -3/(13 cos θ -5) ] dθWe computed I1 already:I1 = [ -2/(13) ln |13 cos θ -5| ]₀^π = -2/13 [ln|13 cos π -5| - ln|13 cos 0 -5|] = -2/13 [ln(18) - ln(8)] = -2/13 ln(18/8) = -2/13 ln(9/4)Then for I2, we expressed it as:3/13 ∫₀^π dθ + 15/169 ∫₀^π 1/(13 cos θ -5) dθ = 3/13 [π -0] + 15/169 ∫₀^π 1/(13 cos θ -5) dθ = 3π/13 + 15/169 JWhere J = ∫₀^π 1/(13 cos θ -5) dθSimilarly, I3 is -3 ∫₀^π 1/(13 cos θ -5) dθ = -3JTherefore, combining I2 and I3:I2 + I3 = 3π/13 +15/169 J -3J = 3π/13 + (15/169 - 3) JConvert 3 to 507/169:15/169 -507/169 = -492/169Thus, I2 + I3 = 3π/13 - (492/169) JTherefore, the entire integral I is:I = I1 + I2 + I3 = (-2/13 ln(9/4)) + 3π/13 - (492/169) JSo now, we need to compute J = ∫₀^π 1/(13 cos θ -5) dθSo the crux is evaluating J. Let me focus on that.J = ∫₀^π 1/(13 cos θ -5) dθIntegrals of the form ∫1/(a cos θ + b) dθ can be tricky. One method is to use the substitution t = tan(θ/2), which transforms the integral into a rational function. Let's try that.Weierstrass substitution: Let t = tan(θ/2). Then, dθ = 2 dt/(1 + t²), cos θ = (1 - t²)/(1 + t²). Also, when θ=0, t=0; when θ=π, t=∞.Therefore, J becomes:∫_{t=0}^{t=∞} [1 / (13*( (1 - t²)/(1 + t²) ) -5 ) ] * [2 dt/(1 + t²)]Simplify the denominator:13*(1 - t²)/(1 + t²) -5 = [13(1 - t²) -5(1 + t²)] / (1 + t²) = [13 -13 t² -5 -5 t²]/(1 + t²) = [8 -18 t²]/(1 + t²)Therefore, the integrand becomes:[1 / ( (8 -18 t²)/(1 + t²) ) ] * [2 dt/(1 + t²)] = [ (1 + t²)/(8 -18 t²) ) ] * [2/(1 + t²) ] dt = 2/(8 -18 t²) dt = 2/( -18 t² +8 ) dt = 2/(8 -18 t²) dtFactor out 2 from denominator:2/(2*(4 -9 t²)) dt = 1/(4 -9 t²) dtSo J = ∫₀^∞ [1/(4 -9 t²)] dtBut we have to be careful here. The substitution t = tan(θ/2) goes from 0 to ∞ as θ goes from 0 to π. But the integral ∫₀^∞ 1/(4 -9 t²) dt may have a problem if the denominator is zero. Let's check when 4 -9 t²=0. That is when t=±2/3. Since our integral is from 0 to ∞, t=2/3 is a point in the interval. Therefore, the integral is improper at t=2/3. Wait, but originally, the integral J had a denominator 13 cos θ -5, which we saw earlier has a zero at θ = arccos(5/13). But we already established that the integrand has a removable discontinuity there because numerator and denominator both zero. However, after substitution, we have a different expression. Wait, perhaps the substitution changes the nature? Let me verify.Wait, when θ approaches arccos(5/13), t approaches tan( (arccos(5/13))/2 ). Let me compute that. Let θ0 = arccos(5/13). Then cos θ0 =5/13, so sin θ0=12/13. Then tan(θ0/2) = sin θ0 / (1 + cos θ0) = (12/13)/(1 +5/13)= (12/13)/(18/13)=12/18=2/3. Therefore, t=2/3 is the point corresponding to θ= arccos(5/13). Therefore, in the substitution, t approaches 2/3 as θ approaches θ0. Therefore, the integral J in terms of t is:J = ∫₀^∞ [1/(4 -9 t²)] dt = ∫₀^{2/3} [1/(4 -9 t²)] dt + ∫_{2/3}^∞ [1/(4 -9 t²)] dtBut since there's a discontinuity at t=2/3, we have to take the limit as t approaches 2/3 from the left and right. However, since the original integral J is proper (due to the removable discontinuity), this improper integral in terms of t should also converge. Therefore, even though 4 -9 t² is zero at t=2/3, the integral converges. Let's check.Near t=2/3, let's let t =2/3 - ε, then 4 -9 t²=4 -9*(4/9 - (4/3)ε + ε²)=4 -4 +12 ε -9 ε²=12 ε -9 ε²≈12 ε. So as ε approaches 0 from the left, 4 -9 t²≈12 ε, so the integrand ≈1/(12 ε). But the integral near t=2/3 would be like ∫ 1/(12 ε) dε, which diverges. Wait, but that contradicts our earlier conclusion that the original integral J is proper. Therefore, there must be an error in the substitution process.Wait, but earlier we saw that when θ approaches θ0=arccos(5/13), the integrand 1/(13 cos θ -5) has a removable discontinuity because the numerator in the original integrand (from the whole problem) was zero there. Wait, but wait: in the original problem, the integrand was (2 sin θ +3 cos θ -3)/(13 cos θ -5). However, in J, which is part of the problem, we are integrating 1/(13 cos θ -5). But in the original problem, that term is multiplied by coefficients and combined with other terms. However, when we split the integral into I1, I2, I3, we separated J = ∫1/(13 cos θ -5) dθ. But in the original integrand, even though at θ0=arccos(5/13), there is a removable discontinuity, when we split into J, we are integrating a function that has a simple pole there. Therefore, J itself is actually divergent? Wait, but that contradicts the original problem's integrand being continuous. Wait, perhaps splitting the integral was invalid? Because the original integral is proper, but splitting into J which is improper? That suggests that even though the original integral is proper, the individual integrals I1, I2, I3 might not be, but their combination is.Therefore, perhaps splitting the integral into I1, I2, I3 is problematic because J is divergent. But that can't be, since the original integral is convergent, so the combination of I1, I2, I3 must result in cancellation of the divergences. Therefore, even though J is divergent, when combined with other terms, the divergences cancel, leading to a finite result. Therefore, perhaps we need to compute the combination of terms involving J without computing J separately.Alternatively, perhaps there's a smarter substitution. Let me think. Maybe combining the terms before integrating.Alternatively, let's go back to the original integral:I = ∫₀^π [ (2 sin θ +3 cos θ -3)/(13 cos θ -5) ] dθWe can write the numerator as A*(13 cos θ -5) + B*sin θ + C. Let me check if that's possible.Let me suppose:2 sin θ +3 cos θ -3 = A*(13 cos θ -5) + B sin θ + CWait, but why add a constant C? The original numerator is 2 sin θ +3 cos θ -3. So maybe:Let me write it as A*(13 cos θ -5) + B sin θ + C.Expanding:13 A cos θ -5 A + B sin θ + CComparing to original numerator: 2 sin θ +3 cos θ -3Therefore, equate coefficients:13 A =3 => A=3/13B=2-5 A + C =-3 => -5*(3/13) + C =-3 => -15/13 + C= -3 => C= -3 +15/13= -39/13 +15/13= -24/13Therefore,2 sin θ +3 cos θ -3 = (3/13)(13 cos θ -5) +2 sin θ -24/13Therefore, the integrand becomes:[ (3/13)(13 cos θ -5) +2 sin θ -24/13 ] / (13 cos θ -5 ) = 3/13 + [2 sin θ -24/13]/(13 cos θ -5)Therefore, the integral I becomes:∫₀^π [3/13 + (2 sin θ -24/13)/(13 cos θ -5) ] dθ = 3/13 ∫₀^π dθ + ∫₀^π [2 sin θ -24/13]/(13 cos θ -5) dθCompute the first integral: 3/13 * πThe second integral: ∫₀^π [2 sin θ -24/13]/(13 cos θ -5) dθLet me split this into two parts:2 ∫₀^π [ sin θ / (13 cos θ -5) ] dθ - (24/13) ∫₀^π [1/(13 cos θ -5) ] dθ = 2 I1' - (24/13) JWait, but I1' is ∫ sin θ / (13 cos θ -5) dθ, which is similar to the I1 we computed earlier. Wait, but in the original split, I1 was ∫ [2 sin θ / (13 cos θ -5) ] dθ. So here, 2 times I1' would be 2 times ∫ [ sin θ / (13 cos θ -5) ] dθ, which is similar.Earlier, we computed I1 as -2/13 ln |13 cos θ -5 | evaluated from 0 to π. So that integral is:2 ∫ [ sin θ / (13 cos θ -5) ] dθ = 2 * [ -1/13 ln |13 cos θ -5 | ] + C = -2/13 ln |13 cos θ -5 | + CWait, but in the previous calculation, we had for the original I1, which was ∫ [2 sin θ / (13 cos θ -5) ] dθ = -2/13 ln |13 cos θ -5 | + C. So in this new expression, 2 ∫ [ sin θ / (13 cos θ -5) ] dθ is the same as I1, which is -2/13 ln |13 cos θ -5 | evaluated from 0 to π. So we can write this as:2 I1' = I1 = -2/13 ln(9/4)Therefore, the second integral becomes:2 I1' - (24/13) J = -2/13 ln(9/4) - (24/13) JThus, the entire integral I is:3π/13 + [ -2/13 ln(9/4) - (24/13) J ]But we need to find J. However, we saw that J is ∫₀^π [1/(13 cos θ -5) ] dθ, which through substitution became ∫₀^∞ [1/(4 -9 t²) ] dt. But this integral is divergent unless we take principal value or something. However, in the original integral, the combination of terms must cancel any divergence. Wait, but this seems conflicting. Let me check again.Wait, when we did the substitution for J, we saw that t=2/3 is a point where the denominator becomes zero, which would make the integral divergent. But since the original integral I is proper, the combination of terms involving J must cancel the divergence. Let's see.In the expression for I:I = 3π/13 -2/13 ln(9/4) - (24/13) JBut if J is divergent, how can I be finite? Therefore, perhaps my approach is wrong here.Wait, but earlier we saw that J is actually convergent because when θ approaches arccos(5/13), the integrand 1/(13 cos θ -5) has a simple pole, but the numerator in the original integrand (2 sin θ +3 cos θ -3) also approaches zero, leading to a removable discontinuity. However, when we split the integral into I1, I2, I3, the term J is separated from the other terms. Therefore, the divergence in J must be canceled by another term. Wait, but in our current expression, I is written as 3π/13 -2/13 ln(9/4) - (24/13) J. So if J is divergent, then I would be divergent unless the coefficient of J is zero, which it's not (-24/13). Therefore, this suggests that my method is flawed.Therefore, perhaps splitting the integral into I1, I2, I3 is not the right approach. Alternatively, maybe I need to consider that when combining all the terms, the divergent parts cancel out. Let me see.Wait, going back to the substitution t = tan(θ/2). Let's compute the entire original integral using this substitution. Maybe that would be better.Original integral:I = ∫₀^π [ (2 sin θ +3 cos θ -3)/(13 cos θ -5) ] dθUsing substitution t = tan(θ/2), as before. Then, θ from 0 to π, t from 0 to ∞.sin θ = 2t/(1 + t²), cos θ = (1 - t²)/(1 + t²), dθ = 2 dt/(1 + t²)Therefore, numerator:2 sin θ +3 cos θ -3 = 2*(2t/(1 + t²)) +3*((1 - t²)/(1 + t²)) -3 = (4t +3(1 - t²) -3(1 + t²))/(1 + t²)Simplify numerator:4t +3 -3t² -3 -3t²=4t -6t²Denominator:13 cos θ -5 =13*((1 - t²)/(1 + t²)) -5= [13(1 - t²) -5(1 + t²)]/(1 + t²)= (13 -13 t² -5 -5 t²)/(1 + t²)= (8 -18 t²)/(1 + t²)Therefore, integrand becomes:(4t -6t²)/( (8 -18 t²)/(1 + t²) ) * (2 dt/(1 + t²))= (4t -6t²)*(1 + t²)/(8 -18 t²) * 2/(1 + t²) dt= (4t -6t²)*2/(8 -18 t²) dt= [ (8t -12 t²) / (8 -18 t²) ] dtFactor numerator and denominator:Numerator: -12 t² +8 t = -4 t (3 t -2 )Denominator: -18 t² +8 = -2(9 t² -4 ) = -2(3t -2)(3t +2 )Therefore, the integrand becomes:[ -4 t (3 t -2 ) ] / [ -2 (3 t -2 )(3 t +2 ) ] dt = [ (-4 t)/ (-2(3t +2)) ] dt = (2 t)/(3 t +2 ) dtSimplify:(2t)/(3t +2) dt = [2/3*(3t +2) -4/3]/(3t +2) dt = 2/3 dt - (4/3)/(3t +2) dtTherefore, integrand simplifies to 2/3 - 4/(3*(3t +2)) dtTherefore, integral I becomes:∫_{t=0}^{t=∞} [2/3 -4/(3*(3t +2)) ] dt = ∫₀^∞ 2/3 dt -4/3 ∫₀^∞ 1/(3t +2) dtCompute each integral:First integral: 2/3 ∫₀^∞ dt. Hmm, this diverges to infinity.Second integral: -4/3 ∫₀^∞ 1/(3t +2) dt. Let u=3t +2, du=3 dt, limits from u=2 to u=∞. So integral becomes -4/3*(1/3) ∫₂^∞ 1/u du= -4/9 [ln u ]₂^∞= -4/9 (∞ -ln2), which diverges to -∞.Therefore, I is divergent? But that's impossible because the original integral is over a finite interval and the integrand has a removable discontinuity. Wait, but this contradicts our earlier analysis. There must be a mistake in the substitution.Wait, but when we did the substitution, we saw that θ from 0 to π translates to t from 0 to ∞. However, near t=2/3, which corresponds to θ= arccos(5/13), there is a removable discontinuity. But in terms of t, the integrand simplifies to 2/3 -4/(3*(3t +2)) dt. However, evaluating this from 0 to ∞ gives divergent integrals. This suggests that the original integral is divergent, which contradicts the initial analysis.But that can't be. The original integrand has a removable discontinuity at θ= arccos(5/13). Therefore, the function is actually continuous on [0, π], hence the integral should be finite.Wait, but when we did the substitution, we transformed the integral into an improper integral from 0 to ∞, which seems to diverge. But that must be due to the substitution not accounting for the removable discontinuity properly.Wait, let's check the transformed integrand:After substitution, integrand became (2t)/(3t +2). Let's analyze this as t approaches ∞. The integrand approaches 2/3, so integrating from 0 to ∞ would indeed lead to a divergent integral. But this contradicts the original integral being finite. Therefore, there must be a mistake in the substitution.Wait, going back through the substitution steps:Numerator:2 sin θ +3 cos θ -3. With substitution, sin θ=2t/(1 +t²), cos θ=(1 -t²)/(1 +t²). So:2*(2t/(1 +t²)) +3*((1 -t²)/(1 +t²)) -3 = (4t)/(1 +t²) + (3 -3t²)/(1 +t²) -3Combine terms:[4t +3 -3t² -3(1 +t²)]/(1 +t²) = [4t +3 -3t² -3 -3t²]/(1 +t²)= [4t -6t²]/(1 +t²)Denominator:13 cos θ -5=13*(1 -t²)/(1 +t²) -5= [13(1 -t²) -5(1 +t²)]/(1 +t²)= [13 -13t² -5 -5t²]/(1 +t²)= (8 -18t²)/(1 +t²)So integrand is [4t -6t²]/[8 -18t²] * [2/(1 +t²)] * (1 +t²) ?Wait, no. The substitution step:Original integrand: numerator/(denominator) * dθNumerator: (4t -6t²)/(1 +t²)Denominator: (8 -18t²)/(1 +t²)Therefore, numerator/denominator = (4t -6t²)/(8 -18t²)Then multiply by dθ=2/(1 +t²) dt:Therefore, integrand becomes (4t -6t²)/(8 -18t²) * 2/(1 +t²) dtWait, but earlier I simplified this to [ (4t -6t²) *2 ] / [8 -18t²] dt, which is correct.Then factoring numerator and denominator:4t -6t² = -6t² +4t = -2t(3t -2 )8 -18t²= -18t² +8= -2(9t² -4 )= -2(3t -2)(3t +2 )Therefore, [ -2t(3t -2 ) ] / [ -2(3t -2)(3t +2 ) ] *2 dtThe -2 cancels out, and (3t -2) cancels out:[ t / (3t +2 ) ] *2 dt= (2t)/(3t +2 ) dtBut here, we canceled (3t -2), which is valid except when 3t -2=0, i.e., t=2/3. But at t=2/3, which corresponds to θ= arccos(5/13), we have a removable discontinuity. Therefore, strictly speaking, the integrand in t is (2t)/(3t +2 ) dt, but with a removable discontinuity at t=2/3. Therefore, the integral from t=0 to t=∞ is actually the same as integrating (2t)/(3t +2 ) dt, excluding the point t=2/3. But integrating this function from 0 to ∞:∫₀^∞ (2t)/(3t +2 ) dtLet me compute this integral. Let's make substitution u=3t +2. Then du=3dt, t=(u -2)/3, dt= du/3. When t=0, u=2; when t=∞, u=∞.Integral becomes:∫_{u=2}^{u=∞} [2*( (u -2)/3 ) / u ] * (du/3 )= ∫₂^∞ [2(u -2)/3u ]*(1/3) du= (2/9) ∫₂^∞ (u -2)/u du= (2/9) ∫₂^∞ [1 -2/u ] du= (2/9)[ ∫₂^∞ 1 du -2 ∫₂^∞ 1/u du ]But these integrals are divergent. The first integral ∫₂^∞ 1 du is divergent, and the second integral ∫₂^∞1/u du is also divergent. Therefore, the integral in terms of t diverges. However, this contradicts the original integral being convergent. Therefore, there must be a mistake in the substitution.Wait, perhaps the mistake is in the substitution step. Let's check again.Original integrand after substitution:[ (4t -6t²)/(1 +t²) ] / [ (8 -18t²)/(1 +t²) ] * [2/(1 +t²) ] dt= [ (4t -6t²) / (8 -18t²) ] * [2/(1 +t²) ] dt= [ (4t -6t²)*2 ] / [ (8 -18t²)(1 +t²) ] dt= [ (8t -12t²) ] / [ (8 -18t²)(1 +t²) ] dtWait, but previously, I canceled out (3t -2) from numerator and denominator, but perhaps there's an error in factoring.Let me factor numerator and denominator again:Numerator:8t -12 t² = -12 t² +8t = -4t(3t -2 )Denominator: (8 -18t²)(1 +t²)= -2(9t² -4)(1 +t²)= -2(3t -2)(3t +2)(1 +t²)Therefore, numerator/denominator= -4t(3t -2)/[ -2(3t -2)(3t +2)(1 +t²) ]= [ -4t / (-2(3t +2)(1 +t²)) ]= [2t/( (3t +2)(1 +t²) ) ]Therefore, integrand becomes [2t/( (3t +2)(1 +t²) ) ] *2 dt=4t/( (3t +2)(1 +t²) ) dtWait, previously, I thought that (3t -2) cancels out, but in reality, denominator after factoring is -2(3t -2)(3t +2)(1 +t²). So after canceling (3t -2), the denominator is -2(3t +2)(1 +t²). Therefore, numerator is -4t(3t -2), denominator is -2(3t -2)(3t +2)(1 +t²). So when we cancel (3t -2), we get numerator: -4t, denominator: -2(3t +2)(1 +t²). The negatives cancel, leading to 4t/[2(3t +2)(1 +t²)]= 2t/[ (3t +2)(1 +t²) ]Therefore, integrand is 2t/[ (3t +2)(1 +t²) ] *2 dt from substitution?Wait, no. Wait, the original expression after substitution is:[ (4t -6t²)/(8 -18t²) ] * [2/(1 +t²) ] dt= [ -4t(3t -2) / (-2(3t -2)(3t +2)) ] * [2/(1 +t²) ] dt= [ (4t)/(2(3t +2)) ] * [2/(1 +t²) ] dt= [ (2t)/(3t +2) ] * [2/(1 +t²) ] dt= [4t]/[ (3t +2)(1 +t²) ] dtAh, I see. So my previous simplification was incorrect. The correct integrand after substitution is 4t/[ (3t +2)(1 +t²) ] dtTherefore, I need to compute:I= ∫₀^∞ [4t]/[ (3t +2)(1 +t²) ] dtThis integral is convergent because as t approaches infinity, the integrand behaves like 4t/(3t * t² )=4/(3 t²), which is integrable at infinity. Similarly, near t=0, the integrand behaves like 4t/(2*1)=2t, which is integrable. Therefore, the integral is convergent. Therefore, my earlier mistake was in the algebra step where I incorrectly simplified the integrand to (2t)/(3t +2) dt. The correct integrand is 4t/[ (3t +2)(1 +t²) ] dt, which is convergent.Therefore, now we need to compute:I= ∫₀^∞ [4t]/[ (3t +2)(1 +t²) ] dtLet me use partial fractions to decompose the integrand.Let me write 4t / [ (3t +2)(t² +1) ] = A/(3t +2 ) + (Bt +C)/(t² +1)Multiply both sides by (3t +2)(t² +1):4t = A(t² +1) + (Bt +C)(3t +2)Expand the right side:A t² +A + Bt*(3t) + Bt*2 + C*3t + C*2= A t² +A +3B t² +2B t +3C t +2CCombine like terms:(A +3B) t² + (2B +3C) t + (A +2C)Set equal to left side 4t:So coefficients must satisfy:A +3B =0 (coefficient of t²)2B +3C =4 (coefficient of t)A +2C=0 (constant term)We have three equations:1) A +3B =02) 2B +3C=43) A +2C=0Let's solve this system.From equation 1: A= -3BFrom equation 3: -3B +2C=0 => 2C=3B => C= (3/2)BSubstitute into equation 2:2B +3*(3/2 B)=4 =>2B +9/2 B=4 => (4/2 +9/2)B=4 =>13/2 B=4 =>B=4*(2/13)=8/13Then, A= -3*(8/13)= -24/13C= (3/2)*(8/13)=24/26=12/13Therefore, partial fractions decomposition:4t / [ (3t +2)(t² +1) ]= (-24/13)/(3t +2 ) + ( (8/13)t +12/13 )/(t² +1 )Therefore, integral I becomes:∫₀^∞ [ (-24/13)/(3t +2 ) + (8t/13 +12/13)/(t² +1 ) ] dt= (-24/13)∫₀^∞ 1/(3t +2 ) dt + (8/13)∫₀^∞ t/(t² +1 ) dt + (12/13)∫₀^∞1/(t² +1 ) dtCompute each integral separately.First integral: (-24/13) ∫₀^∞1/(3t +2 ) dtLet u=3t +2, du=3 dt, when t=0, u=2; t=∞, u=∞.Integral becomes (-24/13)*(1/3) ∫₂^∞1/u du= (-8/13)[ln u ]₂^∞= (-8/13)(∞ -ln2 ) → -∞. But wait, this is a problem. But earlier we saw that the integral I is convergent. Therefore, the divergence must cancel with other terms. Let's check the other integrals.Second integral: (8/13) ∫₀^∞ t/(t² +1 ) dtLet u= t² +1, du=2t dt, so t dt= du/2. Integral becomes (8/13)*(1/2) ∫₁^∞1/u du= (4/13)[ln u ]₁^∞= (4/13)(∞ -0 ) → ∞Third integral: (12/13) ∫₀^∞1/(t² +1 ) dt= (12/13)*(π/2 )= (6π)/13Therefore, combining all three integrals:I= (-8/13)(∞ -ln2 ) + (4/13)(∞) +6π/13This is an indeterminate form ( -∞ + ∞ ). Therefore, the divergence cancels. To compute this properly, we need to consider the integrals as limits.Let's compute each integral as a limit approaching some upper bound T and then take T→∞.First integral:(-24/13) ∫₀^T1/(3t +2 ) dt= (-24/13)*(1/3)[ln(3t +2 )]₀^T= (-8/13)[ln(3T +2 ) -ln2 ]Second integral:(8/13) ∫₀^T t/(t² +1 ) dt= (4/13)[ln(t² +1 )]₀^T= (4/13)[ln(T² +1 ) -ln1 ]= (4/13) ln(T² +1 )Third integral:(12/13) ∫₀^T1/(t² +1 ) dt= (12/13)[ arctan t ]₀^T= (12/13)( arctan T -0 )= (12/13) arctan TTherefore, combining all three:I(T)= (-8/13)[ln(3T +2 ) -ln2 ] + (4/13) ln(T² +1 ) + (12/13) arctan TNow, take the limit as T→∞.Compute each term as T→∞:1. (-8/13)[ln(3T +2 ) -ln2 ] ~ (-8/13) ln(3T ) + constants2. (4/13) ln(T² +1 ) ~ (4/13) ln(T² )= (8/13) ln T3. (12/13) arctan T ~ (12/13)*(π/2 )=6π/13Combine the leading terms:-8/13 ln(3T ) +8/13 ln T +6π/13= -8/13 (ln3 +lnT ) +8/13 lnT +6π/13= -8/13 ln3 -8/13 lnT +8/13 lnT +6π/13= -8/13 ln3 +6π/13The divergent terms cancel out. Therefore, the limit as T→∞ is:I= -8/13 ln3 +6π/13Therefore, the original integral:I=6π/13 -8/13 ln3But let's check the coefficients:Wait, the integral I(T)= (-8/13)[ln(3T +2 ) -ln2 ] + (4/13) ln(T² +1 ) + (12/13) arctan TExpand:= (-8/13) ln(3T +2 ) + (8/13) ln2 + (4/13) ln(T² +1 ) + (12/13) arctan TNow, as T→∞, ln(3T +2 )=ln3 +lnT +ln(1 +2/(3T ))~ln3 +lnTSimilarly, ln(T² +1 )=2 lnT +ln(1 +1/T² )~2 lnTarctan T ~π/2Therefore,I(T)~ (-8/13)(ln3 +lnT ) +8/13 ln2 + (4/13)(2 lnT ) + (12/13)(π/2 )Simplify:= (-8/13 ln3 -8/13 lnT ) +8/13 ln2 +8/13 lnT +6π/13The -8/13 lnT and +8/13 lnT cancel, leaving:= (-8/13 ln3 ) +8/13 ln2 +6π/13Factor out 8/13:=8/13 (-ln3 +ln2 ) +6π/13=8/13 ln(2/3 ) +6π/13Therefore, the original integral I=8/13 ln(2/3 ) +6π/13Alternatively, factor out 2/13:=2/13 [4 ln(2/3 ) +3π ]But perhaps it's better written as:I=(6π)/13 - (8 ln3 -8 ln2 )/13= (6π -8 ln(3/2 ))/13Therefore, the final answer is (6π -8 ln(3/2 )) /13Simplify further if needed. Since the question likely expects a simplified answer, perhaps written as:(6π -8 ln(3/2 )) /13Or factor out 2: 2*(3π -4 ln(3/2 )) /13= (3π -4 ln(3/2 )) *2/13, but I think the first form is acceptable.Alternatively, combine the constants:ln(3/2 )=ln3 -ln2, so -8 ln(3/2 )= -8 ln3 +8 ln2Thus, 6π/13 +8 ln2 /13 -8 ln3 /13= (6π +8 ln2 -8 ln3 )/13= (6π +8(ln2 -ln3 ))/13= (6π -8 ln(3/2 ))/13So both forms are equivalent.Therefore, the value of the integral is (6π -8 ln(3/2 )) /13.Final AnswerThe value of the integral is boxed{dfrac{6pi - 8 lnleft(dfrac{3}{2}right)}{13}}.</think>To compute the integralint_{0}^{pi} frac{2 sin theta + 3 cos theta - 3}{13 cos theta - 5} , dtheta,we start by splitting the integrand into simpler parts and using substitutions.First, we decompose the numerator:2 sin theta + 3 cos theta - 3 = frac{3}{13}(13 cos theta - 5) + 2 sin theta - frac{24}{13}.This allows us to rewrite the integrand as:frac{3}{13} + frac{2 sin theta - frac{24}{13}}{13 cos theta - 5}.Next, we use the substitution ( t = tan(theta/2) ), which transforms the integral into a rational function. After substitution, the integrand simplifies to:frac{4t}{(3t + 2)(1 + t^2)} , dt.Using partial fractions, we decompose:frac{4t}{(3t + 2)(t^2 + 1)} = frac{-24/13}{3t + 2} + frac{(8/13)t + 12/13}{t^2 + 1}.We then integrate each term separately, taking limits as ( t ) approaches infinity to handle the improper integrals. The divergent parts cancel out, leaving us with the finite result.After evaluating the integrals and simplifying, we find the result:int_{0}^{pi} frac{2 sin theta + 3 cos theta - 3}{13 cos theta - 5} , dtheta = frac{6pi - 8 lnleft(frac{3}{2}right)}{13}.The final answer is:boxed{dfrac{6pi - 8 lnleft(dfrac{3}{2}right)}{13}}.,<think>Okay, let me try to solve this integral step by step. The integral given is:int_{0}^{pi} frac{2 sin theta + 3 cos theta - 3}{13 cos theta - 5} , dthetaHmm, integrating a rational function of sine and cosine over 0 to π. These types of integrals can sometimes be tackled with the Weierstrass substitution, but maybe there's a simpler approach here. Let me see.First, let me check if the denominator can be simplified or if the numerator can be expressed in terms of the derivative of the denominator. That might help in breaking the integral into parts. The denominator is 13 cos θ - 5. Let's compute its derivative:d/dθ [13 cos θ - 5] = -13 sin θLooking at the numerator: 2 sin θ + 3 cos θ - 3. There's a 2 sin θ term, which is similar to the derivative of the denominator, except for a coefficient. Let me see if I can split the fraction accordingly.Let me suppose that the numerator can be written as A*(derivative of denominator) + B*(denominator) + C. Let's try that.Let me denote:Numerator = 2 sin θ + 3 cos θ - 3Denominator = 13 cos θ - 5So, suppose:2 sin θ + 3 cos θ - 3 = A*(-13 sin θ) + B*(13 cos θ - 5) + CLet's expand the right-hand side:-13A sin θ + 13B cos θ - 5B + CNow, equate the coefficients of sin θ, cos θ, and the constants:For sin θ:2 = -13A => A = -2/13For cos θ:3 = 13B => B = 3/13For constants:-3 = -5B + CWe already know B = 3/13, so plug that into the constants equation:-3 = -5*(3/13) + CCompute -5*(3/13): -15/13Therefore:-3 = -15/13 + C => C = -3 + 15/13 = -39/13 + 15/13 = -24/13So, we've expressed the numerator as:Numerator = (-2/13)*(-13 sin θ) + (3/13)*(13 cos θ - 5) -24/13Wait, let's check that:-13A sin θ = -13*(-2/13) sin θ = 2 sin θ13B cos θ -5B = 13*(3/13) cos θ -5*(3/13) = 3 cos θ -15/13Then adding C gives:3 cos θ -15/13 -24/13 = 3 cos θ -39/13 = 3 cos θ -3Which matches the numerator: 2 sin θ +3 cos θ -3. Perfect!So, substituting back into the integral:int_{0}^{pi} frac{2 sin theta + 3 cos theta - 3}{13 cos theta - 5} dtheta = int_{0}^{pi} left[ frac{-2/13*(-13 sin theta) + 3/13*(13 cos theta -5) -24/13}{13 cos theta -5} right] dthetaWait, but since we expressed the numerator as:A*(-13 sin θ) + B*(13 cos θ -5) + CThen:frac{Numerator}{Denominator} = frac{A*(-13 sin theta) + B*(13 cos theta -5) + C}{13 cos theta -5}Which simplifies to:A*frac{-13 sin theta}{13 cos theta -5} + B + frac{C}{13 cos theta -5}Therefore, substituting A = -2/13, B = 3/13, C = -24/13:frac{Numerator}{Denominator} = left( -frac{2}{13} right)*frac{-13 sin theta}{13 cos theta -5} + frac{3}{13} + frac{-24/13}{13 cos theta -5}Simplifying term by term:First term:(-2/13)*(-13 sin θ)/(13 cos θ -5) = (2 sin θ)/(13 cos θ -5)Second term: 3/13Third term: (-24/13)/(13 cos θ -5)So overall:frac{2 sin theta}{13 cos theta -5} + frac{3}{13} - frac{24}{13(13 cos theta -5)}Therefore, the integral becomes:int_{0}^{pi} left[ frac{2 sin theta}{13 cos theta -5} + frac{3}{13} - frac{24}{13(13 cos theta -5)} right] dthetaLet's split this into three separate integrals:1. Integral of [2 sin θ / (13 cos θ -5)] dθ from 0 to π2. Integral of 3/13 dθ from 0 to π3. Integral of [-24/(13(13 cos θ -5))] dθ from 0 to πLet me handle each integral separately.First Integral: I1 = ∫ [2 sin θ / (13 cos θ -5)] dθLet’s substitute u = 13 cos θ -5Then du/dθ = -13 sin θ => du = -13 sin θ dθ => sin θ dθ = -du/13Therefore, I1 becomes:2 ∫ [sin θ dθ / (13 cos θ -5)] = 2 ∫ [ (-du/13) / u ] = 2*(-1/13) ∫ (1/u) du = (-2/13) ln|u| + CSo, evaluated from 0 to π:I1 = (-2/13) [ ln|13 cos θ -5| ] from 0 to πCompute at θ = π: 13 cos π -5 = 13*(-1) -5 = -18At θ = 0: 13 cos 0 -5 = 13*1 -5 = 8So,I1 = (-2/13)[ln| -18 | - ln|8|] = (-2/13)[ln(18) - ln(8)] = (-2/13) ln(18/8) = (-2/13) ln(9/4) = (-2/13)(ln9 - ln4) = (-2/13)(2 ln3 - 2 ln2) = (-4/13)(ln3 - ln2)Wait, but maybe we can write it as (-2/13) ln(9/4). Let's keep it as that for now.Second Integral: I2 = ∫ (3/13) dθ from 0 to πThis is straightforward:I2 = (3/13)(π - 0) = 3π/13Third Integral: I3 = ∫ [-24/(13(13 cos θ -5))] dθ from 0 to πSo, I3 = (-24)/(13*13) ∫ [1/(cos θ -5/13)] dθ from 0 to πWait, let me factor out constants:I3 = (-24)/(13*13) ∫ [1/(cos θ -5/13)] dθ from 0 to πHmm, integrating 1/(cos θ - a) dθ. This is a standard integral. Let me recall the integral of 1/(cos θ - a) dθ. Alternatively, maybe use substitution. Let me see. Perhaps use the substitution t = tan(θ/2), the Weierstrass substitution.But since the integral is from 0 to π, the substitution t = tan(θ/2) would map θ from 0 to π into t from 0 to ∞. Let me try that.Using the substitution t = tan(θ/2). Then, cos θ = (1 - t²)/(1 + t²), and dθ = 2 dt/(1 + t²)Therefore, the integral becomes:∫ [1/( (1 - t²)/(1 + t²) - a ) ] * [2 dt/(1 + t²)] Let me compute this step by step. Let's denote a = 5/13.Then, denominator in the integrand becomes:(1 - t²)/(1 + t²) - a = [1 - t² - a(1 + t²)]/(1 + t²) = [1 - t² - a - a t²]/(1 + t²) = [ (1 - a) - t²(1 + a) ]/(1 + t²)Therefore, the integral becomes:∫ [ (1 + t²) / ( (1 - a) - t²(1 + a) ) ] * [2 dt/(1 + t²) ] = ∫ [2 / ( (1 - a) - t²(1 + a) ) ] dtSimplify the denominator:(1 - a) - t²(1 + a) = (1 - a) - (1 + a)t²Factor out (1 + a):= (1 + a)[ ( (1 - a)/(1 + a) ) - t² ]Let me compute (1 - a)/(1 + a):Given a = 5/13, so:(1 - 5/13)/(1 + 5/13) = (8/13)/(18/13) = 8/18 = 4/9Thus, the denominator becomes:(1 + a)[4/9 - t²] = (18/13)(4/9 - t²) = (18/13)( (4 - 9t²)/9 ) = (18/13)( (4 - 9t²)/9 ) = (2/13)(4 - 9t²) = (8 - 18t²)/13Wait, maybe I miscalculated:Wait, (1 + a) = 1 + 5/13 = 18/13Then (1 - a)/(1 + a) = (8/13)/(18/13) = 8/18 = 4/9Therefore, the denominator is (1 + a)*(4/9 - t²) = (18/13)*(4/9 - t²) = (18/13)*(4/9 - t²) = (18/13)*( (4 - 9t²)/9 ) = (2/13)*(4 - 9t²)Therefore, the integral becomes:∫ [2 / ( (2/13)(4 - 9t²) ) ] dt = ∫ [2 * (13/2) / (4 - 9t²) ] dt = 13 ∫ [1/(4 - 9t²) ] dtSo, integral becomes 13 ∫ 1/(4 - 9t²) dtFactor out 4 from denominator:= 13 ∫ 1/[4(1 - (9/4)t²)] dt = (13/4) ∫ 1/[1 - ( (3t/2) )² ] dtLet me substitute u = (3t)/2, so du = (3/2) dt => dt = (2/3) duThen, integral becomes:(13/4) * (2/3) ∫ 1/(1 - u²) du = (13/6) ∫ [1/(1 - u²)] duThe integral of 1/(1 - u²) is (1/2) ln |(1 + u)/(1 - u)| + CTherefore, the integral is:(13/6) * (1/2) ln |(1 + u)/(1 - u)| + C = (13/12) ln |(1 + u)/(1 - u)| + CSubstituting back u = (3t)/2:= (13/12) ln |(1 + (3t/2))/(1 - (3t/2))| + CSimplify:= (13/12) ln |(2 + 3t)/(2 - 3t)| + CBut remember that t = tan(θ/2). Therefore, substituting back:= (13/12) ln |(2 + 3 tan(θ/2))/(2 - 3 tan(θ/2))| + CBut we need to evaluate the definite integral from θ = 0 to θ = π.Let’s compute the bounds:When θ approaches π from below, tan(θ/2) approaches tan(π/2) which goes to +∞. Let's check:At θ = π, tan(π/2) is undefined (infinite). Let me approach θ = π as a limit. Let θ = π - ε, where ε approaches 0+.tan( (π - ε)/2 ) = tan(π/2 - ε/2 ) = cot(ε/2) ≈ 2/ε as ε → 0. So, tan(θ/2) approaches infinity as θ approaches π.Similarly, at θ = 0, tan(0/2) = tan(0) = 0.Therefore, the integral from θ = 0 to θ = π becomes:[ (13/12) ln |(2 + 3t)/(2 - 3t)| ] evaluated from t = 0 to t → ∞.Compute as t → ∞:(2 + 3t)/(2 - 3t) = (3t + 2)/(-3t + 2) ≈ (3t)/(-3t) = -1 as t → ∞. Therefore, the absolute value becomes | -1 | = 1, and ln(1) = 0.Wait, but wait:Wait, let me take the limit as t approaches infinity:(2 + 3t)/(2 - 3t) = (3t + 2)/(-3t + 2) = [t(3 + 2/t)] / [ t(-3 + 2/t) ] = (3 + 0)/(-3 + 0) = -1So, ln | -1 | = ln 1 = 0. So the upper limit evaluates to 0.At t = 0:(2 + 0)/(2 - 0) = 1, so ln(1) = 0.Therefore, the entire definite integral evaluates to 0 - 0 = 0?Wait, that can't be right. Wait, but we had:The integral was:13 ∫ [1/(4 - 9t²) ] dt from t = 0 to t → ∞But when we evaluated the antiderivative, we found that as t approaches infinity, the expression inside the log tends to -1, whose absolute value is 1, so ln(1) = 0. Similarly, at t =0, the expression is 1, so ln(1) =0. Therefore, the entire integral is 0. But this seems suspicious.Wait, but let me check the substitution steps again. Because integrating 1/(cos θ - a) from 0 to π, but using substitution t = tan(θ/2), which might lead to symmetric results. However, maybe the integral is indeed zero? Let me check with another method.Alternatively, recall that the integral of 1/(cos θ - a) over 0 to π. Let me consider the substitution θ = π - φ. Then, when θ = 0, φ = π; θ = π, φ = 0. So,Integral becomes ∫ from π to 0 of 1/(cos(π - φ) -a) (-dφ) = ∫ from 0 to π 1/(-cos φ -a) dφSo, original integral I = ∫0^π [1/(cos θ - a)] dθ = ∫0^π [1/(-cos φ - a)] dφ = -∫0^π [1/(cos φ + a)] dφBut this implies that I = -I', where I' = ∫0^π [1/(cos φ + a)] dφBut unless a = 0, this doesn't directly help. However, unless the integral is zero, which would require I = -I => I =0, but that would be true only if I = I', which would require a = -a, i.e., a =0. But in our case a =5/13 ≠0, so perhaps this approach doesn't imply I=0. Therefore, maybe my earlier conclusion was wrong.Wait, but according to the substitution, we ended up with the integral being zero. That contradicts. Let me check with a specific example. Let’s take a =0. Then ∫0^π [1/(cos θ)] dθ, which diverges, but in our case a =5/13, which is less than 1, so the denominator 13 cos θ -5 =13(cos θ -5/13). The denominator is 13(cos θ -5/13). The question is, does cos θ -5/13 ever equal zero in [0, π]?cos θ =5/13. Since 5/13 ≈0.3846, which is achievable in [0, π], since cos θ decreases from 1 to -1. So θ = arccos(5/13) ≈1.176 radians, which is approximately 67.38 degrees. So the denominator crosses zero here. Wait, but wait: denominator is 13 cos θ -5. So cos θ =5/13 is the point where denominator is zero. So, θ = arccos(5/13) ≈1.176, which is in [0, π]. Therefore, the integrand has a singularity at θ ≈1.176. Therefore, the integral is improper and needs to be evaluated as a principal value or it diverges.Wait, but the original problem didn't mention anything about principal values, so maybe the integral is actually convergent? Let me check.Near θ = arccos(5/13), the denominator behaves like 13(cos θ -5/13) ≈13*(-sin(arccos(5/13)))(θ - arccos(5/13)). Let’s compute sin(arccos(5/13)): sqrt(1 - (25/169))=sqrt(144/169)=12/13. Therefore, near θ ≈ arccos(5/13), denominator ≈13*(-12/13)(θ - arccos(5/13))= -12(θ - arccos(5/13)). Therefore, the integrand near the singularity behaves like 1/(-12(θ - a)), so the integral has a singularity like 1/(θ -a), which is integrable in the principal value sense, but the integral is actually divergent as an improper integral. But since the problem is given, perhaps the integral is intended to be interpreted as principal value.But this complicates things. However, in our earlier calculation using substitution, we obtained zero, but the integral may actually be divergent. Wait, but hold on. The original integral combines three terms: I1, I2, I3. Let me check if each of them is convergent.First, the original integrand is [2 sin θ +3 cos θ -3]/[13 cos θ -5]. At θ where denominator is zero, let's see the numerator:At θ = arccos(5/13), compute numerator: 2 sin θ +3*(5/13) -3. sin θ =12/13, so numerator=2*(12/13)+15/13 -3=24/13 +15/13 -39/13=0. Therefore, the numerator is zero at the same point where denominator is zero. Therefore, the integrand has a removable discontinuity at θ = arccos(5/13). Therefore, the integral is actually convergent.Therefore, the integrand is continuous on [0, π], since the problematic point is a removable discontinuity. Therefore, the integral is proper and converges. Therefore, our earlier substitution leading to zero might be correct. Wait, but how?Wait, when we did the substitution t = tan(θ/2), which led to an antiderivative that seemed to evaluate to zero. But let me verify.Wait, if the integral is proper, then the result should be a finite number. Let's check again.Wait, when we did the substitution for the third integral I3, we had:I3 = (-24)/(13*13) * ∫ [1/(cos θ -5/13)] dθ from 0 to πBut after substitution, we found that the integral of [1/(cos θ -5/13)] dθ from 0 to π is zero? But that seems odd.Alternatively, perhaps my substitution missed something. Let me check again.Wait, the integral was:∫ [1/(cos θ - a)] dθ from 0 to π, where a =5/13.But using substitution θ = π - φ, as before:I = ∫0^π [1/(cos θ - a)] dθ = ∫0^π [1/(-cos φ - a)] dφ = -∫0^π [1/(cos φ + a)] dφBut if I denote this as -J, where J = ∫0^π [1/(cos φ + a)] dφBut unless J = -I, which would imply I = -J = I, which would require I=0. But that's only possible if J = -I and I = -J, so I + J =0, but unless a=0, which it isn't.But since a ≠0, this approach doesn't directly give I =0, unless...Wait, maybe there's symmetry. Let me compute I + J:I + J = ∫0^π [1/(cos θ -a) + 1/(cos θ +a) ] dθ = ∫0^π [ (2 cos θ)/(cos² θ -a²) ] dθBut not sure if that helps. Alternatively, perhaps integrate using a different substitution.Alternatively, use the standard integral formula for ∫ dθ / (b + c cos θ). The formula is:∫0^π [dθ / (b + c cos θ)] = π / sqrt(b² -c²) if b > |c|But in our case, for the integral I3, we have ∫0^π [1/(cos θ -a)] dθ. Let's write this as ∫0^π [1/( -a + cos θ ) ] dθ. Comparing with the standard integral ∫0^π [dθ / (b + c cos θ)].So, set b = -a, c =1. Then the integral becomes ∫0^π [dθ / (b + c cos θ)] with b = -a, c =1. The formula is π / sqrt(b² -c²) if b > |c|. However, here b = -a = -5/13 ≈ -0.3846, c=1. Since |c|=1, and b=-0.3846 <1, so the formula doesn't apply. The standard formula is for ∫0^{2π} [dθ / (b + c cos θ)] = 2π / sqrt(b² -c²) when b > |c|. But in our case, the integral is from 0 to π, and b < |c|. Therefore, the integral diverges in the usual sense, but since our integrand has a removable discontinuity, the integral is actually convergent.Wait, but according to our previous analysis, the integral I3 is part of the expression, and the original integral is proper. Therefore, maybe the substitution gives zero because the positive and negative areas cancel out. However, given that the original integrand is real and the integral is over an interval where the function is continuous except at a removable discontinuity, the integral should result in a real number. But how?Wait, perhaps I made a mistake in the substitution steps. Let me check again.Original substitution for I3:I3 = (-24)/(13*13) ∫ [1/(cos θ -5/13)] dθ from 0 to πLet me compute this integral ∫ [1/(cos θ -5/13)] dθ from 0 to π.Let me use the substitution t = tan(θ/2). Then, as before, cos θ = (1 - t²)/(1 + t²), dθ = 2 dt/(1 + t²)Thus, the integral becomes:∫ [1 / ( (1 - t²)/(1 + t²) -5/13 ) ] * [2 dt/(1 + t²) ]Simplify the denominator:(1 - t²)/(1 + t²) -5/13 = [13(1 - t²) -5(1 + t²)] / [13(1 + t²)]Compute numerator:13(1 - t²) -5(1 + t²) =13 -13 t² -5 -5 t² =8 -18 t²Therefore, denominator becomes [8 -18 t²]/[13(1 + t²)]Thus, the integral becomes:∫ [ (13(1 + t²))/(8 -18 t²) ] * [2 dt/(1 + t²) ] = ∫ [13 *2 / (8 -18 t²) ] dt = 26 ∫ dt/(8 -18 t²)Factor denominator:8 -18 t² = 2*(4 -9 t²) =2*(2^2 - (3 t)^2)Thus,26 ∫ dt/(2*(4 -9 t²)) =13 ∫ dt/(4 -9 t²)Which is:13*(1/4) ∫ dt/(1 - ( (3t)/2 )² ) =13/4 ∫ dt/(1 - ( (3t/2 )² )Let me substitute u =3t/2, then du =3/2 dt => dt =2 du/3Thus,13/4 * (2/3) ∫ du/(1 -u² ) =13/6 ∫ du/(1 -u² )Integral of 1/(1 -u² ) is (1/2) ln |(1 + u)/(1 - u)| + CTherefore,13/6 *1/2 [ ln |(1 + u)/(1 - u)| ] +C =13/12 ln |(1 +u)/(1 -u)| +CSubstituting back u=3t/2,=13/12 ln |(1 + 3t/2)/(1 -3t/2)| + C=13/12 ln |(2 +3t)/(2 -3t)| + CBut t=tan(θ/2). Therefore, the antiderivative is:13/12 ln |(2 +3 tan(θ/2))/(2 -3 tan(θ/2))| evaluated from θ=0 to θ=π.At θ=0, tan(0)=0, so the expression becomes ln(2/2)=ln1=0.At θ approaching π, tan(θ/2) approaches infinity. Let's compute the limit as t→infty of (2 +3t)/(2 -3t):Divide numerator and denominator by t:(2/t +3)/(2/t -3) → (0 +3)/(0 -3)=3/(-3)= -1Therefore, the expression becomes ln| -1 | = ln1=0.Thus, the definite integral is 13/12 (0 -0 )=0.Therefore, the integral I3 is (-24/(13*13)) *0=0.Wait, so I3=0? Then the entire third integral is zero?But then, the original integral is:I1 + I2 + I3 = I1 + I2 +0 = (-2/13) ln(9/4) + 3π/13Wait, but let's confirm this.First Integral I1:(-2/13)[ln|13 cos θ -5|] from 0 to πAt θ=π: 13 cos π -5= -13 -5= -18At θ=0:13 cos 0 -5=13 -5=8Therefore:I1= (-2/13)(ln| -18| -ln8)= (-2/13)(ln18 -ln8)= (-2/13)ln(18/8)= (-2/13)ln(9/4)Yes.So I1= -2/13 * ln(9/4)I2=3π/13I3=0Therefore, the total integral is:-2/13 * ln(9/4) +3π/13Simplify ln(9/4)=2 ln(3/2)Therefore:-2/13 *2 ln(3/2) +3π/13= -4/13 ln(3/2) +3π/13But the problem might want the answer in terms of ln(9/4) or something else. Let me check if we can write this as (3π -4 ln(3/2))/13 or combine terms.Alternatively, express ln(9/4) as ln9 -ln4=2 ln3 -2 ln2, so:-2/13*(2 ln3 -2 ln2)= -4/13(ln3 -ln2)So total integral:-4/13(ln3 -ln2) +3π/13= (3π -4 ln3 +4 ln2)/13Which can be written as [3π +4 ln(2/3)]/13Alternatively, [3π -4 ln(3/2)]/13Either form is acceptable. Let me see the possible answer forms.The problem says "Compute the integral", and unless specified, both forms are correct. But perhaps the answer is expressed in terms of pi and logarithms. Let me check if my calculation is correct.Wait, let me verify again:Original integrand decomposed into three parts, leading to three integrals:1. Integral of [2 sin θ/(13 cos θ -5)] dθ = (-2/13) ln|13 cos θ -5| from 0 to π = (-2/13)(ln18 -ln8)= (-2/13)ln(18/8)= (-2/13)ln(9/4)2. Integral of 3/13 dθ from 0 to π =3π/133. Integral of [-24/(13*(13 cos θ -5))] dθ from 0 to π = (-24/169)*0=0Therefore, total integral= (-2/13)ln(9/4)+3π/13+0= (3π -2 ln(9/4))/13But ln(9/4)=2 ln(3/2), so:= (3π -4 ln(3/2))/13Alternatively, factor out 2:= (3π -2*2 ln(3/2))/13= same as before.Alternatively, write ln(9/4) as ln9 -ln4=2 ln3 -2 ln2, so:= (3π -2*(2 ln3 -2 ln2))/13= (3π -4 ln3 +4 ln2)/13Which can be expressed as:= (3π +4 ln2 -4 ln3)/13= (3π +4(ln2 -ln3))/13= (3π -4 ln(3/2))/13Either form is acceptable. Let me check numerically to verify.Compute approximate value:Compute 3π ≈9.4248ln(3/2)≈0.4055, so 4 ln(3/2)≈1.622Thus, 3π -4 ln(3/2)≈9.4248 -1.622≈7.8028Divide by13: ≈7.8028 /13≈0.6002Alternatively, compute original integral numerically.Let me compute the original integral numerically using approximate methods to check if the answer is approximately0.6.Compute the integrand f(θ)= [2 sin θ +3 cos θ -3]/[13 cos θ -5]Let me check at several points:At θ=0:Numerator=0 +3*1 -3=0Denominator=13*1 -5=8So f(0)=0/8=0At θ=π:Numerator=0 +3*(-1) -3= -6Denominator=13*(-1)-5= -18f(π)= (-6)/(-18)=1/3≈0.333At θ=π/2:Numerator=2*1 +0 -3= -1Denominator=0 -5= -5f(π/2)= (-1)/(-5)=0.2At θ=arccos(5/13)≈1.176, the denominator is zero, but numerator is also zero. Let's compute near θ≈1.176:Let θ=1.176, compute cos θ≈5/13≈0.3846. Let’s take θ=1.176, compute f(θ):Numerator=2 sin θ +3 cos θ -3. sin θ=sqrt(1 -25/169)=12/13≈0.9231So numerator=2*(12/13)+3*(5/13)-3=24/13 +15/13 -39/13=0/13=0Therefore, at θ=1.176, the function is 0/0, which is the removable discontinuity.Therefore, the integrand is continuous everywhere on [0, π], with a hole at θ≈1.176. Therefore, the integral is proper.Numerically integrating this function from 0 to π:But to check if the analytical result ≈0.6 is correct, let's perform a rough numerical integral.Divide the interval [0, π] into four intervals: [0, π/4], [π/4, π/2], [π/2, 3π/4], [3π/4, π]. Compute the function at midpoints and use midpoint rule.First interval: midpoint θ=π/8≈0.3927Compute f(π/8):cos(π/8)≈0.9239, sin(π/8)≈0.3827Numerator≈2*0.3827 +3*0.9239 -3≈0.7654 +2.7717 -3≈0.5371Denominator≈13*0.9239 -5≈12.0107 -5≈7.0107f≈0.5371/7.0107≈0.0766Second interval: midpoint θ=3π/8≈1.178 (close to arccos(5/13)=1.176)Wait, actually midpoint of [π/4, π/2] is 3π/8≈1.178. But arccos(5/13)≈1.176, so very close. So this midpoint is near the singularity.Compute f(3π/8):cos(3π/8)≈0.3827, sin(3π/8)≈0.9239Numerator≈2*0.9239 +3*0.3827 -3≈1.8478 +1.1481 -3≈0.0 (approximately)Denominator≈13*0.3827 -5≈5 -5≈0. So near the removable discontinuity. But since we need to avoid exactly that point, but as it's a midpoint, the function value here is undefined, but since it's a removable discontinuity, we can assign the value 0. However, for numerical integration, it's better to take points away from the singularity.Alternatively, pick a slightly different θ. Let's take θ=1.2 (slightly past arccos(5/13)=1.176)cos(1.2)=0.3624, sin(1.2)=0.9320Numerator=2*0.9320 +3*0.3624 -3≈1.864 +1.0872 -3≈-0.0488Denominator=13*0.3624 -5≈4.7112 -5≈-0.2888f≈-0.0488/-0.2888≈0.169Third interval midpoint θ=5π/8≈1.9635cos(5π/8)=cos(π -3π/8)=-cos(3π/8)≈-0.3827, sin(5π/8)=sin(3π/8)≈0.9239Numerator=2*0.9239 +3*(-0.3827) -3≈1.8478 -1.1481 -3≈-2.3003Denominator=13*(-0.3827) -5≈-5 -5≈-10f≈-2.3003/-10≈0.23Fourth interval midpoint θ=7π/8≈2.7489cos(7π/8)=cos(π -π/8)= -cos(π/8)≈-0.9239, sin(7π/8)=sin(π/8)≈0.3827Numerator=2*0.3827 +3*(-0.9239) -3≈0.7654 -2.7717 -3≈-5.0063Denominator=13*(-0.9239) -5≈-12.0107 -5≈-17.0107f≈-5.0063/-17.0107≈0.294Now, using midpoint rule with four intervals:Each interval has width Δθ=π/4≈0.7854Approximate integral≈ Δθ*(0.0766 +0.169 +0.23 +0.294)=0.7854*(0.769)≈0.7854*0.769≈0.604Which is close to the analytical result≈0.6002. Therefore, the analytical result seems correct.Therefore, the answer is (3π -4 ln(3/2))/13. However, let's present it in the simplest form.Alternatively, as:frac{3pi - 4 ln frac{3}{2}}{13}Or using logarithmic properties:frac{3pi}{13} - frac{4}{13} ln frac{3}{2} = frac{3pi}{13} + frac{4}{13} ln frac{2}{3}Either way is acceptable, but the problem might prefer combining the terms over a common denominator. Let me check if there's a more simplified form.Alternatively, recall that ln(9/4) =2 ln(3/2), so:-2/13 * ln(9/4)= -4/13 ln(3/2)Thus, the integral is:(3π)/13 - (4/13) ln(3/2)= (3π -4 ln(3/2))/13Alternatively, factor out 1/13:frac{1}{13} left(3pi -4 ln frac{3}{2}right)This seems like a concise form. Therefore, the final answer is:boxed{dfrac{3pi - 4 ln dfrac{3}{2}}{13}}Alternatively, if we want to rationalize the logarithm terms, since ln(3/2) can be written as ln3 - ln2, so:boxed{dfrac{3pi - 4 (ln 3 - ln 2)}{13}} = boxed{dfrac{3pi - 4 ln 3 + 4 ln 2}{13}}But the problem might prefer the compact form with ln(3/2). Let me check standard integral answers; often, combining the logs is preferred. Since ln(9/4) was part of the initial expression, but we simplified it to 2 ln(3/2), so either form is acceptable. However, the problem might expect the answer in terms of ln(9/4). Let me re-express:Original I1 was (-2/13) ln(9/4), and the total integral is (-2/13) ln(9/4) +3π/13, so:frac{3pi -2 ln frac{9}{4}}{13}This is also a valid expression. Let me compute which form is simpler. Since 9/4 is (3/2)^2, so ln(9/4)=2 ln(3/2). Therefore, the two forms are equivalent. The answer can be written either way, but perhaps the problem expects the answer in the form with ln(9/4). Let me check:If we write it as (3π -2 ln(9/4))/13, that's also correct.But in the step-by-step calculation, I found:-2/13 ln(9/4) +3π/13= (3π -2 ln(9/4))/13Which is equivalent.Alternatively, in terms of ln(3/2):(3π -4 ln(3/2))/13Both are correct. The question is, which one is preferred. Since ln(9/4) is directly from the integral I1, maybe that's acceptable. However, often, simplifying the logarithm to have the smallest argument inside is preferred, so ln(3/2) instead of ln(9/4). Therefore, I would go with:boxed{dfrac{3pi - 4 ln dfrac{3}{2}}{13}}But just to confirm, let's see what the exact value is numerically:3π ≈9.424777964 ln(3/2)≈4*0.4054651081≈1.621860432Subtracting:9.42477796 -1.621860432≈7.802917528Divide by13:≈0.600224425Which matches the numerical approximation earlier. So this is correct.Alternatively, if written as (3π -2 ln(9/4))/13:ln(9/4)=0.81093021622 ln(9/4)=1.621860432Same result.Therefore, both forms are acceptable. However, since the problem might prefer the form with the least number of terms, the form with ln(3/2) is better. Therefore, I will present the answer as:boxed{dfrac{3pi - 4 ln dfrac{3}{2}}{13}}Final Answerboxed{dfrac{3pi - 4 ln dfrac{3}{2}}{13}}</think>To compute the integralint_{0}^{pi} frac{2 sin theta + 3 cos theta - 3}{13 cos theta - 5} , dtheta,we start by expressing the numerator in terms of the derivative of the denominator and the denominator itself. We decompose the numerator as follows:1. Decompose the numerator: - The numerator (2 sin theta + 3 cos theta - 3) is expressed as (A cdot (-13 sin theta) + B cdot (13 cos theta - 5) + C). - Solving for coefficients (A), (B), and (C), we find (A = -frac{2}{13}), (B = frac{3}{13}), and (C = -frac{24}{13}).2. Rewrite the integrand: - The integrand is split into three parts: [ frac{2 sin theta}{13 cos theta - 5} + frac{3}{13} - frac{24}{13(13 cos theta - 5)}. ]3. Evaluate each part separately: - First integral: [ int_{0}^{pi} frac{2 sin theta}{13 cos theta - 5} , dtheta = -frac{2}{13} left[ ln|13 cos theta - 5| right]_{0}^{pi} = -frac{2}{13} left( ln 18 - ln 8 right) = -frac{2}{13} ln frac{9}{4}. ] - Second integral: [ int_{0}^{pi} frac{3}{13} , dtheta = frac{3}{13} pi. ] - Third integral: [ int_{0}^{pi} frac{-24}{13(13 cos theta - 5)} , dtheta = 0 quad text{(due to symmetry and removable discontinuity)}. ]4. Combine results: - Combining the results of the three integrals, we get: [ -frac{2}{13} ln frac{9}{4} + frac{3pi}{13}. ] - Simplifying (ln frac{9}{4}) as (2 ln frac{3}{2}), the integral becomes: [ frac{3pi}{13} - frac{4}{13} ln frac{3}{2}. ]Thus, the final answer is:boxed{dfrac{3pi - 4 ln dfrac{3}{2}}{13}}]

🔑:Answer:frac{3 pi}{13}-frac{4}{13} log frac{3}{2}We havebegin{aligned}int_{0}^{pi} frac{2 sin theta+3 cos theta-3}{13 cos theta-5} mathrm{~d} theta & =2 int_{0}^{pi / 2} frac{2 sin 2 x+3 cos 2 x-3}{13 cos 2 x-5} mathrm{~d} x & =2 int_{0}^{pi / 2} frac{4 sin x cos x-6 sin ^{2} x}{8 cos ^{2} x-18 sin ^{2} x} mathrm{~d} x & =2 int_{0}^{pi / 2} frac{sin x(2 cos x-3 sin x)}{(2 cos x+3 sin x)(2 cos x-3 sin x)} mathrm{d} x & =2 int_{0}^{pi / 2} frac{sin x}{2 cos x+3 sin x}end{aligned}To compute the above integral we want to write sin x as a linear combination of the denominator and its derivative:begin{aligned}2 int_{0}^{pi / 2} frac{sin x}{2 cos x+3 sin x} & =2 int_{0}^{pi / 2} frac{-frac{1}{13}[-3(2 cos x+3 sin x)+2(3 cos x-2 sin x)]}{2 cos x+3 sin x} & =-frac{2}{13}left[int_{0}^{pi / 2}(-3)+2 int_{0}^{pi} frac{-2 sin x+3 cos x}{2 cos x+3 sin x}right] & =-frac{2}{13}left[-frac{3 pi}{2}+left.2 log (3 sin x+2 cos x)right|_{0} ^{pi / 2}right] & =-frac{2}{13}left[-frac{3 pi}{2}+2 log frac{3}{2}right] & =frac{3 pi}{13}-frac{4}{13} log frac{3}{2} .end{aligned}