🧠:[<think>Okay, so there's this problem where two identical rectangles are placed obliquely on top of each other, and the overlapping area X is one eighth of the total shaded area. We need to find what fraction X is of one rectangle's area. The answer choices are given from A to E. Let me try to visualize this setup first.Since the rectangles are identical and placed obliquely, the overlapping region is probably a smaller parallelogram or another rectangle, depending on how they're rotated. But since they are rectangles, if they are rotated, the overlap might be a more complex shape. Hmm, maybe a diamond shape? Wait, actually, when two rectangles overlap at an angle, the overlapping area is generally a convex polygon, sometimes a hexagon or an octagon, but in simpler cases, like when they're just slightly rotated, it might be a quadrilateral. But the problem mentions that the overlapping region is shaded more darkly, and X is one eighth of the total shaded area. Wait, the total shaded area must be the union of both rectangles, but since they are overlapping, the total shaded area would be the area of both rectangles minus the overlapping area. But the problem says X is one eighth of the total shaded area. So, X = (1/8)*(Total Shaded Area). Then, Total Shaded Area = 2A - X, where A is the area of one rectangle. So substituting, X = (1/8)*(2A - X). Let's write that equation down:X = (1/8)*(2A - X)Multiply both sides by 8:8X = 2A - XBring X to the left side:8X + X = 2A => 9X = 2A => X = (2/9)AWait, so X is 2/9 of the area of one rectangle? Then the answer would be D, 2/9. But wait, let me check if my reasoning is correct.Total shaded area is the union of the two rectangles. Since they are overlapping, union = area of first + area of second - overlap. So union = A + A - X = 2A - X. The problem states X is 1/8 of the total shaded area. So X = (1/8)*(2A - X). Solving for X gives 8X = 2A - X => 9X = 2A => X = 2A/9. Since we need X as a fraction of one rectangle's area (which is A), X/A = 2/9. So the answer is D. That seems straightforward, but let me make sure there's no trick here.Wait, maybe I misunderstood the problem. It says "the area X of the overlapping region (shaded more darkly) is one eighth of the total shaded area." So total shaded area includes both rectangles, but overlapped region is counted once? Or is the total shaded area referring to both overlapped and non-overlapped parts? But in typical terms, the total shaded area would be the union, which is 2A - X. So X is part of that union? Wait, but X is the overlapping part. So if you have two rectangles overlapping, the total shaded area would be the union, which is indeed 2A - X. So the problem states that X = (1/8)*(2A - X). Solving this gives X = 2A/9, so X is 2/9 of A, which is the area of one rectangle. So the answer is D, 2/9.But wait, this seems too straightforward. The problem mentions that the rectangles are placed obliquely. Does that affect anything? Maybe the overlapping area calculation is different because of the angle? But in the problem statement, they just refer to the area X of the overlapping region, regardless of how the rectangles are placed. So even if the overlapping region is a different shape, the area is still X, and the relationship given is X is 1/8 of the total shaded area. So the setup might just require solving the equation regardless of the angle. Then maybe the answer is indeed D. But let me think again.Alternatively, maybe the problem is referring to the total shaded area as both the overlapped and the non-overlapped parts. But when you place two rectangles on top of each other, the total shaded area would be the union, which is 2A - X. So if X is 1/8 of that, then the equation holds. Therefore, solving gives X = 2A/9, so 2/9 of one rectangle's area. So answer D.Wait, but let me check with a different approach. Suppose each rectangle has area A. The overlapping area is X. The total shaded area when you put them together is 2A - X. According to the problem, X = (1/8)*(2A - X). So 8X = 2A - X => 9X = 2A => X = 2A/9. So yes, the fraction is 2/9. Therefore, answer D. Hmm, maybe that's it. So despite the oblique placement, the key is the relationship between X and the total shaded area. So the answer is D. I think that's correct.But just to be thorough, let me imagine a concrete example. Suppose each rectangle has area 9. Then X would be 2, since 2/9 of 9 is 2. The total shaded area would be 2*9 - 2 = 16. Then X is 2, which is 1/8 of 16. That works. So in this case, 2 is 1/8 of 16, and 2 is 2/9 of 9. So the math checks out. Therefore, the answer should be D.Final Answerboxed{D}</think>Given two identical rectangles placed obliquely on top of each other, the area ( X ) of the overlapping region is one eighth of the total shaded area. We need to determine what fraction of the area of one rectangle ( X ) represents.1. Let the area of one rectangle be ( A ).2. The total shaded area (the union of the two rectangles) is given by ( 2A - X ).3. According to the problem, ( X ) is one eighth of the total shaded area: [ X = frac{1}{8} times (2A - X) ]4. Solving the equation: [ 8X = 2A - X ] [ 9X = 2A ] [ X = frac{2A}{9} ]5. Therefore, the fraction of the area of one rectangle that ( X ) represents is: [ frac{X}{A} = frac{2}{9} ]The final answer is (boxed{D}).,<think>Okay, let me try to figure out this problem. So, there are two identical rectangles placed obliquely on top of each other, and the overlapping area X is one eighth of the total shaded area. We need to find what fraction of one rectangle's area is X. The options are given, so maybe that can help check my answer later.First, let me visualize this. If two identical rectangles are overlapping at an angle, the overlapping region is probably a smaller quadrilateral, maybe a parallelogram or another rectangle? Hmm. The problem says the overlapping region is shaded more darkly, and X is one eighth of the total shaded area. Wait, the total shaded area would be the union of both rectangles? But since they are overlapping, the total shaded area is the area of both rectangles minus the overlapping area. But the problem states that X (the overlapping area) is 1/8 of the total shaded area. So:X = (1/8) * (Total Shaded Area)But Total Shaded Area = Area of Rectangle 1 + Area of Rectangle 2 - XSince the rectangles are identical, let's let the area of one rectangle be A. Then:Total Shaded Area = A + A - X = 2A - XGiven that X = (1/8)(2A - X), so substituting:X = (1/8)(2A - X)Let me write that equation again:X = (2A - X)/8Multiply both sides by 8:8X = 2A - XBring X to the left side:8X + X = 2A9X = 2ATherefore, X = (2/9)ASo X is 2/9 of the area of one rectangle. Hmm, that seems straightforward, but let me verify if I interpreted the problem correctly.Wait, the problem says "the area X of the overlapping region (shaded more darkly) is one eighth of the total shaded area." So the total shaded area would include both rectangles but subtracting the overlap once. So yes, that's correct: Total Shaded Area = 2A - X. Then X = (1/8)(2A - X), leading to X = 2/9 A. So the answer should be D, 2/9.But let me make sure there's no trick here. Maybe the total shaded area is considered differently? For example, if one rectangle is on top of the other, perhaps the total shaded area is the area of the top rectangle plus the non-overlapping part of the bottom rectangle? But that would still be the same as 2A - X.Alternatively, maybe the problem is referring to the total shaded area as both the overlapping and non-overlapping parts, but in different shades? The overlapping region is shaded more darkly, so perhaps the total shaded area is both the dark and light regions. Then the dark region is X, and the total shaded area would be X (dark) plus (2A - 2X) (light), since each rectangle's non-overlapping part is A - X. Wait, that might complicate things.Wait, if the overlapping region is shaded more darkly, then the total shaded area could be considered as the union, which is 2A - X. But the problem says "the area X of the overlapping region... is one eighth of the total shaded area". So X = 1/8 of (2A - X). So as before. So solving that gives X = 2/9 A.But let me check with numbers. Suppose each rectangle has area 9. Then according to the answer, X would be 2. Total shaded area would be 2*9 - 2 = 16. Then X is 2, which is 1/8 of 16. That works. So that seems correct.Alternatively, if X was 1/4 of A, which is 2.25 in this case, then total shaded area would be 18 - 2.25 = 15.75. Then X would be 2.25, which is 2.25/15.75 = 0.1428... which is 1/7, not 1/8. So that's not matching.If X was 2/7 of A, then with A=7, X=2. Total shaded area would be 14 - 2 = 12. Then X=2 is 1/6 of 12, which is not 1/8.If X was 1/3, with A=9, X=3. Total shaded area=18-3=15. Then 3 is 1/5 of 15, not 1/8.If X=1/5 of A=5, then X=1. Total shaded area=10 -1=9. 1 is 1/9 of 9, not 1/8. Wait, no, if X=1/5 of 5? Wait, perhaps A=5, then X=1, total shaded area=10 -1=9, which would make X=1/9 of total shaded area, which is not 1/8. So that doesn't work either.So the only one that works is 2/9. Let me confirm again with A=9, X=2. Total shaded area=18-2=16, so 2 is 1/8 of 16. Yes, that's correct.So unless there's some different interpretation, I think D is the correct answer.But maybe the problem is presented differently? For instance, if the two rectangles are overlapping such that the overlapping area is a specific shape, but given that they are identical and placed obliquely, the overlapping area's proportion could be different. But the problem states that X is 1/8 of the total shaded area, which gives us an equation that leads to X=2/9 A. So unless there's some geometric consideration I'm missing here, perhaps the answer is 2/9.Alternatively, maybe the overlapping area is calculated based on the angle of the oblique placement. But the problem doesn't give any specific angles or dimensions, so perhaps it's a general case where the relation is purely algebraic based on areas.Therefore, based on the given information and the equations derived, the answer should be D, 2/9.Final Answerboxed{D}</think>Given two identical rectangles placed obliquely on top of each other, the area ( X ) of the overlapping region is one eighth of the total shaded area. We need to determine what fraction of the area of one rectangle ( X ) represents.1. Let the area of one rectangle be ( A ).2. The total shaded area (the union of both rectangles) is given by: [ text{Total Shaded Area} = 2A - X ]3. According to the problem, the overlapping area ( X ) is one eighth of the total shaded area: [ X = frac{1}{8} times (text{Total Shaded Area}) ]4. Substituting the expression for the total shaded area into the equation: [ X = frac{1}{8} times (2A - X) ]5. Solving for ( X ): [ 8X = 2A - X ] [ 9X = 2A ] [ X = frac{2}{9}A ]Thus, the fraction of the area of one rectangle that ( X ) represents is (boxed{D}).]

🔑:21. D Let the area of each rectangle be Y. Then the total shaded area is 2(Y-X)+X=2 Y-X. Therefore X=frac{1}{8}(2 Y-X). So 8 X=2 Y-X, that is 9 X=2 Y. Therefore frac{X}{Y}=frac{2}{9}.

🧠:[<think>Okay, let's see. The problem is about an arithmetic sequence of positive numbers, {a_n}. We are given that the product of the first term and the 20th term is 100, and we need to find the minimum value of the sum of the 7th and 14th terms. Hmm, arithmetic sequences, right? So, let's recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference. So, the nth term can be written as a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference.Alright, so given that a_1 * a_{20} = 100. Let me write down the expressions for a_1 and a_{20} first. The first term is just a_1. The 20th term would be a_1 + (20 - 1)d = a_1 + 19d. So, their product is a_1*(a_1 + 19d) = 100. Got that.Now, we need to find the minimum value of a_7 + a_{14}. Let's write expressions for these terms as well. The 7th term is a_1 + 6d, and the 14th term is a_1 + 13d. So, the sum a_7 + a_{14} would be (a_1 + 6d) + (a_1 + 13d) = 2a_1 + 19d. Wait, that simplifies to 2a1 +19d. Interesting. So, the sum we need to minimize is 2a1 +19d.But we have a constraint: a1*(a1 +19d) = 100. Let me note that down. Let's denote S = 2a1 +19d, which is the sum we need to minimize. So, we need to minimize S subject to the constraint a1*(a1 +19d) = 100, with a1 and d being positive numbers (since all terms in the sequence are positive).Hmm, so this seems like an optimization problem with a constraint. I remember that in calculus, we can use Lagrange multipliers, but since all variables are positive and it's a single-variable problem, maybe substitution would work here. Let me try that.Let's express S in terms of a1 and d, but since we have a relationship between a1 and d, perhaps we can express d in terms of a1, or vice versa, and substitute into S. Let me see.From the constraint, a1*(a1 +19d) = 100. Let's expand that: a1^2 + 19a1*d = 100. Hmm. So, if we can write 19a1*d in terms of a1^2, maybe. But S is 2a1 +19d. Wait, S has 19d, and in the constraint, we have 19a1*d. Let's see.Alternatively, maybe we can let x = a1, then 19d = (100/x - x). Wait, let me check. From the constraint: x*(x + 19d) = 100. So, x^2 +19x*d =100. Therefore, 19x*d=100 -x^2. Therefore, 19d= (100 -x^2)/x. So, 19d = 100/x -x. Therefore, S =2x +19d = 2x + (100/x -x) = 2x +100/x -x =x +100/x.Oh! That's nice. So, S simplifies to x + 100/x, where x is a1, which is positive. Therefore, the problem reduces to finding the minimum value of x + 100/x for x >0. That's a standard calculus problem. Alternatively, by AM-GM inequality, the minimum occurs when x=100/x, so x=10. Therefore, the minimum value is 10 +10=20. But let's verify that.Wait, let me confirm. So, S = x +100/x. To find its minimum, take derivative: dS/dx =1 -100/x^2. Set derivative to zero: 1 -100/x^2 =0 => x^2=100 =>x=10. Since x>0, x=10. Then the minimum value is S=10 +100/10=20. Therefore, the minimum value of a7 +a14 is 20. But wait, let's check if this is feasible with the original constraint.If a1=10, then from the constraint: a1*(a1 +19d)=100 =>10*(10 +19d)=100 =>10 +19d=10 =>19d=0 =>d=0. But d=0 would mean the sequence is constant, which is technically an arithmetic sequence, but all terms are 10. Then, a7 +a14=10 +10=20. However, the problem states that it's an arithmetic sequence of positive numbers. If d=0, is that allowed? Well, usually, in arithmetic sequences, the common difference can be zero, making it a constant sequence. Since the problem didn't specify that the common difference has to be positive, just that the numbers are positive, which they are (all terms are 10). So, that should be acceptable.But wait a minute, if d=0, then all terms are 10, so the product a1*a20=10*10=100, which matches the given condition. Then, a7 +a14=10 +10=20, which would be the minimum. But I need to check if d=0 is allowed. The problem says "arithmetic sequence of positive numbers". An arithmetic sequence with common difference 0 is still an arithmetic sequence. So, yes, I think that's acceptable. Hence, the minimum value is 20.But wait, let me check another way. Suppose d is not zero. Let's pick another value. For example, suppose a1=5. Then, from the constraint: 5*(5 +19d)=100 =>5 +19d=20 =>19d=15 =>d=15/19≈0.789. Then, a7 +a14=2a1 +19d=2*5 +19*(15/19)=10 +15=25, which is higher than 20. Similarly, if a1=20, then 20*(20 +19d)=100 =>20 +19d=5 =>19d=-15. But d would be negative here, which would make the sequence eventually have negative terms. But since all terms are positive, d cannot be negative. So, in that case, a1 cannot be 20. So, a1 must be less than or equal to 10? Wait, when a1=10, d=0. If a1 is larger than 10, then 19d would have to be negative to make a1*(a1 +19d)=100, but since d is the common difference and the terms must remain positive, we cannot have d negative. Because if d is negative, eventually the terms will become negative. Let's check.Suppose a1=11, then 11*(11 +19d)=100 =>11 +19d=100/11≈9.09 =>19d≈9.09 -11≈-1.91 =>d≈-0.1005. Then, the next terms would decrease. But we need all terms to be positive. The 20th term is 100/11≈9.09, which is positive, but let's check the 21st term: a21= a1 +20d≈11 +20*(-0.1005)=11 -2.01≈8.99, which is still positive. But wait, if d is negative, then eventually, as n increases, the terms will become negative. But since the problem doesn't specify a finite sequence, we need to ensure that all terms are positive. However, for an infinite arithmetic sequence with negative common difference, it will eventually have negative terms. Therefore, if d is negative, even if a1 and a20 are positive, beyond some term, the sequence will become negative. Therefore, to have all terms positive, d cannot be negative. So, d must be non-negative. Therefore, in this problem, since the sequence is of positive numbers, d must be non-negative, so 19d ≥0. Therefore, from the constraint a1*(a1 +19d)=100, if a1 >10, then 19d =100/a1 -a1, which would be negative since 100/a1 <10 (as a1>10) and hence 100/a1 -a1 <0. Therefore, d would be negative, which is not allowed. Therefore, a1 cannot exceed 10. Similarly, if a1 <10, then 19d=100/a1 -a1, which would be positive since 100/a1 >10 (as a1<10). Therefore, d would be positive, which is allowed. So, in this case, a1 can be between 0 and10, with d positive. But when a1=10, d=0. So, the domain of a1 is (0,10], and d is non-negative.Therefore, when a1=10, d=0, which gives S=20, and when a1<10, d positive, which gives S=a1 +100/a1. Wait, but earlier we saw that S=x +100/x where x=a1. So, the minimum of S is 20 when x=10. However, when x approaches 0, S approaches infinity, and when x approaches 10 from below, S approaches 10 +100/10=20. Therefore, the minimum value is 20, achieved when a1=10, d=0. Therefore, the answer is 20.But let's just confirm once again with another example. Take a1=5, then as before, d=15/19≈0.789. Then, a7 +a14=25. If a1=2, then 2*(2 +19d)=100 =>2 +19d=50 =>19d=48 =>d≈2.526. Then, a7 +a14=2*2 +19*2.526≈4 +48≈52. That's way higher. If a1=10, then d=0, sum is 20. If a1=8, then 8*(8 +19d)=100 =>8 +19d=12.5 =>19d=4.5 =>d≈0.2368. Then, a7 +a14=2*8 +19*0.2368≈16 +4.5≈20.5. So, which is more than 20. Similarly, if a1=9, 9*(9 +19d)=100 =>9 +19d≈11.111 =>19d≈2.111 =>d≈0.111. Then, a7 +a14=2*9 +19*0.111≈18 +2.111≈20.111, still more than 20. Therefore, it seems that as a1 approaches 10, S approaches 20, and when a1=10, S=20. Therefore, the minimum is indeed 20.So, conclusion: the minimum value of a7 +a14 is 20, achieved when a1=10 and d=0. So, the answer is 20.Final AnswerThe minimum value of (a_7 + a_{14}) is boxed{20}.</think>Given an arithmetic sequence of positive numbers ({a_n}), we know that (a_1 cdot a_{20} = 100). We need to find the minimum value of (a_7 + a_{14}).First, we express the terms (a_7) and (a_{14}) in terms of the first term (a_1) and the common difference (d):[a_7 = a_1 + 6d][a_{14} = a_1 + 13d]Thus, the sum (a_7 + a_{14}) is:[a_7 + a_{14} = (a_1 + 6d) + (a_1 + 13d) = 2a_1 + 19d]We are given the constraint (a_1 cdot a_{20} = 100). The 20th term (a_{20}) can be written as:[a_{20} = a_1 + 19d]Thus, the constraint becomes:[a_1 cdot (a_1 + 19d) = 100]Let (x = a_1). Then, the constraint can be rewritten as:[x(x + 19d) = 100]Expanding and solving for (19d):[x^2 + 19xd = 100 implies 19xd = 100 - x^2 implies 19d = frac{100 - x^2}{x}]Substituting (19d) back into the expression for (a_7 + a_{14}):[2a_1 + 19d = 2x + frac{100 - x^2}{x} = 2x + frac{100}{x} - x = x + frac{100}{x}]We need to minimize the function (S = x + frac{100}{x}) for (x > 0). Using the AM-GM inequality:[x + frac{100}{x} geq 2sqrt{x cdot frac{100}{x}} = 2 times 10 = 20]The minimum value is achieved when (x = frac{100}{x}), which gives (x = 10).Substituting (x = 10) back into the constraint, we find (d = 0) (since (19d = frac{100 - 10^2}{10} = 0)). Therefore, the sequence is constant with all terms equal to 10, and the sum (a_7 + a_{14} = 10 + 10 = 20).Thus, the minimum value of (a_7 + a_{14}) is (boxed{20}).,<think>Okay, so I have this problem here about an arithmetic sequence. Let me read it again: Given an arithmetic sequence of positive numbers {a_n}, if a₁ ⋅ a₂₀ = 100, find the minimum value of a₇ + a₁₄. Hmm, arithmetic sequences. Let me recall. An arithmetic sequence is one where each term increases by a constant difference, right? So, the nth term can be written as a₁ + (n-1)d, where d is the common difference. Since all the terms are positive, that means the common difference d must be such that all terms stay positive. If the sequence is increasing, d is positive; if decreasing, d is negative. But since all terms are positive, if it's decreasing, the first term has to be large enough so that the terms never become negative. But since the problem says it's an arithmetic sequence of positive numbers, I guess d can be positive or negative, as long as all terms stay positive. So, given that a₁ multiplied by a₂₀ is 100, we need to find the minimum value of a₇ + a₁₄. Hmm. Let me write down the expressions for these terms. First, let's express a₁ and a₂₀. Since it's an arithmetic sequence:a₁ = a₁ (just the first term)a₂₀ = a₁ + (20 - 1)d = a₁ + 19dSimilarly, a₇ = a₁ + 6da₁₄ = a₁ + 13dSo, the problem states that a₁ * a₂₀ = 100, and we need to minimize a₇ + a₁₄. Let me write these equations:a₁ * (a₁ + 19d) = 100and we need to find the minimum value of (a₁ + 6d) + (a₁ + 13d) = 2a₁ + 19d.So, the expression to minimize is 2a₁ + 19d, given that a₁(a₁ + 19d) = 100, with a₁ > 0 and a₁ + 19d > 0 (since all terms are positive).Wait, but since the sequence is of positive numbers, all terms must be positive. So, a₁ > 0, and for the 20th term, a₁ + 19d > 0. If d is positive, then all terms after a₁ will be larger, so they'll automatically be positive. If d is negative, then we need to make sure that a₁ + 19d > 0, so that the 20th term is still positive. So, d can be negative, but not too negative.So, our constraints are:1. a₁ > 02. a₁ + 19d > 03. a₁(a₁ + 19d) = 100And we need to minimize 2a₁ + 19d.Hmm, this seems like an optimization problem with a constraint. Maybe I can use substitution. Let me denote S = a₁ + 19d, so that a₁ * S = 100. Then the expression to minimize is 2a₁ + (S - a₁) = a₁ + S. Wait, because 2a₁ + 19d = 2a₁ + (S - a₁) = a₁ + S. So, actually, the expression to minimize is a₁ + S, where a₁ * S = 100. That's interesting. So, we have two variables a₁ and S, with a₁ * S = 100, and we need to minimize a₁ + S. Oh, wait, this is the classic problem where for two positive numbers with a fixed product, their sum is minimized when they are equal. So, by AM ≥ GM inequality, a₁ + S ≥ 2√(a₁ * S) = 2√100 = 20. Equality holds when a₁ = S. So, the minimum sum is 20, achieved when a₁ = S = 10. But wait, hold on. Let me check. S is a₁ + 19d. So, if a₁ = S = 10, then S = 10 = a₁ + 19d. But a₁ is 10, so 10 + 19d = 10, which implies 19d = 0, so d = 0. But if d = 0, then all terms in the arithmetic sequence are equal to a₁ = 10. So, the 20th term is also 10, so a₁ * a₂₀ = 10 * 10 = 100, which satisfies the condition. Then a₇ + a₁₄ = 10 + 10 = 20. So, that would be the minimum. But is this valid? Wait, if d can be zero, then yes. But the problem says it's an arithmetic sequence. But an arithmetic sequence with d = 0 is still an arithmetic sequence, right? It's a constant sequence. So, the minimum value is 20, achieved when d = 0. But wait, wait. Let me think again. Is there a case where d is negative, and a₁ is bigger, such that a₁ + S is smaller? But according to AM-GM inequality, the sum a₁ + S is minimized when a₁ = S. So, if they are equal, that gives the minimal sum. So, unless there's some constraint that prevents a₁ and S from being equal. Wait, but S is a₁ + 19d, which would require that d = (S - a₁)/19. If a₁ = S, then d = 0. So, that's allowed. So, that's feasible. So, in that case, the minimal sum is indeed 20. But let me check with an example. Suppose d is not zero. Let's say d is positive. Let's take d = 1. Then a₁ * (a₁ + 19) = 100. So, a₁² + 19a₁ - 100 = 0. Solving this quadratic: a₁ = [-19 ± sqrt(361 + 400)]/2 = [-19 ± sqrt(761)]/2. sqrt(761) is approx 27.6. So, positive solution is ( -19 + 27.6 ) / 2 ≈ 8.6 / 2 ≈ 4.3. So, a₁ ≈4.3, a₂₀ ≈4.3 +19≈23.3. Then a₇ + a₁₄ = 2a₁ +19d ≈2*4.3 +19*1≈8.6 +19=27.6. Which is higher than 20. Alternatively, take d negative. Let's say d = -1. Then a₁ +19*(-1) = a₁ -19. Since a₁ -19 must be positive (as a₂₀ is positive), so a₁ >19. Then a₁*(a₁ -19)=100. So, a₁² -19a₁ -100=0. Solving: a₁ = [19 ± sqrt(361 +400)]/2 = [19 ± sqrt(761)]/2. sqrt(761)≈27.6. So, positive solution is (19 +27.6)/2≈46.6/2≈23.3. Then a₇ + a₁₄ =2a₁ +19d =2*23.3 +19*(-1)=46.6 -19≈27.6. Same as before, which is again higher than 20. So, if we take d=0, a₁=10, then a₇ +a₁₄=20. So, in that case, the sum is 20. So, that must be the minimum. But wait, just to confirm, maybe if d is something else, but not 1 or -1. Let me try another approach. Let me set d as a variable. Let me let a₁ = x, so that a₁ +19d = y. Then x*y = 100, and we need to minimize x + y. Then as per AM-GM, the minimum is 20. But is there a constraint that y = x +19d. So, if x and y are variables with x*y=100, then x + y is minimized when x=y=10, but in this problem, y =x +19d. So, if we set x = y=10, then y =x +19d => 10=10 +19d => d=0. So, which is valid, as the sequence is allowed to have d=0. Alternatively, maybe if d is not zero, but allows x and y to be unequal but x*y=100, then x + y would be more than 20. Therefore, the minimal sum is indeed 20. Therefore, the answer should be 20. But let me double-check. Suppose we consider d not equal to zero, but some other value. Let me parametrize the problem. Let me denote the common difference as d. Then, given a₁ * (a₁ +19d)=100, we can express a₁ in terms of d, or vice versa. Let's express a₁ in terms of d. So, a₁² +19d a₁ -100=0. Solving for a₁:a₁ = [ -19d ± sqrt( (19d)^2 + 400 ) ] / 2But since a₁ must be positive, we take the positive root:a₁ = [ -19d + sqrt(361d² + 400 ) ] / 2Then, the expression to minimize is 2a₁ +19d. Substitute a₁:2a₁ +19d = [ -19d + sqrt(361d² + 400 ) ] +19d = sqrt(361d² + 400 )Therefore, the expression to minimize is sqrt(361d² + 400 ). Hmm, interesting. So, we need to find the minimum value of sqrt(361d² + 400 ). Since the square root is a monotonically increasing function, this is equivalent to minimizing 361d² + 400.So, 361d² + 400. Since 361 is positive, this is a quadratic in d², which is minimized when d² is as small as possible, i.e., when d=0. Then the minimum value is 400, so sqrt(400)=20. So, indeed, the minimum value is 20 when d=0. Therefore, the answer is 20.Therefore, regardless of how we approach it, the minimum value of a₇ +a₁₄ is 20, achieved when d=0, i.e., the sequence is constant with all terms equal to 10. But wait, just to make sure, let me check when d is non-zero. Let's say d is very small but non-zero. Let's take d approaching zero. Then, a₁ is approaching 10. Then, a₇ + a₁₄ would be 2a₁ +19d ≈20 +19d. If d is approaching zero from positive side, then this sum approaches 20 from above. If d is approaching zero from negative side, then 19d approaches zero from below, so the sum approaches 20 from below? Wait, but if d is negative, even a tiny bit negative, then a₁ must be greater than 10, because a₁ * (a₁ +19d)=100. If d is negative, then a₁ +19d < a₁. So, if d is negative, say d = -k where k>0, then a₁*(a₁ -19k)=100. So, if k is very small, then a₁ must be slightly larger than 10. For example, let d = -0.001. Then, a₁*(a₁ -19*(-0.001))= a₁*(a₁ +0.019)=100. If a₁ is approximately 10, then 10*(10.019)=100.19, which is slightly more than 100. So, to get back to 100, a₁ needs to be slightly less than 10. Wait, that seems contradictory. Wait, if d is negative, then a₁ +19d = a₁ -19k, where k = |d|. So, if d is negative, the term a₁ +19d must still be positive, so a₁ >19k. So, let's take d = -k, k>0. Then, a₁*(a₁ -19k)=100. Let's suppose k is very small, then a₁ is approximately sqrt(100)=10. So, plugging in a₁≈10, we have 10*(10 -19k)=100 -190k. But we need this to equal 100, so 100 -190k=100 => k=0. Therefore, only when k=0 do we have equality. So, if d is negative, then to satisfy a₁*(a₁ +19d)=100, a₁ must be greater than 10, but then a₁ +19d would be less than 10. Therefore, their product remains 100. For example, if a₁ = 20, then a₁ +19d =5, so 20*5=100. Then, d=(5 -20)/19= (-15)/19≈-0.789. Then, a₇ +a₁₄=2a₁ +19d=40 + (-15)=25. Which is higher than 20. So, even with d negative, the sum is higher. So, indeed, the minimal sum is achieved when d=0. Therefore, the answer is 20. Final AnswerThe minimum value of (a_7 + a_{14}) is (boxed{20}).</think>Given an arithmetic sequence of positive numbers ({a_n}), we need to find the minimum value of (a_7 + a_{14}) given that (a_1 cdot a_{20} = 100).First, we express the terms involved in the problem:- The 20th term (a_{20}) is (a_1 + 19d).- The 7th term (a_7) is (a_1 + 6d).- The 14th term (a_{14}) is (a_1 + 13d).We need to minimize the expression (a_7 + a_{14}), which simplifies to:[a_7 + a_{14} = (a_1 + 6d) + (a_1 + 13d) = 2a_1 + 19d]Given the constraint (a_1 cdot (a_1 + 19d) = 100), we let (S = a_1 + 19d). This transforms the constraint into (a_1 cdot S = 100). We need to minimize the expression (2a_1 + 19d), which can be rewritten as (a_1 + S).By the AM-GM inequality, for two positive numbers (a_1) and (S) with a fixed product, their sum is minimized when they are equal. Therefore, the minimum value of (a_1 + S) is achieved when (a_1 = S). Given (a_1 cdot S = 100), setting (a_1 = S) gives (a_1 = S = 10).This implies:- (a_1 = 10)- (S = a_1 + 19d = 10), leading to (19d = 0) or (d = 0)Thus, the arithmetic sequence is a constant sequence where all terms are 10. Therefore, (a_7 + a_{14} = 10 + 10 = 20).Hence, the minimum value of (a_7 + a_{14}) is (boxed{20}).]

🔑:Since {a_n} is an arithmetic sequence of positive numbers and a_1 cdot a_{20} = 100, we havea_1 cdot a_{20} = 100 = a_7 cdot a_{14}By the arithmetic mean-geometric mean inequality, we havea_7 + a_{14} geq 2 sqrt{a_7 cdot a_{14}} = 20with equality if and only if a_7 = a_{14}. Therefore, the minimum value of a_7 + a_{14} is boxed{20}.To see why this is true, note that since {a_n} is an arithmetic sequence of positive numbers, we havefrac{a_{20}}{a_1} = left(frac{a_{7}}{a_1}right)^3 = left(frac{a_{14}}{a_7}right)^3Taking the cube root of both sides, we getfrac{a_{7}}{a_1} = frac{a_{14}}{a_7}Rearranging, we obtaina_7^2 = a_{14} cdot a_1Substituting this into our original equation, we geta_1 cdot a_{20} = a_7^2 cdot a_{14} = (a_7 cdot a_{14})^2 = 100Taking the square root of both sides, we geta_7 cdot a_{14} = 10Now, by the arithmetic mean-geometric mean inequality, we havea_7 + a_{14} geq 2 sqrt{a_7 cdot a_{14}} = 20with equality if and only if a_7 = a_{14}. Therefore, the minimum value of a_7 + a_{14} is boxed{20}.

🧠:[<think>Okay, let's see. So we have a regression line equation given as (bar{y} = frac{1}{3}x + bar{a}). Wait, the notation here is a bit confusing. Usually, the regression line is written as (hat{y} = bx + a), where (b) is the slope and (a) is the y-intercept. But here they've used (bar{y}) and (bar{a}). Hmm, maybe (bar{y}) is the mean of the y-values? Or is it the predicted y-value? Wait, no, in regression, the line is used to predict y based on x, so (hat{y}) is the predicted value. The notation here uses a bar, which typically denotes a mean. So maybe this is a different way of writing the regression equation? Or perhaps there's a typo or misunderstanding in the problem statement?Wait, let me check the problem again. It says "the regression line equation is (bar{y}= frac{1}{3}x+ bar{a})". Hmm. Maybe they actually mean (hat{y} = frac{1}{3}x + a), where (a) is the intercept, and there's a typo with the bar? Or maybe (bar{a}) is a constant? This is a bit confusing. Let's read the rest of the problem to see if we can figure it out.The problem gives that the sum of all x_i from 1 to 8 is equal to 6, and twice the sum of all y_i from 1 to 8 is equal to 6. So, (x_1 + x_2 + ... + x_8 = 6) and (2(y_1 + y_2 + ... + y_8) = 6), which means the sum of the y_i's is 3. So, (sum y_i = 3). The question is asking for the value of the real number (a), which is part of the regression equation. The options are fractions like 1/16, 1/8, etc.In linear regression, the equation is usually (hat{y} = b(x - bar{x}) + bar{y}), where (b) is the slope, (bar{x}) is the mean of x, and (bar{y}) is the mean of y. Alternatively, it can be written as (hat{y} = bx + a), where (a = bar{y} - bbar{x}). So maybe in this problem, they're writing the regression equation in the form (hat{y} = frac{1}{3}x + a), and the value of (a) is what we need to find.But in the problem statement, the equation is written as (bar{y} = frac{1}{3}x + bar{a}). Maybe this is a notation issue. If (bar{y}) is the mean of y, then that equation would have to hold for all x, which doesn't make sense. So perhaps it's a misnotation, and it should be (hat{y} = frac{1}{3}x + a). Alternatively, maybe (bar{a}) is a constant term. Wait, but the problem says "the value of the real number a", so maybe (bar{a}) is supposed to be the intercept, denoted by (a) here. Maybe the bar over the a is just a typo or a formatting error.Assuming that the regression equation is meant to be (hat{y} = frac{1}{3}x + a), then we need to find the intercept (a). To find the intercept, we can use the fact that the regression line passes through the point ((bar{x}, bar{y})), where (bar{x}) is the mean of the x-values and (bar{y}) is the mean of the y-values. So, according to this property, (bar{y} = frac{1}{3}bar{x} + a). Therefore, (a = bar{y} - frac{1}{3}bar{x}).So, we need to compute (bar{x}) and (bar{y}). Given that the sum of all x_i is 6, and there are 8 data points, the mean (bar{x} = frac{6}{8} = frac{3}{4}). Similarly, the sum of all y_i is 3 (since twice the sum is 6, so sum is 3), so the mean (bar{y} = frac{3}{8}). Therefore, plugging into the equation for (a):(a = frac{3}{8} - frac{1}{3} times frac{3}{4} = frac{3}{8} - frac{1}{4} = frac{3}{8} - frac{2}{8} = frac{1}{8}).So the value of (a) is (frac{1}{8}), which is option B.Wait, let me verify again. The regression line must pass through the mean of x and mean of y. So if the equation is (hat{y} = frac{1}{3}x + a), then when (x = bar{x}), (hat{y} = bar{y}). Therefore:(bar{y} = frac{1}{3}bar{x} + a)Solving for (a):(a = bar{y} - frac{1}{3}bar{x})Yes, that's correct. Then compute the means:Sum of x is 6, n=8, so (bar{x} = 6/8 = 3/4)Sum of y is 3, n=8, so (bar{y} = 3/8)Thus,(a = 3/8 - (1/3)(3/4) = 3/8 - 1/4 = 3/8 - 2/8 = 1/8)So yes, the answer should be B: 1/8.Alternatively, if the original equation was supposed to be (bar{y} = frac{1}{3}x + bar{a}), but that notation is confusing. If they meant that the average of y is equal to (1/3)x plus the average of a, that doesn't make much sense. So I think the correct interpretation is that the regression equation is (hat{y} = (1/3)x + a), and using the fact that it passes through ((bar{x}, bar{y})), leading to the calculation above. So the answer is 1/8, which is option B.Final Answerboxed{B}</think>Given a set of observed data ((x_i, y_i)) for (i = 1, 2, ldots, 8) with a linear relationship, the regression line equation is given as (bar{y} = frac{1}{3}x + bar{a}). We need to find the value of the real number (a).First, we calculate the means of the x-values and y-values. The sum of the x-values is given as 6, and the sum of the y-values is 3 (since twice the sum of the y-values is 6). Therefore, the means are:[bar{x} = frac{x_1 + x_2 + cdots + x_8}{8} = frac{6}{8} = frac{3}{4}][bar{y} = frac{y_1 + y_2 + cdots + y_8}{8} = frac{3}{8}]The regression line must pass through the point ((bar{x}, bar{y})). Using the regression line equation (hat{y} = frac{1}{3}x + a), we substitute (bar{x}) and (bar{y}) into the equation:[bar{y} = frac{1}{3}bar{x} + a]Substituting the calculated means:[frac{3}{8} = frac{1}{3} left(frac{3}{4}right) + a]Solving for (a):[frac{3}{8} = frac{1}{4} + a][a = frac{3}{8} - frac{1}{4} = frac{3}{8} - frac{2}{8} = frac{1}{8}]Thus, the value of (a) is (boxed{B}).,<think>Okay, so I need to find the value of 'a' in the regression line equation given some data. The regression equation is written as (bar{y} = frac{1}{3}x + bar{a}). Wait, actually, hold on—is that the equation for the regression line? Because usually, a regression line is something like (hat{y} = bx + a), where (b) is the slope and (a) is the y-intercept. The notation here uses (bar{y}) and (bar{a}), which might be related to averages? Hmm, maybe the problem is using a different notation. Let me read the problem again carefully.The problem states: "For a set of observed data ((x_i, y_i)) for (i = 1,2,...,8) of variables (x), (y) that have a linear relationship, the regression line equation is (bar{y}= frac{1}{3}x+ bar{a}). Given that (x_1+x_2+x_3+...+x_8=2(y_1+y_2+y_3+...+y_8)=6), the value of the real number (a) is (    )."Wait, so the regression line is given as (bar{y} = frac{1}{3}x + bar{a}). But in regression, the line should pass through the point ((bar{x}, bar{y})), right? So maybe the equation is written in terms of deviations from the mean? Or perhaps there's a typo in the notation? Because typically, (bar{y}) is the mean of the y-values. But here, they have (bar{y}) as the predicted y for a given x? That seems confusing. Let me check the original problem again.Wait, maybe the bar over the y and a is just part of the notation for the regression line, not indicating an average. Maybe it's like (hat{y} = frac{1}{3}x + bar{a}), where (hat{y}) is the predicted y. But in the problem statement, it's written as (bar{y}). Hmm. Alternatively, maybe it's a different type of regression. Let me see.Given that the sum of all x's is 6, and twice the sum of all y's is 6, which means the sum of all y's is 3. So (sum x_i = 6) and (sum y_i = 3). There are 8 data points, so the means would be (bar{x} = 6/8 = 3/4) and (bar{y} = 3/8).In linear regression, the regression line passes through the point ((bar{x}, bar{y})). So if the equation is (hat{y} = frac{1}{3}x + a), then plugging in (bar{x}) into the equation should give (bar{y}). Therefore, (bar{y} = frac{1}{3}bar{x} + a). So substituting the values of (bar{x}) and (bar{y}), we can solve for a.Let me compute that. So (bar{y} = frac{1}{3} cdot frac{3}{4} + a). That would be (frac{1}{3} cdot frac{3}{4} = frac{1}{4}). So (frac{3}{8} = frac{1}{4} + a). Then solving for a: (a = frac{3}{8} - frac{1}{4} = frac{3}{8} - frac{2}{8} = frac{1}{8}). So the value of a is 1/8, which is option B.Wait, but the problem mentions (bar{a}) in the regression equation. Is that supposed to be the intercept term? Maybe the notation is different here. If the regression equation is (bar{y} = frac{1}{3}x + bar{a}), where (bar{a}) is the intercept, then using the point ((bar{x}, bar{y})) gives (bar{y} = frac{1}{3}bar{x} + bar{a}). Then solving for (bar{a}) gives (bar{a} = bar{y} - frac{1}{3}bar{x}). Then substituting the values of (bar{y}) and (bar{x}), which are 3/8 and 3/4 respectively, we get (bar{a} = 3/8 - (1/3)(3/4) = 3/8 - 1/4 = 1/8. So that would make (bar{a} = 1/8), so the answer is B: 1/8.But wait, the question asks for the value of the real number 'a', not (bar{a}). Wait, the regression equation is given as (bar{y} = frac{1}{3}x + bar{a}). Maybe there's a confusion in notation here. If it's a standard regression line, the equation should be (hat{y} = bx + a), where (b) is the slope and (a) is the intercept. But in this problem, the equation is written with (bar{y}) on the left side and (bar{a}) on the right. That could be a typo or a notation difference.Alternatively, maybe the problem is referring to the regression of y on x, with the line expressed in terms of the mean of y. Wait, but in standard regression, the line passes through the mean of x and mean of y. So regardless, substituting the mean values into the equation should hold. Let me confirm that.In linear regression, the equation is (hat{y}_i = b x_i + a), and we have that (hat{bar{y}} = b bar{x} + a). But since the regression line is designed such that (hat{bar{y}} = bar{y}), that equation must hold. Therefore, even if they denote the regression equation as (bar{y} = frac{1}{3}x + bar{a}), when x is the variable, then substituting (bar{x}) into the equation should give (bar{y}). So:(bar{y} = frac{1}{3}bar{x} + bar{a})Therefore, solving for (bar{a}):(bar{a} = bar{y} - frac{1}{3}bar{x})Given that (sum x_i =6), so (bar{x} =6/8 = 3/4). And (sum y_i =3) (since 2 times sum y_i is 6, so sum y_i is 3), so (bar{y} =3/8).Substituting these in:(bar{a} = 3/8 - (1/3)(3/4) = 3/8 - 1/4 = 3/8 - 2/8 = 1/8).Therefore, the value of (bar{a}) is 1/8, which is option B. But the problem asks for the value of the real number 'a'. Wait, in the problem statement, the regression equation is written with (bar{a}), but the options are just 'a'. Maybe it's a typo and (bar{a}) is the intercept term, which is just 'a'. So then the answer is 1/8, option B.Alternatively, maybe the problem uses (bar{a}) to denote the intercept, and they just call it 'a' in the options. So likely, the answer is B: 1/8.Wait, but let me check again the problem statement. The regression line is given as (bar{y} = frac{1}{3}x + bar{a}). Maybe the notation here is that (bar{y}) is the predicted value, so it's equivalent to (hat{y}), and (bar{a}) is the intercept. Then, as in standard regression, the intercept is calculated as (bar{y} - b bar{x}). So that would be exactly what I did above. Therefore, the intercept (bar{a} = 1/8), so the answer is B.Therefore, I think the correct answer is option B: 1/8.Final Answerboxed{B}</think>Given a set of observed data ((x_i, y_i)) for (i = 1, 2, ldots, 8) with a linear relationship, the regression line equation is (bar{y} = frac{1}{3}x + bar{a}). We know that the sum of all (x_i) is 6 and twice the sum of all (y_i) is 6, which means the sum of all (y_i) is 3. First, we calculate the means of (x) and (y):[bar{x} = frac{sum_{i=1}^8 x_i}{8} = frac{6}{8} = frac{3}{4}][bar{y} = frac{sum_{i=1}^8 y_i}{8} = frac{3}{8}]The regression line must pass through the point ((bar{x}, bar{y})). Substituting (bar{x}) and (bar{y}) into the regression equation:[bar{y} = frac{1}{3} bar{x} + bar{a}][frac{3}{8} = frac{1}{3} cdot frac{3}{4} + bar{a}][frac{3}{8} = frac{1}{4} + bar{a}]Solving for (bar{a}):[bar{a} = frac{3}{8} - frac{1}{4} = frac{3}{8} - frac{2}{8} = frac{1}{8}]Thus, the value of the real number (a) is (boxed{B}).]

🔑:AnalysisThis problem examines the linear regression equation. The key to solving the problem is that the linear regression line must pass through the sample mean point, which is one of the steps in solving the linear regression equation.SolutionGiven: x_1+x_2+x_3+…+x_8=2(y_1+y_2+y_3+…+y_8)=6, Therefore, bar{x} = frac{6}{8}, bar{y} = frac{3}{8}, Thus, the coordinates of the sample mean point are (frac{6}{8}, frac{3}{8}), Substituting into the regression line equation, we get frac{3}{8} = frac{1}{3} times frac{6}{8} +a, Therefore, a= frac{1}{8}. Hence, the correct choice is boxed{B}.

🧠:[<think>Okay, so I need to solve this problem about the sequence {a_n} and then another part involving an arithmetic sequence {b_n}. Let's start with part (1).First, the problem states that the sequence {a_n} has its first term a_1 = t, and the recursive relation a_{n+1} = 2S_n + 1, where S_n is the sum of the first n terms. We need to find the value of t such that {a_n} is a geometric sequence.Hmm, let me recall that a geometric sequence has each term multiplied by a common ratio. So, if {a_n} is geometric, then a_{n+1} = r * a_n for some constant ratio r. Let me try to use the given recursion relation and see if I can express it in terms of a geometric sequence.Given a_{n+1} = 2S_n + 1. But if the sequence is geometric, then S_n would be the sum of the first n terms of a geometric series, right? The formula for the sum of the first n terms of a geometric sequence is S_n = a_1*(r^n - 1)/(r - 1). Since a_1 = t, then S_n = t*(r^n - 1)/(r - 1).But also, from the recursive formula, a_{n+1} = 2S_n + 1. But if the sequence is geometric, a_{n+1} should also equal r*a_n = r*(t*r^{n-1}) = t*r^n. So, setting these equal:t*r^n = 2*S_n + 1 = 2*(t*(r^n - 1)/(r - 1)) + 1.So, the equation is:t*r^n = 2t*(r^n - 1)/(r - 1) + 1.Hmm, this seems a bit complex. Maybe there's another approach. Let's compute the first few terms and see if we can find a pattern or spot the common ratio.Given a_1 = t.Then, S_1 = a_1 = t.So, a_2 = 2*S_1 + 1 = 2t + 1.Now, if the sequence is geometric, then a_2 should be r*a_1. So, r*t = 2t + 1.Similarly, let's compute S_2 = a_1 + a_2 = t + (2t + 1) = 3t + 1.Then, a_3 = 2*S_2 + 1 = 2*(3t + 1) + 1 = 6t + 2 + 1 = 6t + 3.But if the sequence is geometric, a_3 should be r^2*a_1 = r^2*t. So, r^2*t = 6t + 3.So now we have two equations:1) r*t = 2t + 12) r^2*t = 6t + 3We can try solving these equations for r and t.From equation 1: r = (2t + 1)/t, provided t ≠ 0.Substitute this into equation 2:[(2t + 1)/t]^2 * t = 6t + 3Simplify left side:( (2t + 1)^2 / t^2 ) * t = ( (4t^2 + 4t + 1) / t^2 ) * t = (4t^2 + 4t + 1)/tSo:(4t^2 + 4t + 1)/t = 6t + 3Multiply both sides by t:4t^2 + 4t + 1 = 6t^2 + 3tBring all terms to one side:4t^2 +4t +1 -6t^2 -3t = 0Combine like terms:-2t^2 + t +1 =0Multiply both sides by -1:2t^2 - t -1 =0Solve the quadratic equation:2t^2 - t -1 =0Using quadratic formula: t = [1 ± sqrt(1 + 8)] /4 = [1 ± 3]/4So, t = (1 +3)/4 = 4/4 =1 or t=(1 -3)/4 = -2/4 = -1/2So t can be 1 or -1/2. Let's check these solutions to see if they work.First, t=1:Check if the sequence is geometric.a_1=1a_2=2*S_1 +1=2*1 +1=3a_3=2*S_2 +1=2*(1+3)+1=8 +1=9a_4=2*S_3 +1=2*(1+3+9)+1=26 +1=27So the sequence is 1, 3, 9, 27,... which is geometric with common ratio 3. So this works.Now, check t= -1/2:a_1= -1/2a_2=2*S_1 +1=2*(-1/2) +1= -1 +1=0But then a_2=0. Then, a_3=2*S_2 +1=2*(-1/2 +0) +1=2*(-1/2) +1= -1 +1=0So the sequence becomes -1/2, 0, 0, 0,... which is not a geometric sequence because after the first term, all terms are zero, but the ratio is undefined (division by zero). So t= -1/2 is invalid.Therefore, the only valid solution is t=1.So answer to part (1) is t=1.Now part (2): Under the condition of (1), which means t=1, so the sequence {a_n} is geometric with a1=1 and ratio 3, so the terms are 1, 3, 9, 27, etc.Now, given that the sum T_n of the first n terms of an arithmetic sequence {b_n} has a maximum value, and T_3=15. Also, a1 + b1, a2 + b2, a3 + b3 form a geometric sequence. We need to find T_n.First, since {b_n} is an arithmetic sequence, its sum T_n is given by n/2 [2b1 + (n-1)d], where d is the common difference.But since T_n has a maximum value, that implies that the arithmetic sequence {b_n} has a negative common difference. Because the sum of an arithmetic sequence will have a maximum only if the terms eventually become negative, so the sum will increase up to a certain point and then start decreasing. Hence, the common difference d must be negative.Alternatively, the formula for T_n is a quadratic in n: T_n = (d/2)n^2 + (b1 - d/2)n. Since the coefficient of n^2 is d/2, for the quadratic to have a maximum, the coefficient of n^2 must be negative. Therefore, d < 0.So we know that d < 0.Given that T_3=15. Let's compute T_3.T_3 = 3/2 [2b1 + 2d] = 3(b1 + d) =15Thus, 3(b1 + d)=15 => b1 + d=5.So equation (1): b1 + d =5.Also, the terms a1 + b1, a2 + b2, a3 + b3 form a geometric sequence. Let's denote these terms as c1, c2, c3. So,c1 = a1 + b1 =1 + b1c2 = a2 + b2 =3 + (b1 + d)c3 = a3 + b3 =9 + (b1 + 2d)Since these form a geometric sequence, the ratio c2/c1 should equal c3/c2.So,(3 + b1 + d)/(1 + b1) = (9 + b1 + 2d)/(3 + b1 + d)Cross multiplying:(3 + b1 + d)^2 = (1 + b1)(9 + b1 + 2d)Let me write this equation. Let's denote b1 as b for simplicity, and since from equation (1), we have b + d=5, so d=5 - b. Therefore, we can substitute d=5 - b into the equation.So replacing d with 5 - b:Left side: (3 + b + (5 - b))^2 = (3 +5)^2 =8^2=64Right side: (1 + b)(9 + b + 2*(5 - b))= (1 + b)(9 + b +10 -2b)= (1 + b)(19 - b)So equation becomes:64 = (1 + b)(19 - b)Expand the right side:64 = 19 - b +19b -b^2= -b^2 +18b +19Bring all terms to left:b^2 -18b -19 +64=0 => b^2 -18b +45=0Solve quadratic equation: b^2 -18b +45=0Using quadratic formula: b = [18 ± sqrt(324 -180)] /2 = [18 ± sqrt(144)] /2 = [18 ±12]/2Thus, b=(18 +12)/2=30/2=15 or b=(18 -12)/2=6/2=3So b=15 or b=3. Then, since d=5 - b, so:If b=15, then d=5 -15= -10If b=3, then d=5 -3=2But earlier, we established that d must be negative for the sum T_n to have a maximum. So d=2 is invalid because positive, so d=-10 is acceptable.Thus, b1=15, d=-10.Therefore, the arithmetic sequence {b_n} has first term 15 and common difference -10.Now, find T_n, the sum of the first n terms.The formula for T_n is n/2 [2b1 + (n -1)d]Substitute b1=15, d=-10:T_n = n/2 [30 + (n -1)(-10)] = n/2 [30 -10(n -1)] = n/2 [30 -10n +10] = n/2 [40 -10n] = n/2 *10*(4 -n)=5n(4 -n)= -5n(n -4)= -5n^2 +20nAlternatively, T_n= -5n² +20nWe can check if this has a maximum. Since the coefficient of n² is -5, which is negative, the parabola opens downward, so it indeed has a maximum. The maximum occurs at n= -b/(2a) for quadratic an² +bn +c. Here, a=-5, b=20, so n= -20/(2*(-5))= -20/-10=2. So maximum at n=2. But the problem states that "the sum T_n has a maximum value", which is consistent with our calculation.Also, check T_3= -5*(9) +20*3= -45 +60=15, which matches given.Now, we need to find T_n. So the answer is T_n= -5n² +20n. But let me write it in standard form.Alternatively, T_n=5n(4 -n)=20n -5n². Either form is acceptable, but maybe factorized form is better. However, the problem might expect the answer in a specific form. But since it's a quadratic, writing as T_n = -5n² +20n is also fine.But let me check the problem statement again: "find T_n". Since {b_n} is an arithmetic sequence, T_n is the sum, so in terms of n, the expression is correct.Alternatively, perhaps the answer is expected in terms of the arithmetic sequence formula. Let's see:Since {b_n} is arithmetic with first term 15 and common difference -10, the nth term is b_n=15 + (n-1)(-10)=15 -10(n -1)=15 -10n +10=25 -10nThen, the sum T_n= n/2*(b1 +b_n)=n/2*(15 +25 -10n)=n/2*(40 -10n)=20n -5n², which matches.Yes, so T_n= -5n² +20n, or equivalently T_n=20n -5n². Either form is correct. But since the problem might prefer the factored form, maybe T_n=5n(4 -n). But both are equivalent. Let me check the standard way of presenting such answers. I think either form is okay, but perhaps the expanded form is better here.Alternatively, if the problem expects a formula in terms of arithmetic sequence sum, which is n/2[2b1 + (n-1)d], so plugging in the values:T_n= n/2[2*15 + (n -1)(-10)] =n/2[30 -10(n -1)] =n/2[30 -10n +10] =n/2[40 -10n]=20n -5n². So same result.Thus, the answer is T_n= -5n² +20n, which can be written as boxed{-5n^2 + 20n} or boxed{20n -5n^2}.Alternatively, if the problem expects a more factored form, maybe boxed{5n(4 - n)}. But since the question says "find T_n", which is the sum, and there are multiple equivalent expressions, but in most cases, the expanded quadratic form is standard. Let's check the initial terms.For n=1, T_1=20*1 -5*1=15. Which is correct, since b1=15.For n=2, T_2=20*2 -5*4=40 -20=20.For n=3, T_3=20*3 -5*9=60 -45=15. Correct.For n=4, T_4=20*4 -5*16=80 -80=0.For n=5, T_5=20*5 -5*25=100 -125= -25.So the sum peaks at n=2 with T_2=20, then decreases. So the formula makes sense.Therefore, the answer is T_n= -5n² +20n. So boxed as boxed{-5n^2 + 20n}Final Answer(1) The value of ( t ) is boxed{1}.(2) The sum ( T_n ) is boxed{-5n^2 + 20n}.</think>Given the sequence ({a_n}) with (a_1 = t) and (a_{n+1} = 2S_n + 1) where (S_n) is the sum of the first (n) terms:1. To determine the value of (t) such that ({a_n}) is a geometric sequence: - Compute the first few terms: - (a_1 = t) - (a_2 = 2S_1 + 1 = 2t + 1) - (S_2 = t + (2t + 1) = 3t + 1) - (a_3 = 2S_2 + 1 = 6t + 3) - For ({a_n}) to be geometric, (a_{n+1} = r cdot a_n). This leads to the equations: - (r cdot t = 2t + 1) - (r^2 cdot t = 6t + 3) - Solving these equations, we find (t = 1) (since (t = -frac{1}{2}) leads to a non-geometric sequence).2. Under the condition (t = 1), the sequence ({a_n}) is geometric with (a_1 = 1) and common ratio 3. For the arithmetic sequence ({b_n}) with (T_3 = 15) and the terms (a_1 + b_1), (a_2 + b_2), (a_3 + b_3) forming a geometric sequence: - Given (T_3 = 15), we find (b_1 + d = 5) where (d) is the common difference. - The terms (1 + b_1), (3 + (b_1 + d)), (9 + (b_1 + 2d)) form a geometric sequence. Solving the quadratic equation derived from the geometric sequence condition, we find (b_1 = 15) and (d = -10). - The sum (T_n) of the arithmetic sequence ({b_n}) is given by (T_n = -5n^2 + 20n). Final Answer(1) The value of ( t ) is (boxed{1}).(2) The sum ( T_n ) is (boxed{-5n^2 + 20n}).,<think>Okay, so I have this problem about sequences. Let me try to understand what's being asked here. There are two parts. Let me start with part (1).The sequence {a_n} has its first term a1 = t, and the next term is defined by a_{n+1} = 2S_n + 1, where S_n is the sum of the first n terms. They want to know for what value of t the sequence {a_n} is a geometric sequence.Hmm. So, first, let me recall that a geometric sequence is one where each term is a constant multiple of the previous term. That is, a_{n+1} = r * a_n for some common ratio r. So, if the entire sequence is geometric, then this recursive relation must hold for all n.But here, the given recursive formula is a_{n+1} = 2S_n + 1. So, maybe I can express S_n in terms of the geometric sequence and set up equations to find t.Let me start by writing down the first few terms of the sequence {a_n} based on the given recursion. Let's see:Given a1 = t.Then a2 = 2S1 + 1 = 2a1 + 1 = 2t + 1.Then S2 = a1 + a2 = t + (2t + 1) = 3t + 1.Then a3 = 2S2 + 1 = 2*(3t + 1) + 1 = 6t + 2 + 1 = 6t + 3.Similarly, S3 = S2 + a3 = (3t + 1) + (6t + 3) = 9t + 4.Then a4 = 2S3 + 1 = 2*(9t + 4) + 1 = 18t + 8 + 1 = 18t + 9.Wait, let me check that again. 2*(9t + 4) is 18t + 8, plus 1 is 18t + 9. Yes.So, the terms so far are:a1 = ta2 = 2t + 1a3 = 6t + 3a4 = 18t + 9If the sequence is geometric, then the ratio between consecutive terms should be constant. Let me check the ratios:a2 / a1 = (2t + 1)/ta3 / a2 = (6t + 3)/(2t + 1) = [3(2t + 1)]/(2t + 1) = 3a4 / a3 = (18t + 9)/(6t + 3) = [9(2t + 1)]/[3(2t + 1)] = 3Wait, so from a2 to a3, the ratio is 3, and from a3 to a4, it's also 3. But from a1 to a2, it's (2t + 1)/t. For the sequence to be geometric, all these ratios must be equal. So, (2t + 1)/t should equal 3.So, setting (2t + 1)/t = 3.Solving this equation:2t + 1 = 3tSubtract 2t both sides:1 = tSo, t = 1.Wait, let's check if this works. If t = 1, then:a1 = 1a2 = 2*1 + 1 = 3a3 = 6*1 + 3 = 9a4 = 18*1 + 9 = 27So, the sequence is 1, 3, 9, 27,... which is geometric with common ratio 3. That works. So, t must be 1.Wait, but let me verify if this is consistent with the given recursion. Let me check a3. If S2 = 1 + 3 = 4, then a3 = 2*4 + 1 = 9. Which matches. Then S3 = 4 + 9 = 13, so a4 = 2*13 + 1 = 27. Wait, 2*13 is 26 +1 is 27. Correct. So, yes, when t=1, the sequence is geometric with ratio 3.Therefore, the answer to part (1) is t=1.Okay, that seems straightforward. Let me confirm again. If t=1, the first term is 1, and each subsequent term is three times the previous one, which satisfies the given recursion. So, part (1) is solved.Now moving to part (2). Under the condition of (1), meaning t=1, so {a_n} is a geometric sequence with a1=1 and ratio 3. So, the terms are 1, 3, 9, 27, etc.Now, we have an arithmetic sequence {b_n}, whose sum of the first n terms is T_n, and T_n has a maximum value. Also, T_3 =15. Additionally, the sequence formed by a1 + b1, a2 + b2, a3 + b3 is a geometric sequence. We need to find T_n.First, let's recall that in an arithmetic sequence {b_n}, the nth term is given by b_n = b1 + (n-1)d, where d is the common difference. The sum of the first n terms, T_n, is given by T_n = n/2 [2b1 + (n-1)d]. Alternatively, T_n can be written as n*b1 + n(n-1)d/2.Now, the problem states that T_n has a maximum value. Since T_n is a quadratic function in n (because it's of the form An^2 + Bn + C), for it to have a maximum value, the coefficient of n^2 must be negative. That coefficient comes from the term involving d: in T_n = n*b1 + n(n-1)d/2 = (d/2)n^2 + (b1 - d/2)n. Therefore, the coefficient of n^2 is d/2. For this quadratic to have a maximum, d/2 must be negative, so d < 0.So, the common difference d must be negative. That means the arithmetic sequence is decreasing.Also, T_3 =15. Let's write T_3:T_3 = 3/2 [2b1 + 2d] = 3*(b1 + d) = 15. Therefore, b1 + d = 5. So, b1 + d = 5.Additionally, the terms a1 + b1, a2 + b2, a3 + b3 form a geometric sequence. Let's write down these terms.Given that {a_n} is 1, 3, 9, 27,... so:a1 =1, a2=3, a3=9.For {b_n}, which is arithmetic, b1 = b1, b2 = b1 + d, b3 = b1 + 2d.Therefore:a1 + b1 = 1 + b1a2 + b2 = 3 + (b1 + d)a3 + b3 = 9 + (b1 + 2d)These three terms must form a geometric sequence. Let's denote them as x, y, z. So:x = 1 + b1y = 3 + b1 + dz = 9 + b1 + 2dFor x, y, z to be a geometric sequence, the ratio between y/x should equal z/y. Therefore:(y)^2 = x*zSo, [3 + b1 + d]^2 = (1 + b1)(9 + b1 + 2d)We also know from T_3 =15 that 3*(b1 + d) =15, so b1 + d =5. Hence, d =5 - b1.Let me substitute d =5 - b1 into the equation from the geometric sequence.First, let's compute each term:x =1 + b1y =3 + b1 + d =3 + b1 + (5 - b1) =3 +5 =8Wait, that simplifies nicely. Because the d =5 - b1, so when you substitute into y, the b1 terms cancel:y=3 + b1 +5 - b1=8Similarly, z=9 + b1 +2d=9 + b1 +2*(5 -b1)=9 +b1 +10 -2b1=19 -b1So, x=1 + b1, y=8, z=19 - b1.So, the three terms are 1 + b1, 8, 19 - b1. These should form a geometric sequence. Therefore, the square of the middle term should equal the product of the first and third terms.Therefore:8^2 = (1 + b1)(19 - b1)Compute 8^2=64.So, (1 + b1)(19 - b1) =64Let's expand the left side:(1)(19) +1*(-b1) +b1*19 +b1*(-b1)=19 -b1 +19b1 -b1^2=19 +18b1 -b1^2Set equal to 64:19 +18b1 -b1^2=64Rearranging:-b1^2 +18b1 +19 -64=0Simplify:-b1^2 +18b1 -45=0Multiply both sides by -1:b1^2 -18b1 +45=0Now, solving this quadratic equation for b1.Use quadratic formula:b1 = [18 ± sqrt(324 - 180)] / 2 = [18 ± sqrt(144)] / 2 = [18 ±12]/2So, sqrt(144)=12, so:b1=(18 +12)/2=30/2=15or b1=(18 -12)/2=6/2=3Therefore, b1=15 or b1=3.Now, since d=5 -b1, if b1=15, then d=5 -15= -10If b1=3, then d=5 -3=2But earlier, we established that the sum T_n has a maximum value. The sum T_n is given by T_n = (d/2)n^2 + (b1 - d/2)n.For T_n to have a maximum, the coefficient of n^2 must be negative. The coefficient is d/2. Therefore, d must be negative.So, check the two cases:Case 1: b1=15, d=-10. Then d/2=-5 <0. Okay, that's valid, as it will create a maximum.Case 2: b1=3, d=2. Then d/2=1 >0. So, the quadratic would open upwards, meaning T_n has a minimum, not a maximum. But the problem states that T_n has a maximum. Therefore, this case is invalid.Therefore, only b1=15 and d=-10 is valid.So, the arithmetic sequence {b_n} has b1=15 and common difference d=-10.Thus, the sum of the first n terms, T_n, is given by:T_n = n/2 [2*15 + (n -1)*(-10)] =n/2 [30 -10(n -1)] =n/2 [30 -10n +10]=n/2 [40 -10n]=n/2 *10*(4 -n)=5n(4 -n)= -5n(n -4)Alternatively, expanding:T_n =5n(4 -n)=20n -5n^2= -5n^2 +20nWhich is a quadratic function in n with coefficient of n^2 being -5, which indeed has a maximum.So, T_n = -5n^2 +20nBut maybe the problem wants the expression in terms of n. Let me check.Alternatively, since {b_n} is arithmetic with b1=15 and d=-10, then T_n can be written as:T_n = n/2 [2b1 + (n -1)d] =n/2 [30 + (n -1)(-10)] =n/2 [30 -10n +10] =n/2 [40 -10n] =20n -5n^2.Yes, same result.So, the answer for part (2) is T_n = -5n^2 +20n, which can be written as T_n =20n -5n^2.Alternatively, factored as T_n =5n(4 -n). Any of these forms is acceptable, but maybe the standard form is T_n = -5n² +20n. But let me check if the problem expects it in a specific form.The problem says "find T_n". Since T_n is the sum of the first n terms of the arithmetic sequence, and arithmetic sequences' sums are usually expressed as quadratic functions. So, either form is okay, but perhaps we can present it as T_n = -5n² +20n.But let me verify once again. If b1=15 and d=-10, then:b1=15, b2=15 -10=5, b3=5 -10=-5, etc.Sum of first 3 terms:15 +5 + (-5)=15, which matches T_3=15. Good.Then, the terms a1 +b1=1 +15=16, a2 +b2=3 +5=8, a3 +b3=9 +(-5)=4. So, the sequence 16, 8, 4. Is this a geometric sequence?Yes, the common ratio is 0.5. 8/16=0.5, 4/8=0.5. So, that works. Therefore, our solution is correct.Therefore, the answer for part (2) is T_n= -5n² +20n.But let me check if they need it in another form, like factored form. 5n(4 -n) is also correct. But perhaps the standard form is more likely. Either way, both are correct. But maybe the problem expects the answer in the form with parentheses, like Tn = 5n(4 - n). Alternatively, Tn = -5n² +20n. Let me see if the problem specifies, but it just says "find T_n". Either form is acceptable. However, since they mention the sum has a maximum value, which occurs at the vertex of the parabola. The vertex is at n = -b/(2a) where the quadratic is an² + bn + c. Here, a=-5, b=20, so vertex at n = -20/(2*(-5))= -20/-10=2. So, the maximum occurs at n=2. So, T_2 is the maximum sum.But regardless, the question just asks for T_n, so either form is okay. Let me check if the answer is expected in a specific notation. Since in the first part they wanted t, here they might want T_n expressed as a formula, which could be written as T_n =20n -5n². Alternatively, using the arithmetic sequence sum formula, which could be written as Tn = n/2 [2*15 + (n-1)(-10)] =n/2 [30 -10(n-1)] =n/2 [30 -10n +10] =n/2 [40 -10n] =20n -5n². So, that's consistent.Alternatively, since the problem is in Chinese (given the user's instruction mentions "the sum of the first n terms of the arithmetic sequence {b_n}, denoted as T_n", but the solution is in English. Either way, the answer is T_n = -5n² +20n.But in the box, we need to present it. Let me check the problem statement again. It says "find T_n". So, they probably want the expression for T_n. So, boxed as boxed{-5n^2 + 20n}.Alternatively, if they prefer factored form, boxed{5n(4 - n)}. Either is acceptable. Let me check standard textbooks. Usually, the expanded form is more common for such answers. So, I think -5n² +20n is better. Also, in the problem, T_n is "denoted as T_n", so they just want the formula. Therefore, boxed as boxed{-5n^2 + 20n}.But to confirm, let's check with the given information. If n=3, then T_3= -5*(9)+20*3= -45 +60=15, which matches. And for n=1, T_1= -5 +20=15? Wait, but b1=15, so T_1 should be 15. Yes, T_1=15. Wait, T_n= -5n² +20n. For n=1, T1= -5 +20=15, which is correct. For n=2, T2= -20 +40=20, which is the maximum. For n=3, 15. For n=4, T4= -5*16 +80= -80 +80=0. So, yes, that seems correct.So, the answer is T_n = -5n² +20n. Boxed.Final Answer(1) boxed{1}(2) boxed{-5n^2 + 20n}</think>Given the sequence {a_n} with a_1 = t and a_{n+1} = 2S_n + 1 where S_n is the sum of the first n terms:1. For what value of t is the sequence {a_n} a geometric sequence? We start by calculating the first few terms of the sequence: - a_1 = t - a_2 = 2S_1 + 1 = 2t + 1 - S_2 = a_1 + a_2 = t + (2t + 1) = 3t + 1 - a_3 = 2S_2 + 1 = 2(3t + 1) + 1 = 6t + 3 - S_3 = S_2 + a_3 = (3t + 1) + (6t + 3) = 9t + 4 - a_4 = 2S_3 + 1 = 2(9t + 4) + 1 = 18t + 9 For the sequence to be geometric, the ratio between consecutive terms must be constant. Checking the ratios: - frac{a_2}{a_1} = frac{2t + 1}{t} - frac{a_3}{a_2} = frac{6t + 3}{2t + 1} = 3 - frac{a_4}{a_3} = frac{18t + 9}{6t + 3} = 3 Setting frac{2t + 1}{t} = 3 gives 2t + 1 = 3t implies t = 1. Verifying, when t = 1, the sequence is 1, 3, 9, 27, ldots, which is geometric with common ratio 3. Thus, the value of t is boxed{1}.2. Under the condition of (1), find T_n if T_3 = 15, the sum T_n has a maximum value, and a_1 + b_1, a_2 + b_2, a_3 + b_3 form a geometric sequence. Given t = 1, the sequence {a_n} is 1, 3, 9, ldots. For the arithmetic sequence {b_n} with T_3 = 15: - T_3 = frac{3}{2} [2b_1 + 2d] = 3(b_1 + d) = 15 implies b_1 + d = 5 implies d = 5 - b_1 - The terms 1 + b_1, 3 + (b_1 + d), 9 + (b_1 + 2d) form a geometric sequence. Substituting d = 5 - b_1: - x = 1 + b_1 - y = 8 - z = 19 - b_1 For these to form a geometric sequence, 8^2 = (1 + b_1)(19 - b_1): - 64 = 19 + 18b_1 - b_1^2 implies b_1^2 - 18b_1 + 45 = 0 - Solving, b_1 = 15 or b_1 = 3. Since T_n must have a maximum, d must be negative, so b_1 = 15 and d = -10. The sum T_n is: - T_n = frac{n}{2} [2 cdot 15 + (n - 1)(-10)] = frac{n}{2} [30 - 10(n - 1)] = frac{n}{2} [40 - 10n] = 20n - 5n^2 Thus, the sum T_n is boxed{-5n^2 + 20n}.]

🔑:(1) From a_{n+1}=2S_n+1 (①), we can derive a_n=2S_{n-1}+1 (②) for n geq 2.Subtracting these two equations, we obtain a_{n+1}-a_n = 2a_n Rightarrow a_{n+1}=3a_n.Since {a_n} is a geometric sequence, a_2=2S_1+1=2a_1+1=3a_1 Rightarrow a_1=t=1.Thus, sequence {a_n} is a geometric sequence with the first term 1 and a common ratio of 3.Hence, a_n=3^{n-1}.(2) Let the common difference of the arithmetic sequence {b_n} be d.From T_3=15 Rightarrow b_1+b_2+b_3=15 Rightarrow b_2=5,we can set b_1=5-d and b_3=5+d.Given a_1=1, a_2=3, and a_3=9, the problem states (5-d+1)(5+d+9)=(5+3)^2. Solving this equation, we obtain d=-10 or d=2.Since the sum of the first n terms of the arithmetic sequence {b_n}, denoted as T_n, has a maximum value and b_2=5, we take d=-10.Solving for b_1, we get b_1=15.Thus, T_{n}=15n+frac{n(n-1)}{2}times(-10)=boxed{20n-5n^2}.

🧠:[<think>Okay, let's see. I need to find the value of ( y_1^2 + a y_2^2 ) given these equations. Hmm. The problem gives me three equations with variables ( x_1, x_2, y_1, y_2 ), and constants ( a, b, c ), right?So the first equation is ( x_1^2 + a x_2^2 = b ). That looks like a quadratic form. The second equation is ( x_2 y_1 - x_1 y_2 = a ), and the third is ( x_1 y_1 + a x_2 y_2 = c ). And I need to find ( y_1^2 + a y_2^2 ). I wonder if there's a way to relate these equations. Maybe through substitution or elimination? Let me think.Let me list the equations again:1. ( x_1^2 + a x_2^2 = b )2. ( x_2 y_1 - x_1 y_2 = a )3. ( x_1 y_1 + a x_2 y_2 = c )We have two equations involving ( y_1 ) and ( y_2 ), maybe I can solve for ( y_1 ) and ( y_2 ) in terms of ( x_1 ) and ( x_2 ), then substitute into ( y_1^2 + a y_2^2 ).So equations 2 and 3 are linear equations in ( y_1 ) and ( y_2 ). Let me write them as a system:[begin{cases}x_2 y_1 - x_1 y_2 = a quad (1) x_1 y_1 + a x_2 y_2 = c quad (2)end{cases}]Let me try to solve this system. Let's denote equation (1) as the first equation and equation (2) as the second. Let's solve for ( y_1 ) and ( y_2 ).From equation (1): ( x_2 y_1 - x_1 y_2 = a ). Let's solve for ( y_1 ):( x_2 y_1 = a + x_1 y_2 )( y_1 = frac{a + x_1 y_2}{x_2} )Then plug this into equation (2):( x_1 left( frac{a + x_1 y_2}{x_2} right) + a x_2 y_2 = c )Multiply out:( frac{x_1 a}{x_2} + frac{x_1^2 y_2}{x_2} + a x_2 y_2 = c )Let me combine the terms with ( y_2 ):( y_2 left( frac{x_1^2}{x_2} + a x_2 right) + frac{x_1 a}{x_2} = c )Factor ( y_2 ):( y_2 left( frac{x_1^2 + a x_2^2}{x_2} right) + frac{a x_1}{x_2} = c )Wait, from the first original equation, ( x_1^2 + a x_2^2 = b ), so substitute that in:( y_2 left( frac{b}{x_2} right) + frac{a x_1}{x_2} = c )Multiply both sides by ( x_2 ):( b y_2 + a x_1 = c x_2 )Solve for ( y_2 ):( b y_2 = c x_2 - a x_1 )( y_2 = frac{c x_2 - a x_1}{b} )Now substitute this back into the expression for ( y_1 ):( y_1 = frac{a + x_1 y_2}{x_2} )Plugging ( y_2 ):( y_1 = frac{a + x_1 left( frac{c x_2 - a x_1}{b} right)}{x_2} )Simplify numerator:( a + frac{c x_1 x_2 - a x_1^2}{b} = frac{a b + c x_1 x_2 - a x_1^2}{b} )Therefore:( y_1 = frac{a b + c x_1 x_2 - a x_1^2}{b x_2} )So now, expressions for ( y_1 ) and ( y_2 ) in terms of ( x_1, x_2 ). Now, we need to compute ( y_1^2 + a y_2^2 ).Let me compute each term.First, compute ( y_1^2 ):( y_1^2 = left( frac{a b + c x_1 x_2 - a x_1^2}{b x_2} right)^2 )That's going to be messy. Maybe expand it:Numerator squared:( (a b - a x_1^2 + c x_1 x_2)^2 )Denominator squared:( (b x_2)^2 )Similarly, compute ( y_2^2 ):( y_2^2 = left( frac{c x_2 - a x_1}{b} right)^2 )So ( a y_2^2 = a left( frac{c x_2 - a x_1}{b} right)^2 )So combining both terms:( y_1^2 + a y_2^2 = frac{(a b - a x_1^2 + c x_1 x_2)^2}{b^2 x_2^2} + a left( frac{c x_2 - a x_1}{b} right)^2 )This looks complicated, but maybe we can simplify it. Let me try expanding the first numerator:First term numerator: ( (a b - a x_1^2 + c x_1 x_2)^2 )Let me factor out 'a' from the first two terms: ( a(b - x_1^2) + c x_1 x_2 )But ( b - x_1^2 = a x_2^2 ), from the first original equation: ( x_1^2 + a x_2^2 = b implies a x_2^2 = b - x_1^2 ). So substitute that:( a(a x_2^2) + c x_1 x_2 = a^2 x_2^2 + c x_1 x_2 )Therefore, the first numerator becomes ( (a^2 x_2^2 + c x_1 x_2)^2 )So:( y_1^2 = frac{(a^2 x_2^2 + c x_1 x_2)^2}{b^2 x_2^2} )Similarly, the second term:( a left( frac{c x_2 - a x_1}{b} right)^2 = a cdot frac{c^2 x_2^2 - 2 a c x_1 x_2 + a^2 x_1^2}{b^2} )So let's write both terms with denominator ( b^2 ):First term: ( frac{(a^2 x_2^2 + c x_1 x_2)^2}{b^2 x_2^2} = frac{(a^2 x_2^2 + c x_1 x_2)^2}{b^2 x_2^2} )Second term: ( frac{a(c^2 x_2^2 - 2 a c x_1 x_2 + a^2 x_1^2)}{b^2} )Hmm. Let me compute the first term's numerator:( (a^2 x_2^2 + c x_1 x_2)^2 = a^4 x_2^4 + 2 a^2 c x_1 x_2^3 + c^2 x_1^2 x_2^2 )Divided by ( x_2^2 ), we get:( a^4 x_2^2 + 2 a^2 c x_1 x_2 + c^2 x_1^2 )So the first term becomes:( frac{a^4 x_2^2 + 2 a^2 c x_1 x_2 + c^2 x_1^2}{b^2} )The second term is:( frac{a c^2 x_2^2 - 2 a^2 c x_1 x_2 + a^3 x_1^2}{b^2} )Now add them together:Total numerator:( a^4 x_2^2 + 2 a^2 c x_1 x_2 + c^2 x_1^2 + a c^2 x_2^2 - 2 a^2 c x_1 x_2 + a^3 x_1^2 )Let's combine like terms:- Terms with ( x_2^2 ):( a^4 x_2^2 + a c^2 x_2^2 = x_2^2(a^4 + a c^2) )- Terms with ( x_1 x_2 ):( 2 a^2 c x_1 x_2 - 2 a^2 c x_1 x_2 = 0 )- Terms with ( x_1^2 ):( c^2 x_1^2 + a^3 x_1^2 = x_1^2(c^2 + a^3) )So total numerator:( x_2^2(a^4 + a c^2) + x_1^2(c^2 + a^3) )Therefore, ( y_1^2 + a y_2^2 = frac{x_2^2(a^4 + a c^2) + x_1^2(c^2 + a^3)}{b^2} )Hmm, maybe factor out ( a ) and ( c^2 ):Wait, let's see:( a^4 + a c^2 = a(a^3 + c^2) )Similarly, ( c^2 + a^3 = a^3 + c^2 )So numerator becomes:( a(a^3 + c^2) x_2^2 + (a^3 + c^2) x_1^2 )Factor out ( (a^3 + c^2) ):( (a^3 + c^2)(a x_2^2 + x_1^2) )But from the first original equation, ( x_1^2 + a x_2^2 = b ). So substitute that:Numerator becomes ( (a^3 + c^2) cdot b )Therefore, ( y_1^2 + a y_2^2 = frac{(a^3 + c^2) cdot b}{b^2} = frac{a^3 + c^2}{b} )Wait, so the answer is ( frac{a^3 + c^2}{b} )?Let me check again to make sure I didn't make a mistake.Starting from the system of equations, solving for ( y_1 ) and ( y_2 ), substituting into ( y_1^2 + a y_2^2 ), expanding, simplifying, and then using the first equation ( x_1^2 + a x_2^2 = b ). It seems correct. Let me check the steps:1. Solved equations 2 and 3 for ( y_1 ) and ( y_2 ), which should be okay.2. Then substituted those expressions into ( y_1^2 + a y_2^2 ), expanded, combined terms. The cross terms canceled out, which is nice.3. Factored the numerator, utilized the first equation to substitute ( x_1^2 + a x_2^2 = b ), leading to simplification.Yes, seems correct. Therefore, the value of ( y_1^2 + a y_2^2 ) is ( frac{a^3 + c^2}{b} ). So boxed answer is ( boxed{dfrac{a^3 + c^2}{b}} ).Final Answerboxed{dfrac{a^3 + c^2}{b}}</think>Given real numbers (a, b, c) with (b neq 0), and real numbers (x_1, x_2, y_1, y_2) that satisfy the equations:1. (x_1^2 + a x_2^2 = b)2. (x_2 y_1 - x_1 y_2 = a)3. (x_1 y_1 + a x_2 y_2 = c)We need to find the value of (y_1^2 + a y_2^2).First, we solve the system of linear equations (2) and (3) for (y_1) and (y_2):From equation (2):[ x_2 y_1 - x_1 y_2 = a ]Solving for (y_1):[ y_1 = frac{a + x_1 y_2}{x_2} ]Substitute this into equation (3):[ x_1 left( frac{a + x_1 y_2}{x_2} right) + a x_2 y_2 = c ]Simplifying and using (x_1^2 + a x_2^2 = b) from equation (1), we find:[ y_2 = frac{c x_2 - a x_1}{b} ]Substitute (y_2) back into the expression for (y_1):[ y_1 = frac{a b + c x_1 x_2 - a x_1^2}{b x_2} ]Next, we compute (y_1^2 + a y_2^2):For (y_1^2):[ y_1^2 = left( frac{a b + c x_1 x_2 - a x_1^2}{b x_2} right)^2 ]Simplifying using (x_1^2 + a x_2^2 = b), we get:[ y_1^2 = frac{(a^2 x_2^2 + c x_1 x_2)^2}{b^2 x_2^2} ]For (a y_2^2):[ a y_2^2 = a left( frac{c x_2 - a x_1}{b} right)^2 ]Combining both terms:[ y_1^2 + a y_2^2 = frac{(a^4 x_2^2 + 2 a^2 c x_1 x_2 + c^2 x_1^2) + (a c^2 x_2^2 - 2 a^2 c x_1 x_2 + a^3 x_1^2)}{b^2} ]Simplifying the numerator:[ (a^3 + c^2)(x_1^2 + a x_2^2) ]Using (x_1^2 + a x_2^2 = b) again, we get:[ y_1^2 + a y_2^2 = frac{(a^3 + c^2) cdot b}{b^2} = frac{a^3 + c^2}{b} ]Thus, the value of (y_1^2 + a y_2^2) is:[boxed{dfrac{a^3 + c^2}{b}}],<think>Okay, let's see. I have this problem with real numbers a, b, c, and b is not zero. There are these variables x1, x2, y1, y2 that satisfy three equations:1. ( x_1^2 + a x_2^2 = b )2. ( x_2 y_1 - x_1 y_2 = a )3. ( x_1 y_1 + a x_2 y_2 = c )And I need to find the value of ( y_1^2 + a y_2^2 ).Hmm. So, three equations and four variables, but we are supposed to find an expression in terms of a, b, c. Maybe we can manipulate these equations to express y1 and y2 in terms of x1, x2, and the constants, then substitute back into the expression we need.Let me start by writing down the equations again:1. ( x_1^2 + a x_2^2 = b )2. ( x_2 y_1 - x_1 y_2 = a )3. ( x_1 y_1 + a x_2 y_2 = c )We need to find ( y_1^2 + a y_2^2 ).Hmm. Equations 2 and 3 are linear in y1 and y2. Maybe we can solve for y1 and y2 from these two equations. Let's treat equations 2 and 3 as a system of linear equations.Let me write equations 2 and 3 as:Equation 2: ( x_2 y_1 - x_1 y_2 = a )Equation 3: ( x_1 y_1 + a x_2 y_2 = c )So, this is a system of two equations with two variables y1 and y2. Let's write this in matrix form to solve for y1 and y2.The coefficient matrix is:[begin{bmatrix}x_2 & -x_1 x_1 & a x_2end{bmatrix}]And the constants on the right-hand side are [a; c]. So, the system is:[begin{bmatrix}x_2 & -x_1 x_1 & a x_2end{bmatrix}begin{bmatrix}y_1 y_2end{bmatrix}=begin{bmatrix}a cend{bmatrix}]To solve this system, we can use Cramer's Rule or find the inverse of the coefficient matrix. Let me compute the determinant of the coefficient matrix first.The determinant D is:( D = x_2 cdot (a x_2) - (-x_1) cdot x_1 = a x_2^2 + x_1^2 )Wait, from equation 1, we have ( x_1^2 + a x_2^2 = b ). So, the determinant D is exactly equal to b. Since b ≠ 0, the determinant is non-zero, which means the system has a unique solution. Great, that's good news.So, using Cramer's Rule:( y_1 = frac{D_1}{D} ), ( y_2 = frac{D_2}{D} ), where D1 and D2 are the determinants obtained by replacing the first and second columns with the constants, respectively.Compute D1: replace first column with [a; c]:[begin{bmatrix}a & -x_1 c & a x_2end{bmatrix}]Determinant D1 = a*(a x_2) - (-x_1)*c = a^2 x_2 + c x_1Similarly, D2: replace second column with [a; c]:[begin{bmatrix}x_2 & a x_1 & cend{bmatrix}]Determinant D2 = x_2*c - a*x_1Therefore:( y_1 = frac{a^2 x_2 + c x_1}{b} )( y_2 = frac{c x_2 - a x_1}{b} )Hmm, let me check that again.Wait, D1 is determinant of the matrix where the first column is replaced by [a; c], so:First column: a, cSecond column: -x1, a x2So D1 = a*(a x2) - (-x1)*c = a^2 x2 + c x1Yes.Similarly, D2 is determinant of the matrix where the second column is replaced by [a; c]:First column remains x2, x1Second column: a, cSo D2 = x2*c - x1*a = c x2 - a x1Therefore, y2 = D2 / D = (c x2 - a x1)/bSo, we have expressions for y1 and y2 in terms of x1, x2, a, c, and b.Now, we need to compute ( y_1^2 + a y_2^2 ). Let's substitute the expressions for y1 and y2 into this.First, compute y1^2:( y_1^2 = left( frac{a^2 x_2 + c x_1}{b} right)^2 = frac{(a^2 x_2 + c x_1)^2}{b^2} )Similarly, compute a y2^2:( a y_2^2 = a left( frac{c x_2 - a x_1}{b} right)^2 = frac{a (c x_2 - a x_1)^2}{b^2} )Therefore, ( y_1^2 + a y_2^2 = frac{(a^2 x_2 + c x_1)^2 + a (c x_2 - a x_1)^2}{b^2} )Now, we need to expand the numerator and simplify.Let's compute the numerator:First term: (a²x₂ + c x₁)² = a⁴x₂² + 2 a² c x₁ x₂ + c² x₁²Second term: a (c x₂ - a x₁)² = a (c² x₂² - 2 a c x₁ x₂ + a² x₁²) = a c² x₂² - 2 a² c x₁ x₂ + a³ x₁²So, adding these two terms:First term + Second term:= (a⁴ x₂² + 2 a² c x₁ x₂ + c² x₁²) + (a c² x₂² - 2 a² c x₁ x₂ + a³ x₁²)Let's combine like terms:For x₁² terms: c² x₁² + a³ x₁² = x₁² (c² + a³)For x₂² terms: a⁴ x₂² + a c² x₂² = x₂² (a⁴ + a c²)For cross terms: 2 a² c x₁ x₂ - 2 a² c x₁ x₂ = 0So, the numerator simplifies to:x₁² (c² + a³) + x₂² (a⁴ + a c²)Factor out common terms:x₁² (c² + a³) + x₂² a (a³ + c²)Wait, let me check:Wait, x₂² (a⁴ + a c²) can be written as x₂² a (a³ + c²)/a? Wait, no:Wait, a⁴ + a c² = a(a³ + c²). Hmm, actually:a⁴ + a c² = a³ * a + a c² = a(a³ + c²). Wait, no:Wait, a⁴ is a^4, and a c² is a c², so factoring out an a gives a(a³ + c²). Yes.So, numerator becomes:x₁² (c² + a³) + x₂² a (a³ + c²) = (a³ + c²)(x₁² + a x₂²)But from equation 1, x₁² + a x₂² = b. Therefore, numerator is (a³ + c²) * bTherefore, numerator = b(a³ + c²)Therefore, ( y_1^2 + a y_2^2 = frac{b(a³ + c²)}{b²} = frac{a³ + c²}{b} )So, the value is ( frac{a^3 + c^2}{b} )Wait, but let me check again.Wait, the numerator was:x₁² (c² + a³) + x₂² a (a³ + c²)So, factor out (a³ + c²):(a³ + c²)(x₁² + a x₂²)But x₁² + a x₂² = b, so numerator is (a³ + c²) * bTherefore, the expression is (a³ + c²) * b / b² = (a³ + c²)/bYes. So, the answer is (a³ + c²)/b. Hmm.But let me check with a simple case to see if this makes sense. Let's choose some numbers where we can compute everything.Suppose a = 1, b = 2, c = 3.Then equation 1: x₁² + x₂² = 2Equation 2: x₂ y₁ - x₁ y₂ = 1Equation 3: x₁ y₁ + x₂ y₂ = 3We can choose x1 and x2 such that x1² + x2² = 2. Let's pick x1 = 1, x2 = 1. Then 1 + 1 = 2, which satisfies equation 1.Now, solve equations 2 and 3 with x1=1, x2=1:Equation 2: 1 * y1 - 1 * y2 = 1 → y1 - y2 = 1Equation 3: 1 * y1 + 1 * y2 = 3 → y1 + y2 = 3Solving these two equations:From equation 2: y1 = y2 + 1Substitute into equation 3:(y2 + 1) + y2 = 3 → 2 y2 + 1 = 3 → 2 y2 = 2 → y2 = 1Then y1 = 1 + 1 = 2So y1 = 2, y2 = 1Then compute y1² + a y2² = 2² + 1 * 1² = 4 + 1 = 5Now, according to our formula, (a³ + c²)/b = (1 + 9)/2 = 10/2 = 5. Perfect, matches. So that's correct.Another test case. Let's take a = 2, b = 3, c = 4.Compute expected value: (8 + 16)/3 = 24/3 = 8.Now, we need to find x1, x2 such that x1² + 2 x2² = 3. Let's choose x1=1, then 1 + 2 x2² = 3 ⇒ x2² = 1 ⇒ x2=1 or -1. Let's take x2=1.Then equations 2 and 3:Equation 2: 1 * y1 - 1 * y2 = 2 → y1 - y2 = 2Equation 3: 1 * y1 + 2 * 1 * y2 = 4 → y1 + 2 y2 = 4Solve these two equations:From equation 2: y1 = y2 + 2Substitute into equation 3:(y2 + 2) + 2 y2 = 4 ⇒ 3 y2 + 2 = 4 ⇒ 3 y2 = 2 ⇒ y2 = 2/3Then y1 = 2/3 + 2 = 8/3Compute y1² + 2 y2² = (8/3)^2 + 2*(2/3)^2 = 64/9 + 2*(4/9) = 64/9 + 8/9 = 72/9 = 8, which matches (a³ + c²)/b = 8 + 16 /3? Wait, wait:Wait a=2, c=4. So (2³ + 4²)/3 = (8 + 16)/3 = 24/3 = 8. Correct. So yes, works.Therefore, the answer is (a³ + c²)/b, which is boxed{dfrac{a^3 + c^2}{b}}.Final Answerboxed{dfrac{a^3 + c^2}{b}}</think>Given real numbers (a, b, c) with (b neq 0), and real numbers (x_1, x_2, y_1, y_2) that satisfy the equations:1. (x_1^2 + a x_2^2 = b)2. (x_2 y_1 - x_1 y_2 = a)3. (x_1 y_1 + a x_2 y_2 = c)We need to find the value of (y_1^2 + a y_2^2).First, we solve the system of linear equations (2) and (3) for (y_1) and (y_2). The system can be written in matrix form as:[begin{bmatrix}x_2 & -x_1 x_1 & a x_2end{bmatrix}begin{bmatrix}y_1 y_2end{bmatrix}=begin{bmatrix}a cend{bmatrix}]The determinant of the coefficient matrix is (D = x_1^2 + a x_2^2 = b) (from equation 1). Using Cramer's Rule, we find:[y_1 = frac{a^2 x_2 + c x_1}{b}, quad y_2 = frac{c x_2 - a x_1}{b}]Next, we substitute these expressions into (y_1^2 + a y_2^2):[y_1^2 + a y_2^2 = left( frac{a^2 x_2 + c x_1}{b} right)^2 + a left( frac{c x_2 - a x_1}{b} right)^2]Expanding and simplifying the numerator:[(a^2 x_2 + c x_1)^2 + a (c x_2 - a x_1)^2 = a^4 x_2^2 + 2a^2 c x_1 x_2 + c^2 x_1^2 + a c^2 x_2^2 - 2a^2 c x_1 x_2 + a^3 x_1^2]Combining like terms:[x_1^2 (c^2 + a^3) + x_2^2 (a^4 + a c^2) = (a^3 + c^2)(x_1^2 + a x_2^2)]Using equation 1, (x_1^2 + a x_2^2 = b), the numerator becomes:[(a^3 + c^2) b]Thus, the value of (y_1^2 + a y_2^2) is:[frac{(a^3 + c^2) b}{b^2} = frac{a^3 + c^2}{b}]The final answer is:[boxed{dfrac{a^3 + c^2}{b}}]]

🔑:5. frac{c^{2}+a^{3}}{b}.From the given, we have left(x_{2} y_{1}-x_{1} y_{2}right)^{2}=a^{2}, that isbegin{array}{l}2 x_{1} x_{2} y_{1} y_{2}+a^{2}=x_{2}^{2} y_{1}^{2}+x_{1}^{2} y_{2}^{2} . text { Hence } bleft(y_{1}^{2}+a y_{2}^{2}right)=left(x_{1}^{2}+a x_{2}^{2}right)left(y_{1}^{2}+a y_{2}^{2}right) =x_{1}^{2} y_{1}^{2}+aleft(x_{1}^{2} y_{2}^{2}+x_{2}^{2} y_{1}^{2}right)+a^{2} x_{2}^{2} y_{2}^{2} =x_{1}^{2} y_{1}^{2}+aleft(2 x_{1} x_{2} y_{1} y_{2}+a^{2}right)+a^{2} x_{2}^{2} y_{2}^{2} =left(x_{1} y_{1}+a x_{2} y_{2}right)^{2}+a^{3}=c^{2}+a^{3} .end{array}Since b neq 0, we have y_{1}^{2}+a y_{2}^{2}=frac{c^{2}+a^{3}}{b}.

🧠:[<think>Okay, let's try to figure out this problem step by step. So, we have Annemiek and Bart, each with three different positive integers on their papers. There's exactly one common number between them. Also, if you take any two different numbers from Annemiek's paper and add them, the result is always on Bart's paper. Additionally, Annemiek's favorite number multiplied by 3 is on Bart's paper, and Bart's favorite number is 25. We need to find Annemiek's favorite number.First, let me parse the problem again to make sure I didn't miss any details. Both have three different positive integers. One number is common to both. All pairwise sums from Annemiek's numbers are on Bart's paper. Annemiek has a favorite number, which when multiplied by 3 is on Bart's paper. Bart's favorite number is 25. So, 25 is one of Bart's numbers.Let me denote Annemiek's numbers as A = {a, b, c}, all different positive integers, and Bart's numbers as B = {d, e, f}, also different positive integers. They share exactly one number, so one of a, b, c is equal to one of d, e, f. The sums a+b, a+c, b+c must all be in B. Also, one of Annemiek's numbers, say a (her favorite), when multiplied by 3 gives another number in B. And Bart's favorite number is 25, so 25 is in B.So, Bart's numbers include 25 and 3a (if a is the favorite). But wait, the problem states that Annemiek's favorite number multiplied by 3 is on Bart's paper. So, 3*(favorite) is in B. Also, since there is exactly one common number between A and B, that common number is one of A's numbers and one of B's numbers. But 3*(favorite) could be another number in B, unless the favorite number itself is the common one. So, we need to consider that possibility.Let me outline the constraints:1. A ∩ B = {x}, where x is one number present in both A and B.2. All pairwise sums of A are in B. So, a+b, a+c, b+c ∈ B.3. 3*(favorite) ∈ B, where favorite is one of a, b, c.4. 25 ∈ B, and it's Bart's favorite number.Also, Bart has three numbers, so B has three elements. Therefore, B must contain 25, 3*(favorite), and possibly some other numbers. But since all pairwise sums of A are in B, and B has only three numbers, those sums must be exactly the three numbers in B. However, Bart's numbers also include the common number x. Wait, this seems conflicting.Wait, let's check again. The three numbers in B must include:- The common number x (which is in both A and B)- The three pairwise sums of A: a+b, a+c, b+c- The 3*(favorite) number- 25But B can only have three numbers. So, there must be overlaps here. So, the three numbers in B must be x, 25, and 3*(favorite), but also all pairwise sums of A must be in B. Wait, unless the pairwise sums are exactly the three numbers in B. But if the three pairwise sums are the three numbers in B, then B would consist of a+b, a+c, b+c. But at the same time, B must include x (the common number) and 25, Bart's favorite. So, this implies that one of the pairwise sums is x, and another is 25, and the third is 3*(favorite). Or perhaps x is one of the pairwise sums?Wait, no. Because x is a number that is on both A and B. So, x is an element of A, which is a single number, and also in B, which has three numbers. The pairwise sums of A must all be in B. So, if A has numbers a, b, c, then the sums a+b, a+c, b+c must all be in B. But B can only have three numbers. So, unless those three sums are exactly the three numbers in B. But B also has 25 and 3*(favorite). Therefore, the three sums (a+b, a+c, b+c) must be 25, 3*(favorite), and the common number x. But wait, the common number x is in both A and B. So, x is one of a, b, c, and also in B.But if the sums a+b, a+c, b+c are in B, then unless x is one of the sums. But x is an original number in A, not a sum. Wait, but x is in both A and B. So, x is one of a, b, c, and also in B. But B's numbers are the three sums and 25 and 3*(favorite). Hmm, this is getting confusing. Let me try to model this.Suppose A = {x, y, z}, where x is the common number in B. Then, B must contain x, and also the sums x+y, x+z, y+z. But B can only have three numbers. So, B must be {x, x+y, x+z, y+z} but that's four numbers, which is impossible. Therefore, there must be overlaps. The three sums must include x and 25 and 3*(favorite). Wait, maybe not. Let's think again.Wait, B is Bart's numbers, which have three elements. The three elements of B must include:- The common number x (since A and B share exactly one number)- 25 (Bart's favorite)- The three sums of A's numbers: a+b, a+c, b+c- 3*(favorite) where favorite is in ABut since B only has three numbers, these must overlap. So, the three numbers in B are x, 25, and 3*(favorite), but also must include the three sums. Therefore, this is only possible if the three sums are exactly x, 25, and 3*(favorite). However, x is an element of A, so it's one of a, b, c. The sums a+b, a+c, b+c must equal x, 25, and 3*(favorite). But since x is in A, and the sums are of two numbers from A, which are positive integers, so the sums would be larger than any individual number in A (assuming all numbers are positive). Wait, but if x is one of the sums, that would mean that x = a + b, but x is also in A. So, x is equal to one of the numbers in A, which is also a sum of two numbers in A. But that would require that x is equal to the sum of two numbers in A, which are both smaller than x. But all numbers are positive integers. Wait, but if x is the sum of two smaller numbers, then x can't be one of those smaller numbers. So, this leads to a contradiction. Therefore, the common number x cannot be equal to any of the sums. Therefore, the three sums must be 25, 3*(favorite), and another number, but since B can only have three numbers, which are x, 25, 3*(favorite). Therefore, the three sums must be 25, 3*(favorite), and x. But since x is in A, but the sums are a+b, a+c, b+c, which are all greater than any individual number in A (assuming all numbers are distinct positive integers). So, unless x is the largest number in A, but even then, the sum of two numbers would be larger than x, which would mean x is not in B. Wait, this is getting really tangled.Let me try to approach this step by step.First, since all pairwise sums of Annemiek's numbers are in Bart's numbers. Bart has three numbers. So, there are three sums (a+b, a+c, b+c) that must be the three numbers in Bart's list. However, Bart also has 25 as his favorite number and 3*(favorite) from Annemiek. But since Bart can only have three numbers, the three sums must include 25 and 3*(favorite). Also, Bart's numbers include the common number x. Therefore, the common number x must be one of the three sums. Wait, but x is in A and B. If x is in A, then x is one of a, b, c, and also in B. But the sums a+b, a+c, b+c are all in B. Therefore, x must be equal to one of the sums? But that's impossible because if x is in A, then x is one of a, b, c, but the sums a+b, etc., are larger than each of a, b, c (since they are positive integers). Therefore, x cannot be equal to any of the sums. Therefore, there's a contradiction here unless one of the numbers in A is equal to a sum of the other two. Wait, but they are distinct positive integers. If a + b = c, for example, then c is in A and the sum a + b is equal to c. But then c is in A, and the sum a + b is equal to c, which would mean that c is in B as well. So, in this case, x would be c. Therefore, c is the common number. Then, Bart's numbers would be a + b (=c), a + c, b + c. But Bart's numbers must also include 25 and 3*(favorite). Wait, so if c is the common number, then Bart's numbers are c, a + c, b + c. But Bart's numbers also must include 25 and 3*(favorite). Therefore, either c = 25 or one of a + c or b + c is 25, and similarly for 3*(favorite). Let's suppose c is the common number, so c is in both A and B. Then, Bart's numbers are c, a + c, b + c. But Bart's numbers must include 25 and 3*(favorite). So, either c = 25, or a + c = 25, or b + c = 25. Similarly, 3*(favorite) must be one of c, a + c, b + c. Since favorite is one of a, b, c. Let's consider possible cases.Case 1: c is the common number, and c = 25. Then Bart's numbers are 25, a + 25, b + 25. But Bart's favorite number is 25, so that's okay. Also, 3*(favorite) must be in Bart's numbers. So, 3*(favorite) is either 25, a + 25, or b + 25. If favorite is one of a, b, c (which is 25). If favorite is 25, then 3*25 = 75. So, 75 must be in Bart's numbers. Therefore, either a + 25 = 75 or b + 25 = 75. So, a or b would be 50. But then, a and b are different positive integers, and along with c =25, they must all be distinct. So, if a =50, then Annemiek's numbers are 50, b, 25. Then the sums would be 50 + b, 50 +25=75, b +25. These are Bart's numbers. So Bart's numbers would be 25, 75, 50 + b, and b +25. But Bart can only have three numbers. So, 50 + b and b +25 must be equal to 25, 75, or another number. Wait, but 25 is already in Bart's numbers. If 50 + b is 75, then b =25, but then Annemiek's numbers would be 50,25,25, which are not distinct. So that's invalid. If b +25 =75, then b=50, but then Annemiek's numbers would be 50,50,25, again duplicates. So this case doesn't work. Therefore, c cannot be 25.Case 2: c is the common number, and a + c =25. So, Bart's numbers are c, 25, b + c. Also, 3*(favorite) must be in Bart's numbers. So, favorite is either a, b, or c. If favorite is a, then 3a must be in Bart's numbers: c, 25, b + c. So, 3a is either c, 25, or b + c. Similarly for favorite being b or c.If favorite is c, then 3c must be in B. So, 3c is either c, 25, or b + c. 3c cannot be c unless c=0, which is invalid. 3c =25 would mean c=25/3, not integer. 3c = b + c => 2c = b. So, b=2c. Then, since a + c =25, a=25 -c. So, Annemiek's numbers are a=25 -c, b=2c, c. All must be distinct positive integers. So, 25 -c, 2c, c must be distinct. So, 25 -c ≠ 2c => 25 ≠ 3c => c ≠25/3. Similarly, 25 -c ≠c => 25 ≠2c => c≠12.5. And 2c ≠c => c≠0. So, c must be an integer such that c <25 (since a=25 -c must be positive), and c must not be 12.5. So c is integer from 1 to24. But also, Bart's numbers are c,25,b +c =2c +c=3c. So, Bart's numbers are c,25,3c. But Bart's numbers must be distinct. So c ≠25, which is already satisfied, c≠3c => c≠0, which is okay, and 3c≠25. So, 3c ≠25 => c≠25/3 ≈8.333. So c must not be 8.333. So c is integer from1 to24, not equal to 25/3.But also, Annemiek's numbers are 25 -c, 2c, c. These must all be distinct and positive. Let's check if c=8: then a=25-8=17, b=16, c=8. So A={17,16,8}. All distinct. Then Bart's numbers are 8,25,24. So B={8,24,25}. The common number is 8. Then, check if 3*(favorite) is in B. If favorite is c=8, then 3*8=24, which is in B. So that works. Then, favorite number is 8. So is this the answer? Wait, let's check if all pairwise sums of A are in B.A={17,16,8}. The pairwise sums are 17+16=33, 17+8=25, 16+8=24. So the sums are 33,25,24. Bart's numbers are 8,24,25. So 24 and25 are in B, but 33 is not. Instead, 8 is in B. So the sums 24,25,33 should be in B, but B only has 8,24,25. So 33 is missing. Therefore, this is invalid. Therefore, our assumption is wrong.Therefore, this case doesn't work. So maybe favorite is a or b.If favorite is a, then 3a must be in B. B is {c,25,3c} (if we take favorite as c earlier, but we saw that doesn't work). Wait, in this case, favorite is a. So 3a is in B. B is {c,25,b + c}. So 3a must equal c, 25, or b + c. Let's see:Given that a =25 -c, and b=2c (from previous assumption where favorite was c). Wait, but if favorite is a, then 3a must be in B. So 3a =3*(25 -c). So 3*(25 -c) must equal c,25, or b +c=3c.So, 3*(25 -c)=c => 75 -3c =c =>75=4c =>c=75/4=18.75, not integer.3*(25 -c)=25 =>75 -3c=25 =>3c=50 =>c=50/3≈16.666, not integer.3*(25 -c)=3c =>75 -3c=3c =>75=6c =>c=12.5, which is not integer.Therefore, no solution here.Similarly, if favorite is b, then 3b must be in B. B={c,25,3c} (if b=2c). So 3b=6c. So 6c must equal c,25, or3c. 6c=c=>c=0 invalid. 6c=25=>c=25/6≈4.166 invalid. 6c=3c=>3c=0 invalid. Thus, no solution.Therefore, this case where a +c=25 and c is common number leads to no solution. Similarly, if instead, b +c=25, we can check.Case 3: c is the common number, and b +c=25. Then, Bart's numbers are c, a +c,25. Also, 3*(favorite) must be in B. So favorite is one of a,b,c. Let's see.If favorite is c, then 3c must be in B={c, a +c,25}. So 3c is either c (invalid), a +c, or25. If 3c = a +c, then a=2c. If 3c=25, then c=25/3≈8.333, invalid.If a=2c, then Annemiek's numbers are a=2c, b, c. But b +c=25 =>b=25 -c. So A={2c,25 -c,c}. These must be distinct. 2c ≠25 -c =>3c≠25 =>c≠25/3. 2c≠c =>c≠0. 25 -c ≠c =>25≠2c =>c≠12.5. So c must be integer, 1 ≤c≤24, c≠25/3≈8.333, c≠12.5. So c is integer. Then, Bart's numbers are c, a +c=3c,25. So B={c,3c,25}. Also, 3*(favorite)=3c must be in B, which it is. If favorite is c, then 3c is in B. So, this works if favorite is c. But let's check the sums of A.A={2c,25 -c,c}. The pairwise sums are:2c + (25 -c)=25 +c2c +c=3c(25 -c) +c=25So the sums are 25 +c,3c,25. But Bart's numbers are c,3c,25. So the sum 25 +c must equal one of c,3c,25. 25 +c can only equal 25 if c=0, invalid. 25 +c=3c =>25=2c =>c=12.5, invalid. 25 +c=c =>25=0, invalid. Therefore, the sum 25 +c is not in B. Therefore, invalid. So this case doesn't work.If favorite is a, which is2c, then 3*(2c)=6c must be in B={c,3c,25}. 6c can only equal 3c or25. 6c=3c =>c=0 invalid. 6c=25 =>c=25/6≈4.166 invalid. Similarly, favorite being b=25 -c: 3*(25 -c)=75 -3c must be in B={c,3c,25}. So 75 -3c equals c,3c, or25. If 75 -3c=c =>75=4c =>c=18.75 invalid. If 75 -3c=3c =>75=6c =>c=12.5 invalid. If 75 -3c=25 =>3c=50 =>c≈16.666 invalid. No solution here.Therefore, this case also doesn't work. So maybe the common number isn't c. Maybe it's a or b. Let's try that.Suppose the common number is a. So a is in both A and B. Then, Bart's numbers are a, and the three sums a+b, a+c, b+c. But Bart can only have three numbers, so two of the sums must be equal to a, which is impossible because a+b >a (since b is positive). Similarly, if the common number is b, same issue. Therefore, the common number cannot be a or b because the sums would be larger than a or b, hence cannot be equal. Therefore, the common number must be c, as we considered before, but that led to contradictions. So maybe our initial assumption is wrong.Wait, maybe the three sums are the three numbers in B, and the common number is one of those sums. But the common number is in A, which is a single number. If the common number is a sum of two numbers in A, then as before, that sum would have to be equal to one of the numbers in A, which is impossible because the sum is larger. Therefore, the common number cannot be one of the sums. Therefore, Bart's three numbers must be the three sums plus the common number. But since Bart can only have three numbers, the common number must be one of the sums. But we just saw that's impossible. Therefore, contradiction.Wait, perhaps the problem is that B contains the common number x, 25, and 3*(favorite), and the sums are overlapping with these. Let's consider that B = {x, 25, 3*(favorite)}, and these three numbers must also include all pairwise sums from A. So, the three sums must be x, 25, 3*(favorite). Therefore, the pairwise sums of A are exactly x,25,3*(favorite). Since A has three numbers, the sums are a+b, a+c, b+c. So:a + b = xa + c =25b + c =3*(favorite)But x is the common number, so x is in A. Therefore, x is one of a, b, c. Let's suppose x =a. Then:a + b =a =>b=0, invalid.If x =b:a + b =b =>a=0, invalid.If x =c:a + b =ca + c =25b + c =3*(favorite)Also, Annemiek's favorite number is one of a, b, c. Let's suppose favorite is a:Then 3a must be in B, which is {c,25,3a} (since x=c). So B={c,25,3a}. But from the sums:a + b =ca + c =25 =>a + (a + b)=25 =>2a +b=25b + c=3a =>b + (a + b)=3a =>a +2b=3a =>2b=2a =>b=aBut Annemiek's numbers are a, b=a, c=a +b=2a. But they must be distinct. So b=a is invalid. Therefore, favorite cannot be a.If favorite is b:3b must be in B={c,25,3b}From the sums:a + b =ca + c =25 =>a +a + b=25 =>2a +b=25b + c =3b =>b +a +b=3b =>a +2b=3b =>a =bAgain, duplicate numbers, invalid.If favorite is c:3c must be in B={c,25,3c}From the sums:a + b =ca + c =25 =>a +c=25b + c =3c =>b=2cFrom a +c=25 and a +b =c, substitute b=2c:a +2c =c =>a= -c. But a must be positive. Contradiction.Therefore, this approach also leads to contradictions.Alternatively, maybe the three sums are 25,3*(favorite), and another number, which is the common number x. Wait, but we thought x can't be a sum. Hmm. Let's try this.Let’s denote the three sums as S1, S2, S3. Then B = {x,25,3*(favorite)}, and the three sums must be these three numbers. Therefore:S1 =xS2=25S3=3*(favorite)But S1 is a sum of two numbers from A. If x is in A, then S1 =x =a +b (assuming S1 is the sum of a and b). But then x must be one of the numbers in A, which is equal to a +b. But since a and b are positive integers, a +b >a and a +b >b. Therefore, x cannot be equal to a or b. So x must be the third number c. Therefore, c =a +b. Then, the sums would be:S1 =a +b =cS2 =a +c =a + (a +b)=2a +bS3 =b +c =b + (a +b)=a +2bSo B = {c,25,3*(favorite)}. So S1, S2, S3 must be c,25,3*(favorite). Therefore, we have:c =c (duh)One of S2 or S3 is 25, and the other is 3*(favorite). Let's say S2 =2a +b =25 and S3 =a +2b =3*(favorite). Or vice versa.Also, favorite is one of a, b, c. Let's suppose favorite is c. Then, 3*(favorite)=3c =3(a +b). So S3 =a +2b =3(a +b). Then:a +2b =3a +3b =>0=2a +b. But a and b are positive integers. Contradiction.If favorite is a, then 3a must be equal to S3 or S2. Suppose S3=3a. Then:a +2b =3a =>2b=2a =>b=a. But then Annemiek's numbers are a, a, c=2a. Duplicates, invalid.If favorite is b, then 3b must be equal to S2 or S3. Suppose S2=3b. Then:2a +b=3b =>2a=2b =>a=b. Again duplicates.Alternatively, if S3=3b. Then a +2b=3b =>a=b. Again duplicates.Alternatively, if S2=25 and S3=3*(favorite). Let's assume S2=25=2a +b and S3=3*(favorite)=a +2b. Then:From S2: 2a +b=25From S3: a +2b=3*(favorite)Also, Annemiek's numbers are a, b, c=a +b. Favorite is one of a, b, c.Case 1: favorite is a.Then, 3a =a +2b =>2a =2b =>a =b. Duplicates. Invalid.Case 2: favorite is b.Then, 3b =a +2b =>a =b. Duplicates. Invalid.Case3: favorite is c.Then, 3c=3(a +b)=a +2b =>3a +3b =a +2b =>2a +b=0. Impossible.Therefore, no solution.Alternatively, maybe S2=3*(favorite) and S3=25.So 2a +b=3*(favorite)a +2b=25Again, favorite is a,b,c.If favorite is a:3a=2a +b =>a =b. Then from a +2b=25 =>3a=25 =>a=25/3≈8.333. Not integer.If favorite is b:3b=2a +b =>2a=2b =>a=b. Then from a +2b=25 =>3a=25 =>a≈8.333 invalid.If favorite is c:3c=2a +b. But c=a +b. So 3(a +b)=2a +b =>3a +3b=2a +b =>a +2b=0. Impossible.Therefore, no solution here.This is getting frustrating. Maybe our initial approach is wrong. Let's think differently.Since Bart's numbers include 25 and 3*(favorite). Let's suppose that 3*(favorite) is 25. Then favorite=25/3≈8.333, not integer. So 3*(favorite) cannot be 25. Therefore, 25 and 3*(favorite) are distinct numbers in B. Also, B includes the common number x. So B={x,25,3f}, where f is the favorite. Also, all pairwise sums from A must be in B. Therefore, the three sums must be x,25,3f. So the three sums are x,25,3f. Therefore, the three sums of A are x,25,3f. Also, A has three numbers, say a,b,c. Let's assume the sums are:a +b =xa +c=25b +c=3fBut x is one of a,b,c. Suppose x=a. Then:a +b =a =>b=0 invalid.If x=b:a +b =b =>a=0 invalid.If x=c:a +b =cThen the other sums are a +c=25 and b +c=3f.But Annemiek's numbers are a,b,c. And favorite is one of them. Let's say favorite is a:Then 3a must be in B={c,25,3a}But from the sums:a +c=25 =>c=25 -ab +c=3a =>b +25 -a=3a =>b=4a -25Also, since c =a +b (from a +b =c), then:c= a +b= a +4a -25=5a -25But c=25 -a, so:5a -25=25 -a =>6a=50 =>a=50/6≈8.333, not integer.If favorite is b:Then 3b must be in B={c,25,3b}From sums:a +c=25 =>c=25 -ab +c=3b =>25 -a +b=3b =>25 -a=2b =>a=25 -2bAlso, from a +b =c=25 -a =>a +b=25 -a =>2a +b=25But a=25 -2b, so substitute:2*(25 -2b) +b=25 =>50 -4b +b=25 =>50 -3b=25 =>3b=25 =>b≈8.333 invalid.If favorite is c:Then 3c must be in B={c,25,3c}From sums:a +c=25b +c=3c =>b=2cFrom a +b =c =>a +2c =c =>a= -c. Invalid.Therefore, no solution.Alternatively, maybe the sums are arranged differently. Let's suppose:a +b=25a +c=3fb +c=xAnd x is the common number. So B={x,25,3f}Then, since x is in A, x is one of a,b,c.If x=a:From b +c=aBut a +b=25 =>a=25 -bSo b +c=25 -b =>c=25 -2bAlso, a +c=3f =>(25 -b) +c=3f =>25 -b +25 -2b=3f =>50 -3b=3f =>f=(50 -3b)/3Since f must be an integer, 50 -3b must be divisible by 3. So 50 ≡3b mod3 =>50≡0 mod3 =>50≡2 mod3. So 2≡0 mod3? No. Contradiction.If x=b:From b +c=b =>c=0 invalid.If x=c:From b +c=c =>b=0 invalid.So this arrangement also doesn't work.Another arrangement:a +b=3fa +c=25b +c=xB={x,25,3f}Again, x is in A. If x=a:Then b +c=aFrom a +c=25 =>c=25 -aThen b +25 -a =a =>b=2a -25From a +b=3f =>a +2a -25=3f =>3a -25=3f =>f=a -25/3Not integer unless 25 divisible by3, which it's not.If x=b:From b +c=b =>c=0 invalid.If x=c:From b +c=c =>b=0 invalid.Another arrangement:a +b=3fa +c=xb +c=25B={x,25,3f}x is in A.If x=a:From a +c=a =>c=0 invalid.If x=b:From a +c=bFrom b +c=25 =>a +c +c=25 =>a +2c=25From a +b=3f =>a + (a +c)=3f =>2a +c=3fBut b=a +cSo Annemiek's numbers are a, a +c, c. Wait, but c must be different from a and a +c.Wait, Annemiek's numbers are a, b, c. If x=b, then b is in both A and B. From a +c =b. From b +c=25 =>(a +c) +c=25 =>a +2c=25. From a +b=3f =>a + (a +c)=2a +c=3f. So:From a +2c=25 =>a=25 -2cSubstitute into 2a +c=3f:2*(25 -2c) +c=50 -4c +c=50 -3c=3f =>f=(50 -3c)/3Again, 50 -3c must be divisible by3. 50 ≡2 mod3, so 50 -3c ≡2 -0≡2 mod3. So 2≡0 mod3? No. Therefore, no solution.If x=c:From a +c=c =>a=0 invalid.This is really tricky. Maybe there's another approach.Since Bart's numbers are 25, 3f, and x (the common number). The three sums from A must be these three numbers. Let's assume that the largest sum is 3f, and another sum is25. Let's suppose the numbers on Annemiek's paper are such that two of them add up to25, and another two add up to3f, and the third add up tox. And x is the common number.Let’s denote Annemiek's numbers as p, q, r, with p < q < r. Then the sums are p+q, p+r, q+r. These must be the three numbers in B, which are x,25,3f. So the three sums are x,25,3f. Since the sums are ordered: p+q < p+r < q+r. So the smallest sum is x, next is25, largest is3f. So:p +q =xp +r=25q +r=3fAlso, x is one of p, q, r.If x is p:Then p +q =p =>q=0 invalid.If x is q:p +q =q =>p=0 invalid.If x is r:p +q =rBut since p < q < r, r is the largest number. However, p +q =r, but p and q are smaller than r. Therefore, r =p +q. Then the sums are:r,25,3fAlso, Annemiek's numbers are p, q, r=p +q. The other sums are p +r=25 and q +r=3f.So:p +r =25 =>p +p +q=25 =>2p +q=25q +r=3f =>q +p +q=3f =>p +2q=3fAlso, Bart's numbers are r=p +q,25,3f. Since r must be one of Bart's numbers and also in A. Since r is in A and B. So Bart's numbers are {p +q,25,3f}.Also, 3*(favorite) is in B. Favorite is one of p, q, r. So 3*(favorite) must be one of p +q,25,3f.If favorite is p:3p must be in B. So 3p is either p +q,25, or3f.If 3p =p +q =>2p=qFrom 2p +q=25 =>2p +2p=25 =>4p=25 =>p=6.25 invalid.If 3p=25 =>p=25/3≈8.333 invalid.If 3p=3f =>p=f. Then favorite=p, so3p=3f =>f=p. So from p +2q=3f =>p +2q=3p =>2q=2p =>q=p. But p < q, so invalid.If favorite is q:3q must be in B={p +q,25,3f}If3q =p +q =>2q=p. From2p +q=25 =>2*(2q) +q=5q=25 =>q=5, thenp=10. But p < q would mean10 <5, which is invalid.If3q=25 =>q≈8.333 invalid.If3q=3f =>q=f. Fromp +2q=3f =>p +2q=3q =>p=q. Invalid.If favorite is r:3r must be in B={r,25,3f}So3r =r =>r=0 invalid.3r=25 =>r≈8.333 invalid.3r=3f =>r=f. So favorite=r=f. Fromp +2q=3f=3r. But r=p +q. So:p +2q=3(p +q) =>p +2q=3p +3q =>0=2p +q. Invalid.Alternatively, maybe the sums are ordered differently. Suppose the middle sum is x. So:p +q =xp +r=3fq +r=25But x is one of p, q, r.If x is r:p +q=rFrom q +r=25 =>q +p +q=25 =>p +2q=25From p +r=3f =>p +p +q=2p +q=3fAnnemiek's numbers are p, q, r=p +qFavorite is one of them. 3*(favorite) must be in B={x=r,3f,25}If favorite is p:3p in B. So3p=r,3f,25If3p=r=p +q. Then3p =p +q =>q=2pFrom p +2q=25 =>p +4p=25 =>5p=25 =>p=5, q=10, r=15So A={5,10,15}B={15,3f,25}The sums are p +q=15, p +r=20, q +r=25. So B should be {15,20,25}, but Bart's numbers are {15,3f,25}. Therefore, 3f must be20 =>f=20/3≈6.666 invalid.If3p=3f =>p=f. Then from2p +q=3f=3p =>q=p. But p < q invalid.If3p=25 =>p≈8.333 invalid.If favorite is q:3q in B. So3q=15,3f,25If3q=15 =>q=5. But q=10 above. Invalid.If3q=3f =>q=f. Fromp +2q=25 and2p +q=3f=3q =>2p +q=3q =>2p=2q =>p=q. Invalid.If3q=25 =>q≈8.333 invalid.If favorite is r=15:3*15=45 must be in B. But B={15,3f,25}. So45 must be one of them. But25 is already there. If45=3f =>f=15. So B={15,45,25}. But the sums are15,20,25. So15,20,25 must be B, but according to this, B is15,45,25. Missing20. Contradiction.Alternatively, maybe the sums are arranged as p +q=25, p +r=3f, q +r=x. Then B={25,3f,x}. x is in A.If x is p:q +r=p. But p < q < r, so q +r >q >p. Contradiction.If x is q:q +r=q =>r=0 invalid.If x is r:q +r=r =>q=0 invalid.This approach isn't working. Maybe we need to consider that one of the sums is 3f and another sum is25, and the third sum is the common number x. But since x is in A, which is one of the numbers, and sums are greater than the numbers, this leads to a problem.Wait, let's consider that the common number x is the smallest number in A. Suppose A={x, a, b}, and x is the common number. Then the sums arex +a, x +b, a +b. These must be in B. B contains x,25,3f. Therefore, the three sums must be x,25,3f. Therefore:x +a =x =>a=0 invalid.x +a=25x +b=3fa +b=xBut from a +b=x andx +a=25:x +a=25 =>(a +b) +a=25 =>2a +b=25Fromx +b=3f =>(a +b) +b= a +2b=3fAlso, favorite is one of x, a, b. Let's say favorite is x:Then3x must be in B={x,25,3f}So3x is eitherx,25, or3f.If3x=x =>x=0 invalid.If3x=25 =>x≈8.333 invalid.If3x=3f =>x=f. So3x=3f =>x=f. But fromx +b=3f =>x +b=3x =>b=2xFrom2a +b=25 andb=2x:Buta +b=x =>a +2x=x =>a= -x invalid.If favorite is a:3a must be in B={x,25,3f}So3a=x,25, or3f.If3a=x:Fromx +a=25 =>3a +a=4a=25 =>a=6.25 invalid.If3a=25 =>a≈8.333 invalid.If3a=3f =>a=f. Fromx +b=3f=3a =>x +b=3aFromx +a=25 =>x=25 -aFroma +b=x =>a +b=25 -a =>b=25 -2aSubstitute intox +b=3a:(25 -a) + (25 -2a)=3a =>50 -3a=3a =>50=6a =>a≈8.333 invalid.If favorite is b:3b must be in B={x,25,3f}So3b=x,25,3f.If3b=x:Fromx +a=25 =>3b +a=25Froma +b=x=3b =>a=2bSubstitute into3b +a=25 =>3b +2b=5b=25 =>b=5, a=10, x=15Then A={x=15,a=10,b=5} but must be in order x <a <b, but x=15, a=10, b=5, which is not ordered. Invalid.If3b=25 =>b≈8.333 invalid.If3b=3f =>b=f. Fromx +b=3f=3b =>x=2bFromx +a=25 =>2b +a=25Froma +b=x=2b =>a=bBut a and b must be distinct. Invalid.This is really challenging. Maybe we need to think of actual numbers.Bart has numbers: 25,3f, andx. Annemiek has numbers: x, a, b. The sums arex +a, x +b, a +b. These sums must be the three numbers in Bart's list:25,3f,x.Assuming the sums are25,3f,x. Let's arrange them in order. Let's say the smallest sum isx, then25, then3f. So:x +a =x =>a=0 invalid.So the next possibility: x +a=25, x +b=3f, a +b=x.From a +b=x, substitute intox +a=25: (a +b) +a=25 =>2a +b=25Fromx +b=3f =>(a +b) +b= a +2b=3fSo we have:2a +b=25a +2b=3fAlso, Annemiek's numbers arex=a +b, a, b. Favorite is one of these. So 3*(favorite) must be in B={x,25,3f}.If favorite isx:3x must be in B. So3x =x,25,3f. 3x=x =>x=0 invalid. 3x=25 =>x≈8.333 invalid.3x=3f =>x=f. So fromx +b=3f =>x +b=3x =>b=2xFromx=a +b= a +2x =>a= -x invalid.If favorite isa:3a must be in B={x,25,3f}So3a isx,25, or3f.If3a=x:From2a +b=25 andx=a +b, sox= a +b= a + (25 -2a)=25 -aBut3a=x=25 -a =>4a=25 =>a=6.25 invalid.If3a=25 =>a≈8.333 invalid.If3a=3f =>a=f. Froma +2b=3f=3a =>a +2b=3a =>2b=2a =>b=a. But a and b must be distinct. Invalid.If favorite isb:3b must be in B={x,25,3f}So3b isx,25, or3f.If3b=x:Fromx=a +b=3b =>a=2bFrom2a +b=25 =>4b +b=5b=25 =>b=5, a=10, x=15So A={15,10,5}Check sums:15+10=2515+5=2010+5=15So the sums are25,20,15. B should be {x=15,25,3f}. But 3f must be20 =>f=20/3≈6.666 invalid.If3b=25 =>b≈8.333 invalid.If3b=3f =>b=f. Froma +2b=3f=3b =>a +2b=3b =>a=b. Invalid.This seems like there's no solution. But the problem states that there is a solution. Maybe I missed something.Wait, perhaps the common number is not one of the sums. But how? Because Bart's numbers must include the common number and the three sums. But Bart can only have three numbers. So if the common number is not one of the sums, then the three sums must be the other two numbers in B, but B has three numbers: the common number and two others. But the three sums must be in B, which has three numbers. Therefore, the three sums must be the three numbers in B. But B also contains the common number. So this requires that the common number is one of the sums, which we saw is impossible. Therefore, there must be an overlap where the common number is one of the sums. But as we saw earlier, this leads to contradictions unless the numbers are not distinct. But the problem states they are distinct positive integers.Wait, maybe the sums are not all distinct. But Bart's numbers are three different positive integers. Therefore, the sums must be distinct. Therefore, the three sums must be three distinct numbers. But Bart's numbers also include the common number. So unless the common number is one of the sums. But as we saw, that's impossible.Is there a possibility that Bart's numbers are the three sums, and the common number is one of the sums? But then the common number is in both A and B. So A contains a sum of two of its numbers. For example, A has numbers a, b, c, and c =a +b. Then, the sums are a +b =c, a +c, b +c. B is {c, a +c, b +c}. But B must also contain 25 and 3f. So, this would mean that B has four numbers: c, a +c, b +c,25,3f. But B can only have three. Therefore, two of these must be equal. For example, maybe a +c=25 and b +c=3f. Then B would be {c,25,3f}. So:If A={a, b, c}, with c =a +b.B={c,25,3f}, which must include the sums c, a +c, b +c.But a +c and b +c are 25 and3f.So:a +c =25 =>a +a +b=25 =>2a +b=25b +c =3f =>b +a +b=3f =>a +2b=3fAlso, favorite is one of a, b, c. So 3*(favorite) must be in B={c,25,3f}If favorite is a:3a must be in B. So3a =c,25,3f.If3a =c=a +b, then:3a =a +b =>2a =bFrom2a +b=25 =>2a +2a=25 =>4a=25 =>a=6.25 invalid.If3a=25 =>a≈8.333 invalid.If3a=3f =>a=f. Froma +2b=3f=3a =>a +2b=3a =>2b=2a =>b=a. Duplicate, invalid.If favorite is b:3b in B={c,25,3f}If3b=c=a +b =>3b=a +b =>2b=aFrom2a +b=25 =>2*(2b) +b=5b=25 =>b=5, a=10, c=15So A={10,5,15}Check sums:10+5=1510+15=255+15=20So sums are15,25,20B should be{15,25,20}But Bart's favorite number is25, which is there. Also, 3*(favorite)=3b=15 if favorite is5. But15 is in B. So favorite is5.But Bart's numbers are15,25,20. The common number is15, which is in both A and B. So A={10,5,15}, B={15,25,20}Check if all sums are in B:10+5=15 ∈ B10+15=25 ∈ B5+15=20 ∈ BYes. Also,3*(favorite)=3*5=15 ∈ B. And Bart's favorite is25 ∈ B.This seems to satisfy all conditions. Therefore, Annemiek's favorite number is5.Wait, but Annemiek's numbers are10,5,15. Her favorite number is5, which when multiplied by3 gives15, which is in Bart's numbers. Bart's numbers are15,25,20. The common number is15. Yes, this works.So the answer is5.Final Answerboxed{5}</think>Given the problem, we need to determine Annemiek's favorite number based on the conditions provided. Let's break down the solution step by step.1. Define the sets: - Annemiek's numbers: ( A = {a, b, c} ) - Bart's numbers: ( B = {d, e, f} )2. Key conditions: - There is exactly one common number between ( A ) and ( B ). - All pairwise sums of ( A ) are in ( B ). - Annemiek's favorite number multiplied by 3 is in ( B ). - Bart's favorite number is 25.3. Analyze the sums: - The sums of Annemiek's numbers ( a+b, a+c, b+c ) must be in Bart's numbers ( B ). - Bart's numbers must include 25 and the product of Annemiek's favorite number multiplied by 3.4. Common number and sums: - The common number ( x ) must be one of Annemiek's numbers and also in Bart's numbers. - The sums ( a+b, a+c, b+c ) must be the three numbers in Bart's list.5. Determine the numbers: - Suppose Annemiek's numbers are ( {a, b, c} ) with ( c = a + b ). - The sums are ( a+b = c ), ( a+c = a + (a + b) = 2a + b ), and ( b+c = b + (a + b) = a + 2b ).6. Check conditions: - Bart's numbers must be ( {c, 25, 3f} ) where ( f ) is Annemiek's favorite number. - Solving the equations ( 2a + b = 25 ) and ( a + 2b = 3f ).7. Find valid integers: - Solving ( 2a + b = 25 ) and ( a + 2b = 3f ) with ( f ) as one of ( a, b, c ): - If ( a = 10 ), ( b = 5 ), then ( c = 15 ). - The sums are ( 15, 25, 20 ). - Bart's numbers are ( {15, 25, 20} ). - Annemiek's favorite number ( b = 5 ) (since ( 3 times 5 = 15 ) is in Bart's numbers).Thus, Annemiek's favorite number is (boxed{5}).,<think>Okay, let's try to figure out this problem step by step. So, we have Annemiek and Bart, each with three different positive integers. They have exactly one number in common. Also, if you take any two different numbers from Annemiek's paper and add them together, that sum is always on Bart's paper. Additionally, Annemiek's favorite number multiplied by 3 is on Bart's paper. Bart's favorite number is 25. We need to find Annemiek's favorite number.First, let's break down the information:1. Both Annemiek (A) and Bart (B) have three distinct positive integers.2. There's exactly one common number between A and B.3. All pairwise sums of A's numbers are present on B's list.4. Three times Annemiek's favorite number is on B's list.5. Bart's favorite number is 25.So, let's denote Annemiek's numbers as a, b, c (all different positive integers) and Bart's numbers as d, e, f (also distinct positive integers). The common number between them is, say, x. So x is one of a, b, c and also one of d, e, f.The pairwise sums of A's numbers are a+b, a+c, b+c. Each of these sums must be one of B's numbers (d, e, f). But B only has three numbers, so the three sums from A must map to B's three numbers. However, since there's exactly one common number between A and B, the three sums (which are on B's list) can't include that common number unless one of the sums equals x. Wait, but the problem states that there is exactly one common number. So the three sums from A must be the other two numbers on B's list plus possibly overlapping with the common number? Hmm, maybe not. Let me think.Wait, B has three numbers: one of them is x (the common number), and the other two must be the sums from A. But A's pairwise sums are three numbers. So how can three sums fit into two remaining spots on B's list? That suggests that maybe two of the sums are the same as the other two numbers on B's list, and one of the sums is equal to x. But the problem states that there's exactly one common number, which is x. So if one of the sums equals x, then x would be a sum of two numbers from A, but x is also a number on A's list. So that would mean that one of A's numbers is equal to the sum of the other two. Is that possible?But all numbers are positive integers, so if x is on both A and B's lists, and x is also a sum of two numbers from A, then A's numbers would be two numbers and their sum. But then A's numbers would be, say, a, b, a+b. But the problem states that all three numbers on A's list are different. So that's allowed as long as a, b, and a+b are distinct, which they are if a and b are positive integers.Wait, but then if x is a+b, which is on both A and B's lists, then B's list would have x, and the other two numbers would have to be the other sums from A. But the sums of A's numbers would be a+b (which is x), a + (a+b) = 2a + b, and b + (a+b) = a + 2b. So unless those other sums are duplicates, but since all numbers on B's list must be distinct, those sums must be distinct and equal to the other two numbers on B's list. So B's numbers would be x, 2a + b, a + 2b. Also, Bart's favorite number is 25. So one of B's numbers is 25. Additionally, three times Annemiek's favorite number is on B's list. So if Annemiek's favorite number is, say, a, then 3a must be one of B's numbers (which are x, 2a + b, a + 2b). Similarly, if her favorite is b or x (a+b), then 3b or 3(a+b) is on B's list.But let's try to formalize this.Let’s denote:A’s numbers: a, b, c (distinct positive integers, with one of them being x, the common number with B’s list)B’s numbers: x, p, q (distinct positive integers, with p and q being the other two sums from A’s pairwise sums)So the pairwise sums of A’s numbers are:a + b, a + c, b + c.These sums must be equal to x, p, q in some order. However, since A’s numbers include x, which is one of a, b, c, let's say x = c. Then A’s numbers are a, b, x, and the pairwise sums would be a + b, a + x, b + x. These sums must be the numbers on B’s list: x, p, q. So, either:1. a + b = x, a + x = p, b + x = qOr, another arrangement. But if x is one of A's numbers, let's suppose x = c. Then the pairwise sums are a + b, a + c, b + c. Since x is on both A and B's lists, B's numbers are x, p, q. Therefore, the three sums (a + b, a + x, b + x) must be the three numbers on B's list: x, p, q. But x is already on B's list. So one of the sums must equal x, and the other two sums must be p and q. So, either:Case 1: a + b = x, a + x = p, b + x = qCase 2: a + b = p, a + x = x, which would imply a = 0, but numbers are positive integers. So this case is invalid.Case 3: a + x = x, implies a = 0, invalid.Case 4: Similarly, if b + x = x, then b = 0, invalid.Therefore, the only possible case is that a + b = x, and the other two sums (a + x and b + x) are p and q. So B's numbers are x, a + x, b + x. But B's numbers are three distinct numbers, so x, a + x, b + x must all be distinct, which they are as long as a ≠ b, which they are because all numbers are distinct.Additionally, Bart's favorite number is 25, so one of B's numbers is 25. Therefore, either x = 25, a + x = 25, or b + x = 25.Also, three times Annemiek's favorite number is on B's list. Suppose Annemiek's favorite number is one of a, b, x. Then 3 times that number must be one of B's numbers: x, a + x, b + x.So let's consider possible scenarios.First, let's suppose that x is the common number. So x is on both A and B's lists. If x is 25, then Bart's favorite number is the common number. But then B's numbers are 25, a + 25, b + 25. But Annemiek's numbers are a, b, 25. The other two numbers on B's list are a + 25 and b + 25. Also, three times Annemiek's favorite number is on B's list. If Annemiek's favorite number is 25, then 3*25 = 75 would have to be on B's list, so 75 would be either a + 25 or b + 25. Thus, either a or b would be 50. But then Annemiek's numbers would be 50, b, 25. Then the pairwise sums would be 50 + b, 50 + 25 = 75, and b + 25. But B's numbers are 25, a + 25 = 50 + 25 = 75, and b + 25. So that would mean B's numbers are 25, 75, and b + 25. But Annemiek's favorite number is 25, which when multiplied by 3 is 75, which is on B's list. So this seems possible. However, we need to check if all numbers are distinct.Annemiek's numbers: 50, b, 25. They must all be distinct. So b ≠ 50 and b ≠ 25. Also, B's numbers: 25, 75, b + 25. These must also be distinct, so b + 25 ≠ 25 (which would mean b = 0, invalid), and b + 25 ≠ 75 (so b ≠ 50). So as long as b is not 50, and b is distinct from 25 and 50, this works. But we also need that a + b = x = 25. Wait, hold on. Wait, in this scenario, we assumed x = 25, which is the common number, and we said that a + b = x. So if x = 25, then a + b = 25. But we previously supposed that a = 50. But 50 + b = 25 would imply b = -25, which is invalid. Contradiction. So this scenario is impossible.Therefore, if x = 25, then a + b = 25, but a and b are positive integers. Then Annemiek's numbers are a, b, 25, with a + b = 25. Then the pairwise sums are 25 (a + b), a + 25, b + 25. So B's numbers are 25, a + 25, b + 25. But Bart's favorite number is 25, so that's fine. Now, three times Annemiek's favorite number must be on B's list. Suppose her favorite number is a. Then 3a must be one of 25, a + 25, b + 25. Similarly, if her favorite is b or 25.Let's check:If her favorite is a, then 3a must be 25, a + 25, or b + 25.If 3a = 25, then a = 25/3, which is not an integer. Disregard.If 3a = a + 25, then 2a = 25 ⇒ a = 12.5, not integer.If 3a = b + 25. Since a + b = 25, then b = 25 - a. So substituting, 3a = (25 - a) + 25 ⇒ 3a = 50 - a ⇒ 4a = 50 ⇒ a = 12.5. Again, not integer.If her favorite is b, similar logic:3b = 25 ⇒ invalid.3b = a + 25. Since a = 25 - b, substitute: 3b = (25 - b) + 25 ⇒ 3b = 50 - b ⇒ 4b = 50 ⇒ b = 12.5. Not integer.3b = b + 25 ⇒ 2b = 25 ⇒ b = 12.5. No.If her favorite is 25, then 3*25 = 75 must be on B's list. B's numbers are 25, a + 25, b + 25. So either 75 = a + 25 or 75 = b + 25. Since a + b = 25, then a = 25 - b. So if 75 = a + 25 ⇒ a = 50 ⇒ b = 25 - 50 = -25, invalid. If 75 = b + 25 ⇒ b = 50 ⇒ a = 25 - 50 = -25, invalid. So this is impossible. Therefore, x cannot be 25.So the common number x is not 25. Therefore, Bart's favorite number, 25, is one of the other numbers on his list: p or q. So B's numbers are x, 25, q (or x, p, 25). Also, three times Annemiek's favorite number is on B's list, which is either x, 25, or q.So let's consider that 25 is one of B's numbers, not the common one. So B's numbers are x, 25, and another number. The three pairwise sums from A's numbers must be x, 25, and another number. So let's recap:A's numbers are a, b, x (since x is the common number). The pairwise sums are a + b, a + x, b + x. These correspond to B's numbers: x, 25, and another number, let's say q.So the three sums are x, 25, q. Therefore, we have:Either:1. a + b = x, a + x = 25, b + x = qOr2. a + b = x, a + x = q, b + x = 25Or3. a + b = 25, a + x = x (invalid, since this would mean a = 0), or other permutations.Wait, since x is one of the sums, and 25 is another sum. Let me consider possible assignments.Case 1: a + b = x, a + x = 25, b + x = qFrom a + x = 25 and a + b = x, substitute x into the first equation:a + (a + b) = 25 ⇒ 2a + b = 25Also, b + x = q ⇒ b + (a + b) = q ⇒ a + 2b = qSo in this case, B's numbers are x = a + b, 25, and q = a + 2b.Additionally, three times Annemiek's favorite number is on B's list. Her favorite number is one of a, b, x. So 3 times that number must be in {x, 25, q}.Let’s check each possibility:1. If her favorite is a: 3a ∈ {x, 25, q}But x = a + b, q = a + 2b. So 3a could be 25, but 25 is already in B's numbers. If 3a = 25 ⇒ a = 25/3 ≈ 8.33, not integer. If 3a = x = a + b ⇒ 3a = a + b ⇒ 2a = b. Then from 2a + b = 25, substituting b = 2a gives 2a + 2a = 25 ⇒ 4a = 25 ⇒ a = 6.25, not integer. If 3a = q = a + 2b ⇒ 3a = a + 2b ⇒ 2a = 2b ⇒ a = b. But a and b are distinct, so this is impossible.2. If her favorite is b: 3b ∈ {x, 25, q}Similarly, if 3b = 25 ⇒ b ≈ 8.33, invalid.If 3b = x = a + b ⇒ 3b = a + b ⇒ a = 2b. From 2a + b = 25, substituting a = 2b gives 4b + b = 25 ⇒ 5b = 25 ⇒ b = 5. Then a = 10. So check:a = 10, b = 5, x = a + b = 15, q = a + 2b = 10 + 10 = 20.So A's numbers: 10, 5, 15. B's numbers: 15, 25, 20.Check all conditions:- All numbers are distinct positive integers: yes.- Common number x = 15: yes, 15 is in both A and B's lists.- Pairwise sums of A: 10 + 5 = 15, 10 + 15 = 25, 5 + 15 = 20. These are exactly B's numbers: 15, 25, 20. Correct.- 3 times Annemiek's favorite number: If her favorite is b = 5, then 3*5 = 15, which is in B's list. Yes.- Bart's favorite number is 25, which is on his list. Yes.So this works! Therefore, Annemiek's favorite number is 5.But let's check other possibilities just to be thorough.3. If her favorite is x: 3x ∈ {x, 25, q}If 3x = x ⇒ 2x = 0 ⇒ x=0 invalid.If 3x = 25 ⇒ x ≈ 8.33, invalid.If 3x = q = a + 2b ⇒ 3x = a + 2b. But x = a + b, so 3(a + b) = a + 2b ⇒ 3a + 3b = a + 2b ⇒ 2a + b = 0 ⇒ impossible since a and b are positive. So this case invalid.Therefore, in Case 1, the only valid solution is when Annemiek's favorite number is 5.Case 2: a + b = x, a + x = q, b + x = 25From a + x = q and b + x = 25. Since x = a + b, substitute:a + (a + b) = q ⇒ 2a + b = qb + (a + b) = 25 ⇒ a + 2b = 25So we have two equations:1. a + 2b = 252. x = a + b3. q = 2a + bNow, three times Annemiek's favorite number must be in B's numbers: x, 25, q.Again, check each possibility:1. Favorite number is a: 3a ∈ {x, 25, q} = {a + b, 25, 2a + b}If 3a = a + b ⇒ 2a = b. Substitute into a + 2b = 25:a + 2*(2a) = 25 ⇒ 5a = 25 ⇒ a = 5, then b = 10.Thus, x = 5 + 10 = 15, q = 2*5 + 10 = 20. So B's numbers: 15, 25, 20. Same as before but a and b swapped. A's numbers: 5, 10, 15. Wait, but in this case, Annemiek's favorite number would be a = 5, so 3*5 = 15, which is in B's list. Correct.But this is similar to the previous case, just swapping a and b. However, since the problem states that Annemiek has three different numbers, the order doesn't matter. So this is another valid solution, but Annemiek's favorite number is still 5 (if she chose a) or 10 (if she chose b). Wait, hold on.Wait, in this scenario, if a = 5, b = 10, then A's numbers are 5, 10, 15. If her favorite number is a = 5, then 3*5 = 15 is on B's list. If her favorite number is b = 10, then 3*10 = 30, which is not on B's list (15, 25, 20). If her favorite number is x = 15, then 3*15 = 45, not on B's list. So only if she chooses a = 5 as her favorite, which gives 15 on B's list. So this is the same solution as before, just swapping a and b. Therefore, Annemiek's favorite number is still 5.Another possibility:If 3a = 25 ⇒ a = 25/3 ≈8.33, invalid.If 3a = q = 2a + b ⇒ 3a = 2a + b ⇒ a = b. But a and b must be distinct. Invalid.2. Favorite number is b: 3b ∈ {x, 25, q}If 3b = x = a + b ⇒ 3b = a + b ⇒ a = 2b. Substitute into a + 2b = 25:2b + 2b = 25 ⇒ 4b = 25 ⇒ b = 6.25, invalid.If 3b = 25 ⇒ b ≈8.33, invalid.If 3b = q = 2a + b ⇒ 3b = 2a + b ⇒ 2b = 2a ⇒ a = b. Not allowed.3. Favorite number is x: 3x ∈ {x, 25, q}If 3x = x ⇒ x=0 invalid.If 3x =25 ⇒ x≈8.33, invalid.If 3x = q =2a + b. But x = a + b, so 3(a + b) = 2a + b ⇒ 3a + 3b = 2a + b ⇒ a + 2b =0, invalid.Thus, only valid solution in this case is a=5, b=10, leading to favorite number 5.Therefore, both Case 1 and Case 2 lead to Annemiek's favorite number being 5. So regardless of the order, the answer is 5.Wait, but let's check another scenario where x is not the sum a + b, but one of the other sums. Wait, we considered that x is one of A's numbers, so if x is one of the numbers a, b, c. But if x is the sum a + b, then c is x. Wait, but in the previous reasoning, we assumed that x is one of the numbers, so if x = c, then the sums are a + b, a + c, b + c. But that would be a + b, a + x, b + x. So maybe x can also be equal to one of the sums? But we already considered that.Alternatively, suppose that x is one of the numbers on A's list, say a, and also on B's list. Then the sums from A's list are a + b, a + c, b + c. These sums must be the numbers on B's list: a, p, q. But since x = a is common, then one of the sums must be a, and the other two sums must be p and q. So:Either:1. a + b = a ⇒ b =0 invalid.2. a + c = a ⇒ c=0 invalid.3. b + c = a. So, b + c = a. Then the other sums are a + b and a + c. These would be p and q. Therefore, B's numbers are a (common), a + b, a + c.Also, Bart's favorite number is 25. So either a =25, a + b=25, or a + c=25.Additionally, three times Annemiek's favorite number is on B's list. Her favorite number is either a, b, or c. So 3*(favorite) must be in {a, a + b, a + c}.Let's explore this scenario.Case 3: x = a is the common number. Then, sums of A's numbers: a + b, a + c, b + c. Since x = a is on B's list, one of the sums must be a. From above, the only possibility is b + c = a. Then the other sums are a + b and a + c, which are B's other numbers. So B's numbers are a, a + b, a + c.Bart's favorite is 25, so either a =25, a + b=25, or a + c=25.Also, 3*(Annemiek's favorite) must be in B's list. Let's check possibilities.First, suppose a =25. Then B's numbers are 25, 25 + b, 25 + c. But A's numbers are 25, b, c, with b + c =25. So Annemiek's numbers are 25, b, c, where b + c =25. Then B's numbers are 25, 25 + b, 25 + c. Now, 3*(Annemiek's favorite) must be in B's list.If her favorite is 25, then 3*25=75 must be in B's list. So 75 must be either 25 + b or 25 + c. Let's say 25 + b =75 ⇒ b=50. Then since b + c =25, c=25 -50 =-25 invalid. Similarly, 25 + c=75 ⇒ c=50 ⇒ b=25 -50=-25 invalid. So impossible.If her favorite is b, then 3b must be in {25, 25 + b, 25 + c}. Similarly:3b =25 ⇒ b≈8.33 invalid.3b=25 + b ⇒2b=25⇒b=12.5 invalid.3b=25 + c. Since c=25 -b, then 3b=25 +25 -b ⇒4b=50⇒b=12.5 invalid.Same for favorite number c: same results. So a=25 is invalid.Second, suppose a + b=25. Then since B's numbers are a, a + b=25, a + c. Also, A's numbers are a, b, c with b + c =a. So from b + c =a, we can write c =a -b. Then a + c =a + (a -b)=2a -b. So B's numbers are a,25,2a -b.Bart's favorite number is25, which is already there. Now, 3*(favorite) must be in B's list. Let's check possibilities.If her favorite is a: 3a must be in {a,25,2a -b}.If 3a =a ⇒2a=0⇒invalid.If 3a=25⇒a≈8.33 invalid.If 3a=2a -b⇒a= -b⇒invalid.If her favorite is b: 3b ∈ {a,25,2a -b}.From b + c =a and c =a -b, we can write a =b +c, but since c =a -b, this is consistent. Let's see.We have:a + b =25 ⇒ a =25 -b.Also, c =a -b=25 -b -b=25 -2b.But c must be a positive integer, so 25 -2b >0 ⇒2b <25⇒b <12.5. So b can be from 1 to12.Also, since a, b, c must be distinct positive integers.So let's write a =25 -b, c=25 -2b.Then Annemiek's numbers are a=25 -b, b, c=25 -2b.These must be distinct:25 -b ≠ b ⇒25 ≠2b ⇒b≠12.5 (since b is integer, okay).25 -b ≠25 -2b ⇒ -b ≠ -2b ⇒b≠0, which it isn't.b ≠25 -2b ⇒3b ≠25⇒b≠25/3≈8.33. Since b is integer, b can't be8.33. So for b=8, check:a=25 -8=17, c=25 -16=9. So numbers 17,8,9. All distinct.Similarly, for other b.Now, 3*(favorite) must be in B's numbers: a=25 -b,25,2a -b=2*(25 -b)-b=50 -2b -b=50 -3b.So B's numbers:25 -b,25,50 -3b.So 3*(favorite) is in {25 -b,25,50 -3b}.If her favorite is a=25 -b: 3*(25 -b) must be in B's list. So 75 -3b is in {25 -b,25,50 -3b}.Check if 75 -3b=25 -b⇒75 -3b=25 -b⇒50=2b⇒b=25. But b <12.5. Invalid.If 75 -3b=25⇒75 -3b=25⇒3b=50⇒b≈16.67. Invalid.If 75 -3b=50 -3b⇒75=50⇒invalid.Thus, favorite can't be a.If her favorite is b: 3b must be in {25 -b,25,50 -3b}.Check:3b=25 -b⇒4b=25⇒b=6.25 invalid.3b=25⇒b≈8.33 invalid.3b=50 -3b⇒6b=50⇒b≈8.33 invalid.If her favorite is c=25 -2b: 3*(25 -2b)=75 -6b. Must be in B's list: {25 -b,25,50 -3b}.Check:75 -6b=25 -b⇒75 -6b=25 -b⇒50=5b⇒b=10.Check if b=10 is valid:b=10, so a=25 -10=15, c=25 -20=5. So A's numbers:15,10,5. B's numbers:25 -10=15,25,50 -30=20. So B's numbers:15,25,20. Same as before. So 3*c=3*5=15, which is in B's list (15). So if her favorite is c=5, then 3*5=15 is on B's list. Therefore, Annemiek's favorite number is5.So in this case, even if x is a=15 (the common number), but her favorite number is c=5, which gives 3*5=15 on B's list.Thus, this is another valid scenario leading to favorite number5.Therefore, regardless of whether x is the sum a + b or one of the original numbers, the only possible favorite number is5.Therefore, the answer is5.Final Answerboxed{5}</think>Given the problem, we need to determine Annemiek's favorite number based on the conditions provided. Here is the step-by-step reasoning:1. Identify the common number: Annemiek (A) and Bart (B) each have three different positive integers with exactly one common number, denoted as ( x ).2. Pairwise sums condition: The pairwise sums of Annemiek's numbers must be on Bart's list. Let Annemiek's numbers be ( a, b, x ) (since ( x ) is the common number). The pairwise sums are ( a + b ), ( a + x ), and ( b + x ). These sums must be Bart's numbers.3. Bart's favorite number: Bart's favorite number is 25, so one of his numbers is 25.4. Three times favorite number: Three times Annemiek's favorite number must be on Bart's list.Key steps:- Assume ( x ) is the common number. Bart's numbers are ( x ), 25, and another number.- The pairwise sums of Annemiek's numbers ( a, b, x ) are ( a + b ), ( a + x ), ( b + x ).- Since ( a + b ), ( a + x ), ( b + x ) must be Bart's numbers, we consider different scenarios where these sums include 25.Detailed analysis:- If ( a + x = 25 ) and ( a + b = x ), solving these equations leads to ( 2a + b = 25 ) and ( x = a + b ).- Substituting ( x = a + b ) into ( a + x = 25 ) gives ( 2a + b = 25 ).- Solving for ( a ) and ( b ) with distinct positive integers, we find ( a = 10 ), ( b = 5 ), and ( x = 15 ).- Annemiek's numbers are ( 10, 5, 15 ), and Bart's numbers are ( 15, 25, 20 ).- Three times Annemiek's favorite number (5) is 15, which is on Bart's list.Verification:- All conditions are satisfied: distinct numbers, common number, pairwise sums, and the favorite number condition.Thus, Annemiek's favorite number is (boxed{5}).]

🔑:B3. Suppose we call the three numbers on Annemieke's note a, b, and c and assume that a<b<c. Then the three numbers on Bart's note, in order of size, are a+b, a+c, and b+c. The last two numbers are greater than c and thus greater than any number on Annemieke's note. The number that appears on both notes must therefore be a+b and that must be equal to c.Annemieke thus has the numbers a, b, and a+b, and Bart has a+b, 2a+b, and a+2b. If we now look at 3b and 3(a+b)=3a+3b, we see that these numbers are both greater than all the numbers on Bart's note. Annemieke's favorite number, which appears on Bart's note when multiplied by 3, is therefore a.The number 3a is thus one of the numbers a+b, 2a+b, a+2b on Bart's note, and since a and b are different, it must be the number a+b. In other words: b=2a. Bart thus has the numbers 3a, 4a, and 5a. The only one of these three numbers that can be the number 25 is 5a. We thus see that a=5 and that is Annemieke's favorite number.