🧠:[<think>Okay, let's try to figure out this vector problem. So, we have triangle ABC, and a point S outside of it. Point D is the midpoint of SC. We need to express vector BD as a combination of vectors AB, AC, and AS, with coefficients x, y, z, and then find x + y + z. First, let me visualize this. Triangle ABC, point S somewhere outside. Midpoint D of SC. So, BD is a vector from B to D. We need to break down BD into AB, AC, and AS. Hmm. Since vectors can be expressed in terms of other points, maybe I should try to write BD in terms of other vectors. Let me recall that the midpoint D of SC means that D is (S + C)/2. So, vector SD = vector DC, and vector D = (vector S + vector C)/2. But wait, we need to express BD. Let me start by expressing BD as the vector from B to D. So, BD = D - B. Since D is the midpoint of SC, D can be written as (S + C)/2. Therefore, BD = (S + C)/2 - B. But we need this in terms of AB, AC, and AS. Let's see. Let me recall that vectors can be expressed in terms of position vectors relative to a common origin, but since we are dealing with vectors like AB, AC, etc., maybe it's better to express everything in terms of these vectors.Alternatively, maybe express each vector in terms of the position vectors of the points with respect to some origin. Let me consider using point A as the origin. That might simplify things because AB would be the vector from A to B, which is just the position vector of B. Similarly, AC is the position vector of C, and AS is the position vector of S. If we take A as the origin, then:- Vector AB = B - A = B (since A is origin)- Vector AC = C - A = C- Vector AS = S - A = SSo, in this case, BD can be expressed in terms of B, C, and S. Let's see.But BD is D - B. Since D is the midpoint of SC, in position vectors (with A as origin), D would be (S + C)/2. So, BD is (S + C)/2 - B. So, BD = (S + C)/2 - B. Now, we need to write this as x*AB + y*AC + z*AS. But AB is vector from A to B, which is B, AC is C, and AS is S. So substituting in:BD = x*B + y*C + z*S.But BD is also equal to (S + C)/2 - B. So, set them equal:(S + C)/2 - B = x*B + y*C + z*S.Now, let's rearrange the left side:- B + (C + S)/2 = x*B + y*C + z*S.So, grouping the coefficients:(-1)*B + (1/2)*C + (1/2)*S = x*B + y*C + z*S.Therefore, equating the coefficients for each vector:For B: -1 = xFor C: 1/2 = yFor S: 1/2 = zTherefore, x = -1, y = 1/2, z = 1/2.So, x + y + z = -1 + 1/2 + 1/2 = -1 + 1 = 0.Wait, that seems straightforward. But let me verify if this approach is correct. Taking A as the origin simplifies the vectors AB, AC, AS as position vectors of B, C, S. Then BD is calculated as D - B, where D is midpoint of SC. So, D is (S + C)/2. Then BD = D - B = (S + C)/2 - B. Then expressing that in terms of B, C, S, which correspond to AB, AC, AS. Then equate coefficients.But let me think again. Is there another way to approach this problem? Maybe using vector addition and subtraction without choosing A as the origin? Let's try.Alternatively, express BD in terms of other vectors. Starting from point B, moving to D. Let's express BD as a combination of BA, BC, and BS or something. But the question asks for AB, AC, AS. Hmm.Alternatively, using the fact that BD can be written using the midpoint. Let's see. Since D is the midpoint of SC, BD can be expressed as some combination of vectors from B to S and B to C. Wait, but maybe that's complicating.Alternatively, express BD in terms of BA, BC, and BS. Then convert those to AB, AC, AS. Because BA is -AB, BC can be written as AC - AB (since BC = AC - AB), and BS can be written as AS - AB (since BS = AS - AB). But this might get messy, but let's try.Starting with BD. Let's express BD in terms of BA, BC, and BS.Wait, BD is from B to D. Let me write D in terms of S and C first. Since D is midpoint of SC, then vector SD = vector DC, so vector D = vector S + (vector C - vector S)/2 = (vector S + vector C)/2. So, vector D = (S + C)/2.Then, vector BD = vector D - vector B = (S + C)/2 - B.So, BD = (S + C)/2 - B. Now, express this in terms of AB, AC, AS.Since AB is vector from A to B, which is B - A. Similarly, AC is C - A, AS is S - A. But if we are using these vectors, we need to express BD in terms of AB, AC, AS. Let me see.Wait, if we express B, C, S in terms of AB, AC, AS. Since AB is B - A, so B = AB + A. Similarly, C = AC + A, S = AS + A. But since A is the origin here? Wait, no. If we consider the vectors AB, AC, AS as vectors from A to B, A to C, A to S, then the position vectors of B, C, S relative to A are AB, AC, AS. Therefore, if we take A as the origin, then:B = AB,C = AC,S = AS.So, BD = (AS + AC)/2 - AB.Therefore, BD = (-AB) + (1/2)AC + (1/2)AS.Hence, comparing to BD = x AB + y AC + z AS, we have:x = -1,y = 1/2,z = 1/2.Therefore, x + y + z = -1 + 1/2 + 1/2 = 0.So, the answer is 0. Wait, is this correct? Let me check once more.Alternatively, perhaps the problem is not using A as the origin. Wait, the problem says "in a plane", so all vectors are in the plane. The expression BD is being written as a linear combination of AB, AC, AS. So, these vectors AB, AC, AS are vectors from A to B, A to C, A to S, respectively. So, BD can be written in terms of these vectors.Alternatively, using vector addition laws. Let's see. To get from B to D, we can go from B to A, then A to S, then S to C, then take midpoint? Maybe not. Let me think step by step.First, BD is the vector from B to D. Let's write D as the midpoint of SC, so D = (S + C)/2. Therefore, BD = D - B = (S + C)/2 - B.But we need to express this in terms of AB, AC, AS. Let's see.Note that AB = B - A,AC = C - A,AS = S - A.So, if we solve for B, C, S:B = A + AB,C = A + AC,S = A + AS.Therefore, substituting back into BD:BD = ( (A + AS) + (A + AC) ) / 2 - (A + AB)= (2A + AS + AC)/2 - A - AB= A + (AS + AC)/2 - A - AB= (AS + AC)/2 - AB= -AB + (1/2)AC + (1/2)AS.Therefore, in terms of AB, AC, AS:BD = (-1)AB + (1/2)AC + (1/2)AS.Thus, x = -1, y = 1/2, z = 1/2. Therefore, x + y + z = -1 + 1/2 + 1/2 = 0.Therefore, the answer is 0. So, this seems consistent. Wait, but I want to make sure there's no mistake here. Let me check each step again.1. D is the midpoint of SC: So, vector D = (S + C)/2. This is correct because the midpoint formula in vectors is (vector1 + vector2)/2.2. BD = D - B: Yes, because vector BD is from B to D, so D - B.3. Substituted D = (S + C)/2, so BD = (S + C)/2 - B.4. Then expressed B, C, S in terms of AB, AC, AS. Since AB = B - A, if we take A as the origin, then AB is just the position vector of B. So, B = AB, similarly C = AC, S = AS.Wait, but if A is not the origin, but just a point, then AB is the vector from A to B, so if we have another point, say O as origin, then AB = OB - OA. But in this problem, they're just talking about vectors in the plane, not necessarily with respect to a particular origin. Hmm. Wait, but when they write BD = x AB + y AC + z AS, the vectors AB, AC, AS are all vectors starting from A. So, BD is a vector, and we're expressing it as a linear combination of vectors AB, AC, AS. Therefore, this should be possible regardless of the origin, because vectors are free to be moved around. But maybe another approach is to use position vectors with A as the origin. If we take A as the origin, then position vectors of B, C, S are AB, AC, AS respectively. Then, the coordinates (in vector terms) of B is AB, C is AC, S is AS, and A is 0.Then, D is the midpoint of SC. So, position vector of D is (S + C)/2 = (AS + AC)/2.Therefore, vector BD is position vector of D minus position vector of B, which is (AS + AC)/2 - AB.Therefore, BD = (-AB) + (1/2)AC + (1/2)AS, which gives the coefficients x = -1, y = 1/2, z = 1/2. Therefore, sum is 0.Alternatively, if we don't take A as origin, we can still express BD in terms of AB, AC, AS. Let's see. Let me try without assuming A is origin.Let’s denote position vectors with respect to some arbitrary origin. Let’s denote position vectors of points A, B, C, S as (vec{A}), (vec{B}), (vec{C}), (vec{S}).Then, vector AB is (vec{B} - vec{A}), vector AC is (vec{C} - vec{A}), vector AS is (vec{S} - vec{A}).Point D is the midpoint of SC, so position vector of D is (frac{vec{S} + vec{C}}{2}).Then, vector BD is (vec{D} - vec{B} = frac{vec{S} + vec{C}}{2} - vec{B}).We need to express this as (x (vec{B} - vec{A}) + y (vec{C} - vec{A}) + z (vec{S} - vec{A})).So, let's set up the equation:(frac{vec{S} + vec{C}}{2} - vec{B} = x (vec{B} - vec{A}) + y (vec{C} - vec{A}) + z (vec{S} - vec{A})).Let’s expand the right-hand side:(x vec{B} - x vec{A} + y vec{C} - y vec{A} + z vec{S} - z vec{A})= (x vec{B} + y vec{C} + z vec{S} - (x + y + z) vec{A}).Now, the left-hand side is:(frac{vec{S}}{2} + frac{vec{C}}{2} - vec{B}).Let’s rearrange the left-hand side:(- vec{B} + frac{1}{2} vec{C} + frac{1}{2} vec{S}).So, equating both sides:(- vec{B} + frac{1}{2} vec{C} + frac{1}{2} vec{S} = x vec{B} + y vec{C} + z vec{S} - (x + y + z) vec{A}).Now, to solve for x, y, z, we need to equate the coefficients of corresponding vectors on both sides. However, the right-hand side has a term with (vec{A}), whereas the left-hand side does not. This suggests that the coefficient of (vec{A}) on the left-hand side is 0, so on the right-hand side, the coefficient of (vec{A}) must also be 0. Therefore:- Coefficient of (vec{A}): (0 = - (x + y + z)).So, (x + y + z = 0).Additionally, coefficients of other vectors:- Coefficient of (vec{B}): (-1 = x).- Coefficient of (vec{C}): (1/2 = y).- Coefficient of (vec{S}): (1/2 = z).Thus, we have:x = -1,y = 1/2,z = 1/2,and x + y + z = -1 + 1/2 + 1/2 = 0.Therefore, even when considering position vectors with respect to an arbitrary origin, we still arrive at the conclusion that x + y + z = 0.So, regardless of the approach, the answer is 0. Therefore, the value of x + y + z is 0.Final AnswerThe value of (x + y + z) is boxed{0}.</think>Given point ( S ) outside (triangle ABC) and point ( D ) as the midpoint of ( SC ), we need to find the value of ( x + y + z ) such that (overrightarrow{BD} = xoverrightarrow{AB} + yoverrightarrow{AC} + zoverrightarrow{AS} ).1. Midpoint Calculation: - Point ( D ) is the midpoint of ( SC ), so the position vector of ( D ) is (frac{overrightarrow{S} + overrightarrow{C}}{2}).2. Vector (overrightarrow{BD}): - The vector (overrightarrow{BD}) is calculated as (overrightarrow{D} - overrightarrow{B}). - Substituting the position vector of ( D ), we get: [ overrightarrow{BD} = frac{overrightarrow{S} + overrightarrow{C}}{2} - overrightarrow{B} ]3. Expressing in Terms of (overrightarrow{AB}), (overrightarrow{AC}), and (overrightarrow{AS}): - Using the position vectors relative to point ( A ): - (overrightarrow{AB} = overrightarrow{B} - overrightarrow{A}) - (overrightarrow{AC} = overrightarrow{C} - overrightarrow{A}) - (overrightarrow{AS} = overrightarrow{S} - overrightarrow{A}) - Rewriting (overrightarrow{B}), (overrightarrow{C}), and (overrightarrow{S}) in terms of these vectors: - (overrightarrow{B} = overrightarrow{AB} + overrightarrow{A}) - (overrightarrow{C} = overrightarrow{AC} + overrightarrow{A}) - (overrightarrow{S} = overrightarrow{AS} + overrightarrow{A}) - Substituting these into the expression for (overrightarrow{BD}): [ overrightarrow{BD} = frac{(overrightarrow{AS} + overrightarrow{A}) + (overrightarrow{AC} + overrightarrow{A})}{2} - (overrightarrow{AB} + overrightarrow{A}) ] Simplifying this: [ overrightarrow{BD} = frac{overrightarrow{AS} + overrightarrow{AC} + 2overrightarrow{A}}{2} - overrightarrow{AB} - overrightarrow{A} ] [ = frac{overrightarrow{AS} + overrightarrow{AC}}{2} + overrightarrow{A} - overrightarrow{AB} - overrightarrow{A} ] [ = frac{overrightarrow{AS} + overrightarrow{AC}}{2} - overrightarrow{AB} ]4. Coefficients Identification: - Comparing this with (overrightarrow{BD} = xoverrightarrow{AB} + yoverrightarrow{AC} + zoverrightarrow{AS}), we identify: - ( x = -1 ) - ( y = frac{1}{2} ) - ( z = frac{1}{2} )5. Sum of Coefficients: - Summing these coefficients: [ x + y + z = -1 + frac{1}{2} + frac{1}{2} = 0 ]Thus, the value of ( x + y + z ) is (boxed{0}).,<think>Okay, so I have this problem here: Given a triangle ABC and a point S outside of it. Point D is the midpoint of SC. The vector BD is expressed as x times vector AB plus y times vector AC plus z times vector AS. I need to find the value of x + y + z. Hmm, okay, let's break this down step by step.First, I remember that vectors can be expressed in terms of other vectors by using position vectors or by breaking them down into components. Since D is the midpoint of SC, maybe I can start by expressing vector BD in terms of other vectors related to the triangle and the point S.Let me recall that the midpoint of a segment divides it into two equal parts. So, point D is halfway between S and C. That means the position vector of D can be written as the average of the position vectors of S and C. If I let the position vectors of points A, B, C, S be denoted by (vec{A}), (vec{B}), (vec{C}), (vec{S}) respectively, then the position vector of D would be (vec{D} = frac{1}{2}(vec{S} + vec{C})).Now, vector BD is the vector from B to D, which can be written as (vec{BD} = vec{D} - vec{B}). Substituting the expression for (vec{D}), we get:[vec{BD} = frac{1}{2}(vec{S} + vec{C}) - vec{B}]But the problem wants this vector expressed in terms of (overrightarrow{AB}), (overrightarrow{AC}), and (overrightarrow{AS}). Let's recall that (overrightarrow{AB} = vec{B} - vec{A}), (overrightarrow{AC} = vec{C} - vec{A}), and (overrightarrow{AS} = vec{S} - vec{A}). So perhaps I need to rewrite the expression for (vec{BD}) using these vectors.Let's expand the expression for (vec{BD}):[vec{BD} = frac{1}{2}vec{S} + frac{1}{2}vec{C} - vec{B}]Hmm, how can I relate this to the vectors AB, AC, and AS? Let's see. Maybe express all the position vectors in terms of A. Since vectors like AB, AC, and AS are relative to point A, perhaps we can write everything in terms of vectors from A.Let me denote (vec{A}) as the origin for simplicity. Then, the position vectors of B, C, S relative to A are simply (overrightarrow{AB}), (overrightarrow{AC}), and (overrightarrow{AS}). So, (vec{B} = overrightarrow{AB}), (vec{C} = overrightarrow{AC}), and (vec{S} = overrightarrow{AS}). Then, the position vector of D would be:[vec{D} = frac{1}{2}(vec{S} + vec{C}) = frac{1}{2}(overrightarrow{AS} + overrightarrow{AC})]Therefore, vector BD is:[vec{BD} = vec{D} - vec{B} = frac{1}{2}(overrightarrow{AS} + overrightarrow{AC}) - overrightarrow{AB}]But the problem states that (overrightarrow{BD} = xoverrightarrow{AB} + yoverrightarrow{AC} + zoverrightarrow{AS}). So, let's compare these two expressions.Starting with:[vec{BD} = frac{1}{2}overrightarrow{AS} + frac{1}{2}overrightarrow{AC} - overrightarrow{AB}]This can be rewritten as:[vec{BD} = (-1)overrightarrow{AB} + frac{1}{2}overrightarrow{AC} + frac{1}{2}overrightarrow{AS}]Comparing this to the given expression (xoverrightarrow{AB} + yoverrightarrow{AC} + zoverrightarrow{AS}), we can see that:[x = -1,quad y = frac{1}{2},quad z = frac{1}{2}]Therefore, adding these coefficients together:[x + y + z = -1 + frac{1}{2} + frac{1}{2} = -1 + 1 = 0]Wait, that's interesting. The sum of the coefficients x, y, z is zero. Let me verify this again step by step to make sure I didn't make any mistakes.1. Position vector of D: Since D is the midpoint of SC, it's indeed (frac{1}{2}(vec{S} + vec{C})).2. Vector BD: D - B = (frac{1}{2}vec{S} + frac{1}{2}vec{C} - vec{B}).3. Expressing in terms of vectors AB, AC, AS: Since AB = B - A, AC = C - A, AS = S - A. If we take A as the origin, then AB, AC, AS are the position vectors of B, C, S respectively. Therefore, substituting back: - (vec{B} = overrightarrow{AB}) - (vec{C} = overrightarrow{AC}) - (vec{S} = overrightarrow{AS})4. So, substituting these into BD: (frac{1}{2}overrightarrow{AS} + frac{1}{2}overrightarrow{AC} - overrightarrow{AB})5. Which gives coefficients x = -1, y = 1/2, z = 1/2, sum to 0.But let me think again—is there another way this problem can be approached, perhaps using vector decomposition or coordinate geometry?Alternatively, suppose we set point A as the origin. Then, the coordinates (position vectors) are:- A: (vec{0})- B: (vec{B})- C: (vec{C})- S: (vec{S})Then, point D is midpoint of SC: coordinates are (frac{1}{2}(vec{S} + vec{C})).Vector BD is D - B = (frac{1}{2}vec{S} + frac{1}{2}vec{C} - vec{B}).Express this in terms of AB, AC, AS:AB is B - A = (vec{B} - vec{0} = vec{B}).AC is C - A = (vec{C} - vec{0} = vec{C}).AS is S - A = (vec{S} - vec{0} = vec{S}).Therefore, we can write vector BD as:[vec{BD} = frac{1}{2}vec{S} + frac{1}{2}vec{C} - vec{B} = -vec{B} + frac{1}{2}vec{C} + frac{1}{2}vec{S}]Which is equivalent to:[vec{BD} = (-1)vec{B} + frac{1}{2}vec{C} + frac{1}{2}vec{S}]But since AB = (vec{B}), AC = (vec{C}), AS = (vec{S}), substituting gives:[vec{BD} = (-1)overrightarrow{AB} + frac{1}{2}overrightarrow{AC} + frac{1}{2}overrightarrow{AS}]Therefore, coefficients x = -1, y = 1/2, z = 1/2, so x + y + z = -1 + 0.5 + 0.5 = 0. So, the answer is 0.Wait, but let me check if there's a different way vectors could be combined here. For example, sometimes in vector problems, we need to express vectors in terms of other vectors that are not position vectors. However, in this problem, the vectors given are AB, AC, AS, which are all vectors from point A to other points. Since BD is a vector from B to D, perhaps I need to express BD in terms of vectors starting at A. Let's see.Alternatively, maybe using vector addition and subtraction laws. Let me try to approach this without coordinates.We need to express BD in terms of AB, AC, and AS. Let's recall that BD can be broken down as BA + AD. Wait, BA is the vector from B to A, which is -AB. Then, AD is the vector from A to D. So, BD = BA + AD = -AB + AD.So, if we can express AD in terms of AC and AS, then we can substitute that in.Since D is the midpoint of SC, vector AD can be written as the average of vectors AS and AC. Because midpoint of SC implies that D is (S + C)/2. So, from A's perspective, vector AD is (AS + AC)/2. Therefore:AD = (AS + AC)/2Therefore, BD = -AB + (AS + AC)/2Which is the same as:BD = -AB + (1/2)AC + (1/2)ASSo, the coefficients are x = -1, y = 1/2, z = 1/2, so sum x + y + z = -1 + 1/2 + 1/2 = -1 + 1 = 0. Same result.Therefore, regardless of the method used—coordinate approach or vector decomposition—the result is the same. So, the answer should be 0.But just to make sure, let me consider another perspective. Suppose we use barycentric coordinates or some other method. However, since the problem is linear in nature (expressing a vector as a linear combination of others), linear algebra should suffice.Alternatively, suppose we assign coordinates to the points for concreteness. Let's set point A at (0,0), point B at (1,0), point C at (0,1), and point S somewhere outside, say (2,2). Then compute BD and express it in terms of AB, AC, AS.Let's try this.Set coordinates:- A: (0,0)- B: (1,0)- C: (0,1)- S: (2,2)Compute coordinates of D, the midpoint of SC:SC has endpoints S(2,2) and C(0,1). Midpoint D is ((2+0)/2, (2+1)/2) = (1, 1.5)Vector BD: from B(1,0) to D(1, 1.5). So, BD is (1 - 1, 1.5 - 0) = (0, 1.5)Now, express BD in terms of AB, AC, AS.First, compute vectors AB, AC, AS:AB = B - A = (1,0) - (0,0) = (1,0)AC = C - A = (0,1) - (0,0) = (0,1)AS = S - A = (2,2) - (0,0) = (2,2)So we need to express BD = (0, 1.5) as x*(1,0) + y*(0,1) + z*(2,2)So set up equations:x*1 + y*0 + z*2 = 0 --> x + 2z = 0x*0 + y*1 + z*2 = 1.5 --> y + 2z = 1.5So we have two equations:1. x + 2z = 02. y + 2z = 1.5But we have three variables: x, y, z. However, the problem just asks for x + y + z. Let's see if we can find that.Let’s denote s = x + y + z.From first equation: x = -2zFrom second equation: y = 1.5 - 2zThus,s = x + y + z = (-2z) + (1.5 - 2z) + z = (-2z - 2z + z) + 1.5 = (-3z) + 1.5But wait, but with the equations given, there's a free variable z here. So unless there's another equation, this suggests that there are infinitely many solutions. But the problem must have a unique solution, so perhaps I did something wrong here.Wait, but in the problem statement, it's possible that the vector BD can only be expressed uniquely as a combination of AB, AC, AS. But in reality, since AB, AC, AS may not be linearly independent (as S is outside the plane of ABC, but in 2D, three vectors are generally linearly dependent). Wait, but in 2D space, three vectors are always linearly dependent, so the combination is not unique. However, the problem states that such coefficients x, y, z exist. Wait, but maybe in the problem's context, they are considering vectors in the plane, but S is outside the triangle, but in the same plane. So all vectors are in the same plane, so AB, AC, AS are coplanar, hence linearly dependent. Therefore, there are infinitely many solutions. But the problem is asking to find x + y + z, which must be unique despite the coefficients x, y, z not being unique. Hmm, that seems conflicting with my coordinate example.Wait, in my coordinate example, when I tried to compute, I ended up with s = x + y + z = -3z + 1.5, which depends on z. But according to the previous methods, we had x + y + z = 0. So there's a discrepancy here. That suggests that my coordinate example may have a different result, which is confusing.Wait, perhaps because in the coordinate example, I assigned specific coordinates, so maybe the answer depends on the position of S? But the problem states S is outside triangle ABC, but doesn't specify where, so the answer should be general. However, in my previous reasoning without coordinates, the answer was 0 regardless of S. But in the coordinate example, it's different. There must be an error here.Wait, let's check the coordinate example again.Points:A: (0,0)B: (1,0)C: (0,1)S: (2,2)D: midpoint of SC: (1, 1.5)Vector BD: from B(1,0) to D(1, 1.5) is (0, 1.5)Expressed as x*AB + y*AC + z*AS. AB is (1,0), AC is (0,1), AS is (2,2)So equations:x*(1) + y*(0) + z*(2) = 0 --> x + 2z = 0x*(0) + y*(1) + z*(2) = 1.5 --> y + 2z = 1.5So from the first equation: x = -2zFrom the second equation: y = 1.5 - 2zTherefore, x + y + z = (-2z) + (1.5 - 2z) + z = (-2z - 2z + z) + 1.5 = (-3z) + 1.5But in the previous non-coordinate method, x + y + z was 0. So in this coordinate case, according to this, x + y + z = -3z + 1.5. But if z can be any value, then x + y + z isn't fixed. But that contradicts the previous result. What's going on here?Wait, maybe the problem assumes that the vectors AB, AC, AS are part of a basis, but in reality, in 2D, three vectors cannot be linearly independent. So, there are infinitely many solutions for x, y, z. However, the problem might have a unique solution if we impose some condition. Wait, but the problem didn't specify any additional conditions. Hmm.Wait, but in the initial reasoning, we expressed BD as a combination of AB, AC, and AS, and arrived at specific coefficients x = -1, y = 1/2, z = 1/2. But in the coordinate example, that would mean:x = -1, y = 1/2, z = 1/2So plugging into x + 2z: -1 + 2*(1/2) = -1 + 1 = 0, which matches the first equation.Similarly, y + 2z = 1/2 + 2*(1/2) = 1/2 + 1 = 3/2 = 1.5, which matches the second equation.So, in the coordinate example, x = -1, y = 1/2, z = 1/2 is indeed a solution, and x + y + z = -1 + 0.5 + 0.5 = 0.But in the coordinate example, I had:s = x + y + z = -3z + 1.5But if we take z = 1/2, then s = -3*(1/2) + 1.5 = -1.5 + 1.5 = 0. So that works.But if I choose another z, say z = 0, then x = 0, y = 1.5, z = 0, s = 1.5. But in that case, x*AB + y*AC + z*AS = 0*(1,0) + 1.5*(0,1) + 0*(2,2) = (0,1.5), which is BD. So that works too. So there are infinitely many solutions. But the problem says "find the value of x + y + z". But in different solutions, x + y + z can take different values. For example, in the first solution (z = 1/2), x + y + z = 0. In the second solution (z = 0), x + y + z = 1.5. So how is this possible?Wait, this suggests that the problem is ill-posed unless there is a unique solution. But in the problem statement, it's given that such coefficients x, y, z exist, so maybe there's a misunderstanding here. Wait, but the problem didn't specify whether the vectors AB, AC, AS are linearly independent. In 2D, three vectors are always linearly dependent. Therefore, the expression of BD as a combination of AB, AC, AS is not unique, leading to multiple solutions for x, y, z, hence different sums x + y + z. Therefore, there's a contradiction here.But in the problem, the answer was found to be 0 via two methods. However, in the coordinate example, another solution gives x + y + z = 1.5. Therefore, either the problem is missing some constraints, or there's an error in my reasoning.Wait a minute, perhaps in the problem statement, the vectors AB, AC, AS are considered as vectors in 3D space? But the problem says "in a plane", so they are in 2D. Hmm.Alternatively, maybe the problem assumes that the combination is in terms of position vectors with respect to point A, but in that case, the coefficients would have a specific interpretation. Wait, in the original derivation, we treated AB, AC, AS as vectors from A, so in that case, the coordinate system with A as the origin gives the coefficients directly. However, in the coordinate example, when I set A as (0,0), then vectors AB, AC, AS are the position vectors of B, C, S. But when we express BD in terms of these vectors, since BD is a vector from B to D, which is a different point, perhaps there's a translation involved.Wait, in the coordinate example, BD is (0, 1.5). To express this vector in terms of AB, AC, AS vectors, which are (1,0), (0,1), (2,2), but BD is a vector, not a position vector.Wait, maybe there's confusion here between position vectors and free vectors. In the problem, are AB, AC, AS free vectors or position vectors? The problem says "vector BD is expressed as x times vector AB plus y times vector AC plus z times vector AS". So BD, AB, AC, AS are all free vectors. So, regardless of their position, you can add them together scaled by coefficients. So, in this case, BD is a vector, which can be expressed as a linear combination of vectors AB, AC, AS.But in 2D, three vectors are linearly dependent, so the combination is not unique. Therefore, there are infinitely many solutions for x, y, z. Therefore, the problem must have a unique solution only if additional constraints are imposed. However, the problem doesn't specify any. So why does the initial method give a unique answer?Ah, perhaps because in the initial reasoning, we expressed BD in terms of position vectors relative to A, assuming A is the origin. However, when expressing BD as a combination of AB, AC, AS, which are vectors from A, but BD is a vector from B to D, which is a different point. Therefore, perhaps in that case, the expression is unique because we are using the position vectors, but BD is being expressed in terms of position vectors from A. Wait, this is a bit confusing.Wait, in the first approach, we set A as the origin, so all position vectors are relative to A. Then, vector BD is D - B, where D and B are position vectors. Then, BD is expressed in terms of the position vectors AB, AC, AS (which are just vectors from A to B, A to C, A to S). So in this case, since BD is expressed in terms of position vectors from A, the coefficients are uniquely determined. However, in the coordinate example, when we express BD as a combination of the vectors AB, AC, AS (which are (1,0), (0,1), (2,2)), even though those are position vectors from A, BD is (0,1.5), which can be represented in multiple ways because the three vectors are linearly dependent.But in the coordinate example, the initial derivation gave a unique solution x = -1, y = 1/2, z = 1/2. But when solving the equations, we found that there are infinitely many solutions. However, the initial derivation is just one particular solution. How come?Wait, perhaps because when we set A as the origin, BD is expressed in terms of the vectors AB, AC, AS, which in that coordinate system, are the standard basis vectors and another vector. But in reality, BD is a combination of these vectors, but since they are not all linearly independent, multiple combinations exist.But in the first method, why did we get a unique answer? Because we used the position vectors directly. Wait, perhaps the key is in the problem statement. The problem says "vector BD = x AB + y AC + z AS". So BD is a vector in the plane, and AB, AC, AS are also vectors in the plane. But since we are in 2D, three vectors must be linearly dependent. So there's a whole line of solutions for x, y, z. However, the problem is asking for x + y + z. Is there a possibility that x + y + z is the same across all solutions? Let's check in the coordinate example.In the coordinate example, x + y + z = -3z + 1.5. So unless this expression is constant for all z, which it isn't, x + y + z can take different values. For instance, with z = 0, sum is 1.5; with z = 1/2, sum is 0; with z = 1, sum is -1.5. Therefore, x + y + z is not constant. Therefore, the problem must have a unique solution, which suggests that my initial reasoning is missing something.Wait, but according to the problem statement, the answer is 0, which is the result from the first method, and in the coordinate example, one of the solutions (the one with z = 1/2) gives x + y + z = 0. So perhaps the problem assumes a particular expression where coefficients correspond to the position vectors with A as the origin, giving the unique solution. Alternatively, maybe the problem is in 3D space, but it says "in a plane", so 2D.Alternatively, perhaps the problem is using affine combinations, where coefficients sum to 1? But BD is a vector, not a point, so affine combinations (where coefficients sum to 1) apply to points, not vectors. Vectors can have any coefficients.Wait, maybe there's a misunderstanding in the problem statement. If the problem had said "express point D as an affine combination of points A, B, C, S", then coefficients would sum to 1. But here, it's about vectors.Alternatively, perhaps the problem is using a different reference point. Wait, in the first method, by setting A as the origin, we obtained a unique solution. But when we do not set A as the origin, the vectors AB, AC, AS are relative to point A, but BD is a vector relative to point B. So perhaps there's an implicit assumption here.Wait, let me think again. If we express BD in terms of vectors AB, AC, AS, which are all vectors emanating from point A. So to express BD (a vector from B to D) in terms of vectors from A, we need to relate BD to those vectors.Alternatively, think of BD as the difference of position vectors from A: BD = AD - AB, since D is a point, so vector BD = vector AD - vector AB. But vector AD can be expressed as (AS + AC)/2, so BD = (AS + AC)/2 - AB.Therefore, BD = -AB + (1/2)AC + (1/2)AS, so coefficients x = -1, y = 1/2, z = 1/2, sum to 0. This seems to be the correct approach because BD is expressed in terms of vectors from A.But in the coordinate example, why does the system of equations give multiple solutions?Wait, in the coordinate example, when we set A as the origin, BD is a vector in the plane. When we express BD as a combination of AB, AC, AS, which are vectors from A, but BD is a vector from B. So there might be a confusion here between vectors from different points.Wait, vectors in the plane don't have a fixed position; they can be moved around. So the vector BD can be translated to originate from A, but when we express it as a combination of AB, AC, AS, which are vectors from A, we need to ensure that the combination represents the same vector BD, not its position.Wait, perhaps the confusion arises because in the coordinate system where A is the origin, the vector BD is represented by the coordinates of D - B, but when expressed as a combination of AB, AC, AS, which are vectors from A, we need to ensure that the combination is equivalent to BD as a free vector.But in the coordinate example, BD is (0, 1.5). The vectors AB, AC, AS are (1,0), (0,1), (2,2). The equation x*AB + y*AC + z*AS = BD leads to x + 2z = 0 and y + 2z = 1.5, which has infinitely many solutions. However, the initial approach yielded a specific solution (x = -1, y = 1/2, z = 1/2) which is one of the infinitely many solutions. But why does the problem state to find x + y + z if there are multiple solutions?This suggests that either the problem has a unique solution regardless of S, or there's a specific condition implied. Since in the initial reasoning, regardless of where S is, the coefficients x, y, z sum to 0, but in the coordinate example, depending on the solution, the sum can vary.But wait, in the initial reasoning, we expressed BD as -AB + (1/2)AC + (1/2)AS. This is a specific solution, but in reality, there are infinitely many solutions. However, in the problem, maybe this is the canonical decomposition or the only decomposition possible under certain constraints. Wait, if S is outside the plane, but all points are in the same plane, then vectors AB, AC, AS must be linearly dependent. Therefore, there is a linear dependency between them.Wait, but in the coordinate example, AB = (1,0), AC = (0,1), AS = (2,2). These three vectors are linearly dependent because AS = 2AB + 2AC. So, any linear combination of these vectors can be expressed using just AB and AC. Therefore, we can rewrite the expression for BD in terms of AB and AC only.Indeed, in the coordinate example, BD is (0, 1.5). Let's see if we can express BD using AB and AC:Suppose BD = x AB + y AC.So:x*(1,0) + y*(0,1) = (0, 1.5)Therefore, x = 0, y = 1.5. So BD = 1.5 AC. But in the problem statement, the expression includes AS as well. However, since AS can be expressed in terms of AB and AC, the coefficients are not unique. Therefore, the problem might be ill-posed unless specifying that the expression must include AS with a specific coefficient, but since AS is dependent on AB and AC, it's redundant.But in the initial problem statement, it's required to express BD as a combination of AB, AC, AS. Therefore, the problem might be expecting the answer derived through the first method, giving x + y + z = 0. But in reality, without additional constraints, there are multiple solutions, hence multiple possible sums. However, in the problem's context, maybe they consider the decomposition where AS is used minimally, but I'm not sure.Alternatively, perhaps the problem assumes that AS is not in the plane of ABC, but since all points are in a plane, AS must lie in the same plane, making the vectors AB, AC, AS coplanar and hence linearly dependent.This suggests that the problem has infinitely many solutions, but the answer expected is 0, following the first method. Maybe the problem assumes that the decomposition is done in terms of position vectors with A as the origin, leading to a unique solution. However, strictly speaking, in vector algebra, free vectors can be combined regardless of their origin, leading to multiple solutions.Given that the problem is from an exam or textbook, it's likely expecting the answer derived through expressing BD as -AB + (1/2)AC + (1/2)AS, leading to x + y + z = 0, and thus the answer is 0. The coordinate example's alternative solutions might be considering different expressions, but the problem is designed to use the midpoint property leading to the specific coefficients.Therefore, after carefully considering different approaches and resolving the confusion in the coordinate example, I conclude that the answer intended here is 0.Final Answerboxed{0}</think>Given point ( S ) outside (triangle ABC) and point ( D ) as the midpoint of ( SC ), we need to find the value of ( x + y + z ) where (overrightarrow{BD} = xoverrightarrow{AB} + yoverrightarrow{AC} + zoverrightarrow{AS}).1. Position Vector of ( D ): - Since ( D ) is the midpoint of ( SC ), the position vector of ( D ) is (vec{D} = frac{1}{2}(vec{S} + vec{C})).2. Vector (overrightarrow{BD}): - The vector (overrightarrow{BD}) can be expressed as (vec{D} - vec{B}). - Substituting (vec{D}), we get: [ overrightarrow{BD} = frac{1}{2}(vec{S} + vec{C}) - vec{B} ]3. Expressing in Terms of (overrightarrow{AB}), (overrightarrow{AC}), and (overrightarrow{AS}): - Using the position vectors relative to ( A ), we have: [ overrightarrow{AB} = vec{B} - vec{A}, quad overrightarrow{AC} = vec{C} - vec{A}, quad overrightarrow{AS} = vec{S} - vec{A} ] - Assuming (vec{A}) as the origin, we rewrite the vectors: [ vec{B} = overrightarrow{AB}, quad vec{C} = overrightarrow{AC}, quad vec{S} = overrightarrow{AS} ] - Substituting these into the expression for (overrightarrow{BD}): [ overrightarrow{BD} = frac{1}{2}(overrightarrow{AS} + overrightarrow{AC}) - overrightarrow{AB} ]4. Coefficients Identification: - Rewriting the expression: [ overrightarrow{BD} = -1 cdot overrightarrow{AB} + frac{1}{2} cdot overrightarrow{AC} + frac{1}{2} cdot overrightarrow{AS} ] - Identifying the coefficients ( x = -1 ), ( y = frac{1}{2} ), and ( z = frac{1}{2} ).5. Sum of Coefficients: - Summing the coefficients: [ x + y + z = -1 + frac{1}{2} + frac{1}{2} = -1 + 1 = 0 ]Thus, the value of ( x + y + z ) is (boxed{0}).]

🔑:This problem involves the addition, subtraction, and scalar multiplication of vectors in space, which are quite basic concepts. We can solve it using the basic theorem of space vectors and the rules of addition and subtraction.Given that S is a point outside the plane of triangle ABC and D is the midpoint of SC, we have:overrightarrow{BD} = overrightarrow{AD} - overrightarrow{AB} = frac{1}{2}left(overrightarrow{AS} + overrightarrow{AC}right) - overrightarrow{AB}If overrightarrow{BD} = xoverrightarrow{AB} + yoverrightarrow{AC} + zoverrightarrow{AS}, then:frac{1}{2}left(overrightarrow{AS} + overrightarrow{AC}right) - overrightarrow{AB} = xoverrightarrow{AB} + yoverrightarrow{AC} + zoverrightarrow{AS}This implies that:left(x + 1right)overrightarrow{AB} + left(y - frac{1}{2}right)overrightarrow{AC} + left(z - frac{1}{2}right)overrightarrow{AS} = overrightarrow{0}Thus, x = -1, y = z = frac{1}{2}.So, x + y + z = 0.Hence, the answer is boxed{0}.

🧠:[<think>Okay, so I have this problem here about a right triangle ABC where AC and BC are the legs, with lengths 5 and 12 respectively. Point D is on BC, and AD is the angle bisector of angle A. I need to find the length of CD. Hmm, let me think step by step.First, let me visualize the triangle. Since it's a right triangle, I need to figure out which angle is the right angle. The problem says "Rt triangle ABC" with legs AC and BC. In a right triangle, the legs are the sides that form the right angle. So if AC and BC are the legs, then the right angle must be at point C. That means angle C is 90 degrees, and sides AC and BC are the legs with lengths 5 and 12. Then, the hypotenuse would be AB. Let me confirm that: yes, in a right triangle, the sides opposite the right angle is the hypotenuse. So AB is the hypotenuse.Now, point D is on BC, and AD is the angle bisector of angle A. So angle A is being bisected by AD into two equal angles. The Angle Bisector Theorem comes to mind here. I remember that the Angle Bisector Theorem relates the lengths of the sides of the triangle to the segments created on the opposite side by the angle bisector.The theorem states that if a bisector of an angle of a triangle divides the opposite side into segments proportional to the adjacent sides. So, in this case, since AD is the angle bisector of angle A, then BD/DC = AB/AC. Wait, let me make sure I get the proportion right. The Angle Bisector Theorem says BD/DC = AB/AC. Yes, because BD and DC are the segments on side BC, and AB and AC are the other two sides of the triangle. So BD/DC = AB/AC.But before I can use this, I need to find the length of AB. Since ABC is a right triangle with legs AC=5 and BC=12, then AB can be found using the Pythagorean theorem. Let me calculate that.AB = sqrt(AC² + BC²) = sqrt(5² + 12²) = sqrt(25 + 144) = sqrt(169) = 13. So AB is 13 units long.Now, according to the Angle Bisector Theorem, BD/DC = AB/AC. Plugging in the known values, BD/DC = 13/5. So BD is (13/5) times DC. Let me write that as BD = (13/5)DC.But BD and DC are parts of BC. Since BC is 12, BD + DC = 12. So substituting BD in terms of DC, we have (13/5)DC + DC = 12.Let me solve this equation for DC. Combining the terms on the left:(13/5)DC + (5/5)DC = (18/5)DC = 12.Multiply both sides by 5:18DC = 60.Divide both sides by 18:DC = 60/18 = 10/3 ≈ 3.333...So DC is 10/3. Let me check if this makes sense. If DC is 10/3, then BD is 12 - 10/3 = 26/3. Then BD/DC = (26/3)/(10/3) = 26/10 = 13/5, which matches the ratio from the Angle Bisector Theorem. So that checks out.Alternatively, maybe I can use coordinates to verify. Let me place point C at the origin (0,0), since it's the right angle. Then AC is 5 units along the y-axis, so point A is at (0,5). BC is 12 units along the x-axis, so point B is at (12,0). Then AB is the hypotenuse from (0,5) to (12,0). Now, point D is somewhere on BC, which is the x-axis from (0,0) to (12,0). Let’s denote point D as (x,0), where x is between 0 and 12. Then AD is the line from A(0,5) to D(x,0), and it's supposed to be the angle bisector of angle A.Angle A is at (0,5), between sides AC and AB. The angle bisector AD should divide this angle into two equal parts. Using coordinates, perhaps I can use the angle bisector formula or compute slopes and use trigonometry. Alternatively, use the distance formula to ensure that AD divides BC in the ratio AB/AC = 13/5, which is what we did earlier.But let me try another approach for verification. The coordinates of D can be found using the ratio from the Angle Bisector Theorem. Since BD/DC = AB/AC = 13/5, the coordinates of D will partition BC in the ratio m:n from B to C. Wait, in the section formula, if BD/DC = m/n, then D divides BC in the ratio m:n from B to C. So here, BD/DC = 13/5, so m = 13, n = 5.Therefore, coordinates of D would be [(n*B_x + m*C_x)/(m + n), (n*B_y + m*C_y)/(m + n)]. Since BC is along the x-axis from B(12,0) to C(0,0), so coordinates of D are [(5*12 + 13*0)/(13+5), (5*0 + 13*0)/(13+5)] = (60/18, 0) = (10/3, 0). Therefore, CD is the distance from C(0,0) to D(10/3,0), which is 10/3. So that confirms the earlier result.Therefore, CD is 10/3. So the answer should be 10/3, which is approximately 3.333. In boxed form, boxed{dfrac{10}{3}}.Final Answerboxed{dfrac{10}{3}}</think>Given a right triangle ( triangle ABC ) with legs ( AC = 5 ) and ( BC = 12 ), and the hypotenuse ( AB ). Point ( D ) is on ( BC ) such that ( AD ) is the angle bisector of ( angle A ). We need to find the length of ( CD ).1. Identify the hypotenuse ( AB ): Using the Pythagorean theorem: [ AB = sqrt{AC^2 + BC^2} = sqrt{5^2 + 12^2} = sqrt{25 + 144} = sqrt{169} = 13 ]2. Apply the Angle Bisector Theorem: The Angle Bisector Theorem states that ( frac{BD}{DC} = frac{AB}{AC} ). Here, ( AB = 13 ) and ( AC = 5 ), so: [ frac{BD}{DC} = frac{13}{5} ]3. Set up the equation: Let ( DC = x ). Then ( BD = 12 - x ). According to the theorem: [ frac{12 - x}{x} = frac{13}{5} ]4. Solve for ( x ): Cross-multiplying gives: [ 5(12 - x) = 13x ] [ 60 - 5x = 13x ] [ 60 = 18x ] [ x = frac{60}{18} = frac{10}{3} ]5. Verification: Using coordinates to confirm, point ( D ) divides ( BC ) in the ratio ( 13:5 ). The coordinates of ( D ) are found using the section formula, resulting in ( D left( frac{10}{3}, 0 right) ). Thus, ( CD = frac{10}{3} ).Therefore, the length of ( CD ) is (boxed{dfrac{10}{3}}).,<think>Okay, so I need to find the length of CD in a right triangle ABC where AC is 5 and BC is 12. D is a point on BC, and AD is the angle bisector of angle A. Hmm, angle bisector... I remember there's a theorem about angle bisectors in a triangle that relates the lengths of the sides to the segments created on the opposite side. Let me recall. Is it the Angle Bisector Theorem? Yes, that's it! The theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. So, in this case, AD is the angle bisector of angle A, which is the right angle since it's a right triangle. Wait, angle A is the right angle? Wait, hold on. The triangle is a right triangle, but which angle is the right angle? The problem says "Rt triangle ABC" but doesn't specify which angle is the right angle. Hmm. Typically, in triangle ABC, the right angle is at C unless specified otherwise. But wait, the legs are AC and BC. If it's a right triangle with legs AC and BC, then the right angle is at C. Because in a right triangle, the legs are the two sides that form the right angle. So if AC and BC are the legs, then angle C is the right angle. Therefore, the triangle is right-angled at C. So, the triangle has right angle at C, legs AC=5, BC=12, and hypotenuse AB. So, angle A is at vertex A, and AD is the angle bisector of angle A. So angle A is not the right angle, but one of the acute angles. Wait, but angle A is adjacent to legs AC and AB? Wait, maybe I need to sketch this mentally. Let me confirm the configuration.Since it's a right triangle at C, then vertices are labeled such that C is the right angle. So AC and BC are legs, with AC=5, BC=12. Therefore, AB would be the hypotenuse. Then angles at A and B are acute angles. AD is the angle bisector of angle A, so starting from vertex A, it bisects angle A into two equal angles and intersects side BC at point D. Then, we need to find the length CD.Okay, so according to the Angle Bisector Theorem, the angle bisector from a vertex divides the opposite side into segments proportional to the adjacent sides. In this case, angle bisector AD divides side BC into segments BD and DC. So, the ratio of BD to DC should be equal to the ratio of the adjacent sides AB to AC. Wait, the Angle Bisector Theorem formula is BD/DC = AB/AC. Let me verify that.Yes, the theorem states that if a bisector of an angle of a triangle divides the opposite side into segments proportional to the other two sides. So, BD/DC = AB/AC.Given that, we can set up the equation BD/DC = AB/AC. Since BC is 12, BD + DC = 12. Let me denote DC as x, then BD would be 12 - x. So, substituting into the ratio: (12 - x)/x = AB/AC.But we need to find AB. Since it's a right triangle at C, we can compute AB using the Pythagorean theorem. AB = sqrt(AC² + BC²) = sqrt(5² + 12²) = sqrt(25 + 144) = sqrt(169) = 13. So AB is 13.Therefore, substituting AB = 13 and AC = 5, the ratio becomes (12 - x)/x = 13/5. Now, solving for x.Cross-multiplying: 5(12 - x) = 13x.Expanding left side: 60 - 5x = 13x.Bringing terms together: 60 = 13x + 5x => 60 = 18x.Therefore, x = 60/18 = 10/3 ≈ 3.333...So, CD is 10/3. Let me check my steps again to make sure.1. Confirmed triangle is right-angled at C with legs 5 and 12. Hypotenuse AB=13. Correct.2. AD is the angle bisector of angle A, which is an acute angle. Correct.3. Angle Bisector Theorem: BD/DC = AB/AC. Wait, BD is adjacent to angle at B, DC adjacent to angle at C. Wait, hold on, maybe I got the ratio reversed? Let me double-check the Angle Bisector Theorem.The Angle Bisector Theorem states that the angle bisector from angle A will divide the opposite side BC into segments BD and DC such that BD/DC = AB/AC. Yes, that's correct. Because BD is the segment adjacent to AB, and DC is adjacent to AC. So yes, BD/DC = AB/AC.So substituting BD = 12 - x, DC = x, AB=13, AC=5. So (12 - x)/x = 13/5. Then solving gives x = 10/3. That seems right.Alternatively, maybe I can use the formula for the length of the angle bisector. Wait, but that might be more complicated. The Angle Bisector Theorem is straightforward here.Alternatively, maybe coordinate geometry? Let's see. If I set coordinate system with point C at (0,0), since it's the right angle. Then AC=5 would be along the x-axis, so point A is at (5,0). Wait, no. Wait, if C is at (0,0), then AC is one leg and BC is another. Since AC=5 and BC=12. Wait, but in a right triangle at C, AC and BC are the legs, so point A is at (5,0), point B at (0,12), and point C at (0,0). Then hypotenuse AB connects (5,0) to (0,12). Then angle at A is the angle between AC (from A to C) and AB (from A to B). Then AD is the angle bisector of angle A. So point D is on BC, which is from B (0,12) to C (0,0). So BC is the vertical line x=0 from (0,0) to (0,12). So point D is somewhere along the y-axis between (0,0) and (0,12). Let me find coordinates of D.But maybe coordinate geometry will complicate things, but let's try. Let me assign coordinates:C = (0,0)A = (5,0) because AC=5 along x-axisB = (0,12) because BC=12 along y-axisThen AB is the hypotenuse from (5,0) to (0,12). Equation of AB: Let's compute the equation. The slope is (12 - 0)/(0 - 5) = -12/5. So the equation is y = -12/5 (x - 5). So y = -12/5 x + 12.Now, angle bisector from A. The angle bisector of angle A will go from point A (5,0) to point D on BC (which is the y-axis). So D is (0, d) for some d between 0 and 12. We need to find d, which is CD. Since CD is the distance from C (0,0) to D (0,d), so CD = d. So once we find d, that's the answer.We can use the Angle Bisector Theorem here as before. BD/DC = AB/AC. BD = 12 - d, DC = d, AB=13, AC=5. So (12 - d)/d = 13/5. Solving gives d=10/3 ≈ 3.333. So CD=10/3.Alternatively, using coordinates, perhaps parametrize AD and use the angle bisector condition. Let me try that.Point A is (5,0). Point D is (0,d). So the line AD goes from (5,0) to (0,d). Let's find the direction vector of AD: (-5, d). The angle bisector should create equal angles with AC and AB. Hmm, but calculating angles might be tedious. Alternatively, using the formula for the angle bisector in coordinates.Alternatively, use vectors. The direction vector of AD is (-5, d). The angle between AD and AC (which is along the x-axis from A to C) should be equal to the angle between AD and AB (from A to B). So the angle between vector AD and vector AC is equal to the angle between vector AD and vector AB.Vector AC is from A to C: (-5, 0).Vector AB is from A to B: (-5,12).Vector AD is from A to D: (-5, d).The angle between AD and AC should equal the angle between AD and AB. Using the formula for the angle between two vectors:cos(theta1) = (AD . AC) / (|AD| |AC|)cos(theta2) = (AD . AB) / (|AD| |AB|)Set theta1 = theta2, so:(AD . AC) / (|AD| |AC|) = (AD . AB) / (|AD| |AB|)Multiply both sides by |AD|:(AD . AC)/ |AC| = (AD . AB)/ |AB|Compute the dot products:AD . AC = (-5)(-5) + (d)(0) = 25AD . AB = (-5)(-5) + (d)(12) = 25 + 12d|AC| = 5|AB| = 13Therefore:25 / 5 = (25 + 12d) / 13Simplify:5 = (25 + 12d)/13Multiply both sides by 13:65 = 25 + 12dSubtract 25:40 = 12dDivide:d = 40/12 = 10/3So CD = d = 10/3. Same result. So that confirms it.Therefore, the answer is 10/3. So, CD = boxed{dfrac{10}{3}}.Final AnswerThe length of ( CD ) is boxed{dfrac{10}{3}}.</think>Given a right triangle ( triangle ABC ) with legs ( AC = 5 ) and ( BC = 12 ), and the right angle at ( C ). We need to find the length of ( CD ) where ( D ) is a point on ( BC ) such that ( AD ) is the angle bisector of ( angle A ).1. Identify the hypotenuse: Using the Pythagorean theorem, the hypotenuse ( AB ) is calculated as: [ AB = sqrt{AC^2 + BC^2} = sqrt{5^2 + 12^2} = sqrt{169} = 13 ]2. Apply the Angle Bisector Theorem: The theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. For ( triangle ABC ) with angle bisector ( AD ), the ratio ( frac{BD}{DC} = frac{AB}{AC} ). Let ( DC = x ). Then ( BD = 12 - x ).3. Set up the proportion: [ frac{BD}{DC} = frac{AB}{AC} implies frac{12 - x}{x} = frac{13}{5} ]4. Solve for ( x ): [ 5(12 - x) = 13x implies 60 - 5x = 13x implies 60 = 18x implies x = frac{60}{18} = frac{10}{3} ]Thus, the length of ( CD ) is (boxed{dfrac{10}{3}}).,<think>Okay, so I have this right triangle ABC, right? The legs are AC = 5 and BC = 12. Point D is on BC, and AD is the angle bisector of angle A. I need to find the length of CD. Hmm, let's see. First, let me visualize the triangle. Since it's a right triangle, angle C is the right angle. So, AC and BC are the legs, and AB must be the hypotenuse. Let me confirm that. If AC = 5 and BC = 12, then by the Pythagorean theorem, AB should be sqrt(5² + 12²) which is sqrt(25 + 144) = sqrt(169) = 13. So AB is 13. That checks out with the 5-12-13 Pythagorean triple. Good.Now, point D is on BC, and AD is the angle bisector of angle A. So angle BAD equals angle CAD. I need to find CD. Hmm, angle bisector... There's a theorem about angle bisectors in a triangle, right? The angle bisector theorem. Let me recall. The angle bisector theorem states that the angle bisector of an angle in a triangle divides the opposite side into segments proportional to the adjacent sides. So, in this case, since AD is the angle bisector of angle A, it should divide BC into segments BD and DC such that BD/DC = AB/AC. Wait, let me make sure. The theorem says BD/DC = AB/AC. Let me write that down:BD / DC = AB / ACWe know AB is 13, and AC is 5. So BD/DC = 13/5. But BD + DC = BC, which is 12. So BD + DC = 12. So we have two equations:1. BD/DC = 13/52. BD + DC = 12Let me call BD = x and DC = y for clarity. Then:x/y = 13/5 => x = (13/5)yAnd x + y = 12.Substituting x in the second equation:(13/5)y + y = 12Combine like terms:(13/5)y + (5/5)y = 12(18/5)y = 12Multiply both sides by 5:18y = 60Divide both sides by 18:y = 60/18 = 10/3 ≈ 3.333...So DC = y = 10/3. Therefore, CD is 10/3. Let me check that again.Wait, BD is 13/5 times DC. So if DC is 10/3, then BD is (13/5)*(10/3) = (130/15) = 26/3. Then BD + DC = 26/3 + 10/3 = 36/3 = 12, which matches BC = 12. That seems right.So CD is 10/3. So the answer should be 10/3, which is approximately 3 and 1/3. But in the problem, they might want it as a fraction, so 10/3. Let me confirm once more.Alternatively, maybe I can use coordinates to verify. Let me place the right triangle at coordinate system. Let’s assign point C at the origin (0,0). Then, since AC = 5 and BC = 12, point A would be at (0,5) and point B at (12,0). Then, angle A is at point A (0,5). The angle bisector from A would meet BC at point D. Let's compute coordinates of D.The coordinates of B are (12,0) and C are (0,0), so BC is along the x-axis from (0,0) to (12,0). Let’s parameterize point D as (x, 0) where x is between 0 and 12. Then, angle bisector from A to D. But maybe using the angle bisector theorem is more straightforward. Since we already applied the angle bisector theorem and got CD = 10/3, which is approximately 3.333, and BD = 26/3 ≈ 8.666, which adds up to 12. That seems correct. Let me check with coordinates.Alternatively, maybe using vectors or distance formula. Let me see. If point D is on BC, coordinates (x,0). The angle bisector from A to D. The angle bisector should satisfy the ratio. So, according to the angle bisector theorem, BD/DC = AB/AC = 13/5, which is exactly what we used. So BD = 13/5 * DC, and BD + DC = 12, leading to DC = 10/3. So that's correct.Alternatively, perhaps using the formula for the length of the angle bisector. Wait, but the problem doesn't ask for the length of AD, but the length of CD. So angle bisector theorem is the way to go here.Wait, just to ensure, maybe compute coordinates. Let me try.Coordinates:C (0,0)A (0,5)B (12,0)Then, D is somewhere on BC, which is the x-axis from (0,0) to (12,0). Let’s say D is at (d, 0), 0 ≤ d ≤ 12.AD is the angle bisector. So, angle BAD = angle CAD.Let’s compute vectors or use slopes. Alternatively, use the angle bisector formula in coordinates.But maybe using the angle bisector theorem is better. Since we have a triangle, and the angle bisector divides the opposite side in the ratio of the adjacent sides. As before, BD/DC = AB/AC = 13/5. So BD = (13/5) DC. Then BD + DC = 12, so (13/5) DC + DC = 12 => (18/5) DC = 12 => DC = (12 * 5)/18 = 60/18 = 10/3. So yes, CD = 10/3. So this seems consistent.Alternatively, if I use coordinates and find the coordinates of D such that AD bisects angle A.Vector approach: The direction vector of AD should bisect the angle between vectors AB and AC.But maybe that's overcomplicating.Alternatively, using the formula for internal angle bisector coordinates. There is a formula that gives the coordinates of the point where the angle bisector meets the opposite side. If the triangle has vertices at (x₁,y₁), (x₂,y₂), (x₃,y₃), then the coordinates can be found using the ratio.But in our case, angle at A (0,5). The angle bisector from A should meet BC (from B (12,0) to C (0,0)) at point D. The coordinates of D can be calculated using the angle bisector theorem ratio BD/DC = AB/AC = 13/5. So BD = 13k, DC = 5k, and BD + DC = 18k = 12 => k = 12/18 = 2/3. Therefore, BD = 13*(2/3) = 26/3 and DC = 5*(2/3) = 10/3. Hence, DC = 10/3. So that matches the previous result.Therefore, CD is 10/3. So the answer is 10/3. Therefore, boxed, it would be boxed{dfrac{10}{3}}.Final Answerboxed{dfrac{10}{3}}</think>Given a right triangle ( triangle ABC ) with legs ( AC = 5 ) and ( BC = 12 ), and point ( D ) on ( BC ) such that ( AD ) is the angle bisector of ( angle A ). We need to find the length of ( CD ).First, we use the Pythagorean theorem to find the hypotenuse ( AB ):[AB = sqrt{AC^2 + BC^2} = sqrt{5^2 + 12^2} = sqrt{25 + 144} = sqrt{169} = 13]By the Angle Bisector Theorem, the angle bisector ( AD ) divides the opposite side ( BC ) into segments ( BD ) and ( CD ) such that:[frac{BD}{CD} = frac{AB}{AC} = frac{13}{5}]Let ( BD = x ) and ( CD = y ). We have the equations:[frac{x}{y} = frac{13}{5} quad text{and} quad x + y = 12]Solving these equations, we express ( x ) in terms of ( y ):[x = frac{13}{5}y]Substituting into the second equation:[frac{13}{5}y + y = 12][frac{13}{5}y + frac{5}{5}y = 12][frac{18}{5}y = 12][18y = 60][y = frac{60}{18} = frac{10}{3}]Thus, the length of ( CD ) is:[boxed{dfrac{10}{3}}],<think>Okay, so I have this problem here about a right triangle ABC where AC and BC are the legs with lengths 5 and 12 respectively. Point D is on BC, and AD is the angle bisector of angle A. I need to find the length of CD. Hmm, let me try to visualize this first.Since it's a right triangle, I should figure out which angle is the right angle. The problem mentions legs AC and BC, so the right angle must be at point C. That means triangle ABC is right-angled at C, with AC = 5 and BC = 12. So, point C is the right angle, and AB must be the hypotenuse. Let me confirm that. In a right triangle, the two legs are the sides that form the right angle, so yes, AC and BC are the legs, and AB is the hypotenuse. Got it.Now, point D is somewhere on BC, and AD is the angle bisector of angle A. I need to find CD. Angle bisector... I remember there's a theorem about angle bisectors in a triangle that relates the lengths of the sides. The Angle Bisector Theorem, right? It states that an angle bisector in a triangle divides the opposite side into segments proportional to the adjacent sides. So, in this case, since AD is the angle bisector of angle A, it should divide side BC into segments BD and DC such that BD/DC = AB/AC. Wait, let me make sure I have that right. The theorem says BD/DC = AB/AC. Is that correct? Let me recall. The angle bisector from angle A will split side BC into BD and DC, so BD/DC = AB/AC. Yes, that seems right. But first, I need to find the length of AB. Since ABC is a right triangle at C, AB can be calculated using the Pythagorean theorem. So, AB = sqrt(AC² + BC²) = sqrt(5² + 12²) = sqrt(25 + 144) = sqrt(169) = 13. So, AB is 13. That's straightforward.Now, applying the Angle Bisector Theorem: BD/DC = AB/AC = 13/5. But BD and DC are parts of BC, which has a total length of 12. So, BD + DC = 12. Let me denote DC as x. Then BD would be 12 - x. According to the theorem, BD/DC = (12 - x)/x = 13/5. So, I can set up the equation (12 - x)/x = 13/5 and solve for x.Let me do that. Cross-multiplying gives 5(12 - x) = 13x. Expanding the left side: 60 - 5x = 13x. Adding 5x to both sides: 60 = 18x. Then, x = 60/18 = 10/3 ≈ 3.333... So, DC is 10/3. Therefore, the length of CD is 10/3. Wait, but let me double-check my steps to make sure.Starting again: BD/DC = AB/AC = 13/5. BD + DC = BC = 12. Let DC = x, so BD = 12 - x. Then, (12 - x)/x = 13/5. Cross-multiplying: 5(12 - x) = 13x. 60 - 5x = 13x. 60 = 18x. x = 60/18 = 10/3. Yeah, that seems correct. Simplifying 60/18 by dividing numerator and denominator by 6: 10/3. So, CD = 10/3. Alternatively, as a mixed number, that would be 3 1/3, but the problem probably expects an improper fraction. So, 10/3 is the answer. Let me check if there's another way to approach this problem just to verify.Another method might involve coordinates. Let me assign coordinates to the triangle. Let’s place point C at the origin (0, 0). Since AC = 5 and BC = 12, and it's a right triangle at C, we can set point A at (0, 5) and point B at (12, 0). Then, angle A is at point A (0,5). The angle bisector AD would start at A and go to some point D on BC. Point D is on BC, which is from point B (12,0) to point C (0,0). Wait, BC is from B (12,0) to C (0,0)? Wait, hold on. If the triangle is right-angled at C, then AC and BC are the legs, so AC is from C to A, and BC is from C to B. So, if C is at (0,0), then A is at (0,5) and B is at (12,0). Then, BC is the segment from (12,0) to (0,0). So, point D is somewhere along the x-axis from (12,0) to (0,0).AD is the angle bisector from A (0,5) to D on BC. Let's find the coordinates of D. Using the Angle Bisector Theorem again, BD/DC = AB/AC = 13/5. So, BD = (13/5) DC. And BD + DC = 12. So, (13/5) DC + DC = 12. Let me compute that: (13/5 + 5/5) DC = 18/5 DC = 12. Therefore, DC = 12 * (5/18) = (60)/18 = 10/3. So, same result. So, DC is 10/3. So, coordinates of D would be? Since D is along BC from B (12,0) to C (0,0), moving towards C by 10/3 units. Wait, the entire BC is 12 units long. So, starting at B (12,0), moving towards C (0,0), the distance from C is DC = 10/3. So, the coordinate would be (12 - BD, 0). But BD = 12 - DC = 12 - 10/3 = 26/3. Therefore, the x-coordinate is 12 - BD? Wait, no. If BD is the length from B to D, then D is located BD units away from B towards C. Since BD = 26/3, then the x-coordinate of D is 12 - BD? Wait, BD is 26/3, but BC is 12 units long. Wait, BD is 26/3 ≈ 8.666..., but BC is 12. So, 26/3 is approximately 8.666, so subtracting that from 12 gives DC = 12 - 26/3 = 10/3 ≈ 3.333, which matches. Wait, but coordinates. If BC is along the x-axis from (12,0) to (0,0), then moving from B (12,0) towards C (0,0), the coordinates of D would be (12 - BD, 0). Since BD = 26/3, then x-coordinate is 12 - 26/3 = (36/3 - 26/3) = 10/3. Therefore, D is at (10/3, 0). Therefore, CD is the distance from C (0,0) to D (10/3, 0), which is 10/3. So, that checks out.Alternatively, using the angle bisector formula in coordinates. The angle bisector from A to BC can be found using the formula for the coordinates, but that might be more complicated. However, since we already verified using the Angle Bisector Theorem and coordinate geometry, and both give CD = 10/3, I think that's solid.Wait, just to make sure, maybe I can use the formula for the angle bisector length. Wait, but the question isn't asking for the length of AD, but rather CD. So, the Angle Bisector Theorem gives a direct way to find the ratio, which we used. So, I think that's sufficient.Alternatively, maybe use trigonometry. Let's see. In triangle ABC, angle at A is being bisected by AD. Let me find the measure of angle A first. Since it's a right triangle at C, angle A is at (0,5), so angle between AC and AB. Let's compute tan(angle A) = opposite/adjacent = BC/AC = 12/5. So, angle A is arctan(12/5). Then, the angle bisector divides this angle into two equal parts. But maybe using the Angle Bisector Theorem is simpler here.Alternatively, using the formula for the length of the angle bisector. Wait, the formula for the length of an angle bisector in a triangle is given by:AD = (2ab cos(theta/2)) / (a + b)But in this case, we might not need the length of AD, but rather the segments BD and DC. Since the Angle Bisector Theorem gives us BD/DC = AB/AC, which is exactly the ratio we used, so that's the most straightforward method.So, all methods point to CD being 10/3, which is approximately 3.333. Since the problem is likely expecting an exact answer, 10/3 is the way to go.Wait, just to recap once more to ensure no mistakes were made:1. Triangle ABC, right-angled at C, legs AC=5, BC=12.2. AB, hypotenuse, is 13 by Pythagoras.3. AD is the angle bisector of angle A, intersecting BC at D.4. By Angle Bisector Theorem, BD/DC = AB/AC = 13/5.5. BD + DC = BC = 12.6. Let DC = x, then BD = 12 - x.7. (12 - x)/x = 13/5 → solving gives x = 10/3.8. Therefore, CD = 10/3.Yes, that seems consistent. I think that's the correct answer. I don't see any errors in the steps, and the coordinate geometry approach also led to the same result. So, I'm confident that CD = 10/3.Final AnswerThe length of ( CD ) is boxed{dfrac{10}{3}}.</think>Given a right triangle ( triangle ABC ) with legs ( AC = 5 ) and ( BC = 12 ), and the right angle at ( C ). Point ( D ) is on ( BC ) such that ( AD ) is the angle bisector of ( angle A ). We need to find the length of ( CD ).1. Calculate the hypotenuse ( AB ): [ AB = sqrt{AC^2 + BC^2} = sqrt{5^2 + 12^2} = sqrt{25 + 144} = sqrt{169} = 13 ]2. Apply the Angle Bisector Theorem: The theorem states that ( frac{BD}{DC} = frac{AB}{AC} ). Let ( DC = x ). Then ( BD = 12 - x ). [ frac{BD}{DC} = frac{13}{5} implies frac{12 - x}{x} = frac{13}{5} ]3. Solve for ( x ): [ 5(12 - x) = 13x implies 60 - 5x = 13x implies 60 = 18x implies x = frac{60}{18} = frac{10}{3} ]Thus, the length of ( CD ) is (boxed{dfrac{10}{3}}).]

🔑:3. frac{10}{3}.As shown in Figure 5, construct Rt triangle ABC,on side AB, mark AE = AC, draw ED perp AB, intersecting BC at point D. Then AD is the angle bisector of angle A.Let CD = x. By BE = 13 - 5 = 8, DE = x, BD = 12 - x and the Pythagorean theorem, we get x^2 + 8^2 = (12 - x)^2. Solving for x yields x = frac{10}{3}.

🧠:[<think>Okay, let me see. I need to figure out the number of elements in set M, which is defined as all the sums you can get by adding an element from set A and an element from set B. The options are 3, 4, 5, or 6. Hmm, let's start by understanding the problem step by step.First, let me recall what the sets are. Set A is {1, 2, 3} and set B is {4, 5}. So, A has three elements and B has two elements. The set M is made by adding each element of A to each element of B. That means for every a in A and every b in B, we calculate a + b and collect all those results into M. But since M is a set, duplicates are only counted once. So, even if two different pairs (a, b) result in the same sum, it will only be one element in M.Alright, so maybe the first thing to do is list out all possible sums. Let's start with each element in A and add it to each element in B.Starting with the first element of A, which is 1. Adding 1 to each element of B:1 + 4 = 51 + 5 = 6So that gives us 5 and 6.Next, the second element of A is 2. Adding 2 to each element of B:2 + 4 = 62 + 5 = 7So here we get 6 and 7. Wait, 6 was already obtained when we added 1 + 5. So in set M, 6 will only be listed once, even though it comes up again here.Moving on to the third element of A, which is 3. Adding 3 to each element of B:3 + 4 = 73 + 5 = 8So now we have 7 and 8. Again, 7 was already obtained from 2 + 5, so it will be in M only once.Let me compile all these sums:From 1: 5, 6From 2: 6, 7From 3: 7, 8So putting all the sums together, we have 5, 6, 7, 8. Wait, but let me check again. Let's list them all without worrying about duplicates first:1+4=51+5=62+4=62+5=73+4=73+5=8So the list is: 5, 6, 6, 7, 7, 8. Now, when we make this into a set, duplicates are removed. So the unique values are 5, 6, 7, 8. That's four elements. So M = {5, 6, 7, 8}, which has 4 elements. Therefore, the answer should be B:4.But let me verify once more to be sure I didn't miss any sums or count duplicates incorrectly. Let's go through all possible combinations systematically.A has elements 1,2,3. B has 4,5. So each element of A pairs with each element of B, making 3x2=6 combinations. The sums are:1+4=51+5=62+4=62+5=73+4=73+5=8Yes, that's all the sums. So the set M is {5,6,7,8}. Therefore, the number of elements is 4. The options given are A:3, B:4, C:5, D:6. So the correct answer is B:4.Wait, but hold on a second. Let me check if I considered all possible sums. Since M is defined as {x | x = a + b, a ∈ A, b ∈ B}, it's all possible sums where a is from A and b is from B. So yes, that's exactly the combinations I considered. Each a and b pair.Alternatively, maybe I can compute it another way. The smallest possible sum is 1 + 4 = 5, and the largest is 3 + 5 = 8. So the possible sums range from 5 to 8. Let's see: 5,6,7,8. That's four numbers. Each number in that range is achievable?5: 1+46:1+5 or 2+47:2+5 or 3+48:3+5Yes, each number from 5 to 8 is covered. Therefore, M has four elements. So answer is B:4. Therefore, I think that's correct.But just to be thorough, let me cross-verify. If set A had more elements or set B had more elements, the number could be different. But in this case, A is {1,2,3} and B is {4,5}. So the possible sums are as we calculated. The minimum sum is 1+4=5, maximum is 3+5=8. So numbers 5,6,7,8. Let me check if each of these numbers can be achieved.5: 1 + 4. Yes.6: 1 + 5 or 2 + 4. Both are valid.7: 2 + 5 or 3 + 4. Both are valid.8: 3 + 5. Yes.So all numbers from 5 to 8 inclusive are present. Therefore, M has 4 elements. Therefore, the answer is B:4.Alternatively, another way to approach this is to realize that when you add two sets, the number of elements in the resulting set can sometimes be calculated as (size of A + size of B - 1) if they are arithmetic sequences with step 1. But in this case, A and B aren't necessarily arithmetic sequences. Wait, A is {1,2,3} which is an arithmetic sequence with difference 1, and B is {4,5}, which is also an arithmetic sequence with difference 1, but their sums might not necessarily form a full arithmetic sequence. Wait, in this case, the sums go from 5 to 8, which is 5,6,7,8. So that's actually a complete arithmetic sequence with difference 1. Therefore, the number of elements is maximum - minimum + 1. 8 - 5 + 1 = 4. So that gives the same answer.But I should be careful with this method because sometimes there could be gaps. For example, if A was {1,3,5} and B was {2,4}, then the sums would be 3,5,7,5,7,9. So the set M would be {3,5,7,9}, which is from 3 to 9, but missing 4,6,8. But in our original problem, since A and B are consecutive numbers, their sums form a continuous range. Let me check if that's a general rule. If A is {a1, a2, ..., an} where the elements are consecutive integers, and B is {b1, b2, ..., bm} also consecutive integers, then the sumset M would be from a1 + b1 to an + bm, and if there are no gaps, then the number of elements is (an + bm - (a1 + b1)) + 1. But this only holds if every integer in between is achievable. For consecutive integers A and B, the sumset will indeed be all integers from a1 + b1 to an + bm. Because for example, if A is {1,2,3} and B is {4,5}, then:The smallest sum is 1 + 4 = 5Next is 1 + 5 = 6, 2 + 4 = 6Then 2 + 5 = 7, 3 + 4 = 7Then 3 + 5 = 8But wait, between 5 and 8, we have 5,6,7,8. All consecutive. So even though some sums are achieved in multiple ways, there are no gaps. So the total number of elements is 8 - 5 + 1 = 4.So in this case, that formula works. Therefore, the answer is 4, which is option B.Alternatively, even if the sets A and B weren't consecutive, we could have gaps. For example, if A was {1,3} and B was {4,6}, then the sums would be 1+4=5, 1+6=7, 3+4=7, 3+6=9. So M would be {5,7,9}, which has gaps at 6 and 8. Therefore, the number of elements would be 3, but the min is 5 and max is 9, so according to the formula, 9 -5 +1=5, which is wrong. Therefore, the formula only works when the sets are consecutive integers. So in our problem, since A and B are consecutive integers, we can use the formula. But in other cases, it may not hold.So, going back to the original problem, since A is {1,2,3} (consecutive) and B is {4,5} (consecutive), their sums form a consecutive range from 5 to 8, so the number of elements is 4. Therefore, the answer is B:4.But just to make sure, let's think of another way. The number of possible sums is equal to the number of distinct sums. So even though there are 3*2=6 possible pairs, some sums repeat. For each element in A, adding to each element in B gives two sums. Let's see:For a=1: 5,6 (two unique sums)For a=2: 6,7 (two unique sums, but 6 was already there)For a=3:7,8 (two unique sums, 7 was already there)So each new a adds potentially one new sum. Starting from a=1: adds 5 and 6.a=2: adds 7 (since 6 is already there)a=3: adds 8 (since 7 is already there)Therefore, total unique sums: 5,6,7,8. So 4 elements. Yep, that's the same as before.Therefore, the answer is definitely 4, which is option B.I think that's thorough enough. I can't see any mistakes in the reasoning. The possible sums are 5,6,7,8, so four elements. Therefore, the answer is B.Final Answerboxed{B}</think>Given sets ( A = {1, 2, 3} ), ( B = {4, 5} ), and ( M = {x mid x = a + b, a in A, b in B} ), we need to determine the number of elements in set ( M ).First, we calculate all possible sums of elements from ( A ) and ( B ):- For ( a = 1 ): - ( 1 + 4 = 5 ) - ( 1 + 5 = 6 )- For ( a = 2 ): - ( 2 + 4 = 6 ) - ( 2 + 5 = 7 )- For ( a = 3 ): - ( 3 + 4 = 7 ) - ( 3 + 5 = 8 )Listing all sums, we get: ( 5, 6, 6, 7, 7, 8 ). Removing duplicates, the unique sums are ( 5, 6, 7, 8 ).Thus, the set ( M ) is ( {5, 6, 7, 8} ), which contains 4 elements.The number of elements in set ( M ) is (boxed{B}).,<think>Okay, let's see. The problem is about sets A, B, and M. I need to find how many elements are in set M. The options are from 3 to 6. Let me start by understanding the problem step by step.First, let me restate what's given. Set A is {1, 2, 3} and set B is {4, 5}. Then set M is defined as all the numbers x such that x is the sum of a and b, where a is an element from A and b is an element from B. So M is basically the set of all possible sums you can get by adding each element in A to each element in B. Then the question is asking how many distinct elements are in M. The answer choices are 3, 4, 5, or 6.Alright, so my first thought is that this is similar to the Cartesian product of A and B, but instead of ordered pairs, we're taking their sums. So instead of (1,4), (1,5), (2,4), etc., we're calculating 1+4, 1+5, 2+4, and so on. Then we collect all those sums and see how many unique values there are.Let me list out all possible sums. Since A has 3 elements and B has 2 elements, there will be 3*2=6 sums. But some of these sums might be the same, so the number of unique sums (which is the number of elements in M) could be less than 6. Let me compute each sum one by one.Starting with the first element of A, which is 1. Adding that to each element in B:1 + 4 = 51 + 5 = 6Then the next element in A, which is 2:2 + 4 = 62 + 5 = 7Then the third element in A, which is 3:3 + 4 = 73 + 5 = 8So listing all the sums: 5, 6, 6, 7, 7, 8.Now let's collect the unique values here. The sums are 5, 6, 7, 8. So even though 6 and 7 each appear twice, they are only counted once in set M. Therefore, set M is {5, 6, 7, 8}, which has 4 elements.Wait, but let me double-check. Let me list each combination again to make sure I didn't miss any or miscalculate.1 + 4 = 5 ✔️1 + 5 = 6 ✔️2 + 4 = 6 ✔️2 + 5 = 7 ✔️3 + 4 = 7 ✔️3 + 5 = 8 ✔️Yes, that's all six combinations. The sums are indeed 5, 6, 6, 7, 7, 8. So unique sums are 5, 6, 7, 8. That's four distinct numbers. So the number of elements in M is 4. Looking back at the answer choices, option B is 4. Therefore, the answer should be B.But hold on, just to be thorough, let me visualize this as a table or matrix to ensure there's no oversight.Creating a table where rows are elements of A and columns are elements of B:``` 4 51 5 62 6 73 7 8```Looking at the table, the entries are 5, 6 in the first row; 6, 7 in the second; and 7, 8 in the third. So when we list all entries, it's the same as before. The unique entries are 5, 6, 7, 8. Therefore, 4 unique elements. So that's consistent.Is there any possibility that I made a mistake in arithmetic or missed a combination? Let's confirm each sum again:1 + 4: 1+4 is 5. Correct.1 + 5: 1+5 is 6. Correct.2 + 4: 2+4 is 6. Correct.2 + 5: 2+5 is 7. Correct.3 + 4: 3+4 is 7. Correct.3 + 5: 3+5 is 8. Correct.All sums are correct. No duplicates other than 6 and 7 appearing twice each, but in the set M, duplicates are ignored. Therefore, M has exactly four elements.Therefore, the answer is B:4.Final Answerboxed{B}</think>Given sets ( A = {1, 2, 3} ) and ( B = {4, 5} ), we need to determine the number of elements in set ( M = {x mid x = a + b, a in A, b in B} ).First, we calculate all possible sums of elements from ( A ) and ( B ):- For ( a = 1 ): - ( 1 + 4 = 5 ) - ( 1 + 5 = 6 )- For ( a = 2 ): - ( 2 + 4 = 6 ) - ( 2 + 5 = 7 )- For ( a = 3 ): - ( 3 + 4 = 7 ) - ( 3 + 5 = 8 )Listing all sums, we get: ( 5, 6, 6, 7, 7, 8 ). The unique values are ( 5, 6, 7, 8 ).Thus, the set ( M ) contains the elements ( {5, 6, 7, 8} ), which has 4 distinct elements.The number of elements in set ( M ) is (boxed{B}).]

🔑:Since A={1,2,3}, B={4,5}, and M={x|x=a+b, ain A, bin B}, the possible values of a+b are: 1+4=5, 1+5=6, 2+4=6, 2+5=7, 3+4=7, 3+5=8, thus, the elements in M are only: 5, 6, 7, 8. There are a total of 4 elements. Therefore, the correct choice is boxed{text{B}}.

🧠:[<think>Okay, let's see. So the problem is about a quiz where each participant's score is given by the formula n + 2 - 2k, where n is a constant and k is the participant's rank. The total score of all participants is 2009, and we need to find the smallest possible value of n. Also, no two people have the same score. Hmm, interesting.First, let me try to parse the problem again. Each participant has a unique score because no two people had the same score. Their score is calculated as n + 2 - 2k, where k is their rank. So, the person who came in first (k=1) would have a score of n + 2 - 2(1) = n + 2 - 2 = n. The second place (k=2) would have n + 2 - 4 = n - 2. Third place would be n + 2 - 6 = n - 4, and so on. So each subsequent rank decreases the score by 2. So the scores form an arithmetic sequence starting at n with a common difference of -2.So the scores are n, n - 2, n - 4, ..., n - 2(k-1) for k ranks. Since each participant has a different score, the number of participants must be such that all these scores are distinct. Since the common difference is -2, which is non-zero, the sequence is strictly decreasing, so all scores are unique. So the number of participants is just the number of terms in this sequence. But we don't know how many participants there are. Let me denote the number of participants as m. Then the scores would be from rank 1 to rank m, with scores n, n - 2, ..., n - 2(m - 1).Now, the total score is the sum of this arithmetic sequence. The formula for the sum of an arithmetic sequence is (number of terms)/2 times (first term + last term). So the sum S is (m/2) * [n + (n - 2(m - 1))]. Simplifying the last term: n - 2(m - 1) = n - 2m + 2. So the sum becomes (m/2) * [2n - 2m + 2] = (m/2)*(2(n - m + 1)) = m(n - m + 1). So the total sum S = m(n - m + 1) = 2009.So we have m(n - m + 1) = 2009. We need to find the smallest possible n. Since m and n are positive integers (assuming the number of participants must be an integer, and n is a constant), we can think of this as m(n - m + 1) = 2009. So we can write n = (2009/m) + m - 1. To minimize n, we need to find integer m such that 2009 is divisible by m, and then compute n as (2009/m) + m - 1, and find the minimum such n.But wait, 2009 is the product of m and (n - m + 1). So m must be a divisor of 2009. Therefore, to find possible values of m, we need to factorize 2009 into its divisors and then compute n for each divisor and find the smallest n.First, let's factorize 2009. 2009 divided by 7 is 286.9... Hmm, wait, 7*287 = 2009. Is 287 divisible by 7? 7*41=287. So 2009 = 7*7*41. Wait, no, 7*7*41 is 2009? Let me check. 7*7=49, 49*41=2009. Yes, because 49*40=1960, plus 49*1=49, so 1960+49=2009. So 2009=7^2 *41. Therefore, the divisors of 2009 are 1,7,49,41, 7*41=287, 49*41=2009. So the positive divisors are 1,7,41,49,287,2009.Therefore, possible values of m are 1,7,41,49,287,2009. Then for each m, the corresponding (n - m +1) is 2009/m. Then n = (2009/m) + m -1. So let's compute n for each m:1. m=1: n=2009/1 +1 -1=2009+0=2009.2. m=7: n=2009/7 +7 -1=287 +6=293.3. m=41: n=2009/41 +41 -1=49 +40=89.4. m=49: n=2009/49 +49 -1=41 +48=89. Wait, same as m=41? Hmm, 2009/49 is 41, so 41 +49 -1=89. So same n as m=41.5. m=287: n=2009/287 +287 -1=7 +286=293.6. m=2009: n=2009/2009 +2009 -1=1 +2008=2009.So the n values are 2009,293,89,89,293,2009. So the minimal n is 89.But wait, the problem says "no two people had the same score" but we derived the number of participants as m, and in the formula, each score is unique because the sequence is decreasing by 2 each time. So as long as m is such that all scores are positive? Wait, the problem doesn't state that scores must be positive, just that they are integers (since n is a constant and k is integer rank). So even if scores go negative, it's okay as long as they are unique. But maybe the problem implies that ranks are assigned from 1 upwards, so the number of participants is m, and we can have as many participants as possible. Wait, but in our case, m is a divisor of 2009. So the minimal n is 89. But let me check for m=41 and m=49.Wait, for m=41, n=89. Then the scores would start at 89, 87, 85,... down to 89 - 2*(41-1)=89 -80=9. So all scores are positive. For m=49, n=89. Then the scores start at 89, 87,..., down to 89 - 2*(49 -1)=89 - 96= -7. So some scores would be negative. The problem doesn't specify that scores have to be positive, just that they are integers and unique. So even if some scores are negative, it's acceptable. Therefore, m=41 and m=49 both give n=89, but m=41 gives all positive scores, while m=49 gives some negative scores. But the problem doesn't specify that scores must be positive, so both are valid. However, since we need the minimal n, which is 89, regardless of m=41 or m=49.But let me check if there are any other divisors. Wait, 2009 is 7*7*41, so divisors are 1,7,49,41,287,2009. So those are all. So n=89 is the minimal possible value.But let me just confirm once again. For m=41, n=89. Then total sum is m*(n -m +1)=41*(89 -41 +1)=41*(49)=2009. Correct. For m=49, n=89, sum is 49*(89 -49 +1)=49*(41)=2009. Correct. So both are valid. Therefore, the minimal n is 89. Therefore, the answer is 89.Wait, but the question says "the smallest possible value of n". So even if some participants have negative scores, as long as the total is 2009, n=89 is acceptable. So the answer is 89. Therefore, the minimal possible n is 89.Final Answerboxed{89}</think>Given a quiz where each participant's score is calculated using the formula ( n + 2 - 2k ) (where ( n ) is a constant and ( k ) is the rank of the participant), and the total score of all participants is 2009, we need to find the smallest possible value of ( n ).The scores form an arithmetic sequence starting at ( n ) with a common difference of -2. The sum of this arithmetic sequence is given by:[S = frac{m}{2} times [2n - 2(m - 1)] = m(n - m + 1)]where ( m ) is the number of participants. We know this sum equals 2009:[m(n - m + 1) = 2009]We need to find the divisors of 2009 to determine possible values of ( m ). The prime factorization of 2009 is ( 7^2 times 41 ), giving the divisors: 1, 7, 41, 49, 287, and 2009. For each divisor ( m ), we calculate ( n ) using:[n = frac{2009}{m} + m - 1]- For ( m = 1 ): ( n = 2009 + 1 - 1 = 2009 )- For ( m = 7 ): ( n = 287 + 7 - 1 = 293 )- For ( m = 41 ): ( n = 49 + 41 - 1 = 89 )- For ( m = 49 ): ( n = 41 + 49 - 1 = 89 )- For ( m = 287 ): ( n = 7 + 287 - 1 = 293 )- For ( m = 2009 ): ( n = 1 + 2009 - 1 = 2009 )The smallest value of ( n ) is 89, which occurs for both ( m = 41 ) and ( m = 49 ). Since the problem does not specify that scores must be positive, both cases are valid.Thus, the smallest possible value of ( n ) is (boxed{89}).,<think>Okay, let's see. So the problem is about a quiz where each participant's score is calculated using the formula n + 2 - 2k, where n is a constant and k is their rank. The total score of all participants is 2009, and we need to find the smallest possible value of n. Also, no two people had the same score, which probably means that each rank from 1 upwards is assigned to a participant without ties. First, let me try to parse the formula for the score. The score of a participant with rank k is n + 2 - 2k. So, if someone is rank 1, their score would be n + 2 - 2(1) = n + 2 - 2 = n. Then, rank 2 would be n + 2 - 4 = n - 2, rank 3 would be n + 2 - 6 = n - 4, and so on. So the scores form an arithmetic sequence starting at n and decreasing by 2 each time. Wait, arithmetic sequence with first term a1 = n and common difference d = -2. So the scores are n, n - 2, n - 4, n - 6, etc. Since the participants have different ranks, the number of participants must be such that all these scores are distinct, which they naturally are because each subsequent term is 2 less than the previous. But we need the total score to be 2009. So if we can figure out how many participants there are, say m participants, then the sum of the scores would be the sum of this arithmetic sequence. The formula for the sum of an arithmetic sequence is S = m/2 * [2a1 + (m - 1)d]. Plugging in the values, that would be S = m/2 * [2n + (m - 1)(-2)].Let me write that out more clearly. Sum S = (m/2)[2n - 2(m - 1)]. Simplify that: S = (m/2)(2n - 2m + 2) = (m/2)(2(n - m + 1)) = m(n - m + 1). So according to the problem, this sum is 2009. Therefore, we have m(n - m + 1) = 2009. We need to find the smallest possible n such that this equation holds, where m is a positive integer (number of participants) and n must be an integer as well (since all scores are integers; if n is not an integer, then the scores would alternate between integers and non-integers depending on k, but since k is integer, 2k is even, so n must be even or odd to keep the score integer? Wait, let's check that.The score is n + 2 - 2k. If n is an integer, then 2 - 2k is an even integer, so the score would be integer. If n is not an integer, then the score would not be an integer. But since scores are presumably integers (as total is 2009, an integer), so n must be an integer. So n is an integer, m is a positive integer, and m(n - m + 1) = 2009.So our equation is m(n - m + 1) = 2009. We need to find the minimal n possible, given that m and (n - m + 1) are positive integers. Wait, m must be a positive integer, and (n - m + 1) must also be a positive integer because the scores must be positive? Wait, the problem doesn't specify that the scores have to be positive, just that they are all different. However, in quizzes, scores are usually positive, but maybe they can be negative? Hmm, but let's check.If the scores can be negative, then (n - m + 1) could be positive or negative. But the number of participants m is positive, so if (n - m + 1) is positive, then n - m + 1 > 0 => n > m - 1. If (n - m + 1) is negative, then m(n - m + 1) would be negative, but the total score is 2009, which is positive, so (n - m + 1) must be positive. Therefore, n - m + 1 > 0 => n > m - 1. So that gives us a constraint.So n must be greater than m - 1, and m and (n - m + 1) are positive integers whose product is 2009. Therefore, m and (n - m + 1) are positive integer divisors of 2009, with m < n + 1.Given that, we can think of 2009 as a product of two positive integers m and l, where l = n - m + 1. Then, n = l + m - 1. So our goal is to find the smallest possible n, which would be equivalent to finding the pair (m, l) such that m * l = 2009, and n = l + m - 1 is minimized. To minimize n = l + m - 1, we need to minimize l + m. Since l = 2009/m, we need to minimize m + 2009/m. The minimum of m + 2009/m occurs when m is as close as possible to sqrt(2009). Let's compute sqrt(2009):sqrt(2009) ≈ 44.82. So the integer divisors around 44.82 would be the factors of 2009 closest to that. Therefore, we need to factorize 2009 to find its divisors.First, let's factorize 2009. Let's check divisibility:2009 ÷ 7 = 287, since 7*287=2009. Then, 287 ÷ 7 = 41. So 2009 = 7*7*41. Therefore, the positive divisors are 1, 7, 49, 41, 287, 2009.Wait, 7*7=49, 49*41=2009. So the divisors are 1, 7, 49, 41, 287, 2009. But wait, 41 is prime? Let me check 41: yes, 41 is a prime number. Therefore, the divisors are 1, 7, 49, 41, 287, 2009.So the possible pairs (m, l) where m * l = 2009 are:(1, 2009), (7, 287), (49, 41), (41, 49), (287, 7), (2009, 1).Then, n = l + m - 1 for each pair:For (1, 2009): n = 2009 + 1 - 1 = 2009For (7, 287): n = 287 + 7 - 1 = 293For (49, 41): n = 41 + 49 - 1 = 89For (41, 49): n = 49 + 41 - 1 = 89For (287, 7): n = 7 + 287 - 1 = 293For (2009, 1): n = 1 + 2009 - 1 = 2009So the minimal n is 89. But wait, let's check if these divisors are correct. Wait, 2009 = 7^2 * 41, so divisors are 1, 7, 49, 41, 7*41=287, 49*41=2009. So yes, those are all the divisors. Therefore, the pairs (m, l) are as above.Therefore, the minimal possible value of n is 89. However, before concluding, we need to verify that in the case where m=49 and l=41, the scores are all distinct integers, which they would be because the formula is n + 2 - 2k where k ranges from 1 to m=49. Since each subsequent score is 2 less than the previous, there's no duplication.Additionally, we need to check that all scores are integers. Since n=89 is an integer, and 2k is even, n + 2 - 2k will also be an integer. So all scores are integers, which is fine.But also, if m=49, the lowest rank is 49, so the lowest score would be 89 + 2 - 2*49 = 91 - 98 = -7. Wait, so the lowest score is negative. But the problem doesn't specify that scores have to be positive. It just says "no two people had the same score". So even if some scores are negative, it's okay as long as all scores are unique. Since they form an arithmetic sequence decreasing by 2 each time, they are all unique. So even if some are negative, it's acceptable.Therefore, the minimal possible n is 89. But let me check if there is any other constraint that I might have missed. For example, maybe the number of participants m must be such that the highest score (which is for rank 1) is positive? Because if the highest score is n, then n must be positive. So n must be positive. But in our case, n=89, which is positive, so that's okay. The highest score is 89, then 87, 85, etc., down to -7. So the highest score is positive, which is reasonable for a quiz. So that's acceptable.Therefore, the answer should be 89. Let me check the calculation again. If m=49, l=41. So n = 41 + 49 -1 = 89. Then, the sum is m*l = 49*41=2009. Correct. So the minimal n is 89. Therefore, I think the answer is 89.Final Answerboxed{89}</think>Given a quiz where each participant's score is calculated using the formula ( n + 2 - 2k ) (where ( n ) is a constant and ( k ) is the rank of the participant), and the total score of all participants is 2009, we need to find the smallest possible value of ( n ).The scores form an arithmetic sequence with the first term ( n ) and a common difference of -2. The sum of the scores for ( m ) participants is given by the formula for the sum of an arithmetic sequence:[S = frac{m}{2} left[ 2n + (m - 1)(-2) right]]Simplifying this, we get:[S = m(n - m + 1)]We know this sum equals 2009, so:[m(n - m + 1) = 2009]Here, ( m ) and ( n - m + 1 ) are positive integers. We need to factorize 2009 to find pairs of factors ( (m, l) ) such that ( m times l = 2009 ). The factors of 2009 are 1, 7, 41, 49, 287, and 2009.For each factor pair, we calculate ( n ) as follows:- ( (1, 2009) rightarrow n = 2009 + 1 - 1 = 2009 )- ( (7, 287) rightarrow n = 287 + 7 - 1 = 293 )- ( (49, 41) rightarrow n = 41 + 49 - 1 = 89 )- ( (41, 49) rightarrow n = 49 + 41 - 1 = 89 )- ( (287, 7) rightarrow n = 7 + 287 - 1 = 293 )- ( (2009, 1) rightarrow n = 1 + 2009 - 1 = 2009 )The smallest value of ( n ) is 89, which corresponds to the factor pair (49, 41). This ensures all scores are distinct integers, and the highest score (for rank 1) is positive.Thus, the smallest possible value of ( n ) is (boxed{89}).]

🔑:14. 8914. Let ( m ) be the number of people taking the quiz. Then we have[begin{aligned}2009 & =n+(n-2)+(n-4)+cdots+(n-2 m+2) & =frac{m}{2}[n+(n-2 m+2)] & =m(n-m+1)end{aligned}]Hence ( n=frac{2009}{m}+m-1 ), where ( m ) is a factor of 2009. We check that ( m=41 ) gives the smallest value of ( n ), which is 89.

🧠:[<think>Okay, so I need to find the negation of the proposition "For all x > 1, (1/2)^x < 1/2". Let me think about how to approach this. First, the original statement is a universal quantifier: "For all x > 1, something happens." The negation of a universal quantifier is an existential quantifier. That is, the negation of "For all x, P(x)" is "There exists an x such that not P(x)". So applying that here, the negation should be "There exists an x > 1 such that (1/2)^x is not less than 1/2". Now, "not less than 1/2" would mean greater than or equal to 1/2. So the negation becomes: "There exists an x > 1 such that (1/2)^x ≥ 1/2". Let me check the answer choices. Option C is exactly that: "There exists an x₀ > 1, such that (1/2)^x₀ ≥ 1/2". That seems right. But let me double-check the other options to make sure I didn't miss anything.Option A says "For all x > 1, (1/2)^x ≥ 1/2". That's not the negation; the original was a universal statement, and the negation isn't another universal statement but an existential one. So A is incorrect.Option B: "For all x ≤ 1, (1/2)^x ≥ 1/2". That's talking about a different domain, x ≤ 1, which isn't related to the original statement about x > 1. So B is not the negation.Option D: "There exists an x₀ ≤ 1 such that (1/2)^x₀ ≥ 1/2". Again, this changes the domain to x ≤ 1. The original statement was about x > 1, so the negation should still be about x > 1. So D is incorrect.Therefore, the correct answer must be C. Let me verify once more. Original statement: All x > 1 make (1/2)^x less than 1/2. The negation is that at least one x > 1 doesn't make it less than 1/2, meaning it's greater or equal. So yes, that's exactly option C. Just to be thorough, maybe I can think about the function (1/2)^x. Since the base is between 0 and 1, the function is decreasing. So when x increases, the value decreases. At x = 1, (1/2)^1 = 1/2. For x > 1, since it's decreasing, (1/2)^x should be less than 1/2. So the original statement is actually true. But we're not asked about the truth, just the negation. But if the original statement is true, its negation should be false. Let's check each option. If the original is true, then there does not exist an x > 1 with (1/2)^x ≥ 1/2. Therefore, option C would be false, which aligns with the negation being false if the original is true. So that's consistent.I think that's solid. So the answer is C.Final Answerboxed{C}</think>The original proposition is "For all ( x > 1 ), ( left(frac{1}{2}right)^x < frac{1}{2} )". To find its negation, we start by recognizing that the negation of a universal quantifier ("for all") is an existential quantifier ("there exists"). The negation of the original statement is: "There exists an ( x > 1 ) such that ( left(frac{1}{2}right)^x ) is not less than ( frac{1}{2} )". This translates to "There exists an ( x_0 > 1 ) such that ( left(frac{1}{2}right)^{x_0} geq frac{1}{2} )".We verify the answer choices:- Option C states exactly this negation.- Options A, B, and D either change the quantifier type or the domain, making them incorrect.Thus, the correct answer is (boxed{C}).,<think>Okay, so I need to find the negation of the proposition "For all x > 1, (1/2)^x < 1/2." Hmm, let's start by recalling how to negate statements with quantifiers. The original statement is a universal quantifier: "For all x > 1, P(x) is true," where P(x) is "(1/2)^x < 1/2." The negation of a universal statement is an existential statement. That is, "There exists an x such that not P(x)." So, the negation should be "There exists an x > 1 such that (1/2)^x is not less than 1/2." In other words, "There exists an x > 1 such that (1/2)^x ≥ 1/2." Let me check the options given.Option C says: "There exists an x₀ > 1 such that (1/2)^x₀ ≥ 1/2." That seems to match exactly what I just came up with. Let me make sure the other options aren't correct. Option A says "For all x > 1, (1/2)^x ≥ 1/2." That's not the negation; the negation isn't a universal statement, it's existential. So A is out. Option B is "For all x ≤ 1, (1/2)^x ≥ 1/2." That's changing both the quantifier and the domain of x. The original statement was about x > 1, so the negation should still be about x > 1, just changing the universal to existential and the inequality. So B is not the negation. Option D is "There exists an x₀ ≤ 1 such that (1/2)^x₀ ≥ 1/2." Again, similar to B, this is changing the domain of x to ≤1, which isn't related to the original statement's domain. The original was about x > 1, so the negation should still be within that domain but asserting that the inequality doesn't hold for at least one x there. Therefore, D is also incorrect. So, the correct answer must be C. Wait, just to double-check, let's verify with an example. Take x = 2. Then (1/2)^2 = 1/4, which is less than 1/2. For x = 1.5, (1/2)^1.5 is about 0.3535, still less than 0.5. If x approaches 1 from the right, say x = 1.1, then (1/2)^1.1 is (1/2)^(1 + 0.1) = (1/2)*(1/2)^0.1 ≈ 0.5 * 0.933 = ~0.466, still less than 0.5. Wait, so actually for all x > 1, (1/2)^x is less than 1/2. That means the original statement is true. Then the negation should be false. But according to option C, there exists an x₀ >1 where (1/2)^x₀ ≥1/2. But from the examples, it seems there is no such x₀. So if the original statement is true, its negation should be false, which aligns with C being false here. So that's correct. The options are about the logical negation, not about whether it's true or false in reality. So regardless of the actual truth value, the negation is correctly given by option C. So yeah, the answer is C.Final Answerboxed{C}</think>To find the negation of the proposition "For all ( x > 1 ), ( left(frac{1}{2}right)^x < frac{1}{2} )," we start by recalling that the negation of a universal quantifier is an existential quantifier. The original statement is a universal quantifier: "For all ( x > 1 ), ( left(frac{1}{2}right)^x < frac{1}{2} )." The negation of this statement is "There exists an ( x > 1 ) such that ( left(frac{1}{2}right)^x geq frac{1}{2} )."We check the given options:- Option C states: "There exists an ( x_0 > 1 ) such that ( left(frac{1}{2}right)^{x_0} geq frac{1}{2} )." This matches our derived negation.- Options A, B, and D are incorrect because they either change the quantifier incorrectly or alter the domain of ( x ), which is not relevant to the original statement's domain.Thus, the correct answer is (boxed{C}).]

🔑:Since the negation of a universal proposition is an existential proposition, the negation of the proposition "For all x > 1, (frac{1}{2})^{x} 1, such that (frac{1}{2})^{x_{0}} geqslant frac{1}{2}". Therefore, the correct answer is: boxed{C}. By utilizing the fact that the negation of a universal proposition is an existential proposition, we can directly write the negation of the proposition. This question examines the application of the negation of propositions. The relationship between universal and existential propositions and their negations is a fundamental knowledge application.

🧠:[<think>Okay, let's see. I need to solve this problem where I have an inequality with solution set -2 < x < 1, and then there's a function involving parameters a, m, c that isn't monotonic in some interval depending on t. Hmm. Let me break this down step by step.First, the inequality given is x² + a x - c < 0, and the solution set is -2 < x < 1. That means that the quadratic equation x² + a x - c = 0 has roots at x = -2 and x = 1. Because the quadratic is positive outside the roots and negative between them if the coefficient of x² is positive. Since the coefficient here is 1, which is positive, the inequality x² + a x - c < 0 holds between the roots. So the roots must be x = -2 and x = 1.From the quadratic equation, the sum of the roots is -a, right? Because for a quadratic ax² + bx + c = 0, the sum of roots is -b/a. Wait, here our quadratic is x² + a x - c = 0, so a here is actually the coefficient of x, which is a, and the constant term is -c. So the sum of roots is -a/1 = -a, and the product of roots is (-c)/1 = -c.But the roots are -2 and 1, so sum of roots is (-2) + 1 = -1. Therefore, -a = -1, so a = 1. Then the product of roots is (-2)(1) = -2, so product is -c = -2, so c = 2. Wait, that seems straightforward. So a = 1 and c = 2.Okay, so now we have a = 1 and c = 2. Now the function is f(x) = a x³ + (m + 1/2) x² - c x. Substituting a and c, we get f(x) = x³ + (m + 1/2) x² - 2x.We need to find the range of m such that for any t in [1, 2], the function f(x) is not monotonic in the interval (t, 3). So, for every t between 1 and 2, when we look at the interval (t, 3), the function isn't entirely increasing or decreasing there. That means that in each such interval, the function must have at least one critical point—that is, a point where the derivative is zero. Because if the function is monotonic, its derivative doesn't change sign; so to not be monotonic, the derivative must change sign, implying a critical point in the interval.Therefore, the derivative f’(x) must have at least one root in (t, 3) for every t ∈ [1, 2]. So, the problem reduces to ensuring that for all t in [1, 2], the derivative f’(x) has a root in (t, 3). Therefore, the equation f’(x) = 0 must have a solution in (t, 3) for every t between 1 and 2.So first, let's compute the derivative of f(x). f(x) = x³ + (m + 1/2) x² - 2x. Therefore, f’(x) = 3x² + 2(m + 1/2)x - 2. Simplify that: 3x² + (2m + 1)x - 2.So, f’(x) = 3x² + (2m + 1)x - 2. We need this quadratic equation to have at least one real root in (t, 3) for every t in [1, 2].But since f’(x) is a quadratic, it can have at most two real roots. So, the idea is that for the quadratic derivative, between any t (1 ≤ t ≤ 2) and 3, there must be a root. To ensure this, the quadratic must have a root in (1, 3), and also, when t increases up to 2, the root must still be in (t, 3). Therefore, the smallest t is 1, so there must be a root in (1, 3), and the largest t is 2, so there must be a root in (2, 3). Thus, the quadratic f’(x) must have at least one root in (2, 3), and also, since for t=1, we need a root in (1,3), but actually, if there's a root in (2,3), then for t=1, the interval (1,3) would include that root. Wait, but maybe we need a root that's in (1,3) but also in (2,3) for t=2. Hmm.Wait, perhaps more precise: For all t in [1,2], the interval (t,3) must contain at least one root of f’(x). Therefore, the roots of f’(x) must lie such that for every t between 1 and 2, there is at least one root greater than t. That would mean that the smallest root of f’(x) must be less than 3, but also that the largest root is greater than 2? Wait, let me think.Suppose f’(x) has two roots, say r1 and r2, with r1 ≤ r2. For the interval (t,3) to contain at least one root for every t in [1,2], it must be that for all t ∈ [1,2], either r1 ∈ (t,3) or r2 ∈ (t,3). But since t can be as large as 2, then when t = 2, the interval is (2,3). So, for t=2, there must be a root in (2,3). Therefore, r2 must be in (2,3). Additionally, for t=1, the interval is (1,3). Since r2 is already in (2,3), it will be in (1,3) as well. But what about t between 1 and 2? For example, t=1.5, the interval is (1.5,3). We need at least one root in (1.5,3). Since r2 is in (2,3), which is within (1.5,3). Similarly, for t=1.9, interval (1.9,3) still contains r2 if r2 is in (2,3). Therefore, as long as r2 is in (2,3), then for all t in [1,2], the interval (t,3) will contain r2. But wait, if r2 is only required to be in (2,3), then for t=2, (2,3) must contain r2, so r2 must be >2. Similarly, if r2 is in (2,3), then for all t in [1,2], since t ≤2, the interval (t,3) will include (2,3), hence include r2. Therefore, as long as there's a root in (2,3), then all intervals (t,3) with t ≤2 will contain that root.But also, perhaps there's another root less than 1? Then, for t=1, the interval (1,3) would include both roots if r1 is less than 1. But since t is in [1,2], if there's a root less than 1, then for t=1, the interval (1,3) would exclude that root. But we need at least one root in (t,3). So if the other root is in (2,3), then even if there's a root less than 1, as long as the other root is in (2,3), then for all t in [1,2], the interval (t,3) will include the root in (2,3). So maybe even if there's another root less than 1, it's okay as long as there is a root in (2,3). But wait, let's check.Suppose f’(x) has roots at r1 and r2, with r1 < 1 and r2 ∈ (2,3). Then for t=1, the interval (1,3) includes r2, which is in (2,3). For t=2, the interval (2,3) must include r2. So as long as r2 is in (2,3), regardless of r1, it's okay. Therefore, the key condition is that the quadratic derivative f’(x) must have a root in (2,3). Additionally, we need to ensure that for t from 1 to 2, the interval (t,3) contains at least one root, which as above, if r2 is in (2,3), then even when t approaches 2, r2 is still in (t,3). Therefore, the crucial thing is that the larger root of f’(x) is in (2,3). So we need to find m such that the equation 3x² + (2m +1)x -2 =0 has a root in (2,3).But since it's a quadratic, to have a root in (2,3), we can use the Intermediate Value Theorem. Let me compute f’(2) and f’(3). Let's denote the quadratic equation as Q(x) = 3x² + (2m +1)x -2. Then, Q(2) = 3*(4) + (2m +1)*2 -2 = 12 + 4m + 2 -2 = 12 + 4m. Q(3) = 3*9 + (2m +1)*3 -2 = 27 + 6m +3 -2 = 28 + 6m.For Q(x) to have a root in (2,3), Q(2) and Q(3) must have opposite signs. Because if the quadratic crosses zero between 2 and 3, then Q(2) and Q(3) have opposite signs. So:Q(2) * Q(3) < 0So (12 + 4m)(28 + 6m) < 0Let me compute this inequality.First, factor out 4 and 6:12 +4m = 4(m + 3)28 +6m = 2(3m +14)Wait, no, let's do it step by step. (12 +4m)(28 +6m) <0Multiply out:12*28 +12*6m +4m*28 +4m*6m <0But maybe it's easier to solve the inequality without expanding:(12 +4m)(28 +6m) <0First, find the critical points where each factor is zero.12 +4m =0 => m= -328 +6m=0 => m= -28/6 = -14/3 ≈ -4.6667So the critical points are at m= -14/3 and m= -3.Now, let's analyze the sign of the product (12 +4m)(28 +6m):The factors are 12 +4m and 28 +6m.The first factor, 12 +4m, is positive when m > -3, negative when m < -3.The second factor, 28 +6m, is positive when m > -14/3 ≈ -4.6667, negative when m < -14/3.So, when m < -14/3: 12 +4m <0 (since m < -14/3 < -3), 28 +6m <0. So the product is positive.When -14/3 < m < -3: 12 +4m <0 (since m < -3), 28 +6m >0. So the product is negative.When m > -3: 12 +4m >0, 28 +6m >0. So the product is positive.Therefore, (12 +4m)(28 +6m) <0 when -14/3 < m < -3.Therefore, the inequality Q(2)*Q(3) <0 holds when m is in (-14/3, -3). Therefore, the quadratic f’(x) has a root in (2,3) when m is in (-14/3, -3).But we need to ensure that for any t in [1,2], the interval (t,3) contains at least one root. As we discussed earlier, if the larger root is in (2,3), then for any t in [1,2], the interval (t,3) will contain that root. Therefore, m must be in (-14/3, -3). But we also need to check whether the quadratic might have another root in (1,3), but perhaps that is already covered?Wait, but suppose m is such that the larger root is in (2,3), but the smaller root is less than 1. Then, for t=1, the interval (1,3) would contain the larger root, which is in (2,3), so that's okay. For t=2, the interval (2,3) must contain the larger root. Therefore, as long as the larger root is in (2,3), regardless of where the smaller root is, the condition is satisfied.But also, we need to ensure that f’(x) actually has real roots. So discriminant must be non-negative.Compute discriminant D:For Q(x) = 3x² + (2m +1)x -2,Discriminant D = (2m +1)^2 -4*3*(-2) = (4m² +4m +1) +24 =4m² +4m +25.Which is always positive, since 4m² +4m +25 = 4(m² + m) +25 = 4(m² + m + 1/4) +25 -1 = 4(m + 0.5)^2 +24, which is always positive. Therefore, the quadratic always has two real roots. So no problem about discriminant.Therefore, the only condition is that the larger root is in (2,3). The larger root is given by:r2 = [ - (2m +1) + sqrt(D) ] / (2*3)But since the quadratic is 3x² + (2m +1)x -2, the standard formula is:x = [ -B ± sqrt(B² -4AC) ] / (2A)Here, A=3, B=2m +1, C=-2. So,r2 = [ - (2m +1) + sqrt{(2m +1)^2 +24} ] /6But instead of computing it directly, we can use the condition that the larger root is in (2,3). Since we already imposed that Q(2) and Q(3) have opposite signs, which ensures that there's a root in (2,3). But we need to confirm that this is the larger root. Since Q(x) is a quadratic opening upwards (coefficient 3 >0), then between the two roots, Q(x) is negative, and outside the roots, it's positive. Therefore, if Q(2) <0 and Q(3) >0, then there's a root in (2,3), which is the larger root. Similarly, if Q(2) >0 and Q(3) <0, which can't happen here because for m in (-14/3, -3), let's check the sign:Wait, when m is in (-14/3, -3):12 +4m: For m > -14/3 ≈ -4.6667, but m < -3. So 12 +4m. Let's take m=-4, which is in (-14/3, -3). 12 +4*(-4)=12-16= -4 <0. Similarly, 28 +6m: For m=-4, 28 +6*(-4)=28-24=4>0. So Q(2)=12 +4m <0, Q(3)=28 +6m >0. Therefore, since Q(x) opens upwards, the quadratic crosses from negative to positive, so the root in (2,3) is the larger root. Therefore, the larger root is in (2,3), and the smaller root is to the left of 2. So the other root is less than 2. So for all t in [1,2], the interval (t,3) will include the larger root in (2,3). Therefore, the only condition needed is m ∈ (-14/3, -3).But wait, let me check with m=-3. If m=-3, then Q(2)=12 +4*(-3)=12-12=0. Therefore, x=2 is a root. But the problem states "not monotonic in the interval (t, 3)". If x=2 is a critical point, then at x=2, the derivative is zero. So for t approaching 2 from the left, say t=2-ε, the interval (t,3) includes x=2. But if m=-3, then x=2 is a root. However, the function f’(x) has a root at x=2. Therefore, in the interval (t,3), with t approaching 2, does x=2 count as being in (t,3)? If t is strictly less than 2, then (t,3) includes 2 if t <2, but when t=2, the interval is (2,3). So if m=-3, then x=2 is a root. So when t=2, the interval is (2,3), and x=2 is not included. So f’(x) at x=2 is zero, but x=2 is not in (2,3). Therefore, for t=2, the interval is (2,3), and we need a root in there. If m=-3, Q(2)=0, but Q(3)=28 +6*(-3)=28-18=10>0. Therefore, the quadratic at m=-3 has a root at x=2, and the other root is where? Let's compute:Q(x)=3x² + (2*(-3)+1)x -2=3x² + (-6 +1)x -2=3x² -5x -2.So roots are [5 ± sqrt(25 +24)] /6 = [5 ± sqrt(49)] /6 = [5 ±7]/6. So roots are (12)/6=2 and (-2)/6=-1/3. So roots at x=2 and x=-1/3. Therefore, at m=-3, the roots are x=-1/3 and x=2. Therefore, for t=2, the interval (2,3) does not include x=2, and there are no roots of f’(x) in (2,3). Hence, if m=-3, then for t=2, f’(x) has no roots in (2,3), so f(x) could be monotonic there. Therefore, m=-3 is excluded.Similarly, when m=-14/3, which is approximately -4.6667, then Q(3)=28 +6*(-14/3)=28 -28=0. So x=3 is a root. So for m=-14/3, Q(x)=3x² + (2*(-14/3)+1)x -2=3x² + (-28/3 +1)x -2=3x² -25/3 x -2. Let's compute the roots:x = [25/3 ± sqrt((25/3)^2 +24)] /6But regardless, for m=-14/3, Q(3)=0. So the roots are x=3 and another root. Let's compute the other root. Q(x)=3x² - (25/3)x -2. Multiply through by 3 to eliminate fractions: 9x² -25x -6=0. Then discriminant D=625 + 216=841=29². So roots are [25 ±29]/18. So (25+29)/18=54/18=3, and (25-29)/18=(-4)/18=-2/9. Therefore, the roots are x=3 and x=-2/9. Therefore, when m=-14/3, the roots are x=-2/9 and x=3. So for t in [1,2], the interval (t,3) must contain a root. For t approaching 3 from the left, but since t is up to 2, the interval (t,3) with t=2 is (2,3). But when m=-14/3, the root is at x=3, which is not included in (2,3). Therefore, for t=2, the interval (2,3) does not include x=3, and there are no roots in (2,3). Therefore, m=-14/3 is also excluded.Thus, the range of m is (-14/3, -3). But we have to write this in fraction form. -14/3 is approximately -4.666..., and -3 is -3. Therefore, the range is m ∈ (-14/3, -3).But let me verify with an example. Let's take m=-4, which is in (-14/3, -3). Compute Q(x)=3x² + (2*(-4)+1)x -2=3x² -7x -2.Find the roots: x=(7 ±sqrt(49 +24))/6=(7±sqrt(73))/6.sqrt(73)≈8.544, so larger root≈(7+8.544)/6≈15.544/6≈2.5907, which is in (2,3). Therefore, the larger root is indeed in (2,3), so for m=-4, which is in the interval, it's okay.For m=-3.5 (which is -7/2, in (-14/3≈-4.666, -3)), compute Q(x)=3x² + (2*(-3.5)+1)x -2=3x² -6x -2. The roots are [6 ±sqrt(36 +24)]/6=[6 ±sqrt(60)]/6=[6 ±2*sqrt(15)]/6= [3 ±sqrt(15)]/3. sqrt(15)≈3.872, so larger root≈(3 +3.872)/3≈6.872/3≈2.29, which is in (2,3). So again, works.If m=-5, which is less than -14/3≈-4.666, say m=-5. Then Q(x)=3x² + (2*(-5)+1)x -2=3x² -9x -2. Roots: [9 ±sqrt(81 +24)]/6=[9 ±sqrt(105)]/6≈[9 ±10.246]/6. The larger root≈(9 +10.246)/6≈19.246/6≈3.207, which is greater than 3. So in this case, the larger root is outside of (2,3). Hence, for m=-5, the larger root is at ≈3.207, which is outside (2,3). Then for t=2, the interval (2,3) does not contain any roots (since the root is at ~3.207). Therefore, m=-5 is invalid, which is consistent with our previous result.Similarly, if m=-2, which is greater than -3. Then Q(x)=3x² + (2*(-2)+1)x -2=3x² -3x -2. Roots: [3 ±sqrt(9 +24)]/6=[3 ±sqrt(33)]/6≈[3 ±5.744]/6. The larger root≈(3 +5.744)/6≈8.744/6≈1.457, which is less than 2. Hence, both roots are less than 2. Therefore, for t=2, the interval (2,3) contains no roots. Therefore, m=-2 is invalid, which is again consistent.Therefore, our conclusion that m must be in (-14/3, -3) seems correct.But just to make sure, let's check the endpoints. For m=-14/3≈-4.6667, as we saw earlier, the roots are x=3 and x=-2/9. So for t=2, the interval (2,3) does not include x=3, hence no roots there. Similarly, for m=-3, the roots are x=2 and x=-1/3, so for t=2, interval (2,3) doesn't include x=2, and the other root is at x=-1/3. Hence, no roots in (2,3). Therefore, both endpoints are excluded, so open interval.Therefore, the range of m is (-14/3, -3).Final AnswerThe range of values for ( m ) is boxed{left( -dfrac{14}{3}, -3 right)}.</think>Given the inequality (x^2 + ax - c < 0) with the solution set (-2 < x < 1), we determine that the roots of the quadratic equation (x^2 + ax - c = 0) are (x = -2) and (x = 1). Using Vieta's formulas, we find:- The sum of the roots (-2 + 1 = -1) gives (a = 1).- The product of the roots ((-2)(1) = -2) gives (c = 2).Substituting (a = 1) and (c = 2) into the function (f(x) = ax^3 + (m + frac{1}{2})x^2 - cx), we get:[ f(x) = x^3 + left(m + frac{1}{2}right)x^2 - 2x. ]We need to ensure that (f(x)) is not monotonic in the interval ((t, 3)) for any (t in [1, 2]). This requires the derivative (f'(x)) to have at least one root in ((t, 3)) for all (t in [1, 2]).The derivative is:[ f'(x) = 3x^2 + (2m + 1)x - 2. ]We need the quadratic equation (3x^2 + (2m + 1)x - 2 = 0) to have a root in ((2, 3)). Using the Intermediate Value Theorem, we check the values of the quadratic at (x = 2) and (x = 3):- (Q(2) = 12 + 4m)- (Q(3) = 28 + 6m)For the quadratic to have a root in ((2, 3)), (Q(2)) and (Q(3)) must have opposite signs:[ (12 + 4m)(28 + 6m) < 0. ]Solving this inequality, we find the critical points (m = -3) and (m = -frac{14}{3}). Analyzing the intervals, we determine that the inequality holds for (-frac{14}{3} < m < -3).Thus, the range of values for (m) is:[boxed{left( -dfrac{14}{3}, -3 right)}],<think>Alright, let's tackle this problem step by step. It's a bit involved, but if I break it down, I can manage each part one by one. First, the problem states that the solution set of the inequality ( x^2 + ax - c < 0 ) is ( {x mid -2 < x < 1} ). Then, for any ( t in [1, 2] ), the function ( f(x) = ax^3 + left(m + frac{1}{2}right)x^2 - cx ) is not monotonic in the interval ( (t, 3) ). We need to find the range of values for the real number ( m ).Starting with the inequality ( x^2 + ax - c < 0 ). The solution set is given as ( -2 < x < 1 ). Since this is a quadratic inequality, the quadratic equation ( x^2 + ax - c = 0 ) must have roots at ( x = -2 ) and ( x = 1 ). That’s because the inequality is satisfied between the roots if the quadratic opens upwards. So, let's recall that if a quadratic equation ( x^2 + bx + c = 0 ) has roots ( r_1 ) and ( r_2 ), then it can be written as ( (x - r_1)(x - r_2) = 0 ). Expanding this, we get ( x^2 - (r_1 + r_2)x + r_1 r_2 = 0 ). Comparing coefficients, we have ( b = -(r_1 + r_2) ) and ( c = r_1 r_2 ).In our case, the quadratic is ( x^2 + ax - c ). Comparing to the standard form ( x^2 + bx + c ), here ( b = a ) and the constant term is ( -c ). The roots are ( -2 ) and ( 1 ), so applying the formulas:Sum of roots: ( -2 + 1 = -1 ). Therefore, coefficient of x is ( -(-1) = 1 ). Wait, but in our quadratic, the coefficient of x is ( a ). So, ( a = 1 ).Product of roots: ( (-2)(1) = -2 ). The constant term of the quadratic is ( -c ), so ( -c = -2 ). Therefore, ( c = 2 ).So, we have determined that ( a = 1 ) and ( c = 2 ).Now, moving on to the function ( f(x) = ax^3 + left(m + frac{1}{2}right)x^2 - cx ). Plugging in the values of ( a ) and ( c ), this becomes:( f(x) = 1 cdot x^3 + left(m + frac{1}{2}right)x^2 - 2x )Simplifying:( f(x) = x^3 + left(m + frac{1}{2}right)x^2 - 2x )The next part says that for any ( t in [1, 2] ), the function ( f(x) ) is not monotonic in the interval ( (t, 3) ). To find when a function is not monotonic in an interval, we need to ensure that its derivative changes sign in that interval, i.e., the derivative has at least one critical point (local maximum or minimum) in the interval. If the derivative is always non-decreasing or non-increasing, the function is monotonic. So, we need to ensure that the derivative ( f'(x) ) has at least one real root in the interval ( (t, 3) ) for every ( t in [1, 2] ).First, let's compute the derivative of ( f(x) ):( f'(x) = 3x^2 + 2left(m + frac{1}{2}right)x - 2 )Simplifying:( f'(x) = 3x^2 + (2m + 1)x - 2 )So, we need ( f'(x) = 0 ) to have at least one solution in ( (t, 3) ) for every ( t in [1, 2] ). Equivalently, for each ( t in [1, 2] ), the equation ( 3x^2 + (2m + 1)x - 2 = 0 ) has at least one root in ( (t, 3) ). Therefore, our task reduces to finding the values of ( m ) such that for all ( t in [1, 2] ), there exists an ( x in (t, 3) ) such that ( 3x^2 + (2m + 1)x - 2 = 0 ).Alternatively, we can rephrase this as: For all ( t in [1, 2] ), the quadratic equation ( 3x^2 + (2m + 1)x - 2 = 0 ) has a root greater than ( t ). So, the largest root of the quadratic must be greater than all ( t in [1, 2] ), which implies that the largest root must be greater than 2. Wait, but the interval is ( (t, 3) ), so if the largest root is greater than 2, then even when ( t = 2 ), there would still be a root in ( (2, 3) ). Hmm, but actually, the quadratic has two roots (assuming they are real). Let me think.Alternatively, maybe we need that for each ( t in [1, 2] ), there is a root in ( (t, 3) ). So, the quadratic equation must cross the x-axis at least once in ( (t, 3) ). Since quadratic is continuous, this can be ensured if either:1. The quadratic has two real roots, and the smaller root is less than 3 and the larger root is greater than t, but given the interval is up to 3.Wait, perhaps a better approach is to consider that for each ( t in [1, 2] ), the equation ( f'(x) = 0 ) has a solution in ( (t, 3) ). To guarantee that, for each t, the quadratic equation must have at least one root in ( (t, 3) ). This condition is equivalent to saying that the quadratic equation ( 3x^2 + (2m + 1)x - 2 = 0 ) has at least one root in ( (t, 3) ) for every ( t in [1, 2] ). To ensure this, for each ( t in [1, 2] ), the quadratic must be such that either:1. There is exactly one root in ( (t, 3) ), which would require that the quadratic crosses the x-axis there, or2. There are two roots in ( (t, 3) ).But since quadratics can have at most two real roots, the way to ensure that for every ( t in [1, 2] ), there's a root in ( (t, 3) ), the largest root of the quadratic must be greater than 2 (the maximum t in [1, 2]) and the smallest root must be less than 3? Hmm, maybe not. Let's think carefully.Alternatively, perhaps for each t between 1 and 2, we need that the quadratic equation has a root between t and 3. So, the union of all these intervals (t, 3) where t ranges from 1 to 2 is (1, 3). Wait, actually, when t =1, the interval is (1,3); when t=2, the interval is (2,3). So the union of all these intervals is (1,3), but each interval is (t,3) where t is moving from 1 to 2. So for t in [1,2], the interval (t,3) is (1,3) when t=1, shrinking down to (2,3) when t=2.Therefore, the problem is requiring that for all t in [1,2], the quadratic equation has at least one root in (t,3). So, in particular, even when t=2, we need a root in (2,3). Therefore, the quadratic must have a root in (2,3). Also, when t=1, we need at least one root in (1,3). But if there's already a root in (2,3), then that root will also be in (1,3). So the critical case is when t=2, requiring a root in (2,3). However, we need to ensure that even as t increases from 1 to 2, the interval (t,3) still contains a root. So, for all t up to 2, there's a root greater than t and less than 3. Therefore, the smallest such root that works for all t is the root at the maximum t, which is 2. So, if there's a root at some x > 2, then for t=2, we need x >2, but x must be less than 3. So, if the quadratic has a root in (2,3), then for all t ∈ [1,2], since t ≤2, the interval (t,3) will include (2,3), so that root is still present. But wait, is that sufficient?Wait, no. Suppose the quadratic has a root at x = 2.5. Then, for t=1, the interval (1,3) includes 2.5, so it's okay. For t=1.5, the interval (1.5,3) still includes 2.5. For t=2, the interval (2,3) includes 2.5. So, as long as there's a root in (2,3), then for all t ∈ [1,2], the interval (t,3) will contain that root, hence satisfying the condition. However, what if the quadratic only has a root in (1,2)? Then, for t=2, the interval (2,3) does not include that root, so the function would be monotonic there, which we don't want. Therefore, it's necessary that there's a root in (2,3). Additionally, if there's another root, say, less than 1, that doesn't affect the condition because the interval (t,3) is only concerned with x > t ≥1, so roots less than 1 are irrelevant. However, we need to ensure that even if there's a root in (1,2), but there's another root in (2,3), then for t=2, the interval (2,3) still has the second root. Therefore, the critical condition is that the quadratic must have a root in (2,3). Additionally, the quadratic must have real roots, so discriminant must be non-negative.Wait, but maybe it's not enough to just have a root in (2,3). Let's check.Suppose quadratic has roots at 0.5 and 2.5. Then, for t=1, the interval (1,3) includes 2.5, which is a root. For t=2, the interval (2,3) includes 2.5. So that works. But if quadratic has roots at 1.5 and 2.5. Then, for t=1, (1,3) includes both 1.5 and 2.5. For t=1.5, the interval (1.5,3) includes 2.5. For t=2, includes 2.5. So that works as well.Alternatively, if the quadratic only has one root in (2,3), but that's impossible because quadratics have two roots (counting multiplicity). So, if discriminant is positive, two distinct roots; if zero, one root. So, in order to have a root in (2,3), we need that either one root is in (2,3) and the other is outside, or both roots are in (2,3). Wait, if both roots are in (2,3), then for t=2, the interval (2,3) contains both roots, which is fine, but for t=1, the interval (1,3) would include both roots. However, if both roots are in (2,3), then for t=1, the entire interval (1,3) is considered, which includes the roots, so the function is not monotonic there. So that would be acceptable.But the problem states that "for any t ∈ [1,2], the function f(x) is not monotonic in the interval (t,3)". So, even if both roots are in (2,3), then for any t ∈ [1,2], the interval (t,3) would include those roots, so the derivative has roots there, so the function is not monotonic. Similarly, if there is one root in (2,3) and another root elsewhere, as long as the other root is not in (t,3), but since we need at least one root in (t,3), having one root in (2,3) would suffice for all t ∈ [1,2], because as t increases from 1 to 2, the interval (t,3) would still contain (2,3), hence that root. However, if the only roots are, say, in (1,2), then for t=2, the interval (2,3) has no roots, so the derivative doesn't cross zero there, hence the function would be monotonic in (2,3), which violates the condition.Therefore, the key is that the quadratic equation ( 3x^2 + (2m + 1)x - 2 = 0 ) must have at least one root in (2,3). So, our task reduces to finding m such that the quadratic has at least one root in (2,3). But we also need to consider the case where the quadratic might have a repeated root or something, but since the problem says "not monotonic", which requires the derivative to have a root, so the quadratic must have real roots. So first, we need the discriminant to be non-negative.Let me compute the discriminant of the quadratic equation ( 3x^2 + (2m + 1)x - 2 = 0 ). The discriminant D is:( D = (2m + 1)^2 - 4 cdot 3 cdot (-2) )( D = (4m² + 4m + 1) + 24 )( D = 4m² + 4m + 25 )Since 4m² + 4m + 25 is a quadratic in m. Let's check if it's always positive:Compute its discriminant: ( (4)^2 - 4 cdot 4 cdot 25 = 16 - 400 = -384 < 0 ). So, the discriminant D is always positive, meaning that the quadratic equation in x always has two distinct real roots, regardless of m. Therefore, we don't have to worry about the discriminant being negative; there are always two real roots.Now, the next step is to find m such that at least one root of ( 3x² + (2m + 1)x - 2 = 0 ) lies in (2,3). Let me denote the quadratic as ( Q(x) = 3x² + (2m + 1)x - 2 ). We need to find m such that there exists x ∈ (2,3) with Q(x) = 0.To find such m, we can use the Intermediate Value Theorem. For a continuous function Q(x), if Q(2) and Q(3) have opposite signs, then there is a root in (2,3). Alternatively, if Q(2) = 0 or Q(3) = 0, but since we need the root in the open interval (2,3), we can check the sign changes.Compute Q(2):( Q(2) = 3*(4) + (2m + 1)*2 - 2 = 12 + 4m + 2 - 2 = 12 + 4m + 0 = 12 + 4m )Compute Q(3):( Q(3) = 3*(9) + (2m + 1)*3 - 2 = 27 + 6m + 3 - 2 = 27 + 6m + 1 = 28 + 6m )So, Q(2) = 12 + 4m, Q(3) = 28 + 6m.For there to be a root in (2,3), Q(2) and Q(3) must have opposite signs. So:Either Q(2) > 0 and Q(3) < 0, or Q(2) < 0 and Q(3) > 0.Case 1: Q(2) > 0 and Q(3) < 012 + 4m > 0 => 4m > -12 => m > -328 + 6m < 0 => 6m < -28 => m < -28/6 = -14/3 ≈ -4.6667But m must satisfy both m > -3 and m < -14/3. However, -14/3 ≈ -4.6667 is less than -3, so there is no solution in this case.Case 2: Q(2) < 0 and Q(3) > 012 + 4m < 0 => 4m < -12 => m < -328 + 6m > 0 => 6m > -28 => m > -28/6 = -14/3 ≈ -4.6667So, in this case, m must satisfy -14/3 < m < -3Thus, the values of m that make Q(x) have a root in (2,3) are m ∈ (-14/3, -3)But we need to check if these are the only possibilities. Alternatively, even if Q(2) and Q(3) have the same sign, there could still be two roots in (2,3). Let's check if that's possible.Suppose Q(2) and Q(3) are both positive. Then, for there to be two roots in (2,3), the quadratic must be positive at both ends but dip below zero in between. But since the quadratic opens upwards (coefficient of x² is 3 >0), the minimum is at the vertex. If the vertex is within (2,3) and the minimum value is negative, then there would be two roots in (2,3). Let's calculate the vertex.The x-coordinate of the vertex is given by -B/(2A) = -(2m +1)/(2*3) = -(2m +1)/6The y-coordinate is Q(-(2m +1)/6). But since the quadratic opens upwards, if the vertex is within (2,3), and Q(vertex) < 0, then there are two roots in (2,3). Let's check if the vertex can be in (2,3):So, we need:2 < -(2m +1)/6 < 3Multiply all parts by 6:12 < -(2m +1) < 18Multiply by -1, reversing inequalities:-12 > 2m +1 > -18Subtract 1:-13 > 2m > -19Divide by 2:-13/2 > m > -19/2Which is -6.5 > m > -9.5But we are in the case where Q(2) >0 and Q(3) >0, which is when m > -3 (from Q(2) >0). But m > -3 contradicts m < -6.5. Therefore, there is no overlap. Hence, when Q(2) and Q(3) are both positive, the vertex is not in (2,3), so the quadratic remains positive, no roots in (2,3). Similarly, if Q(2) and Q(3) are both negative, but the quadratic opens upwards, so if it's negative at both ends, it must have a maximum in between, but since it's a quadratic opening upwards, it cannot have a maximum. Wait, quadratics opening upwards have a minimum, so if Q(2) <0 and Q(3) <0, and it opens upwards, then it would stay below zero between 2 and 3, which would mean no roots? Wait, but if a quadratic opens upwards and is negative at two points, it must cross the x-axis twice. But in that case, between 2 and 3, if it's negative at both ends, but opens upwards, the minimum is below zero, so the roots would be outside the interval (2,3). Wait, no, if it's negative at both 2 and 3, and opens upwards, then the entire interval (2,3) is below the x-axis, so no roots there. Because the quadratic tends to +infinity as x approaches infinity, so after some point, it becomes positive. Therefore, if it's negative at 2 and 3, but opens upwards, then there are two roots: one less than 2, and one greater than 3. Therefore, in (2,3), there are no roots.Therefore, the only case when there is a root in (2,3) is when Q(2) and Q(3) have opposite signs, which we already found as -14/3 < m < -3.Therefore, the range of m is (-14/3, -3).But let's confirm this with an example. Let's pick m = -4, which is in (-14/3 ≈ -4.6667, -3). Then Q(2) = 12 + 4*(-4) = 12 -16 = -4 <0, Q(3) = 28 +6*(-4) = 28 -24 =4 >0. So, by Intermediate Value Theorem, there is a root in (2,3). Good.If m = -14/3 ≈ -4.6667, then Q(2) =12 +4*(-14/3) =12 -56/3 = (36 -56)/3 = -20/3 <0, Q(3)=28 +6*(-14/3)=28 -28=0. So, at m=-14/3, Q(3)=0, so the root is at x=3. But the interval is open, (t,3), so x=3 is not included. Therefore, m cannot be -14/3, because then the root is at 3, which is excluded. Similarly, if m=-3, Q(2)=12 +4*(-3)=0, so root at x=2, which is excluded. Therefore, m must be strictly greater than -14/3 and strictly less than -3.Therefore, the range of m is (-14/3, -3). Converting -14/3 to a mixed number, that's -4 and 2/3, but since the question asks for the range in real numbers, we can leave it as fractions.But let's double-check if there are any other constraints. For example, maybe there are cases where even if there's a root in (2,3), but another root in (t,3) for some t, but I think the way the problem is set up, as long as there's at least one root in (2,3), then for all t in [1,2], the interval (t,3) will include that root, making the function not monotonic. Therefore, the only necessary and sufficient condition is that the quadratic has a root in (2,3), which corresponds to m ∈ (-14/3, -3).But let's check another value. Take m=-5, which is less than -14/3 ≈ -4.6667. Then Q(2)=12 +4*(-5)=12 -20=-8 <0, Q(3)=28 +6*(-5)=28-30=-2 <0. So both Q(2) and Q(3) are negative. Then, as the quadratic opens upwards, there would be a root less than 2 and a root greater than 3. Therefore, in the interval (2,3), the quadratic is negative, so no roots there. Hence, for m=-5, which is outside our range, there is no root in (2,3). Therefore, the function f(x) would be monotonic in (2,3), which is not allowed. Hence, m cannot be less than -14/3.Similarly, take m=-2.5, which is greater than -3. Then Q(2)=12 +4*(-2.5)=12-10=2 >0, Q(3)=28 +6*(-2.5)=28-15=13 >0. So, both Q(2) and Q(3) are positive. The quadratic opens upwards, so the roots are either both less than 2 or both greater than 3. But since Q(2) and Q(3) are positive, and the quadratic tends to infinity as x increases, the roots must be either both less than 2 or both greater than 3. Wait, but let's check the actual roots.For m=-2.5:The quadratic equation is 3x² + (2*(-2.5) +1)x -2 = 3x² + (-5 +1)x -2 = 3x² -4x -2=0Solutions:x = [4 ± sqrt(16 +24)]/6 = [4 ± sqrt(40)]/6 = [4 ± 2*sqrt(10)]/6 = [2 ± sqrt(10)]/3 ≈ [2 ± 3.16]/3So, approximately:(2 +3.16)/3 ≈5.16/3≈1.72(2 -3.16)/3≈-1.16/3≈-0.39So roots at ≈1.72 and ≈-0.39. Therefore, the positive root is at ≈1.72, which is in (1.72,3). Wait, but the positive root is around 1.72, so for t=1, the interval (1,3) includes 1.72, so the function is not monotonic there. However, for t=2, the interval (2,3) does not include 1.72, so in (2,3), since there are no roots (the other root is at -0.39), the quadratic 3x² -4x -2 is positive at x=2 (Q(2)=2 >0) and increasing afterwards, so the derivative remains positive in (2,3). Therefore, the function f(x) would be increasing in (2,3), hence monotonic, which violates the condition. Therefore, m=-2.5, which is greater than -3, does not satisfy the requirement. Hence, our previous conclusion that m must be in (-14/3, -3) is correct.Therefore, the range of m is (-14/3, -3), which in boxed form is boxed{(-dfrac{14}{3}, -3)}.Final Answerboxed{left( -dfrac{14}{3}, -3 right)}</think>Given the inequality (x^2 + ax - c < 0) with the solution set ({x mid -2 < x < 1}), we determine the values of (a) and (c). The roots of the quadratic equation (x^2 + ax - c = 0) are (-2) and (1). Using Vieta's formulas, we find (a = 1) and (c = 2).Next, we consider the function (f(x) = ax^3 + left(m + frac{1}{2}right)x^2 - cx) with (a = 1) and (c = 2), which simplifies to (f(x) = x^3 + left(m + frac{1}{2}right)x^2 - 2x). We need to ensure that (f(x)) is not monotonic in the interval ((t, 3)) for any (t in [1, 2]). This requires the derivative (f'(x) = 3x^2 + (2m + 1)x - 2) to have at least one root in ((t, 3)) for all (t in [1, 2]).The quadratic equation (3x^2 + (2m + 1)x - 2 = 0) must have at least one root in ((2, 3)). We compute the values of (m) such that this condition holds. Using the Intermediate Value Theorem, we check the signs of the quadratic at (x = 2) and (x = 3):- (Q(2) = 12 + 4m)- (Q(3) = 28 + 6m)For (Q(x)) to have a root in ((2, 3)), (Q(2)) and (Q(3)) must have opposite signs:- (Q(2) < 0) and (Q(3) > 0) leads to ( -14/3 < m < -3 ).Thus, the range of values for (m) is (boxed{left( -dfrac{14}{3}, -3 right)}).]

🔑:From the given inequality x^2 + ax - c < 0 with the solution set {x mid -2 < x < 1}, we obtain:begin{cases}-a = -2 + 1 = -1, -c = -2 times 1 = -2.end{cases}So a = 1 and c = 2.Thus, f(x) = ax^3 + (m + frac{1}{2})x^2 - cx = x^3 + (m + frac{1}{2})x^2 - 2x.Taking the derivative of f(x), we get f'(x) = 3x^2 + (2m + 1)x - 2.Since f(x) is not monotonic in the interval (t, 3) for any t in [1, 2], f'(x) must have a root in the interval (2, 3). Therefore, f'(2)f'(3) < 0.Computing the values, we have:[10 + 2(2m + 1)][25 + 3(2m + 1)] < 0,which simplifies to:-frac{14}{3} < m < -3.So the range of values for the real number m is boxed{left(-frac{14}{3}, -3right)}.