🧠:[<think>Okay, let's see. The question is asking which of the given events is a certain event. The options are A, B, C, D. First, I need to remember what a certain event is. In probability terms, a certain event is something that will definitely happen, with 100% probability. So, I need to check each option and see which one is guaranteed to occur.Starting with option A: Tossing a coin and getting heads. Well, a standard coin has two sides, heads and tails. When you toss it, there's a 50% chance of heads and 50% of tails. So, it's not certain. It could be either. So, A is not a certain event.Option B: Heating water to 100°C at standard atmospheric pressure, causing it to boil. Hmm. I remember that the boiling point of water at standard atmospheric pressure (which is 1 atmosphere, like at sea level) is indeed 100°C. So, if you heat water to that temperature under those conditions, it should boil. But wait, is there any case where it might not? Like if the water is super pure or something? No, I think that's more about superheating, but that's rare and requires specific conditions. Under normal circumstances, if you heat water to 100°C at standard pressure, it will boil. So B seems like a certain event. But I should check the others just in case.Option C: Meeting an acquaintance while playing by the seaside. Well, that's dependent on a lot of factors. Do you have acquaintances that live nearby? Do they happen to be at the seaside at the same time? It's possible, but not guaranteed. So, this is a possibility, not a certainty. So C isn't certain.Option D: The sun revolving around the earth. Wait a second. From a scientific standpoint, we know that the Earth revolves around the sun, not the other way around. But maybe this is phrased in a geocentric model? But even so, the statement is factually incorrect. So is this event impossible? If so, then it's not a certain event. But wait, the question is about a certain event, which is an event that must happen. If D is impossible, then it's the opposite of certain. So D is impossible, not certain. So that's out.So going back. B is heating water to 100°C at standard pressure causing it to boil. Since under those conditions, water boils at 100°C, this should definitely happen. Therefore, B is a certain event. The others are either probabilistic or impossible. So the answer should be B.</think>To determine which event is certain, we evaluate each option's likelihood:- A: Tossing a coin and getting heads has a 50% probability, not guaranteed. - B: Heating water to (100^{circ}mathrm{C}) at standard atmospheric pressure ensures boiling due to the defined physical properties of water. This is a certain outcome under the given conditions. - C: Meeting an acquaintance at the seaside depends on unpredictable factors, making it uncertain. - D: The Sun revolving around the Earth is scientifically false, making this an impossible event. Thus, the certain event is B. boxed{B},<think>Okay, let's see. The question is asking which of the given events is a certain event. The options are A, B, C, and D. Hmm. Let me think about each one step by step.Starting with option A: Tossing a coin and getting heads. Well, when you toss a coin, there are two possible outcomes—heads or tails. Assuming it's a fair coin, the probability of getting heads is 50%. So, it's not a certain event because there's an equal chance of getting tails. So A isn't certain.Moving on to option B: Heating water to 100°C at standard atmospheric pressure, causing it to boil. From what I remember in science class, the boiling point of water at standard atmospheric pressure (which is 1 atmosphere, or 100 kPa) is indeed 100°C. So if you heat water to that temperature under those conditions, it should boil. That sounds like a sure thing, right? But wait, is there any situation where this might not happen? Maybe impurities in the water? Or if it's in a different container? Hmm, but the question specifies standard atmospheric pressure, so as long as the pressure is standard, even if there are impurities, the boiling point is 100°C. Impurities might affect things like how vigorously it boils or the exact temperature, but generally, at 100°C under standard pressure, water boils. So B seems like a certain event.Option C: Meeting an acquaintance while playing by the seaside. Well, this depends on a lot of factors. How crowded the seaside is, whether your acquaintances happen to be there, how long you stay, etc. It's possible, but not guaranteed. So this is more of a probabilistic event, not certain. So C is not certain.Option D: The sun revolving around the earth. Wait, isn't that an old belief? We know now that the Earth revolves around the Sun, right? So this is actually impossible. The sun doesn't revolve around the Earth; it's the other way around. So D is an impossible event, not a certain one.So, reviewing the options: A is 50-50, B is certain under the given conditions, C is possible but not certain, D is impossible. Therefore, the answer should be B.</think>To determine which event is certain, analyze each option:- A: Tossing a coin has two equally likely outcomes (heads or tails). Not certain. - B: Heating water to (100^{circ}mathrm{C}) at standard atmospheric pressure ensures boiling due to the defined physical properties of water. This is guaranteed under the specified conditions. - C: Meeting an acquaintance depends on chance factors (e.g., their presence at the seaside). Not certain. - D: The Sun does not revolve around the Earth; this is scientifically disproven. Impossible. The certain event is B. boxed{B}]

🔑:To determine which of the following events is a certain event, let's analyze each option:- Event A: Tossing a coin and getting heads. This event is not certain because there is a 50% chance of getting heads and a 50% chance of getting tails. Therefore, it's a random event. - Event B: Heating water to 100^{circ}mathrm{C} at standard atmospheric pressure, causing it to boil. This event is certain because, under standard atmospheric pressure, water always boils at 100^{circ}mathrm{C}. It's a predictable and repeatable outcome. - Event C: Meeting an acquaintance while playing by the seaside. This event is random because it depends on various unpredictable factors, such as whether your acquaintance decides to visit the seaside at the same time as you. - Event D: The sun revolving around the earth. This is an impossible event because, according to our current understanding of astronomy, the Earth revolves around the Sun, not the other way around.Given these analyses:- Event A does not meet the criteria for a certain event.- Event B meets the criteria for a certain event.- Event C does not meet the criteria for a certain event.- Event D does not meet the criteria for a certain event.Therefore, the correct answer is boxed{text{B}}.

🧠:[<think>Okay, so I need to solve this problem where positive integers x and y make the expression xy²/(x + v) a prime number. Then, I have to find how many possible values x + y can have, with the options being 1, 2, 3, or 4. Wait, hold on, the original problem says "x + v" in the denominator? But then the variables mentioned are x and y. Maybe that's a typo? Because the problem mentions x and y as positive integers, but there's a "v" in the denominator. Maybe it's supposed to be "y" instead of "v"? That would make more sense. Let me check again.The problem states: "If positive integers x and y make the value of (xy²)/(x + v) a prime number, then x + y has ( ) possible values." The options are A to D. Since the problem mentions x and y as the variables, but uses v in the denominator, that must be a typo. Probably, it's supposed to be x + y in the denominator. That would make the expression xy²/(x + y) a prime number. Alternatively, maybe the denominator is x + something else, but given the context, it's likely a typo. Let me assume that the denominator is supposed to be x + y. So the expression is (xy²)/(x + y), which is a prime number.So, our goal is to find the number of possible values of x + y, given that x and y are positive integers such that (xy²)/(x + y) is prime. Then, the answer choices are 1, 2, 3, 4. Let me think about how to approach this.First, since the expression (xy²)/(x + y) is a prime number, let's denote that prime as p. So, (xy²)/(x + y) = p, which can be rewritten as xy² = p(x + y). So, we have the equation xy² = p(x + y). We need to find all pairs of positive integers x, y such that this equation holds for some prime p, and then determine how many distinct values x + y can take.Let me rearrange the equation:xy² - px - py = 0But that might not be helpful. Let me think of this as:xy² = p(x + y)Since p is a prime, perhaps we can analyze the possible factors here. Let's consider x and y as variables, and p as a prime that divides the left-hand side, xy². So, since p divides xy², and p is prime, then p divides x or p divides y. So, two cases: p divides x or p divides y.Case 1: p divides x. Then, x = kp for some positive integer k. Substitute back into the equation:kp * y² = p(kp + y)Divide both sides by p:ky² = kp + yRearranged:ky² - y = kpSo, p = (ky² - y)/k = y² - (y/k)But since p must be an integer prime, y/k must be an integer. Therefore, k divides y. Let me write y = mk, where m is a positive integer. Then:p = (k(mk)² - mk)/k = k m² k² - m = k² m² - mWait, no, let me substitute y = mk into the equation:ky² - y = kpk(mk)² - mk = kpk*(m²k²) - mk = kpm²k³ - mk = kpDivide both sides by k:m²k² - m = pSo, p = m²k² - mBut p is a prime. So, m²k² - m must be a prime number. Let's factor this:m(m k² - 1) = pSince p is prime, and m and (m k² - 1) are positive integers (because m, k are positive integers), then one of them must be 1 and the other must be p. So, either:1. m = 1 and m k² - 1 = p, which gives 1 * k² - 1 = p => k² - 1 = p. Since p is prime, k² - 1 must be prime. So, k² - 1 is prime. Let's see, primes of the form k² - 1. For k ≥ 2 (since y = mk = 1 * k, so k must be at least 1, but if k=1, then p=1² -1=0, which isn't prime). So k must be at least 2. For k=2, p=4 -1=3, which is prime. For k=3, p=9 -1=8, which is not prime. k=4: 16-1=15, not prime. k=5: 25-1=24, not prime. So only k=2 gives a prime here. So in this subcase, m=1, k=2, p=3.Then, x = kp = 2*3=6, y = mk =1*2=2. Therefore, x=6, y=2. Then, x+y=8.Alternatively, subcase 2: m k² - 1 =1 and m = p. Then:m k² -1 =1 => m k² = 2. Since m and k are positive integers, m divides 2 and k² divides 2. The possible positive integer solutions for k² dividing 2 is k=1, since 1²=1 divides 2, but k=2 would give k²=4, which does not divide 2. Therefore, k=1. Then, m*1 =2 => m=2. Therefore, m=p=2. So, in this case, m=2, k=1, so x=kp=1*2=2, y=mk=2*1=2. Therefore, x=2, y=2, x+y=4.So, in Case 1, we have two possibilities: (x,y) = (6,2) leading to x+y=8, and (x,y)=(2,2) leading to x+y=4.Case 2: p divides y. Since p divides y, let's write y = lp for some positive integer l. Then, substitute back into the original equation:x*(lp)² = p(x + lp)Simplify:x l² p² = p x + p l pDivide both sides by p:x l² p = x + l pRearrange:x l² p - x = l pFactor x:x(l² p - 1) = l pTherefore,x = (l p)/(l² p -1)Since x must be a positive integer, the denominator (l² p -1) must divide the numerator l p. Therefore, (l² p -1) divides l p.Let me denote d = l² p -1. Then, d divides l p. So, d divides l p, which implies that d divides l p. But d = l² p -1. Let me write that:l² p -1 | l pSince l² p -1 divides l p, then l² p -1 ≤ l p, because any divisor of a positive integer is less than or equal to it. So:l² p -1 ≤ l pRearranged:l² p - l p ≤ 1lp(l -1) ≤1Since l and p are positive integers, and p is prime (so p ≥2), l must be at least 1. Let's check possible values of l:If l=1:Then, lp(l -1) =1*p*0=0 ≤1. So, this holds. So l=1 is possible.If l=2:Then, 2p(2 -1)=2p*1=2p ≤1. But p ≥2, so 2p ≥4, which is not ≤1. Therefore, l=2 is impossible.Similarly, l≥3: even more impossible.Therefore, only l=1 is possible.Therefore, l=1.Substitute back l=1 into x = (l p)/(l² p -1):x = (1*p)/(1² p -1) = p/(p -1)But x must be a positive integer. So p/(p -1) must be integer. Since p is a prime greater than or equal to 2.If p=2: x=2/(2-1)=2/1=2, integer.If p=3: 3/2, not integer.p=5:5/4, not integer. Similarly, other primes won't work. So only p=2 gives x=2.Therefore, l=1, p=2, so y=lp=1*2=2. Then, x=2. So, x=2, y=2. But this is the same solution as in Case 1, subcase 2. So, x+y=4. So, this is not a new solution.Therefore, in Case 2, we only get the solution x=2, y=2, which was already found in Case 1.Therefore, all solutions are:From Case 1:1. (6,2) with x+y=82. (2,2) with x+y=4But wait, in Case 1, when m=1 and k=2, we get x=6, y=2. When m=2 and k=1, we get x=2, y=2. In Case 2, we only get x=2, y=2 again. So, in total, there are two distinct solutions: (6,2) and (2,2), leading to x+y=8 and 4.Wait, but the options given are 1, 2, 3, 4. So, possible values of x+y are 8 and 4. So, two possible values. So the answer should be B) 2.But wait, let me double-check if there are any other possibilities. Let me see if there are other cases or if I missed any.Wait, when I considered Case 1, where p divides x, we had two subcases: m=1 leading to p=3, x=6, y=2, and m=2 leading to p=2, x=2, y=2. But is there another way when p divides x?Wait, in Case 1, when we set x=kp and y=mk, and substituted back, we derived that m(mk² -1)=p. Then, since p is prime, either m=1 and mk² -1=p, or mk² -1=1 and m=p. But is there a possibility that m or (mk² -1) is negative? No, because m and k are positive integers, so mk² -1 is at least m*1 -1 = m -1. Since m ≥1, mk² -1 ≥0. So, the factorization into m and (mk² -1) as positive integers holds. So, only those two possibilities. So, only those two solutions.Wait, but what if m=1, then p=k² -1. We saw that for k=2, p=3. For k=3, p=8, not prime. For k=1, p=0, invalid. So, only k=2 gives a prime. So that's one solution.In the other subcase, when mk² -1=1, so m=2/k². Wait, no. Wait, when we set mk² -1=1, then m=2/k². But m must be a positive integer, so 2 must be divisible by k². The possible k² divides 2, so k=1, as k²=1 divides 2. Then m=2/1=2. So m=2, k=1. Therefore, x=kp=1*p. But since m=2, and m=p in that case (since m(mk² -1)=p => m= p). Wait, wait, in the equation m(mk² -1)=p, we set mk² -1=1 and m=p. Therefore, m=p, and mk² -1=1 => p*k² -1=1 => p*k²=2. Since p is a prime, possible factors of 2 are p=2 and k²=1. So, p=2, k=1. Then m=p=2. Therefore, x=kp=1*2=2, y=mk=2*1=2. So, that's the same solution as before.Therefore, only two solutions: (6,2) and (2,2). So, x+y=8 and 4. So, two possible values, which would be answer B) 2.But wait, let me check if there are other possibilities. For instance, if the denominator was x + v instead of x + y, but since the problem mentions x and y, maybe v is another variable? Wait, but the problem states "positive integers x and y make the value of (xy²)/(x + v) a prime number". Then it's possible that v is another variable, but the problem doesn't mention v. That seems odd. The original problem in Chinese might have a different character, perhaps a typo where "v" is actually "y". Given that, it's more reasonable to assume the denominator is x + y. Alternatively, perhaps "v" is a constant? But the problem doesn't specify. Therefore, given the ambiguity, but in the context of the problem mentioning only x and y as variables, it's more likely a typo, and the denominator is x + y.Alternatively, perhaps the problem is correct as written, with "v" being a typo for "y". So assuming that, the answer would be two possible values. Therefore, the answer is B) 2.Wait, but let me check another approach. Suppose that the denominator is indeed x + v, but since the problem states "positive integers x and y", maybe v is a constant. But since it's not defined, the problem is ill-posed. Therefore, it's almost certain that the denominator is supposed to be x + y. So, moving forward with that, we have two solutions: x+y=4 and 8. So, two possible values. Therefore, the answer is B) 2.But let me think again. Wait, maybe there's another solution. Let me test with small primes.Suppose p=2. Then, (xy²)/(x + y)=2. So, xy²=2(x + y). Let's see if there are solutions other than x=2, y=2.Let's rearrange: xy² -2x -2y =0 => x(y² -2) = 2y => x= 2y/(y² -2). Since x must be a positive integer, denominator y² -2 must divide 2y. So, y² -2 divides 2y. Let's check possible y:y=1: denominator=1 -2=-1, x=2*1/-1=-2, invalid.y=2: denominator=4 -2=2, x=4/2=2. Valid, x=2, y=2. We already have that.y=3: denominator=9 -2=7, x=6/7, not integer.y=4: 16 -2=14, x=8/14=4/7, invalid.y=5:25 -2=23, x=10/23, invalid. So, only y=2 gives a solution here.Next prime p=3: (xy²)/(x + y)=3 => xy²=3(x + y). So, x=3(x + y)/y². Rearranged: x(y² -3)=3y => x=3y/(y² -3). So, x must be positive integer. So y² -3 divides 3y.Check possible y:y=1: denominator=1 -3=-2, x=3*(-1/2)= -3/2 invalid.y=2: denominator=4 -3=1, x=6/1=6. So x=6, y=2. That's the other solution we found.y=3: denominator=9 -3=6, x=9/6=3/2, invalid.y=4:16 -3=13, x=12/13 invalid.y=5:25 -3=22, x=15/22 invalid.So, only y=2 gives x=6.Next prime p=5: (xy²)/(x + y)=5 => xy²=5(x + y). So x=5(x + y)/y² => x(y² -5)=5y => x=5y/(y² -5). Let's check possible y:y=1: denom=1 -5=-4, x=5*(-1/4) invalid.y=2:4 -5=-1, x=10/(-1) invalid.y=3:9 -5=4, x=15/4, not integer.y=4:16 -5=11, x=20/11 invalid.y=5:25 -5=20, x=25/20=5/4, invalid.y=6:36 -5=31, x=30/31 invalid.So, no solutions for p=5.Similarly, higher primes would have even larger denominators, making x non-integer. Therefore, only p=2 and p=3 give solutions, leading to x+y=4 and 8. Thus, two possible values. Therefore, the answer is B) 2.Wait, but the options are (A)1, (B)2, (C)3, (D)4. So, the answer should be B.But let me verify once more. Suppose there's another solution that I missed. For example, maybe when p divides x and y? But p is a prime. If p divides both x and y, then x = p*a, y = p*b. Substitute into the equation:(p*a)(p*b)² / (p*a + p*b) = p*(a*b²) / (a + b) = prime. So, this would mean that (a*b²)/(a + b) must be 1, since the entire expression is equal to a prime number p, so (a*b²)/(a + b) =1, hence a*b² = a + b. Then, solving for a and b positive integers:a*b² -a -b =0 => a(b² -1) = b => a = b/(b² -1). Since a must be positive integer, denominator b² -1 must divide b. So, b² -1 divides b. For b ≥2:b=2: 4 -1=3 divides 2? No. 3 does not divide 2.b=3:9 -1=8 divides 3? No.b=1: denominator=1 -1=0, undefined.Therefore, no solutions here. Therefore, there are no solutions where p divides both x and y.Therefore, all possible solutions are the two found earlier: (2,2) and (6,2), leading to x+y=4 and 8. Therefore, two possible values. So the answer is B)2.Final Answerboxed{B}</think>Given the problem where positive integers (x) and (y) make the value of (frac{xy^2}{x+y}) a prime number, we need to determine the number of possible values of (x + y).First, we rewrite the given condition as:[frac{xy^2}{x+y} = p]where (p) is a prime number. This implies:[xy^2 = p(x + y)]We analyze the equation by considering the cases where (p) divides (x) or (p) divides (y).Case 1: (p) divides (x)- Let (x = kp). Substituting into the equation: [ kp cdot y^2 = p(kp + y) ] Simplifying, we get: [ ky^2 = kp + y implies ky^2 - y = kp ] Let (y = mk), then: [ k(mk)^2 - mk = kp implies m^2k^3 - mk = kp implies m(mk^2 - 1) = p ] Since (p) is prime, (m = 1) or (mk^2 - 1 = 1). Solving these, we find: - (m = 1) gives (k^2 - 1 = p), leading to (k = 2) and (p = 3), so (x = 6) and (y = 2). - (mk^2 - 1 = 1) gives (m = 2) and (k = 1), leading to (x = 2) and (y = 2).Case 2: (p) divides (y)- Let (y = lp). Substituting into the equation: [ x(lp)^2 = p(x + lp) implies xl^2p = x + lp ] Simplifying, we get: [ x(l^2p - 1) = lp implies x = frac{lp}{l^2p - 1} ] For (x) to be an integer, (l = 1) and (p = 2), leading to (x = 2) and (y = 2), which is already found in Case 1.Thus, the solutions are ((x, y) = (6, 2)) and ((2, 2)), leading to (x + y = 8) and (4) respectively. Therefore, there are 2 possible values for (x + y).The answer is (boxed{B}).,<think>Okay, so I have this problem here: If positive integers x and y make the value of (x y²)/(x + v) a prime number, then x + y has ( ) possible values. The options are A)1, B)2, C)3, D)4. Wait, hold on, there's a typo here, right? The denominator is x + v. But the variables mentioned are x and y. Maybe v is supposed to be y? Let me check the original problem again. Hmm, maybe it's a translation error or something. The problem says "positive integers x and y make the value of (x y²)/(x + v) a prime number". Since x and y are the variables, maybe "v" is a typo. Let me assume that it's supposed to be y. So the expression is (x y²)/(x + y). That makes more sense. Otherwise, if it's x + v, then v is another variable, but the problem doesn't mention it. So probably a typo. Let me proceed with that assumption: denominator is x + y.So the problem is: Given positive integers x and y such that (x y²)/(x + y) is a prime number. Then how many possible values does x + y have? The options are 1, 2, 3, or 4. So we need to find the number of possible values of x + y, given that (x y²)/(x + y) is prime.Alright, let's start by analyzing the expression (x y²)/(x + y). Since x and y are positive integers, x + y is also a positive integer, and x y² is a positive integer. So the fraction must simplify to a prime number. Let's denote the prime number as p. So, (x y²)/(x + y) = p. Then, we can write x y² = p (x + y). Let's rearrange this equation to see if we can find some relationships between x and y.First, let's rearrange:x y² = p x + p yBring all terms to one side:x y² - p x - p y = 0Factor terms with x:x (y² - p) - p y = 0Hmm, maybe that's not too helpful. Let's try another approach. Let's solve for x in terms of y and p.Starting from x y² = p (x + y)Bring the terms with x to one side:x y² - p x = p yFactor x:x (y² - p) = p yTherefore,x = (p y)/(y² - p)Since x is a positive integer, the right-hand side must be a positive integer. Therefore, (y² - p) must divide p y. Let's note that y² - p must be a positive integer as well because x is positive. So, y² - p > 0 => y² > p. So, p is a prime number less than y².Also, since x = (p y)/(y² - p), x must be a positive integer, so (y² - p) must divide p y. Let's write that as (y² - p) | p y. Which means that y² - p is a divisor of p y. Let's denote d = y² - p. Then, d divides p y. Since d = y² - p, and p is a prime, perhaps we can use this to find possible y and p.Alternatively, let's think of y as a variable and p as a prime number. Let's try small primes and small y values to see possible solutions.Let me try to set y as a small integer and see if I can find x such that the expression is prime.First, let's try y = 1.If y = 1, then (x * 1²)/(x + 1) = x / (x + 1). But since x and x + 1 are consecutive integers, they are coprime. Therefore, x/(x + 1) is in lowest terms and cannot be an integer unless x + 1 divides x, which is only possible if x + 1 = 1, but x is positive, so x = 0, which is not allowed. Thus, y cannot be 1.Next, y = 2.Compute (x * 2²)/(x + 2) = 4x / (x + 2). We need this to be a prime number p. So, 4x = p(x + 2). Let's solve for x:4x = p x + 2 p4x - p x = 2 px (4 - p) = 2 pTherefore, x = (2 p)/(4 - p)Since x must be a positive integer, 4 - p must divide 2p, and 4 - p must be positive. Therefore, 4 - p > 0 => p < 4. So p can be 2 or 3 (since p is prime).Case 1: p = 2x = (2*2)/(4 - 2) = 4 / 2 = 2. So x = 2. Then x + y = 2 + 2 = 4.Check: (2*2²)/(2 + 2) = (2*4)/4 = 8/4 = 2, which is prime. So this is valid.Case 2: p = 3x = (2*3)/(4 - 3) = 6 / 1 = 6. So x = 6. Then x + y = 6 + 2 = 8.Check: (6*2²)/(6 + 2) = (6*4)/8 = 24/8 = 3, which is prime. Valid.So for y = 2, there are two possible primes (2 and 3), leading to x + y = 4 and 8.Now, let's check y = 3.Compute (x * 3²)/(x + 3) = 9x / (x + 3). Let p be prime, so 9x = p(x + 3)9x = p x + 3p9x - p x = 3px (9 - p) = 3px = (3p)/(9 - p)x must be positive integer. So 9 - p divides 3p, and 9 - p > 0 => p < 9. Primes less than 9 are 2, 3, 5, 7.Check p = 2:x = (3*2)/(9 - 2) = 6/7, not integer.p = 3:x = (3*3)/(9 - 3) = 9/6 = 3/2, not integer.p = 5:x = (3*5)/(9 - 5) = 15/4, not integer.p = 7:x = (3*7)/(9 - 7) = 21/2 = 10.5, not integer.So no solutions for y = 3.Next, y = 4.Compute (x * 4²)/(x + 4) = 16x/(x + 4) = p.So 16x = p(x + 4)16x = p x + 4 p16x - p x = 4 px (16 - p) = 4 px = (4 p)/(16 - p)Need x positive integer, so 16 - p divides 4p, and 16 - p > 0 => p < 16. Primes less than 16: 2,3,5,7,11,13.Check each p:p = 2: x = (8)/(14) = 4/7, not integer.p = 3: x = 12/13, nope.p = 5: x = 20/11, nope.p = 7: x = 28/9, nope.p = 11: x = 44/5 = 8.8, nope.p = 13: x = 52/3 ≈17.33, nope.No solutions for y =4.y =5:Compute (25x)/(x +5)=p.25x = p(x +5)25x -p x =5px(25 -p)=5px=5p/(25 -p)25 - p divides 5p. Also 25 -p >0 => p <25. Primes less than25 are many, but let's check possible ones.But x must be positive integer, so 25 - p must divide 5p.Let's check p such that 25 -p divides 5p.Let’s check primes p:p=2: 25 -2=23, 23 divides 10? No.p=3: 22 divides 15? No.p=5: 20 divides25? 20 divides25? 25/20=5/4, no.p=7: 18 divides35? 35/18 not integer.p=11: 14 divides55? 55/14≈3.928, no.p=13:12 divides65? 65/12≈5.416, no.p=17:8 divides85? 85/8=10.625, nope.p=19:6 divides95? 95/6≈15.833, nope.p=23:2 divides115? 115/2=57.5, nope.No solutions for y=5.y=6:(36x)/(x +6)=p.36x = p(x +6)36x - p x =6px(36 -p)=6px=6p/(36 -p)36 -p divides6p, and 36 -p >0 =>p <36.Possible primes p=2,3,5,7,11,13,17,19,23,29,31.Check each p:p=2: x=12/(34)=6/17, nope.p=3: x=18/33=6/11, nope.p=5: x=30/31, nope.p=7: x=42/29, nope.p=11: x=66/25=2.64, nope.p=13: x=78/23≈3.39, nope.p=17: x=102/19≈5.368, nope.p=19: x=114/17≈6.705, nope.p=23: x=138/13≈10.615, nope.p=29: x=174/7≈24.857, nope.p=31: x=186/5=37.2, nope.No solutions for y=6.Hmm, getting no solutions so far except for y=2. Let me check y=3 again.Wait, for y=3, when I tried p=2,3,5,7, but maybe I made a mistake?Wait, for y=3:x=(3p)/(9 -p). So p must be less than 9. Let me check again.p=2: x=6/7, not integer.p=3: 9/6=3/2, nope.p=5:15/4, nope.p=7:21/2, nope. So no solutions.Ok, let's try y=7.Compute (49x)/(x +7)=p.49x = p(x +7)49x - p x =7px(49 - p)=7px=7p/(49 - p)49 -p divides7p. So 49 -p divides7p. Let's check primes less than49.Possible primes p:Check primes where 49 -p divides7p.Let me check p=2: 49 -2=47. 47 divides14? No.p=3:46 divides21? No.p=5:44 divides35? No.p=7:42 divides49? 49/42≈1.166, nope.Wait, 42 divides49? No.p=11:38 divides77? 77/38≈2.026, nope.p=13:36 divides91? 91/36≈2.527, nope.p=17:32 divides119? 119/32≈3.718, nope.p=19:30 divides133? 133/30≈4.433, nope.p=23:26 divides161? 161/26≈6.192, nope.p=29:20 divides203? 203/20≈10.15, nope.p=31:18 divides217? 217/18≈12.055, nope.p=37:12 divides259? 259/12≈21.583, nope.p=41:8 divides287? 287/8≈35.875, nope.p=43:6 divides301? 301/6≈50.166, nope.p=47:2 divides329? 329/2=164.5, nope.No solutions for y=7.Wait, maybe we need a different approach. Trying each y is tedious. Let's think algebraically.From the equation x = (p y)/(y² - p). Since x is a positive integer, y² - p must divide p y. Also, since y² - p divides p y, let's denote d = y² - p. Then, d divides p y, and d = y² - p. So, d divides p y, and p is prime. Let's write d | p y. Since d = y² - p, and p is prime, there are a few possibilities.Case 1: d divides y. Then, since d divides y, let d = k, where k divides y. Then, y² - p = k. So p = y² - k. Since k divides y, k is a divisor of y. Let y = k m, where m is a positive integer. Then p = (k m)^2 - k = k^2 m² - k = k(k m² - 1). But p is prime, so this expression must be equal to prime. Since k and (k m² -1) are positive integers, the only way their product is prime is if one of them is 1.Therefore, either k=1 and (k m² -1) is prime, or (k m² -1)=1 and k is prime.First, if k=1, then p =1*(1* m² -1) = m² -1. Since p is prime, m² -1 must be prime. m² -1 factors as (m -1)(m +1). So unless (m -1)=1, because otherwise, m² -1 would be composite. If m -1=1, then m=2. So p=2² -1=3, which is prime. So in this case, k=1, m=2. Then y =k m=1*2=2. Then p=3. Then x=(p y)/(y² -p)=(3*2)/(4 -3)=6/1=6. So x=6, y=2, p=3. Which we already found earlier.Alternatively, if (k m² -1)=1, then k m²=2. Since k divides y and k m²=2, possible k and m are (k=1, m²=2) which is not integer, or (k=2, m²=1). Then m=1, k=2. Then y=k m=2*1=2. Then p=k(k m² -1)=2*(2*1 -1)=2*1=2. Then x=(2*2)/(4 -2)=4/2=2. Which is the other solution for y=2. So this gives p=2, x=2, y=2.Therefore, when d divides y, the only possible solutions are when y=2, leading to p=2 and p=3, with x=2 and x=6.Case 2: d divides p. Since d divides p y and p is prime, either d divides p or d divides y. We considered d divides y already. If d divides p, then since p is prime, d is either 1 or p. Let's consider d=1: then y² -p=1, so p=y² -1. Then x=(p y)/1 = p y = y(y² -1). Since x must be positive integer, this is valid. But we need p=y² -1 to be prime. So y² -1 must be prime. As before, y² -1 factors as (y-1)(y+1). For this to be prime, one of the factors must be 1. Therefore, y-1=1 => y=2. Then p=4 -1=3, prime. Then x=3*2=6, which is the same solution as before. So y=2, p=3, x=6.If d=p, then y² -p =p => y²=2p. So p=y²/2. Since p is prime, y² must be twice a prime. Thus, y²=2p. So y must be even. Let y=2k. Then y²=4k²=2p => p=2k². Since p is prime, 2k² must be prime. Therefore, k²=1, so k=1. Thus, p=2*1=2, y=2*1=2. Then x=(2*2)/(4 -2)=4/2=2. Which is again the same solution.Therefore, in Case 2, we only get the same solutions as before: y=2, p=2, x=2 and y=2, p=3, x=6.So combining both cases, the only possible solutions are when y=2, giving x=2 and x=6, leading to x+y=4 and 8. So two possible values for x+y: 4 and 8. Therefore, the number of possible values is 2, which is option B.But wait, the problem didn't specify distinct values. Wait, the question says "possible values". So if x + y can be 4 or 8, that's two different values, so answer B)2.But let me check again to ensure there are no other solutions where d divides neither y nor p. For example, suppose d divides p y but doesn't divide p or y. Since p is prime, if d divides p y, then d is either 1, p, y, or p y. But since d = y² - p, which is greater than 0, and we already considered d=1 and d=p. Let's check d=y.If d=y, then y² -p = y => p=y² - y. Then x=(p y)/y = p. So x=p. Then since x and y are positive integers, p must be prime. So p=y² - y, which needs to be prime. Let's check for y:y=2: p=4 -2=2, prime. Then x=2, so x+y=4.y=3: p=9 -3=6, not prime.y=4: p=16 -4=12, nope.y=5:20, nope. So only y=2 gives prime p=2. Which we already have.Alternatively, d=py. Then y² -p = p y => y² - p y -p =0. Quadratic in y: y² -p y -p=0. For y to be integer, discriminant must be a perfect square. Discriminant D= p² +4p. Must be a perfect square.But p is prime. Let's set D =k². Then p² +4p =k². Let's rearrange:k² =p² +4p =p(p +4). Since p is prime, p divides k², so p divides k. Let k=mp. Then:(m p)^2 =p(p +4)m² p² =p² +4pDivide both sides by p:m² p =p +4Then m² p -p =4p(m² -1)=4Since p is prime, the factors of 4 are 1,2,4. So possible p(m² -1)=4. Let's see possible prime p:Case 1: p=2. Then 2(m² -1)=4 => m² -1=2 => m²=3. Not integer.Case 2: p=2, but that doesn't work. Wait, p must divide 4. So p=2:2(m² -1)=4 => m² -1=2 => m²=3. No.p=2 is only even prime. Other factors, p=1, but 1 is not prime. So no solution here. Therefore, d=py gives no solutions.Therefore, all possible solutions come from y=2 with x=2 and x=6, giving x+y=4 and 8. Thus, there are two possible values. Therefore, the answer is B)2.Final Answerboxed{B}</think>Given positive integers (x) and (y) such that (frac{xy^2}{x+y}) is a prime number, we need to determine the number of possible values of (x + y).We start by setting (frac{xy^2}{x+y} = p), where (p) is a prime number. This leads to the equation:[xy^2 = p(x + y)]Rearranging and factoring, we get:[x(y^2 - p) = py]Thus, (x = frac{py}{y^2 - p}). For (x) to be a positive integer, (y^2 - p) must divide (py). We analyze possible values of (y) and (p) to find solutions.Testing small values of (y):1. For (y = 2): [ frac{4x}{x + 2} = p ] Solving for (x): [ 4x = p(x + 2) implies x = frac{2p}{4 - p} ] - (p = 2): (x = 2), leading to (x + y = 2 + 2 = 4). - (p = 3): (x = 6), leading to (x + y = 6 + 2 = 8).2. For other values of (y) (e.g., 1, 3, 4, 5, 6, 7), no valid solutions were found as the resulting (x) values were not integers or did not yield prime numbers.Further algebraic analysis confirmed that the only valid solutions occur when (y = 2), leading to (x = 2) or (x = 6). Thus, the possible values of (x + y) are 4 and 8.Therefore, the number of possible values of (x + y) is (boxed{B}).]

🔑:BTranslate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.

🧠:[<think>Okay, let's see. The problem is: Given the equation x² - 2x = 1, we need to find the value of 2023 + 6x - 3x². Hmm, okay. So, first, maybe I can start by solving for x in the given quadratic equation and then plug that into the expression we need to evaluate. But wait, solving for x might involve quadratic formula or completing the square, which could give two solutions. Then, substituting both values into the expression might be a bit tedious. Maybe there's a smarter way to approach this without having to find the actual values of x. Let me think.Alternatively, maybe I can manipulate the original equation to express x² in terms of x, and then substitute that into the expression 2023 + 6x - 3x². That sounds better because substitution could simplify the expression directly without needing to find x. Let's try that.Original equation: x² - 2x = 1. Let's solve for x². If I add 2x to both sides, I get x² = 2x + 1. Okay, so x² is equal to 2x + 1. Now, the expression we need is 2023 + 6x - 3x². Let's substitute x² with 2x + 1 in the expression.So, substituting, we get:2023 + 6x - 3*(2x + 1). Let me compute that step by step. First, expand the multiplication: -3*(2x + 1) = -6x - 3. Then, the entire expression becomes 2023 + 6x -6x -3. Wait, the 6x and -6x cancel each other out. That leaves us with 2023 - 3. Which is 2020. So, the value of the expression is 2020? That seems straightforward. Let me verify that again to make sure I didn't make any mistakes.Starting with the original equation: x² - 2x = 1. Solving for x² gives x² = 2x + 1. Then, substituting into the expression: 2023 + 6x - 3x² becomes 2023 + 6x - 3*(2x + 1). Expanding that: 2023 + 6x -6x -3. The 6x and -6x do indeed cancel, so 2023 -3 = 2020. Yeah, that seems correct.Alternatively, if I didn't see that substitution method, maybe I could have rearranged the expression 2023 + 6x -3x² in terms of the original equation. Let's see. Let's take the expression: -3x² +6x +2023. I can factor out a -3 from the first two terms: -3(x² -2x) +2023. Wait a minute, the original equation is x² -2x =1. So, x² -2x is equal to 1. Therefore, substituting that into the expression: -3*(1) +2023 = -3 +2023 = 2020. Oh, that's another way to see it. That's even faster. So, that confirms the answer is 2020.So, both methods lead to the same result. Either substitute x² =2x +1 into the expression or recognize that part of the expression can be expressed as -3*(x² -2x), which is -3*1, and then adding the remaining constant. Either way, the answer is 2020. That gives me confidence that the answer is correct.Wait, just to be thorough, suppose I didn't notice these methods and tried to solve for x first. Let's try that as a check.Original equation: x² -2x -1 =0. Using quadratic formula: x = [2 ± sqrt(4 +4)]/2 = [2 ± sqrt(8)]/2 = [2 ± 2*sqrt(2)]/2 = 1 ± sqrt(2). So, the solutions are x = 1 + sqrt(2) and x = 1 - sqrt(2). Let's plug both into the expression 2023 +6x -3x².First, for x =1 + sqrt(2):Compute 6x: 6*(1 + sqrt(2)) =6 +6sqrt(2).Compute 3x²: 3*(1 + sqrt(2))² = 3*(1 + 2sqrt(2) +2) =3*(3 + 2sqrt(2)) =9 +6sqrt(2).So, 2023 +6x -3x² =2023 + (6 +6sqrt(2)) - (9 +6sqrt(2)) =2023 +6 +6sqrt(2) -9 -6sqrt(2) =2023 + (6 -9) + (6sqrt(2) -6sqrt(2)) =2023 -3 +0 =2020.Similarly, for x=1 - sqrt(2):Compute 6x:6*(1 - sqrt(2))=6 -6sqrt(2).Compute 3x²:3*(1 - sqrt(2))²=3*(1 -2sqrt(2) +2)=3*(3 -2sqrt(2))=9 -6sqrt(2).Then, 2023 +6x -3x²=2023 + (6 -6sqrt(2)) - (9 -6sqrt(2))=2023 +6 -6sqrt(2) -9 +6sqrt(2)=2023 + (6 -9) + (-6sqrt(2) +6sqrt(2))=2023 -3 +0=2020.So, even by plugging in both roots, we get 2020. That's reassuring. So, regardless of the method, the answer is 2020. So, that must be the correct answer. I think that's thorough enough. Tried substitution, factoring, and even solving for x directly. All methods led to the same result. So, the answer is 2020.Final Answerboxed{2020}</think>Given the equation (x^2 - 2x = 1), we need to find the value of (2023 + 6x - 3x^2).First, we solve the given equation for (x^2):[x^2 - 2x = 1 implies x^2 = 2x + 1]Next, substitute (x^2 = 2x + 1) into the expression (2023 + 6x - 3x^2):[2023 + 6x - 3x^2 = 2023 + 6x - 3(2x + 1)]Expanding the expression:[2023 + 6x - 3(2x + 1) = 2023 + 6x - 6x - 3]Simplifying the terms:[2023 + 6x - 6x - 3 = 2023 - 3 = 2020]Thus, the value of (2023 + 6x - 3x^2) is (boxed{2020}).,<think>Okay, so I need to solve the problem: Given that x² - 2x = 1, find the value of 2023 + 6x - 3x². Hmm, let's see. Alright, first, maybe I can start by understanding the given equation and see how it relates to the expression we need to evaluate. The equation is x squared minus 2x equals 1. The expression we need to find is 2023 plus 6x minus 3x squared. Let me write down both to visualize them better:Given:x² - 2x = 1Find:2023 + 6x - 3x²I notice that in the expression we need to evaluate, there's a -3x² term and a +6x term. Maybe if I factor out a -3 from those terms, it would look like part of the given equation. Let me try that.So, 2023 + 6x - 3x² can be rewritten as 2023 - 3x² + 6x. If I factor out -3 from the terms with x, it becomes 2023 -3(x² - 2x). Wait, let's check that:-3(x² - 2x) = -3x² + 6x. Yes, exactly. So the expression 6x - 3x² is equivalent to -3(x² - 2x). But from the given equation, x² - 2x = 1. So substituting that into the expression, we can replace x² - 2x with 1. Therefore, the expression becomes:2023 - 3*(1) = 2023 - 3 = 2020.Wait, is that it? Let me check again step by step to make sure I didn't make a mistake.Original expression: 2023 + 6x - 3x².Factor out -3 from the last two terms: 2023 + (-3)(x² - 2x). Since x² - 2x is given as 1, substitute that in: 2023 + (-3)(1) = 2023 - 3 = 2020.Yes, that seems right. So the value of the expression is 2020. Alternatively, maybe I can solve for x first using the given equation and then plug it into the expression. Let's try that method to verify.Given x² - 2x = 1. Let's solve this quadratic equation for x.First, bring all terms to one side: x² - 2x - 1 = 0.This is a quadratic equation. Using the quadratic formula, x = [2 ± sqrt(4 + 4)] / 2 = [2 ± sqrt(8)] / 2 = [2 ± 2*sqrt(2)] / 2 = 1 ± sqrt(2).So, the solutions are x = 1 + sqrt(2) and x = 1 - sqrt(2). Now, let's plug these values into the expression 2023 + 6x - 3x² and see if we get 2020 in both cases.First, take x = 1 + sqrt(2):Compute 6x: 6*(1 + sqrt(2)) = 6 + 6*sqrt(2)Compute 3x²: 3*(x²). First find x²: (1 + sqrt(2))² = 1 + 2*sqrt(2) + 2 = 3 + 2*sqrt(2). Multiply by 3: 3*(3 + 2*sqrt(2)) = 9 + 6*sqrt(2).So, 2023 + 6x - 3x² becomes 2023 + (6 + 6*sqrt(2)) - (9 + 6*sqrt(2)) = 2023 + 6 + 6*sqrt(2) - 9 - 6*sqrt(2). The 6*sqrt(2) and -6*sqrt(2) cancel out. Then, 2023 + 6 -9 = 2023 -3 = 2020.Similarly, take x = 1 - sqrt(2):Compute 6x: 6*(1 - sqrt(2)) = 6 - 6*sqrt(2)Compute 3x²: 3*(x²). First find x²: (1 - sqrt(2))² = 1 - 2*sqrt(2) + 2 = 3 - 2*sqrt(2). Multiply by 3: 3*(3 - 2*sqrt(2)) = 9 - 6*sqrt(2).So, 2023 + 6x - 3x² becomes 2023 + (6 - 6*sqrt(2)) - (9 - 6*sqrt(2)) = 2023 +6 -6*sqrt(2) -9 +6*sqrt(2). Again, the -6*sqrt(2) and +6*sqrt(2) cancel. Then, 2023 +6 -9 = 2020.So both cases give 2020. Therefore, regardless of the value of x, the expression equals 2020. So the answer is 2020.Wait, so both methods confirm that the answer is 2020. The first method was substituting the expression using the given equation, and the second method was solving for x and plugging it back in. Both lead to the same result, so that gives me confidence that the answer is indeed correct. I think that's thorough. Let me just recap:Given x² -2x =1, need to compute 2023 +6x -3x². Rewriting the expression as 2023 -3(x² -2x) allows substitution of x² -2x with 1, leading to 2023 -3(1)=2020. Alternatively, solving for x gives two solutions, substituting both into the original expression also gives 2020 each time. Hence, the answer is 2020. Yep, that seems solid. I don't see any mistakes here.Final Answerboxed{2020}</think>Given the equation (x^2 - 2x = 1), we need to find the value of (2023 + 6x - 3x^2).First, we rewrite the expression (2023 + 6x - 3x^2) by factoring out (-3) from the terms involving (x):[2023 + 6x - 3x^2 = 2023 - 3(x^2 - 2x)]From the given equation (x^2 - 2x = 1), we substitute (x^2 - 2x) with 1:[2023 - 3(1) = 2023 - 3 = 2020]To verify, we solve the quadratic equation (x^2 - 2x - 1 = 0) using the quadratic formula:[x = frac{2 pm sqrt{4 + 4}}{2} = frac{2 pm 2sqrt{2}}{2} = 1 pm sqrt{2}]Substituting (x = 1 + sqrt{2}) into the expression:[6x = 6(1 + sqrt{2}) = 6 + 6sqrt{2}][3x^2 = 3(1 + sqrt{2})^2 = 3(3 + 2sqrt{2}) = 9 + 6sqrt{2}][2023 + 6x - 3x^2 = 2023 + (6 + 6sqrt{2}) - (9 + 6sqrt{2}) = 2023 + 6 - 9 = 2020]Similarly, substituting (x = 1 - sqrt{2}) into the expression:[6x = 6(1 - sqrt{2}) = 6 - 6sqrt{2}][3x^2 = 3(1 - sqrt{2})^2 = 3(3 - 2sqrt{2}) = 9 - 6sqrt{2}][2023 + 6x - 3x^2 = 2023 + (6 - 6sqrt{2}) - (9 - 6sqrt{2}) = 2023 + 6 - 9 = 2020]Both methods confirm the value is (boxed{2020}).]

🔑:Given the equation x^{2}-2x=1, we are tasked with finding the value of 2023+6x-3x^{2}.First, we recognize that the given equation can be directly applied to simplify the expression we need to find. We start by rewriting the expression in terms of the given equation:[2023+6x-3x^{2} = 2023 - 3(x^{2}-2x)]Substituting the value of x^{2}-2x from the given equation, we have:[= 2023 - 3 times 1]Simplifying the arithmetic:[= 2023 - 3][= 2020]Therefore, the value of 2023+6x-3x^{2} is boxed{2020}.

🧠:[<think>Okay, let me try to figure out this problem step by step. So, we have triangle ABC, and there's this vector equation involving dot products of the vectors AB, AC, BA, BC, CA, and CB. The equation is:AB · AC + 2 BA · BC = 3 CA · CBAnd we need to find the maximum value of sin C. Hmm, okay. Let me recall some vector properties and maybe some trigonometry related to triangles. First, I remember that the dot product of two vectors can be expressed in terms of their magnitudes and the cosine of the angle between them. So, for vectors AB and AC, the dot product AB · AC equals |AB||AC|cos A, where A is the angle at vertex A. Wait, is that right? Wait, AB is the vector from A to B, and AC is the vector from A to C. So the angle between AB and AC is indeed angle A. So yes, AB · AC = |AB||AC|cos A.Similarly, BA · BC would be |BA||BC|cos B, since BA is from B to A, and BC is from B to C, so the angle between them is angle B. Similarly, CA · CB is |CA||CB|cos C. So maybe we can rewrite the given equation in terms of the sides and angles of the triangle.Let me denote the lengths of the sides as follows: let’s say side opposite angle A is a, opposite angle B is b, and opposite angle C is c. But wait, actually, in standard notation, side a is BC, side b is AC, and side c is AB. Hmm, maybe I should confirm that. In triangle ABC, side a is opposite angle A, which is BC. Side b is opposite angle B, which is AC. Side c is opposite angle C, which is AB. So, with that notation, AB is length c, BC is length a, AC is length b.Wait, but the vectors AB, AC, etc., have magnitudes equal to the lengths of the sides. So |AB| is c, |AC| is b, |BA| is also c (since BA is the same vector as AB but in the opposite direction), |BC| is a, |CA| is b, and |CB| is a. So substituting these into the dot products:AB · AC = |AB||AC|cos A = c * b * cos ABA · BC = |BA||BC|cos B = c * a * cos BCA · CB = |CA||CB|cos C = b * a * cos CSo the equation becomes:c * b * cos A + 2 * c * a * cos B = 3 * b * a * cos CSo:cb cos A + 2ca cos B = 3ab cos CHmm. Let me write that again:cb cos A + 2ca cos B = 3ab cos CMaybe I can factor out the common terms. Let's see, each term has a c or a or b. Let's see:First term: cb cos ASecond term: 2ca cos BThird term: 3ab cos CAlternatively, maybe divide both sides by abc to normalize:(cb cos A)/(abc) + (2ca cos B)/(abc) = (3ab cos C)/(abc)Simplifying:(cos A)/a + 2(cos B)/b = 3(cos C)/cSo:(cos A)/a + 2(cos B)/b = 3(cos C)/cHmm, interesting. Now, perhaps we can use the Law of Cosines here? The Law of Cosines relates the sides of a triangle to the cosines of its angles. For example, cos A = (b² + c² - a²)/(2bc), and similarly for the other angles. Let me write those down:cos A = (b² + c² - a²)/(2bc)cos B = (a² + c² - b²)/(2ac)cos C = (a² + b² - c²)/(2ab)So substituting these into the equation:[ (b² + c² - a²)/(2bc) ] / a + 2[ (a² + c² - b²)/(2ac) ] / b = 3[ (a² + b² - c²)/(2ab) ] / cLet me compute each term step by step.First term: [ (b² + c² - a²)/(2bc) ] / a = (b² + c² - a²)/(2bc * a) = (b² + c² - a²)/(2abc)Second term: 2[ (a² + c² - b²)/(2ac) ] / b = 2 * (a² + c² - b²)/(2ac * b) = (a² + c² - b²)/(abc)Third term: 3[ (a² + b² - c²)/(2ab) ] / c = 3(a² + b² - c²)/(2ab * c) = 3(a² + b² - c²)/(2abc)So putting them all together:First term + second term = third termSo:[ (b² + c² - a²) + 2(a² + c² - b²) ] / (2abc) = 3(a² + b² - c²)/(2abc)Multiply both sides by 2abc to eliminate denominators:(b² + c² - a²) + 2(a² + c² - b²) = 3(a² + b² - c²)Simplify the left-hand side:First term: b² + c² - a²Second term: 2a² + 2c² - 2b²Adding them together:(b² - 2b²) + (c² + 2c²) + (-a² + 2a²) =(-b²) + 3c² + a²So left-hand side is a² - b² + 3c²Right-hand side: 3a² + 3b² - 3c²So equation becomes:a² - b² + 3c² = 3a² + 3b² - 3c²Bring all terms to left-hand side:a² - b² + 3c² - 3a² - 3b² + 3c² = 0Combine like terms:(a² - 3a²) + (-b² - 3b²) + (3c² + 3c²) =(-2a²) + (-4b²) + 6c² = 0So:-2a² -4b² +6c² =0Divide both sides by 2:- a² -2b² +3c² =0Rearranged:3c² = a² + 2b²So, 3c² = a² + 2b². Okay, so this is a relation between the sides of the triangle. Now, the problem is to find the maximum value of sin C. So, we need to express sin C in terms of the sides, perhaps, and then maximize it given the constraint 3c² = a² + 2b².Alternatively, maybe use the Law of Sines? Since sin C relates to side c. Let me recall the Law of Sines: a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius. So, perhaps we can express a, b, c in terms of angles.Let me denote 2R as a common value. Let’s say a = 2R sin A, b = 2R sin B, c = 2R sin C. Then, substituting into the equation 3c² = a² + 2b²:3(2R sin C)^2 = (2R sin A)^2 + 2(2R sin B)^2Simplify:3 * 4R² sin² C = 4R² sin² A + 2 * 4R² sin² BDivide both sides by 4R²:3 sin² C = sin² A + 2 sin² BSo, 3 sin² C = sin² A + 2 sin² B. Also, in any triangle, A + B + C = π. So, we can perhaps express angles A and B in terms of C. Since A + B = π - C. Let’s set A = π - C - B, but that might not be helpful directly. Alternatively, perhaps express A and B in terms of variables.Alternatively, let me consider using variables. Let’s denote angles A, B, C. Then, A + B + C = π. So, A + B = π - C. Let’s define variables such that x = A and y = B, so x + y = π - C. Then, our equation is 3 sin² C = sin² x + 2 sin² y, with x + y = π - C. Hmm, this seems a bit complicated, but maybe we can express sin² x and sin² y in terms of C.Alternatively, perhaps use the Law of Cosines again. Since 3c² = a² + 2b², and from Law of Cosines, we have c² = a² + b² - 2ab cos C. Wait, but in our relation, 3c² = a² + 2b². Let me substitute c² from the Law of Cosines into this equation.So, 3(a² + b² - 2ab cos C) = a² + 2b²Multiply out the left side:3a² + 3b² - 6ab cos C = a² + 2b²Bring all terms to left side:3a² + 3b² -6ab cos C -a² -2b² =0Simplify:(3a² - a²) + (3b² -2b²) -6ab cos C =02a² + b² -6ab cos C =0So:6ab cos C =2a² + b²Divide both sides by ab:6 cos C = (2a)/b + b/aSo:cos C = [ (2a)/b + b/a ] /6Hmm, interesting. So cos C is equal to (2a/b + b/a)/6. Let me denote t = a/b. Then, cos C becomes (2t + 1/t)/6. So, cos C = (2t + 1/t)/6. Now, we need to find the maximum value of sin C. Since sin C = sqrt(1 - cos² C). So, to maximize sin C, we need to minimize cos C, given that cos C is expressed in terms of t, where t = a/b is a positive real number (since a and b are lengths).So, let's set f(t) = (2t + 1/t)/6. We need to find the minimum value of f(t) to maximize sin C. Wait, but wait, actually, since sin C increases as cos C decreases in the interval [0, π], so to maximize sin C, we need to minimize cos C. Therefore, the problem reduces to minimizing f(t) = (2t + 1/t)/6 over t > 0.Let’s compute the derivative of f(t) with respect to t to find the critical points. f(t) = (2t + 1/t)/6 = (2t)/6 + (1/(6t)) = t/3 + 1/(6t)So f'(t) = 1/3 - 1/(6t²)Set derivative equal to zero:1/3 - 1/(6t²) =01/3 =1/(6t²)Multiply both sides by 6t²:2t² =1t² =1/2t=1/√2Since t>0, we take t=1/√2.So the minimum occurs at t=1/√2. Let’s compute f(t) at this point:f(1/√2) = (2*(1/√2) + 1/(1/√2))/6Compute 2*(1/√2) = 2/√2 = √21/(1/√2) = √2So f(1/√2) = (√2 + √2)/6 = (2√2)/6 = √2/3 ≈ 0.4714So the minimal value of cos C is √2/3, which would give the maximal sin C as sqrt(1 - (√2/3)^2) = sqrt(1 - 2/9) = sqrt(7/9) = √7/3 ≈ 0.8819Wait, that seems like a possible answer. But let me verify that.Wait, if cos C = √2/3, then sin C = sqrt(1 - (√2/3)^2) = sqrt(1 - 2/9) = sqrt(7/9) = √7 / 3 ≈ 0.8819. So the maximum value of sin C is √7/3. But before finalizing, let me check if this critical point indeed gives a minimum. Since the function f(t) = (2t + 1/t)/6, as t approaches 0, 1/t becomes large, so f(t) approaches infinity. As t approaches infinity, 2t becomes large, so f(t) approaches infinity. So the critical point we found at t=1/√2 must be a minimum. Therefore, this gives the minimal cos C, hence maximal sin C. Therefore, the maximum value of sin C is √7/3.Wait, but let me confirm this with the original equation. Let's see, if t = a/b =1/√2, then a = b/√2. Let's set a =1, then b=√2. Then, from 3c² =a² + 2b², we have 3c²=1 + 2*(2)=1 +4=5, so c²=5/3, c=√(5/3). Then, using the Law of Cosines for angle C: cos C= (a² + b² -c²)/(2ab)= (1 + 2 -5/3)/(2*1*√2)= (3 -5/3)/(2√2)= (4/3)/(2√2)= (2/3)/√2= 2/(3√2)= √2/3, which matches our previous result. Therefore, sin C=√7/3. Therefore, the maximum value of sin C is √7/3. So the answer is √7/3, which in boxed form is boxed{dfrac{sqrt{7}}{3}}.Final AnswerThe maximum value of sin C is boxed{dfrac{sqrt{7}}{3}}.</think>Given the problem in triangle ABC where overrightarrow{AB} cdot overrightarrow{AC} + 2 overrightarrow{BA} cdot overrightarrow{BC} = 3 overrightarrow{CA} cdot overrightarrow{CB}, we need to find the maximum value of sin C.First, we express the dot products in terms of the sides and angles of the triangle:- overrightarrow{AB} cdot overrightarrow{AC} = |AB||AC| cos A = cb cos A- overrightarrow{BA} cdot overrightarrow{BC} = |BA||BC| cos B = ca cos B- overrightarrow{CA} cdot overrightarrow{CB} = |CA||CB| cos C = ab cos CSubstituting these into the given equation:[ cb cos A + 2ca cos B = 3ab cos C ]Dividing both sides by abc:[ frac{cos A}{a} + 2 frac{cos B}{b} = 3 frac{cos C}{c} ]Using the Law of Cosines:- cos A = frac{b^2 + c^2 - a^2}{2bc}- cos B = frac{a^2 + c^2 - b^2}{2ac}- cos C = frac{a^2 + b^2 - c^2}{2ab}Substituting these into the equation and simplifying, we find:[ 3c^2 = a^2 + 2b^2 ]Using the Law of Cosines for cos C:[ cos C = frac{a^2 + b^2 - c^2}{2ab} ]Substituting c^2 = frac{a^2 + 2b^2}{3} into the expression for cos C:[ cos C = frac{a^2 + b^2 - frac{a^2 + 2b^2}{3}}{2ab} = frac{2a^2 + b^2}{6ab} ]Letting t = frac{a}{b}, we get:[ cos C = frac{2t + frac{1}{t}}{6} ]To minimize cos C (and thus maximize sin C), we find the critical points of f(t) = frac{2t + frac{1}{t}}{6}. The minimum occurs at t = frac{1}{sqrt{2}}, leading to:[ cos C = frac{sqrt{2}}{3} ]Thus, the maximum value of sin C is:[ sin C = sqrt{1 - left(frac{sqrt{2}}{3}right)^2} = sqrt{frac{7}{9}} = frac{sqrt{7}}{3} ]The maximum value of sin C is boxed{dfrac{sqrt{7}}{3}}.,<think>Alright, let's try to tackle this problem step by step. The problem states:In triangle ABC, it is known that the dot product of vectors AB and AC plus twice the dot product of vectors BA and BC equals three times the dot product of vectors CA and CB. We need to find the maximum value of sin C.Hmm, okay. First, let's recall some vector properties and how dot products relate to angles and lengths in triangles. The dot product of two vectors u and v is given by u · v = |u||v|cosθ, where θ is the angle between them. So, maybe we can express each of these dot products in terms of the sides and angles of the triangle.Let me denote the sides of the triangle as follows: let’s let AB = c, BC = a, and AC = b, following the standard notation where each side is opposite the corresponding angle. Wait, but actually, in standard triangle notation, side a is opposite angle A, side b opposite angle B, and side c opposite angle C. Hmm, but maybe using vectors complicates that. Alternatively, perhaps using position vectors from a common origin?Wait, maybe it's better to consider vectors AB, AC, BA, etc., as vectors in the plane. Let me try to visualize triangle ABC. Let's fix points A, B, C in the plane. The vectors AB and AC would be vectors from point A to points B and C, respectively. Similarly, BA is the vector from B to A, and BC is the vector from B to C. Similarly, CA and CB are vectors from C to A and C to B.But since we are dealing with dot products, maybe we can express these vectors in terms of the sides of the triangle and the angles between them. Let's see.First, let's write down each dot product:1. AB · AC: This is the dot product of vectors AB and AC. The vectors AB and AC both start at point A. The angle between them is angle A. Wait, angle A is the angle between AB and AC? Wait, actually, angle A is between AB and AC? No, angle A is at vertex A, between sides AB and AC. So yes, the angle between vectors AB and AC is angle A. Therefore, AB · AC = |AB||AC|cos A.Similarly, BA · BC: The vectors BA and BC start at point B. Vector BA is from B to A, and BC is from B to C. The angle between BA and BC is angle at B. Wait, angle at B is between BA and BC? Wait, BA is going from B to A, and BC is from B to C. So the angle between BA and BC is angle at B. So BA · BC = |BA||BC|cos B. Since BA is the same as AB but in the opposite direction, so |BA| = |AB| = c, |BC| = a, so BA · BC = c * a * cos B.Similarly, CA · CB: Vectors CA and CB start at point C. Vector CA is from C to A, and CB is from C to B. The angle between them is angle at C, which is angle C. Therefore, CA · CB = |CA||CB|cos C. |CA| is b, |CB| is a, so CA · CB = b * a * cos C.Therefore, the given equation is:AB · AC + 2 BA · BC = 3 CA · CBTranslating into the dot products:|AB||AC|cos A + 2 |BA||BC|cos B = 3 |CA||CB|cos CBut in standard triangle notation, sides opposite angles A, B, C are a, b, c respectively. Wait, hold on. Wait, side a is opposite angle A, side b opposite angle B, side c opposite angle C. But in our case, AB is a vector from A to B, so the length of AB is equal to side c (opposite angle C), right? Wait, let's clarify.Wait, in standard triangle notation:- Angle A is at vertex A, between sides AB and AC.- The side opposite angle A is BC, which is denoted as a.- The side opposite angle B is AC, denoted as b.- The side opposite angle C is AB, denoted as c.Therefore, |AB| = c, |BC| = a, |AC| = b.Therefore, the previous dot products:AB · AC = |AB||AC|cos A = c * b * cos ABA · BC = |BA||BC|cos B = c * a * cos BCA · CB = |CA||CB|cos C = b * a * cos CTherefore, the equation given is:c * b * cos A + 2 * c * a * cos B = 3 * b * a * cos CSo the equation is:cb cos A + 2ca cos B = 3ab cos CHmm. We need to find the maximum value of sin C. So our goal is to relate this equation to angle C and find the maximum possible sin C given this condition.First, maybe we can use the Law of Cosines to express cos A, cos B, and cos C in terms of the sides. Let's recall that:In triangle ABC,cos A = (b² + c² - a²)/(2bc)cos B = (a² + c² - b²)/(2ac)cos C = (a² + b² - c²)/(2ab)So substituting these into the equation:cb [ (b² + c² - a²)/(2bc) ] + 2ca [ (a² + c² - b²)/(2ac) ] = 3ab [ (a² + b² - c²)/(2ab) ]Let me compute each term step by step.First term: cb * cos A = cb * (b² + c² - a²)/(2bc) = (b² + c² - a²)/2Second term: 2ca * cos B = 2ca * (a² + c² - b²)/(2ac) = (a² + c² - b²)Third term (on RHS): 3ab * cos C = 3ab * (a² + b² - c²)/(2ab) = 3(a² + b² - c²)/2So putting it all together:Left-hand side (LHS): (b² + c² - a²)/2 + (a² + c² - b²) = ?Right-hand side (RHS): 3(a² + b² - c²)/2Let me compute LHS:First term: (b² + c² - a²)/2Second term: (a² + c² - b²) = (a² + c² - b²)/1So to add them, let's get a common denominator:= (b² + c² - a²)/2 + 2(a² + c² - b²)/2= [ (b² + c² - a²) + 2a² + 2c² - 2b² ] / 2Simplify numerator:b² + c² - a² + 2a² + 2c² - 2b²= (b² - 2b²) + (c² + 2c²) + (-a² + 2a²)= (-b²) + 3c² + a²Therefore, LHS = (a² - b² + 3c²)/2RHS = 3(a² + b² - c²)/2Setting them equal:(a² - b² + 3c²)/2 = 3(a² + b² - c²)/2Multiply both sides by 2:a² - b² + 3c² = 3a² + 3b² - 3c²Bring all terms to the left side:a² - b² + 3c² - 3a² - 3b² + 3c² = 0Combine like terms:(a² - 3a²) + (-b² - 3b²) + (3c² + 3c²) = 0(-2a²) + (-4b²) + 6c² = 0Divide both sides by 2:- a² - 2b² + 3c² = 0So,3c² = a² + 2b²So, we have a relationship between the sides: 3c² = a² + 2b²Our goal is to find the maximum value of sin C. Since angle C is opposite side c, and sin C can be expressed using the Law of Sines: sin C = c/(2R), where R is the circumradius, but maybe we can use the relationship between the sides and angles.Alternatively, using the Law of Cosines, we can express cos C in terms of sides, and then sin C = sqrt(1 - cos² C). Then, we can try to express everything in terms of a single variable.But first, let's see if we can express a and b in terms of c, or another variable, using the equation 3c² = a² + 2b².Alternatively, maybe we can relate the sides through some substitution or use trigonometric identities.Alternatively, let's recall that in triangle ABC, by the Law of Sines, we have:a/sin A = b/sin B = c/sin C = 2RSo, a = 2R sin A, b = 2R sin B, c = 2R sin CTherefore, substituting into 3c² = a² + 2b²:3(2R sin C)^2 = (2R sin A)^2 + 2(2R sin B)^2Simplify:3 * 4R² sin² C = 4R² sin² A + 2 * 4R² sin² BDivide both sides by 4R²:3 sin² C = sin² A + 2 sin² BSo, 3 sin² C = sin² A + 2 sin² BBut in a triangle, angles A + B + C = π. So, A + B = π - CWe need to express sin² A and sin² B in terms of angle C.Hmm, this might be a bit complicated, but perhaps we can use some trigonometric identities.Alternatively, let's denote angle C as θ, so θ = C. Then, A + B = π - θ.We can express angles A and B in terms of θ and some other variable. Let’s say let’s set angle A = α, angle B = π - θ - α. Then, we can write sin² A + 2 sin² B in terms of α and θ.So, 3 sin² θ = sin² α + 2 sin² (π - θ - α)But sin (π - θ - α) = sin (θ + α), since sin(π - x) = sin x.Wait, no: sin(π - x) = sin x. Wait, π - θ - α = π - (θ + α), so sin(π - (θ + α)) = sin(θ + α). Hmm, actually, sin(π - x) = sin x, so sin(π - θ - α) = sin(θ + α). Wait, but θ + α is angle (C + A), which may not be directly helpful.Alternatively, since A + B = π - θ, maybe express B in terms of A: B = π - θ - A. Then, sin B = sin(π - θ - A) = sin(θ + A). Wait, that's not correct. Wait, sin(π - x) = sin x, so sin(π - θ - A) = sin(θ + A). Wait, that is correct. Because π - θ - A = π - (θ + A), so sin(π - (θ + A)) = sin(θ + A). So sin B = sin(θ + A). Hmm.But perhaps this is complicating things. Maybe we can use the Law of Cosines relation we had earlier.Wait, we also had the relation 3c² = a² + 2b². Let's see if we can express a and b in terms of c, then relate angles via Law of Sines or Cosines.Alternatively, express a and b as multiples of c, perhaps.Let’s assume c is a constant for now (since in a triangle, sides are relative, we can scale the triangle as needed). Let’s set c = 1 for simplicity. Then, 3(1)^2 = a² + 2b² => 3 = a² + 2b².But in the Law of Sines, a/sin A = b/sin B = c/sin C = 1/sin θ, where θ = angle C. Therefore, a = sin A / sin θ, b = sin B / sin θ.But A + B = π - θ, so B = π - θ - A.Therefore, b = sin(π - θ - A)/sin θ = sin(θ + A)/sin θ.Wait, but this might not be helpful. Alternatively, maybe express a and b in terms of θ and another variable.Alternatively, use the Law of Cosines with 3c² = a² + 2b².Given that, and we also have from Law of Cosines:cos C = (a² + b² - c²)/(2ab)But we need to express sin C, which is sqrt(1 - cos² C). So if we can find cos C in terms of a and b, then perhaps express a and b in terms of each other using 3c² = a² + 2b².But since c is related to a and b via 3c² = a² + 2b², maybe we can write c² = (a² + 2b²)/3.Then, substituting into the expression for cos C:cos C = (a² + b² - c²)/(2ab) = (a² + b² - (a² + 2b²)/3)/(2ab)Let me compute that:= [ (3a² + 3b² - a² - 2b²)/3 ] / (2ab )Simplify numerator:3a² - a² = 2a²3b² - 2b² = b²So, numerator: 2a² + b²Therefore, cos C = (2a² + b²)/(3 * 2ab) = (2a² + b²)/(6ab)So,cos C = (2a² + b²)/(6ab)Our goal is to find sin C maximum. Since sin C = sqrt(1 - cos² C), so we need to minimize cos C to maximize sin C. Wait, but actually, sin C is maximized when cos C is minimized, but angle C is between 0 and π, so sin C is maximized at C = π/2, but depending on constraints, it might not reach π/2. So we need to find the minimum value of cos C given the constraints, which would correspond to the maximum sin C.But the problem is that the sides a, b, c are related by 3c² = a² + 2b², so we need to express cos C in terms of a and b with that constraint, and then optimize.Alternatively, we can use substitution. Let me denote t = a/b. Then, a = tb. Then, substituting into 3c² = a² + 2b²:3c² = t² b² + 2b² = b²(t² + 2)Therefore, c² = (b²(t² + 2))/3 => c = b sqrt( (t² + 2)/3 )Now, express cos C in terms of t:From earlier, cos C = (2a² + b²)/(6ab) = (2t² b² + b²)/(6 t b²) = (2t² + 1)/(6t)Therefore, cos C = (2t² + 1)/(6t)So now, cos C is a function of t, where t = a/b > 0 (since sides are positive). So we can consider t > 0.Our goal is to find the minimum value of cos C, which will maximize sin C. So we need to minimize (2t² + 1)/(6t) over t > 0.Let’s set f(t) = (2t² + 1)/(6t). Let's find the minimum of this function.First, simplify f(t):f(t) = (2t² + 1)/(6t) = (2t²)/(6t) + 1/(6t) = (t/3) + 1/(6t)Therefore, f(t) = (1/3)t + (1/6)t^{-1}To find the minimum, take derivative f’(t):f’(t) = (1/3) - (1/6)t^{-2}Set derivative equal to zero:(1/3) - (1/6)t^{-2} = 0Multiply both sides by 6t² to eliminate denominators:6t²*(1/3) - 6t²*(1/6)t^{-2} = 0Simplify:2t² - 1 = 0 => 2t² = 1 => t² = 1/2 => t = 1/√2Since t > 0, we take t = 1/√2.Now, check the second derivative to confirm it's a minimum.Second derivative f''(t):f''(t) = (0) + (1/3) * 0 + (1/6)*(2)t^{-3} = (1/3)t^{-3}Wait, let's recalculate:Wait, f’(t) = (1/3) - (1/6)t^{-2}Then, f''(t) = 0 + (1/3)t^{-3}Wait, derivative of - (1/6)t^{-2} is (1/6)*2 t^{-3} = (1/3) t^{-3}So f''(t) = (1/3) t^{-3}At t = 1/√2, t > 0, so f''(t) = (1/3)( (1/√2)^{-3} ) = (1/3)( (√2)^3 ) = (1/3)(2√2) = (2√2)/3 > 0Since the second derivative is positive, this critical point is indeed a minimum.Therefore, the minimal value of cos C is f(1/√2):f(1/√2) = (2*(1/2) + 1)/(6*(1/√2)) = (1 + 1)/(6/√2) = 2/(6/√2) = 2*(√2/6) = √2/3 ≈ 0.4714Therefore, the minimal cos C is √2/3, so the maximum sin C is sqrt(1 - (√2/3)^2) = sqrt(1 - 2/9) = sqrt(7/9) = √7 / 3 ≈ 0.8819Therefore, the maximum value of sin C is √7 / 3.Wait, but let me verify this calculation step by step to be sure.First, compute f(t) at t = 1/√2:t = 1/√2Numerator: 2t² + 1 = 2*(1/2) + 1 = 1 + 1 = 2Denominator: 6t = 6*(1/√2) = 6/√2 = 3√2Therefore, cos C = 2/(3√2) = √2/3 (rationalizing the denominator: 2/(3√2) = (2√2)/(3*2) = √2/3 )Yes, that's correct.Then, sin C = sqrt(1 - (√2/3)^2) = sqrt(1 - 2/9) = sqrt(7/9) = sqrt7 / 3Yes, that's correct. So the maximum value of sin C is sqrt7 / 3.But let me check if this is possible in the triangle.We need to ensure that angle C with cos C = √2/3 is valid. Since √2/3 ≈ 0.4714, which is positive, so angle C is acute. The maximum sin C occurs when angle C is as close to 90 degrees as possible, but due to the constraint given in the problem, the maximum sin C is sqrt7 / 3 ≈ 0.8819, which is less than 1, so this is valid.Alternatively, could there be a case where angle C is obtuse, leading to a higher sin C? Wait, sin C is maximum at 90 degrees, and sin(180 - x) = sin x, so sin C is maximum at 90 degrees. But here, angle C is forced to be acute with cos C = √2/3, so sin C = sqrt7 / 3.Hence, the maximum possible value is sqrt7 / 3.Therefore, the answer should be sqrt7 / 3, which can be written as boxed{dfrac{sqrt{7}}{3}}.Final AnswerThe maximum value of sin C is boxed{dfrac{sqrt{7}}{3}}.</think>Given the problem in triangle ( triangle ABC ) where the equation ( overrightarrow{AB} cdot overrightarrow{AC} + 2 overrightarrow{BA} cdot overrightarrow{BC} = 3 overrightarrow{CA} cdot overrightarrow{CB} ) holds, we need to find the maximum value of ( sin C ).First, we express the dot products in terms of the sides and angles of the triangle:- ( overrightarrow{AB} cdot overrightarrow{AC} = cb cos A )- ( overrightarrow{BA} cdot overrightarrow{BC} = ca cos B )- ( overrightarrow{CA} cdot overrightarrow{CB} = ab cos C )Substituting these into the given equation, we get:[ cb cos A + 2ca cos B = 3ab cos C ]Using the Law of Cosines, we express ( cos A ), ( cos B ), and ( cos C ) in terms of the sides ( a ), ( b ), and ( c ):- ( cos A = frac{b^2 + c^2 - a^2}{2bc} )- ( cos B = frac{a^2 + c^2 - b^2}{2ac} )- ( cos C = frac{a^2 + b^2 - c^2}{2ab} )Substituting these into the equation, we simplify to find:[ 3c^2 = a^2 + 2b^2 ]Next, we express ( cos C ) in terms of ( a ) and ( b ):[ cos C = frac{2a^2 + b^2}{6ab} ]To maximize ( sin C ), we minimize ( cos C ). Let ( t = frac{a}{b} ), then:[ cos C = frac{2t^2 + 1}{6t} ]Minimizing this expression, we find the critical point ( t = frac{1}{sqrt{2}} ). Substituting ( t = frac{1}{sqrt{2}} ) back, we get:[ cos C = frac{sqrt{2}}{3} ]Thus, the maximum value of ( sin C ) is:[ sin C = sqrt{1 - left( frac{sqrt{2}}{3} right)^2} = sqrt{frac{7}{9}} = frac{sqrt{7}}{3} ]The maximum value of ( sin C ) is boxed{dfrac{sqrt{7}}{3}}.]

🔑:begin{array}{l}overrightarrow{A B} cdot overrightarrow{A C}+2 overrightarrow{B A} cdot overrightarrow{B C}=3 overrightarrow{C A} cdot overrightarrow{C B} Rightarrow frac{b^{2}+c^{2}-a^{2}}{2}+a^{2}+c^{2}-b^{2}=frac{3left(a^{2}+b^{2}-c^{2}right)}{2} Rightarrow b^{2}+c^{2}-a^{2}+2left(a^{2}+c^{2}-b^{2}right)=3left(a^{2}+b^{2}-c^{2}right) Rightarrow a^{2}+2 b^{2}=3 c^{2} .end{array}Thus, cos C=frac{a^{2}+b^{2}-c^{2}}{2 a b}=frac{3left(a^{2}+b^{2}right)-left(a^{2}+2 b^{2}right)}{6 a b}=frac{2 a^{2}+b^{2}}{6 a b} geqslant frac{sqrt{2}}{3},equality holds when left{begin{array}{l}2 a^{2}=b^{2}, a^{2}+2 b^{2}=3 c^{2}end{array} Rightarrowleft{begin{array}{l}b=sqrt{2} a, c=frac{sqrt{15}}{3} a text {. }end{array}right.right.At this point, sin C=sqrt{1-left(frac{sqrt{2}}{3}right)^{2}}=frac{sqrt{7}}{3}, which means the maximum value of sin C is frac{sqrt{7}}{3}.

🧠:[<think>Okay, so I need to find the range of possible values for side ( c ) in an acute triangle ( triangle ABC ) where sides ( a = 1 ) and ( b = 2 ). Hmm, let's start by recalling some triangle properties. Since it's a triangle, the sides must satisfy the triangle inequality. Also, since it's acute, all angles must be less than 90 degrees. First, let's remember the triangle inequality. For any triangle, the sum of any two sides must be greater than the third side. So, applying that here, for sides ( a = 1 ), ( b = 2 ), and ( c ):1. ( a + b > c ) → ( 1 + 2 > c ) → ( 3 > c ) or ( c < 3 )2. ( a + c > b ) → ( 1 + c > 2 ) → ( c > 1 )3. ( b + c > a ) → ( 2 + c > 1 ) → ( c > -1 ) (which is always true since side lengths are positive)So, from the triangle inequality alone, ( c ) must be between ( 1 ) and ( 3 ). But the problem specifies it's an acute triangle, so the range will be narrower. Now, for a triangle to be acute, the square of each side must be less than the sum of the squares of the other two sides. That comes from the Law of Cosines: if ( c^2 < a^2 + b^2 ), then angle opposite ( c ) is acute. Since all angles need to be acute, we have three inequalities:1. ( a^2 + b^2 > c^2 )2. ( a^2 + c^2 > b^2 )3. ( b^2 + c^2 > a^2 )Let me write these down with the given values ( a = 1 ), ( b = 2 ):1. ( 1^2 + 2^2 > c^2 ) → ( 1 + 4 > c^2 ) → ( 5 > c^2 ) → ( c < sqrt{5} )2. ( 1^2 + c^2 > 2^2 ) → ( 1 + c^2 > 4 ) → ( c^2 > 3 ) → ( c > sqrt{3} )3. ( 2^2 + c^2 > 1^2 ) → ( 4 + c^2 > 1 ) → ( c^2 > -3 ) (which is always true)So, combining these acute conditions with the triangle inequalities:From triangle inequality: ( 1 < c < 3 )From acute conditions:- ( c > sqrt{3} ) (≈1.732)- ( c < sqrt{5} ) (≈2.236)Therefore, combining both, the range for ( c ) is ( sqrt{3} < c < sqrt{5} ).Wait, let me check if that's correct. Let's verify each angle.Suppose ( c ) is the largest side. Then angle opposite ( c ) would be the largest angle. For it to be acute, ( c^2 < a^2 + b^2 ). But if ( c ) isn't the largest side, then the largest side is ( b = 2 ), so angle opposite ( b ) must be acute, which requires ( b^2 < a^2 + c^2 ). Similarly, angle opposite ( a ) must be acute, but since ( a = 1 ) is the smallest side, that condition would be automatically satisfied if the other two are.Wait, maybe I need to check all three conditions:1. When is ( c ) the largest side? When ( c > 2 ). Because the given sides are ( a = 1 ), ( b = 2 ). So if ( c > 2 ), then ( c ) is the largest side, so the condition is ( c^2 < 1 + 4 = 5 ), so ( c < sqrt{5} ).If ( c leq 2 ), then the largest side is ( b = 2 ), so we need ( 2^2 < 1^2 + c^2 ), which is ( 4 < 1 + c^2 ), so ( c^2 > 3 ), ( c > sqrt{3} ).But if ( c leq 2 ), then ( c ) must be greater than ( sqrt{3} approx 1.732 ), which is possible because ( sqrt{3} approx 1.732 < 2 ).Therefore, combining these:If ( c > 2 ), then ( c < sqrt{5} approx 2.236 ). So ( 2 < c < sqrt{5} ).If ( c leq 2 ), then ( c > sqrt{3} approx 1.732 ).Therefore, overall, the range is ( sqrt{3} < c < sqrt{5} ).But wait, does this cover both cases? Let me see. Since ( sqrt{3} approx 1.732 ) and ( 2 ) is in between ( sqrt{3} ) and ( sqrt{5} approx 2.236 ), so the total range is ( sqrt{3} < c < sqrt{5} ).Yes, because when ( c ) is between ( sqrt{3} ) and 2, the largest side is ( b = 2 ), and we need ( 2^2 < 1^2 + c^2 ).When ( c ) is between 2 and ( sqrt{5} ), the largest side is ( c ), and we need ( c^2 < 1 + 4 ).Therefore, combining these two intervals, the total interval for ( c ) is ( sqrt{3} < c < sqrt{5} ).So, the answer should be ( sqrt{3} < c < sqrt{5} ).But let me double-check. Suppose ( c = 2 ). Then the triangle has sides 1, 2, 2. Let's check if it's acute.Compute the angles:Using Law of Cosines:For angle opposite ( a = 1 ):( cos A = (b^2 + c^2 - a^2)/(2bc) = (4 + 4 - 1)/ (2*2*2) = 7/8 ≈ 0.875. So angle A is acute.For angle opposite ( b = 2 ):( cos B = (a^2 + c^2 - b^2)/(2ac) = (1 + 4 - 4)/(2*1*2) = 1/4 = 0.25. So angle B is acute.For angle opposite ( c = 2 ):Same as angle B, since sides are 1, 2, 2. So angle opposite the other 2 is also angle C, which is the same as angle B. So that's also acute. So triangle with sides 1,2,2 is acute. Thus, c=2 is allowed. So the upper limit is sqrt(5) ≈ 2.236, but when c=2, which is less than sqrt(5), it is allowed. Wait, but in the previous reasoning, when c is between sqrt(3) and sqrt(5), which includes c=2, so that's okay.Another test: suppose c=1.8, which is between sqrt(3)≈1.732 and 2. Check if it's acute.Check if b=2 is the largest side, so need 2² < 1² + c² → 4 < 1 + 3.24 → 4 < 4.24, which is true. So angle opposite b is acute. Then check other angles. For angle opposite c=1.8:( cos C = (a² + b² - c²)/(2ab) = (1 + 4 - 3.24)/(2*1*2) = (1.76)/4 = 0.44. So angle C is acute. And angle A is:( cos A = (b² + c² - a²)/(2bc) = (4 + 3.24 -1)/(2*2*1.8) = (6.24)/(7.2) ≈0.866, which is acute. So all angles acute. So c=1.8 is okay.Now, check c= sqrt(3) ≈1.732. Then:If c=sqrt(3), then the triangle sides are 1, 2, sqrt(3). Check if it's acute.The largest side is 2. So check if 2² < 1² + (sqrt(3))² → 4 < 1 + 3 → 4 < 4, which is not true. So equality holds, which is the case for a right triangle. Therefore, when c=sqrt(3), the triangle is right-angled, not acute. Therefore, c must be greater than sqrt(3).Similarly, if c= sqrt(5) ≈2.236, check the triangle. Then the sides are 1, 2, sqrt(5). Then the largest side is sqrt(5). Check if sqrt(5)^2 < 1² + 2² → 5 < 1 +4 → 5 <5, which is again equality, so right-angled. Hence, c must be less than sqrt(5). Therefore, the interval is open: (sqrt(3), sqrt(5)).Therefore, the range of c is sqrt(3) < c < sqrt(5).Hence, the answer is ( sqrt{3} < c < sqrt{5} ).Final AnswerThe range of values for side ( c ) is boxed{sqrt{3} < c < sqrt{5}}.</think>To determine the range of values for side ( c ) in an acute triangle ( triangle ABC ) with ( a = 1 ) and ( b = 2 ), we need to consider both the triangle inequality and the conditions for an acute triangle.First, applying the triangle inequality:1. ( a + b > c ) implies ( 1 + 2 > c ) or ( c < 3 )2. ( a + c > b ) implies ( 1 + c > 2 ) or ( c > 1 )3. ( b + c > a ) is always true since ( c > 0 )Thus, the triangle inequality gives ( 1 < c < 3 ).Next, considering the conditions for an acute triangle (using the Law of Cosines):1. ( a^2 + b^2 > c^2 ) implies ( 1 + 4 > c^2 ) or ( c^2 < 5 ) hence ( c < sqrt{5} )2. ( a^2 + c^2 > b^2 ) implies ( 1 + c^2 > 4 ) or ( c^2 > 3 ) hence ( c > sqrt{3} )3. ( b^2 + c^2 > a^2 ) is always true since ( c > 0 )Combining these conditions with the triangle inequality, we get:- ( sqrt{3} < c < sqrt{5} )Verification for specific values (e.g., ( c = 2 ), ( c = 1.8 )) confirms that the triangle remains acute within this range. The endpoints ( c = sqrt{3} ) and ( c = sqrt{5} ) correspond to right triangles, hence excluded.Thus, the range of values for side ( c ) is (boxed{sqrt{3} < c < sqrt{5}}).,<think>Okay, so I need to find the range of possible values for side c in an acute triangle ABC where sides a=1 and b=2. Hmm, let's think. First, I remember that in any triangle, the sum of any two sides must be greater than the third side. That's the triangle inequality theorem. So, applying that here, the sides must satisfy:1. a + b > c2. a + c > b3. b + c > aGiven that a=1 and b=2, substituting these into the inequalities:1. 1 + 2 > c ⇒ 3 > c ⇒ c < 32. 1 + c > 2 ⇒ c > 13. 2 + c > 1 ⇒ c > -1 (which is redundant since side lengths are positive)So from the triangle inequality alone, c must be between 1 and 3. But the problem specifies that the triangle is acute. So, I need to apply the conditions for an acute triangle. An acute triangle has all angles less than 90 degrees. From the converse of the Pythagorean theorem, for a triangle with sides a, b, c (where c is the longest side), the triangle is acute if a² + b² > c². So, we need to ensure that for all angles, the squares of the sides opposite are less than the sum of the squares of the other two sides.But first, let me recall that in any triangle, the longest side is opposite the largest angle. Since we want all angles to be acute, we need to make sure that even if c is the longest side, it doesn't make the angle opposite to it obtuse, and similarly if one of the other sides is the longest. Wait, but a=1, b=2. So, the sides are a=1, b=2, c. Depending on the value of c, the longest side could be either b=2 or c. If c is greater than 2, then c becomes the longest side. If c is less than or equal to 2, then b=2 remains the longest side.So, to ensure the triangle is acute, we have two cases to consider:1. When c is the longest side (c > 2). Then, the angle opposite c (which is angle C) must be acute. So, we need a² + b² > c².2. When b is the longest side (c ≤ 2). Then, the angle opposite b (which is angle B) must be acute. So, we need a² + c² > b².Additionally, even if a=1 is not the longest side, we need to ensure that the angle opposite a is also acute. But since a is the shortest side (since a=1, b=2, and c must be greater than 1), the angle opposite a (angle A) will be the smallest angle, which is automatically acute if the other two angles are acute. But maybe I need to check that as well? Wait, no. If the triangle is acute, all angles must be less than 90 degrees. However, if we ensure that the largest two angles are acute, then the smallest angle will naturally be acute as well. Because in a triangle, the sum of angles is 180. If the two larger angles are less than 90 each, then the third angle must be less than 90 as well. For example, if two angles are 80 each, the third is 20. So, perhaps I just need to check the conditions for the two potentially largest angles, which correspond to the two longest sides.Therefore, the conditions we need are:- If c is the longest side (c > 2), then a² + b² > c².- If b is the longest side (c ≤ 2), then a² + c² > b².So let's compute these inequalities.First, if c > 2, then the condition is 1² + 2² > c² ⇒ 1 + 4 > c² ⇒ 5 > c² ⇒ c < √5 ≈ 2.236. But since we assumed c > 2 here, combining these gives 2 < c < √5.Second, if c ≤ 2, then the longest side is b=2, so the condition is a² + c² > b² ⇒ 1 + c² > 4 ⇒ c² > 3 ⇒ c > √3 ≈ 1.732. But since c ≤ 2 in this case, combining gives √3 < c ≤ 2.Therefore, combining both cases, the range of c is √3 < c < √5.But we also need to remember the triangle inequality, which gives c >1 and c <3. However, √3 is approximately 1.732, which is greater than 1, so the lower bound from the acute condition is more restrictive. Similarly, √5 is approximately 2.236, which is less than 3, so the upper bound from the acute condition is more restrictive.Therefore, combining the triangle inequality and the acute angle conditions, c must satisfy √3 < c < √5.But wait, let me verify this. Let's check each condition:1. Triangle inequality: 1 < c < 3.2. Acute conditions: - If c >2: c < √5 ≈2.236 - If c ≤2: c > √3 ≈1.732Therefore, the combined range is √3 < c < √5.But let's double-check. Suppose c is exactly √3. Then, the triangle would have sides 1, 2, √3. Let's check if it's acute. The largest side is 2. Then, we check 1² + (√3)^2 = 1 +3=4, which is equal to 2². So, when c=√3, the triangle is right-angled, not acute. Similarly, if c=√5, then sides are 1,2,√5. The largest side is √5. Then, 1² + 2² =1+4=5, which is equal to (√5)^2=5. Again, a right-angled triangle. Therefore, to have an acute triangle, c must be strictly less than √5 and strictly greater than √3.So, the range is √3 < c < √5.But let me also check a value in between. Let's take c=2, which is within √3 (~1.732) and √5 (~2.236). Wait, c=2 is the upper bound of the second case. When c=2, the sides are 1,2,2. The largest side is 2. Check if the triangle is acute. So, check 1² + 2² > 2². That would be 1 +4 >4 ⇒5>4, which is true. So, c=2 is allowed. Wait, but earlier when c=2, which is equal to the upper bound in the second case, but since the condition was c >√3 (≈1.732) and c ≤2, so at c=2, the triangle is still acute? Let me verify.For c=2, the triangle with sides 1,2,2. The angles opposite the sides of length 2 will be equal. Let's compute the angles. Using the Law of Cosines for angle B (opposite side b=2):Wait, sides are a=1, b=2, c=2. Wait, actually, side a=1 is opposite angle A, side b=2 is opposite angle B, and side c=2 is opposite angle C. Wait, but sides b and c are both 2, so angles B and C are equal. Let's check angle B (opposite side b=2). Wait, no, wait: in a triangle, the sides opposite the angles are labeled accordingly. So, if the sides are a=1, b=2, c=2, then sides b and c are both 2, so angles B and C are equal. Let's compute angle A (opposite side a=1):Using the Law of Cosines:cos A = (b² + c² - a²)/(2bc) = (4 +4 -1)/(2*2*2) = (7)/8 ≈0.875So angle A is arccos(7/8) ≈29 degrees, which is acute.Angles B and C are equal. Let's compute angle B:cos B = (a² + c² - b²)/(2ac) = (1 +4 -4)/(2*1*2) = (1)/4 =0.25So angle B = arccos(0.25) ≈75.5 degrees, which is acute.Therefore, when c=2, all angles are acute. So, c=2 is allowed. Similarly, at c=√3, the triangle is right-angled. So, the lower bound is c>√3.Similarly, when c approaches √5 from below, say c=2.2, then check if the triangle is acute. Let's see:If c=2.2, then the largest side is c. Check if a² + b² > c²:1 +4 =5, c²=4.84. 5 >4.84, so yes, angle opposite c is acute. So, it's an acute triangle. Similarly, check the other angles. For example, angle opposite b=2. Using Law of Cosines:cos B = (a² + c² - b²)/(2ac) = (1 +4.84 -4)/(2*1*2.2) = (1.84)/(4.4) ≈0.418. So angle B ≈65.3 degrees, acute. So, yes, c=2.2 is okay.But when c=√5 (~2.236), then a² + b² =5, which equals c², making it right-angled. So, c must be less than √5.Similarly, when c=√3 (~1.732), the triangle is right-angled, so c must be greater than √3.Thus, combining these, the range of c is √3 < c < √5.But let me check another point. Take c=1.5, which is between 1 and √3 (~1.732). So, c=1.5. Then, the sides are 1, 2, 1.5. The largest side is 2. Check if the triangle is acute. Use the condition a² + c² > b²:1² + 1.5² =1 +2.25=3.25. b²=4. 3.25 <4, so the angle opposite b=2 is obtuse. Therefore, the triangle is obtuse, which is not allowed. Hence, c=1.5 is invalid. So, we do need c>√3.Similarly, take c=2.5, which is between √5 (~2.236) and 3. Then, the sides are 1,2,2.5. The largest side is 2.5. Check a² + b² =1 +4=5. Compare to c²=6.25. 5 <6.25, so angle opposite c is obtuse. Therefore, c=2.5 is invalid, hence c must be less than √5.Therefore, indeed, the range of c is √3 < c < √5.So, the final answer is that c must satisfy √3 < c < √5, which can be written using inequalities as:√3 < c < √5Or, in LaTeX:boxed{sqrt{3} < c < sqrt{5}}Final AnswerThe range of values for side ( c ) is boxed{sqrt{3} < c < sqrt{5}}.</think>To determine the range of values for side ( c ) in an acute triangle (triangle ABC) with ( a = 1 ) and ( b = 2 ), we need to consider both the triangle inequality theorem and the conditions for an acute triangle.1. Triangle Inequality Theorem: - ( a + b > c ) implies ( 1 + 2 > c ) or ( c < 3 ). - ( a + c > b ) implies ( 1 + c > 2 ) or ( c > 1 ). - ( b + c > a ) implies ( 2 + c > 1 ), which is always true since ( c > 0 ). Combining these, the range for ( c ) from the triangle inequality is ( 1 < c < 3 ).2. Acute Triangle Conditions: - For a triangle to be acute, the square of each side must be less than the sum of the squares of the other two sides. - If ( c ) is the longest side (( c > 2 )), then ( a^2 + b^2 > c^2 ): [ 1^2 + 2^2 > c^2 implies 5 > c^2 implies c < sqrt{5} approx 2.236 ] Combining with ( c > 2 ), we get ( 2 < c < sqrt{5} ). - If ( b ) is the longest side (( c leq 2 )), then ( a^2 + c^2 > b^2 ): [ 1^2 + c^2 > 2^2 implies 1 + c^2 > 4 implies c^2 > 3 implies c > sqrt{3} approx 1.732 ] Combining with ( c leq 2 ), we get ( sqrt{3} < c leq 2 ).Combining both cases and considering the triangle inequality, the range of ( c ) for an acute triangle is:[sqrt{3} < c < sqrt{5}]Thus, the range of values for side ( c ) is (boxed{sqrt{3} < c < sqrt{5}}).]

🔑:Since in an acute triangle triangle ABC, a=1, b=2,If c is the largest side, then 0 0,which implies c 0, which meansfrac{a^2 + c^2 - b^2}{2ac} > 0,thus c > sqrt{3}.Therefore, the answer is:boxed{sqrt{3} < c < sqrt{5}}.

🧠:[<think>Okay, so I need to convert -1125° into the form k·360° + α where α is between 0° and 360°, and k is an integer. The options are A, B, C, D. Let me think about how to approach this.First, I remember that angles can be converted to their equivalent within a 0° to 360° range by adding or subtracting multiples of 360°. Since the given angle is negative, I need to add multiples of 360° until the result is positive and less than 360°. Alternatively, I can divide the angle by 360° and find the remainder.Wait, let me recall the exact method. For any angle θ, the angle α in the range [0°, 360°) is given by θ mod 360°, which is the remainder when θ is divided by 360°. Since θ here is -1125°, I need to compute -1125° mod 360°.But maybe it's easier to think of how many times 360° goes into 1125°, and then adjust for the negative sign. Let me see. 360° times 3 is 1080°, which is 3·360°=1080°. Then 1125° - 1080°=45°. So 1125° is 3·360° +45°, right? But since it's -1125°, that would be -3·360° -45°. But -45° is not between 0° and 360°, so I need to adjust this.Alternatively, maybe I should add 360° repeatedly to -1125° until the angle is positive. Let's try that.First, add 360° to -1125°: -1125° + 360° = -765°. Still negative. Add another 360°: -765° + 360° = -405°. Still negative. Add another 360°: -405° + 360° = -45°. Still negative. Add another 360°: -45° + 360° = 315°. Now that's positive and less than 360°. So α is 315°, and how many times did I add 360°? Four times. So that means -1125° + 4·360° = 315°, so rearranged, -1125° = -4·360° + 315°. So that would be option C: -4×360° +315°.But let me verify this. Let's check option C: -4×360° +315°. Calculating that: -4×360 is -1440°, plus 315° is -1440 +315 = -1125°, which matches. So that seems correct.But let me check the other options too to make sure.Option A: -3×360° -315°. Let's compute that: -1080° -315° = -1395°, which is not -1125°. So A is wrong.Option B: -9×180° -45°. That's -1620° -45° = -1665°, which is not -1125°, so B is wrong.Option D: -3×360° +45°. That's -1080° +45° = -1035°, not -1125°, so D is wrong.Therefore, the correct answer must be C. But let me think again if there's another way.Alternatively, using division. Let's compute -1125 divided by 360. So, -1125 / 360. Let's do that. 360 goes into 1125 three times (3×360=1080), so 3 with a remainder of 45. But since it's -1125, the division would be -1125 = (-4)×360 + ( -1125 - (-4)×360 ). Let's compute:(-4)×360 = -1440. Then, -1125 - (-1440) = -1125 +1440 = 315. So yes, -1125 = -4×360 +315°, which is option C. So that confirms the answer is C.But wait, another way to think about modulus operations. The formula for angle conversion is α = θ + 360°×k, where k is an integer such that α is in [0°, 360°). So solving for k: k = floor(θ / 360°). But θ is negative here. Let's see.θ = -1125°. θ / 360° = -1125 / 360 ≈ -3.125. So floor(-3.125) is -4. Therefore, k = -4. Then α = θ - k×360° = -1125° - (-4)×360° = -1125° +1440°= 315°, which is correct. So indeed, θ = k×360° + α = -4×360° +315°, so option C.I think that's thorough. So the answer is C.Final Answerboxed{C}</think>To convert -1125° into the form ( k cdot 360° + alpha ) where ( 0° leq alpha < 360° ) and ( k in mathbb{Z} ):1. Add multiples of 360° to -1125° until the result is within the range [0°, 360°): - -1125° + 360° = -765° (still negative) - -765° + 360° = -405° (still negative) - -405° + 360° = -45° (still negative) - -45° + 360° = 315° (positive and within the range)2. Determine the number of times 360° was added: - 360° was added 4 times, so ( k = -4 ).3. Verify the calculation: - -1125° = -4 cdot 360° + 315° - Calculate: -4 cdot 360° = -1440°, then -1440° + 315° = -1125°4. Check other options: - Options A, B, and D do not yield -1125° when calculated.Thus, the correct answer is (boxed{C}).,<think>Okay, so I have this problem here where I need to convert -1125° into the form of k·360° + α, where α is between 0° and 360°, and k is an integer. The options are A, B, C, D. Let me think step by step how to approach this.First, I remember that angles can be converted to their equivalent within a specific range by adding or subtracting multiples of 360°. Since the angle given here is negative, -1125°, I need to add multiples of 360° until the result is between 0° and 360°. Alternatively, maybe subtract, but since it's negative, adding positive multiples might make it less negative or even positive.Let me start by figuring out how many times 360° goes into 1125°, because if I can find how many full rotations are in 1125°, that might help me break it down. But since it's negative, maybe I should approach it differently. Let me think.If the angle is -1125°, I need to represent this as k·360° + α. So, k has to be an integer, and α must be in [0°, 360°). Essentially, we want to find an angle coterminal with -1125° that's within one full rotation (0° to 360°). To do this, we can add 360° repeatedly until we get into the desired range.But since -1125° is a large negative angle, adding 360° multiple times might be tedious. Maybe there's a formula or a division method to find k and α.Let me recall that to find α, we can compute -1125° modulo 360°. The modulo operation here will give us the equivalent positive angle. But how does modulo work with negative numbers?In many programming languages, the modulo operation can give a negative result if the dividend is negative, but in mathematics, modulo should return a non-negative result. So, perhaps I need to compute:α = (-1125°) mod 360°To compute this, we can find how many times 360° divides into -1125°, take the floor of that quotient, multiply by 360°, and subtract from -1125° to get the remainder.Alternatively, maybe just divide -1125 by 360 and find the remainder.Wait, let me try:First, divide 1125 by 360 to see how many times 360 goes into 1125.1125 ÷ 360. Let's see, 360*3 = 1080, which is 1125 - 1080 = 45 less. So 1125 = 3*360 + 45. Therefore, 1125° is 3 full rotations plus 45°, meaning that -1125° is -3*360° -45°. But that might not be in the form they want, since α has to be between 0° and 360°. Hmm.Wait, but -1125° can be written as k·360° + α. If we take k as a negative integer, then adding α (which is positive) would get us to the negative angle. Wait, maybe I need to adjust.Alternatively, let's think of it as starting at -1125°, and adding 360° until we reach an angle in 0°-360°.So, how many times do we need to add 360° to -1125° to get into 0° to 360°? Let's calculate:-1125 + 360 = -765° (still negative)-765 + 360 = -405° (still negative)-405 + 360 = -45° (still negative)-45 + 360 = 315° (positive, within 0°-360°)So, we added 360° four times. Therefore, -1125° + 4*360° = 315°. Therefore, rearranged, -1125° = -4*360° + 315°. So that would mean k = -4 and α = 315°, which is option C. Hmm, let me check the options.Option C is -4×360° + 315°, yes. So that's one way. But let me verify.Alternatively, using division:Compute -1125 ÷ 360. Let's see:-1125 / 360 = -3.125. So the integer division would give -4 (since we take the floor of -3.125, which is -4). Then, the remainder is -1125 - (-4)*360 = -1125 + 1440 = 315°. So that gives α = 315°, k = -4, so -4×360 + 315°, which is option C. So that seems to check out.But let me check the options again.A: -3×360° -315°: This would be -1080 -315 = -1395°, which is not -1125°.B: -9×180° -45°: This is -1620 -45 = -1665°, not -1125°.C: -4×360° +315°: This is -1440 +315 = -1125°, which matches.D: -3×360° +45°: This is -1080 +45 = -1035°, not -1125°.So, option C is the correct one. But let me just make sure I didn't make any mistake in the calculation.Alternatively, another way to think about it is:Since -1125° is the angle, adding 360° four times gives:-1125 + 360*4 = -1125 + 1440 = 315°, which is correct. So, to express -1125° as k*360 + α, we have k = -4 and α = 315°, which is exactly option C.Therefore, the answer should be C.But just to double-check, maybe there's another way someone could approach this and get confused.For example, if someone tried to factor out 180°, which is what option B does. Let's see, option B is -9×180° -45°. Let's compute that: -9×180 is -1620, minus 45 is -1665°, which is way more negative than -1125°, so that's wrong.Option A: -3×360 -315. -3×360 is -1080, minus 315 is -1395, which is not -1125.Option D: -3×360 +45 is -1080 +45 = -1035, still not -1125.So, only option C gives -1125°.Therefore, the correct answer is C.Final Answerboxed{C}</think>To convert -1125° into the form ( k cdot 360° + alpha ) where ( 0° leq alpha < 360° ) and ( k in mathbb{Z} ):1. Determine the number of full rotations (360°) needed to bring -1125° into the range [0°, 360°): - Start with -1125° and add 360° repeatedly until the result is within the desired range. - Calculations: - -1125° + 360° = -765° (still negative) - -765° + 360° = -405° (still negative) - -405° + 360° = -45° (still negative) - -45° + 360° = 315° (positive, within range) - This required adding 360° four times: -1125° + 4 times 360° = 315°.2. Express -1125° in the form ( k cdot 360° + alpha ): - Rearranging the equation from above: -1125° = -4 times 360° + 315°. - Here, ( k = -4 ) and ( alpha = 315° ).3. Verify the options: - Option C: (-4 times 360° + 315°) matches the calculated values.Thus, the correct answer is (boxed{C}).]

🔑:Solution: -1125° = -1440° + 315° = -4 times 360° + 315°, Therefore, the correct option is boxed{C}. Given that the angle is negative, we use a negative multiple of 360° whose absolute value is greater than the absolute value of the given negative angle, and then add a positive angle within a full rotation to get the result. This problem demonstrates the concept of angles with the same terminal side. The key to solving this problem is to outline the general range of the angle and then determine how much needs to be added to balance the change in the angle. This is a basic problem.