answer：Okay, so I need to find the greatest common divisor (GCD) of 24 and 36. Hmm, I remember that the GCD is the largest number that divides both numbers without leaving a remainder. But I'm not entirely sure how to approach this systematically. Maybe I can start by listing the factors of each number and see which ones they have in common.Let me think about the factors of 24 first. Factors are numbers that multiply together to give the original number. So, starting from 1, I can divide 24 by each number to see if it divides evenly.1 times 24 is 24, so 1 and 24 are factors.2 times 12 is 24, so 2 and 12 are factors.3 times 8 is 24, so 3 and 8 are factors.4 times 6 is 24, so 4 and 6 are factors.5 doesn't divide evenly into 24 because 5 times 4.8 is 24, which isn't a whole number.6 times 4 is 24, which I've already listed.7 doesn't divide evenly either.8 times 3 is 24, already listed.9 doesn't divide evenly.10 doesn't divide evenly.11 doesn't divide evenly.12 times 2 is 24, already listed.13 and above won't divide into 24 evenly.So, the factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24.Now, let's do the same for 36.1 times 36 is 36, so 1 and 36 are factors.2 times 18 is 36, so 2 and 18 are factors.3 times 12 is 36, so 3 and 12 are factors.4 times 9 is 36, so 4 and 9 are factors.5 doesn't divide evenly into 36 because 5 times 7.2 is 36, which isn't a whole number.6 times 6 is 36, so 6 is a factor.7 doesn't divide evenly.8 doesn't divide evenly.9 times 4 is 36, already listed.10 doesn't divide evenly.11 doesn't divide evenly.12 times 3 is 36, already listed.13 and above won't divide into 36 evenly.So, the factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36.Now, I need to find the common factors between 24 and 36. Let's list them out:From 24: 1, 2, 3, 4, 6, 8, 12, 24.From 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.Comparing these, the common factors are: 1, 2, 3, 4, 6, 12.Out of these, the greatest one is 12. So, the greatest common divisor of 24 and 36 should be 12.Wait, let me double-check to make sure I didn't miss anything. Maybe I can use another method, like the Euclidean algorithm, to confirm.The Euclidean algorithm involves dividing the larger number by the smaller one and then replacing the larger number with the smaller one and the smaller number with the remainder, repeating this until the remainder is zero. The last non-zero remainder is the GCD.So, starting with 36 and 24:36 divided by 24 is 1 with a remainder of 12 (since 24 times 1 is 24, and 36 minus 24 is 12).Now, take 24 and divide by the remainder 12:24 divided by 12 is 2 with a remainder of 0.Since the remainder is now 0, the last non-zero remainder is 12, which confirms that the GCD is indeed 12.Okay, that matches what I found earlier by listing the factors. So, I'm confident that the greatest common divisor of 24 and 36 is 12.

answer：Alright, so I have this problem about an arithmetic sequence. It says that in the arithmetic sequence {a_n}, if a_8 equals 8, then what is the value of S_{15}? The options are A: 120, B: 60, C: 15, D: 30.Okay, let me recall what an arithmetic sequence is. It's a sequence where each term increases by a constant difference. So, the nth term can be expressed as a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference.They gave me that a_8 = 8. Let me write that down: a_8 = a_1 + 7d = 8. So, that's one equation involving a_1 and d.Now, they're asking for S_{15}, which is the sum of the first 15 terms of this arithmetic sequence. The formula for the sum of the first n terms of an arithmetic sequence is S_n = n/2 * (a_1 + a_n). So, for S_{15}, it would be S_{15} = 15/2 * (a_1 + a_{15}).Hmm, I need to find a_1 and a_{15} to compute this sum. But I only have one equation: a_1 + 7d = 8. That's not enough to find both a_1 and d. Maybe there's another way to express a_{15} in terms of a_8?Let me think. Since a_{15} is the 15th term, it can be written as a_1 + 14d. Similarly, a_8 is a_1 + 7d. So, a_{15} is a_8 plus 7d more. So, a_{15} = a_8 + 7d. But I don't know what d is.Wait, maybe I can express a_1 in terms of a_8. From a_8 = a_1 + 7d, I can solve for a_1: a_1 = a_8 - 7d. So, a_1 = 8 - 7d.Then, a_{15} = a_1 + 14d = (8 - 7d) + 14d = 8 + 7d.So, a_1 + a_{15} = (8 - 7d) + (8 + 7d) = 16. Oh, interesting! The d terms cancel out. So, a_1 + a_{15} = 16.Therefore, S_{15} = 15/2 * 16 = 15 * 8 = 120.Wait, that seems straightforward. So, the sum is 120, which is option A.But let me double-check to make sure I didn't make a mistake. Maybe I can think of it another way. In an arithmetic sequence, the sum of the first n terms can also be expressed as S_n = n * average of the first and nth term. So, S_{15} = 15 * (a_1 + a_{15}) / 2, which is the same as what I did before.Alternatively, since a_8 is the middle term when n is 15, because 15 is odd, the 8th term is the middle one. So, the average of the first and 15th term is equal to a_8. Therefore, S_{15} = 15 * a_8 = 15 * 8 = 120.Yeah, that makes sense too. So, regardless of the common difference d, the sum S_{15} is 15 times the middle term, which is a_8. So, 15 * 8 is 120.I think that's solid. I don't see any mistakes in my reasoning. So, the answer should be A: 120.**Final Answer**boxed{A}

answer：Okay, so I have this equation here: 5x + 3 equals bx minus 22. And I'm told that x is 5. I need to find out what b is. Alright, let's break this down step by step.First, I'll write down the equation again to make sure I have it right: 5x + 3 = bx - 22. And x is 5. So, I guess I need to substitute x with 5 in the equation. That should help me solve for b.Let me do that substitution. So, replacing x with 5, the equation becomes: 5 times 5 plus 3 equals b times 5 minus 22. Let me calculate 5 times 5 first. That's 25. Then, adding 3 to that gives me 28. So, the left side of the equation simplifies to 28.Now, the right side of the equation is b times 5 minus 22. So, that's 5b minus 22. So, putting it all together, I have 28 equals 5b minus 22.Hmm, I need to solve for b. To do that, I think I should get all the constants on one side and the terms with b on the other. So, maybe I can add 22 to both sides to get rid of the minus 22 on the right side.Adding 22 to both sides: 28 plus 22 equals 5b minus 22 plus 22. That simplifies to 50 equals 5b. Okay, that looks better.Now, I have 50 equals 5b. To find b, I need to isolate it. So, I'll divide both sides by 5. That should give me b by itself.Dividing both sides by 5: 50 divided by 5 equals 5b divided by 5. That simplifies to 10 equals b. So, b is 10.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting with the substitution: 5 times 5 is 25, plus 3 is 28. On the right side, 5b minus 22. Then, adding 22 to both sides: 28 plus 22 is 50, and 5b minus 22 plus 22 is just 5b. Dividing both sides by 5 gives me 10 equals b. Yeah, that seems correct.I think I've got it right. So, b is 10.

answer：First, I notice that the equation ( a^2 + frac{a^2}{4} = 5 ) has two terms with ( a^2 ). To simplify, I'll combine these like terms.Combining ( a^2 ) and ( frac{a^2}{4} ) gives me ( frac{5a^2}{4} ). This simplifies the equation to ( frac{5a^2}{4} = 5 ).Next, I'll eliminate the fraction by multiplying both sides of the equation by 4, resulting in ( 5a^2 = 20 ).Now, I'll isolate ( a^2 ) by dividing both sides by 5, which gives ( a^2 = 4 ).Finally, to find the value of ( a ), I'll take the square root of both sides. This yields two possible solutions: ( a = 2 ) and ( a = -2 ).

answer：Okay, so I have this triangle problem to solve. It says in triangle ABC, the sides opposite angles A, B, and C are a, b, and c respectively. There's an equation given: (√3 / 3) * b * sin C + c * cos B = a. Part (1) asks me to find the area of triangle ABC when a = 2 and b = 1. Hmm, okay. Let me think about how to approach this.First, I remember that in any triangle, the Law of Sines relates the sides and angles: a / sin A = b / sin B = c / sin C. Maybe that can help here. Also, the Law of Cosines might come into play: c² = a² + b² - 2ab cos C. But I'm not sure yet.Looking at the given equation: (√3 / 3) * b * sin C + c * cos B = a. Since a, b, c are sides opposite angles A, B, C, maybe I can express everything in terms of angles using the Law of Sines.Let me rewrite the equation using the Law of Sines. Since a = 2, b = 1, and c is unknown. Let me denote the angles as A, B, C.From the Law of Sines: a / sin A = b / sin B = c / sin C. So, 2 / sin A = 1 / sin B = c / sin C. Let me denote this common ratio as k. So, 2 = k sin A, 1 = k sin B, c = k sin C.So, sin A = 2 / k, sin B = 1 / k, sin C = c / k.But maybe instead of introducing k, I can express sin C and cos B in terms of the other sides.Wait, let's see. The given equation is (√3 / 3) * b * sin C + c * cos B = a. Substituting b = 1 and a = 2, we get:(√3 / 3) * 1 * sin C + c * cos B = 2.So, (√3 / 3) sin C + c cos B = 2.Hmm, okay. Maybe I can relate sin C and cos B using the Law of Sines or Cosines.I know that in a triangle, angles sum to π, so C = π - A - B. Maybe that can help.Alternatively, using the Law of Sines, sin C = c / (2R), where R is the circumradius. But I don't know R yet.Wait, maybe I can express sin C in terms of angle B. Since C = π - A - B, and A can be expressed in terms of B using the Law of Sines.From the Law of Sines: a / sin A = b / sin B, so 2 / sin A = 1 / sin B. Therefore, sin A = 2 sin B.So, A = arcsin(2 sin B). Hmm, but arcsin has restrictions. Since sin A must be ≤ 1, 2 sin B ≤ 1, so sin B ≤ 1/2. Therefore, angle B ≤ π/6 or ≥ 5π/6. But since in a triangle, angles are less than π, so B ≤ π/6.So, angle B is at most π/6. That might be useful later.Now, going back to the equation: (√3 / 3) sin C + c cos B = 2.I need to express sin C and cos B in terms of B or something else.Since C = π - A - B, and A = arcsin(2 sin B), maybe I can express sin C as sin(π - A - B) = sin(A + B). Because sin(π - x) = sin x.So, sin C = sin(A + B). Let's compute that:sin(A + B) = sin A cos B + cos A sin B.We know sin A = 2 sin B, so let's substitute that:sin C = (2 sin B) cos B + cos A sin B.Hmm, okay. Now, cos A can be found using sin² A + cos² A = 1. Since sin A = 2 sin B, then cos A = sqrt(1 - (2 sin B)^2) = sqrt(1 - 4 sin² B).So, sin C = 2 sin B cos B + sqrt(1 - 4 sin² B) * sin B.That seems complicated, but maybe manageable.Now, let's substitute sin C back into the equation:(√3 / 3) sin C + c cos B = 2.But c is another side. From the Law of Sines, c = (sin C) * (a / sin A) = sin C * (2 / sin A). Since sin A = 2 sin B, c = sin C * (2 / (2 sin B)) = sin C / sin B.So, c = sin C / sin B.Therefore, the equation becomes:(√3 / 3) sin C + (sin C / sin B) cos B = 2.Let me factor sin C:sin C [ (√3 / 3) + (cos B / sin B) ] = 2.Simplify the term in brackets:(√3 / 3) + (cos B / sin B) = (√3 / 3) + cot B.So, sin C [ (√3 / 3) + cot B ] = 2.But earlier, we had sin C = sin(A + B) = 2 sin B cos B + sqrt(1 - 4 sin² B) sin B.This is getting quite involved. Maybe there's a better approach.Wait, let's think about the given equation again: (√3 / 3) b sin C + c cos B = a.We can write this as:(√3 / 3) * 1 * sin C + c cos B = 2.So, (√3 / 3) sin C + c cos B = 2.But from the Law of Sines, c = (sin C / sin B) * b = (sin C / sin B) * 1 = sin C / sin B.So, c = sin C / sin B.Therefore, substitute c into the equation:(√3 / 3) sin C + (sin C / sin B) cos B = 2.Factor sin C:sin C [ (√3 / 3) + (cos B / sin B) ] = 2.Which is the same as before.So, sin C [ (√3 / 3) + cot B ] = 2.Hmm, maybe I can express cot B in terms of angle B.Alternatively, let's consider using the Law of Cosines on angle B.Law of Cosines: cos B = (a² + c² - b²) / (2ac).But we don't know c yet. Hmm.Wait, maybe we can express everything in terms of angle B.Given that sin A = 2 sin B, and A + B + C = π.Also, from the Law of Sines, c = sin C / sin B.So, perhaps we can write everything in terms of angle B.Let me denote angle B as θ for simplicity. So, θ = B.Then, sin A = 2 sin θ.Since A = arcsin(2 sin θ), which implies that angle A is determined by θ.Also, angle C = π - A - θ.So, sin C = sin(π - A - θ) = sin(A + θ).Which is sin A cos θ + cos A sin θ.As before, sin A = 2 sin θ, cos A = sqrt(1 - 4 sin² θ).So, sin C = 2 sin θ cos θ + sqrt(1 - 4 sin² θ) sin θ.So, sin C = 2 sin θ cos θ + sin θ sqrt(1 - 4 sin² θ).Therefore, plugging back into the equation:sin C [ (√3 / 3) + cot θ ] = 2.So, [2 sin θ cos θ + sin θ sqrt(1 - 4 sin² θ)] [ (√3 / 3) + (cos θ / sin θ) ] = 2.Simplify the term in the second bracket:(√3 / 3) + (cos θ / sin θ) = (√3 / 3) + cot θ.So, the equation becomes:[2 sin θ cos θ + sin θ sqrt(1 - 4 sin² θ)] [ (√3 / 3) + cot θ ] = 2.This is quite complicated. Maybe I can make a substitution.Let me set t = sin θ. Then, cos θ = sqrt(1 - t²).So, sin C becomes:2 t sqrt(1 - t²) + t sqrt(1 - 4 t²).And the term [ (√3 / 3) + cot θ ] becomes:(√3 / 3) + (sqrt(1 - t²) / t).So, the equation is:[2 t sqrt(1 - t²) + t sqrt(1 - 4 t²)] [ (√3 / 3) + (sqrt(1 - t²) / t) ] = 2.This seems very messy. Maybe there's a better approach.Wait, perhaps instead of trying to solve for θ, I can consider specific values or look for a way to simplify the equation.Alternatively, maybe using the area formula. The area of triangle ABC is (1/2)ab sin C. Since a = 2, b = 1, the area is (1/2)*2*1*sin C = sin C.So, if I can find sin C, I can find the area.From the given equation: (√3 / 3) sin C + c cos B = 2.But c = sin C / sin B, as before.So, (√3 / 3) sin C + (sin C / sin B) cos B = 2.Factor sin C:sin C [ (√3 / 3) + (cos B / sin B) ] = 2.So, sin C [ (√3 / 3) + cot B ] = 2.Hmm, maybe I can express cot B in terms of angle B.Alternatively, let's think about the triangle. If I can find angle C, then I can find sin C.Wait, maybe using the fact that in any triangle, the sum of angles is π. So, A + B + C = π.From the Law of Sines, a / sin A = b / sin B = c / sin C.Given a = 2, b = 1, so 2 / sin A = 1 / sin B. Therefore, sin A = 2 sin B.So, A = arcsin(2 sin B). But as before, sin B ≤ 1/2, so B ≤ π/6.So, angle B is at most π/6, which is 30 degrees.So, angle B is acute, and angle A is also acute or obtuse? Since sin A = 2 sin B, and sin B ≤ 1/2, so sin A ≤ 1, which is okay.But if sin A = 2 sin B, and B ≤ π/6, then sin A ≤ 1, so A can be up to π/2, since sin(π/2) = 1.So, A is between arcsin(2 sin B) where B is between 0 and π/6.Wait, maybe I can consider specific angles.Alternatively, maybe I can use the given equation to find angle C.Wait, let's think about the given equation again:(√3 / 3) sin C + c cos B = 2.But c = sin C / sin B, so:(√3 / 3) sin C + (sin C / sin B) cos B = 2.Factor sin C:sin C [ (√3 / 3) + (cos B / sin B) ] = 2.So, sin C [ (√3 / 3) + cot B ] = 2.Let me denote this as sin C * K = 2, where K = (√3 / 3) + cot B.So, sin C = 2 / K.But sin C must be ≤ 1, so 2 / K ≤ 1, which implies K ≥ 2.So, (√3 / 3) + cot B ≥ 2.So, cot B ≥ 2 - √3 / 3.Compute 2 - √3 / 3 ≈ 2 - 0.577 ≈ 1.423.So, cot B ≥ 1.423.But cot B = cos B / sin B.So, cos B / sin B ≥ 1.423.Which implies tan B ≤ 1 / 1.423 ≈ 0.702.So, tan B ≤ 0.702, which implies B ≤ arctan(0.702) ≈ 35 degrees.But earlier, we had B ≤ π/6 ≈ 30 degrees.So, combining both, B ≤ 30 degrees.So, angle B is at most 30 degrees.Hmm, maybe I can find angle B.Let me set angle B = θ, so θ ≤ π/6.Then, sin A = 2 sin θ.Also, angle C = π - A - θ.So, sin C = sin(A + θ) = sin A cos θ + cos A sin θ.As before, sin C = 2 sin θ cos θ + sqrt(1 - 4 sin² θ) sin θ.So, plugging back into the equation:sin C [ (√3 / 3) + cot θ ] = 2.So, [2 sin θ cos θ + sin θ sqrt(1 - 4 sin² θ)] [ (√3 / 3) + (cos θ / sin θ) ] = 2.Let me compute this step by step.First, compute the first bracket: 2 sin θ cos θ + sin θ sqrt(1 - 4 sin² θ).Factor sin θ: sin θ [2 cos θ + sqrt(1 - 4 sin² θ)].Second bracket: (√3 / 3) + (cos θ / sin θ).So, the entire equation is:sin θ [2 cos θ + sqrt(1 - 4 sin² θ)] * [ (√3 / 3) + (cos θ / sin θ) ] = 2.Let me compute [2 cos θ + sqrt(1 - 4 sin² θ)] * [ (√3 / 3) + (cos θ / sin θ) ].Let me denote this as:[2 cos θ + sqrt(1 - 4 sin² θ)] * [ (√3 / 3) + (cos θ / sin θ) ].Let me compute this product:First term: 2 cos θ * (√3 / 3) = (2√3 / 3) cos θ.Second term: 2 cos θ * (cos θ / sin θ) = 2 cos² θ / sin θ.Third term: sqrt(1 - 4 sin² θ) * (√3 / 3) = (√3 / 3) sqrt(1 - 4 sin² θ).Fourth term: sqrt(1 - 4 sin² θ) * (cos θ / sin θ) = (cos θ / sin θ) sqrt(1 - 4 sin² θ).So, combining all terms:(2√3 / 3) cos θ + 2 cos² θ / sin θ + (√3 / 3) sqrt(1 - 4 sin² θ) + (cos θ / sin θ) sqrt(1 - 4 sin² θ).This is getting really complicated. Maybe there's a better approach.Wait, perhaps instead of trying to solve for θ, I can consider specific values.Given that angle B is at most π/6, let's try θ = π/6 (30 degrees).Compute sin θ = 1/2, cos θ = √3 / 2.Then, sin A = 2 sin θ = 1, so A = π/2 (90 degrees).Then, angle C = π - A - θ = π - π/2 - π/6 = π/3 (60 degrees).So, sin C = sin(π/3) = √3 / 2.Now, let's check the given equation:(√3 / 3) sin C + c cos B = 2.Compute c: from Law of Sines, c = sin C / sin B * b = (√3 / 2) / (1/2) * 1 = √3.So, c = √3.Now, compute (√3 / 3) sin C + c cos B:(√3 / 3)(√3 / 2) + √3 * (√3 / 2) = (3 / 6) + (3 / 2) = (1/2) + (3/2) = 2.Yes, it satisfies the equation.So, when angle B = π/6, angle A = π/2, angle C = π/3, sides a = 2, b = 1, c = √3.Therefore, the area is (1/2)ab sin C = (1/2)*2*1*(√3 / 2) = √3 / 2.So, the area is √3 / 2.Wait, but is this the only solution? Because I assumed angle B = π/6, but maybe there are other angles that satisfy the equation.But given the constraints, when angle B = π/6, it satisfies the equation, and since angle B is at most π/6, this might be the only solution.Therefore, the area is √3 / 2.So, for part (1), the area is √3 / 2.Now, moving on to part (2): If c = 2, determine the range of possible values for the perimeter of triangle ABC.So, c = 2. We need to find the possible perimeters, which is a + b + c = a + b + 2.From the given equation: (√3 / 3) b sin C + c cos B = a.Given c = 2, so:(√3 / 3) b sin C + 2 cos B = a.We need to express a and b in terms of angles or find a relationship between them.Again, using the Law of Sines: a / sin A = b / sin B = c / sin C = 2 / sin C.So, a = (2 / sin C) sin A, b = (2 / sin C) sin B.Also, since A + B + C = π, we can express A = π - B - C.So, sin A = sin(π - B - C) = sin(B + C).Using the sine addition formula: sin(B + C) = sin B cos C + cos B sin C.So, sin A = sin B cos C + cos B sin C.Therefore, a = (2 / sin C)(sin B cos C + cos B sin C) = 2 (sin B cos C / sin C + cos B).Simplify: a = 2 sin B cot C + 2 cos B.Similarly, b = (2 / sin C) sin B.So, b = 2 sin B / sin C.Now, from the given equation:(√3 / 3) b sin C + 2 cos B = a.Substitute a and b:(√3 / 3) * (2 sin B / sin C) * sin C + 2 cos B = 2 sin B cot C + 2 cos B.Simplify:(√3 / 3) * 2 sin B + 2 cos B = 2 sin B cot C + 2 cos B.So, (2√3 / 3) sin B + 2 cos B = 2 sin B cot C + 2 cos B.Subtract 2 cos B from both sides:(2√3 / 3) sin B = 2 sin B cot C.Divide both sides by 2 sin B (assuming sin B ≠ 0, which it isn't in a triangle):(√3 / 3) = cot C.So, cot C = √3 / 3, which implies tan C = 3 / √3 = √3.Therefore, angle C = π/3 (60 degrees).So, angle C is fixed at π/3.Therefore, in this case, angle C = 60 degrees.So, now, we can express the sides in terms of angles A and B.Since angle C = π/3, and A + B + C = π, so A + B = 2π/3.From the Law of Sines:a / sin A = b / sin B = c / sin C = 2 / sin(π/3) = 2 / (√3 / 2) = 4 / √3 = (4√3) / 3.So, a = (4√3 / 3) sin A, b = (4√3 / 3) sin B.Since A + B = 2π/3, we can write B = 2π/3 - A.So, sin B = sin(2π/3 - A) = sin(2π/3) cos A - cos(2π/3) sin A = (√3 / 2) cos A - (-1/2) sin A = (√3 / 2) cos A + (1/2) sin A.Therefore, b = (4√3 / 3) [ (√3 / 2) cos A + (1/2) sin A ] = (4√3 / 3)(√3 / 2 cos A + 1/2 sin A).Simplify:= (4√3 / 3)(√3 / 2 cos A) + (4√3 / 3)(1/2 sin A)= (4√3 * √3) / 6 cos A + (4√3) / 6 sin A= (12) / 6 cos A + (4√3) / 6 sin A= 2 cos A + (2√3 / 3) sin A.So, b = 2 cos A + (2√3 / 3) sin A.Similarly, a = (4√3 / 3) sin A.Therefore, the perimeter P = a + b + c = (4√3 / 3) sin A + 2 cos A + (2√3 / 3) sin A + 2.Combine like terms:= [ (4√3 / 3 + 2√3 / 3) sin A ] + 2 cos A + 2= (6√3 / 3) sin A + 2 cos A + 2= 2√3 sin A + 2 cos A + 2.So, P = 2√3 sin A + 2 cos A + 2.We can write this as P = 2(√3 sin A + cos A) + 2.Now, to find the range of P, we need to find the range of √3 sin A + cos A.Recall that expressions of the form M sin A + N cos A can be written as R sin(A + φ), where R = sqrt(M² + N²) and tan φ = N / M.So, let's compute R:R = sqrt( (√3)^2 + 1^2 ) = sqrt(3 + 1) = sqrt(4) = 2.And tan φ = 1 / √3, so φ = π/6.Therefore, √3 sin A + cos A = 2 sin(A + π/6).So, P = 2 * 2 sin(A + π/6) + 2 = 4 sin(A + π/6) + 2.Now, we need to find the range of P as A varies.But we have constraints on angle A.Since A + B = 2π/3, and both A and B must be positive and less than π.So, A must be greater than 0 and less than 2π/3.But also, from the Law of Sines, sin A must be positive, which it is.But more specifically, since angle B = 2π/3 - A, and angle B must be positive, so A < 2π/3.Also, from the earlier part, when c = 2, angle C = π/3, and from the Law of Sines, sides a and b are positive.But we also have to ensure that the triangle inequality holds.But perhaps the main constraint is on angle A.So, A ∈ (0, 2π/3).But let's see the expression for P: 4 sin(A + π/6) + 2.We need to find the range of sin(A + π/6) as A varies.Since A ∈ (0, 2π/3), then A + π/6 ∈ (π/6, 5π/6).So, sin(A + π/6) ∈ (1/2, 1].Therefore, sin(A + π/6) ranges from 1/2 to 1.Thus, 4 sin(A + π/6) ranges from 2 to 4.Therefore, P = 4 sin(A + π/6) + 2 ranges from 2 + 2 = 4 to 4 + 2 = 6.But wait, when A + π/6 = π/2, sin(A + π/6) = 1, which is the maximum.When A + π/6 approaches π/6, sin(A + π/6) approaches 1/2, but A can't be zero, so sin(A + π/6) approaches just above 1/2.Similarly, when A approaches 2π/3, A + π/6 approaches 5π/6, sin(5π/6) = 1/2.Wait, but sin(5π/6) is 1/2, so sin(A + π/6) approaches 1/2 from above as A approaches 2π/3.Wait, actually, when A approaches 0, A + π/6 approaches π/6, sin(π/6) = 1/2.When A approaches 2π/3, A + π/6 approaches 5π/6, sin(5π/6) = 1/2.So, sin(A + π/6) ranges from 1/2 to 1, but actually, when A is in (0, 2π/3), A + π/6 is in (π/6, 5π/6), and sin(A + π/6) is in (1/2, 1].Wait, but when A + π/6 = π/2, sin(A + π/6) = 1, which is the maximum.So, sin(A + π/6) ∈ (1/2, 1].Therefore, 4 sin(A + π/6) ∈ (2, 4].Thus, P = 4 sin(A + π/6) + 2 ∈ (4, 6].But wait, when A approaches 0, P approaches 4*(1/2) + 2 = 2 + 2 = 4.But when A approaches 0, angle B approaches 2π/3, which is 120 degrees, and side a approaches 0, which is not possible in a triangle.Similarly, when A approaches 2π/3, angle B approaches 0, which is also not possible.Therefore, the actual range of P is when A is such that both A and B are positive and less than π.But more precisely, since A must be greater than 0 and less than 2π/3, but also, from the earlier part, when c = 2, angle C = π/3, and from the Law of Sines, sides a and b must satisfy a > 0 and b > 0.But perhaps the perimeter can approach 4 and 6, but not including 4 and 6.Wait, but when A approaches π/3, A + π/6 = π/2, sin(A + π/6) = 1, so P approaches 6.When A approaches 0, P approaches 4, but A can't be 0, so P approaches just above 4.Similarly, when A approaches 2π/3, P approaches 4*(1/2) + 2 = 4.But wait, when A approaches 2π/3, A + π/6 approaches 5π/6, sin(5π/6) = 1/2, so P approaches 4*(1/2) + 2 = 4.But in reality, when A approaches 2π/3, angle B approaches 0, which is not possible, so P can't actually reach 4.Similarly, when A approaches 0, angle B approaches 2π/3, which is allowed, but side a approaches 0, which is not possible.Therefore, the perimeter P must be greater than 4 and less than or equal to 6.Wait, but when A = π/3, P = 6.When A approaches 0 or 2π/3, P approaches 4.But in reality, A must be such that both A and B are positive and less than π.But since A + B = 2π/3, and both A and B must be greater than 0, so A ∈ (0, 2π/3).Therefore, sin(A + π/6) ∈ (1/2, 1], so P ∈ (4, 6].But wait, when A = π/3, P = 6.When A approaches 0 or 2π/3, P approaches 4.But in reality, A can't be 0 or 2π/3, so P can't be exactly 4, but can approach it.Therefore, the range of P is (4, 6].But let me check when A = π/6.Then, A + π/6 = π/3, sin(π/3) = √3 / 2 ≈ 0.866.So, P = 4*(√3 / 2) + 2 = 2√3 + 2 ≈ 2*1.732 + 2 ≈ 3.464 + 2 ≈ 5.464.Which is between 4 and 6.Similarly, when A = π/2, which is greater than π/3, but wait, A + B = 2π/3, so if A = π/2, then B = 2π/3 - π/2 = π/6.So, A = π/2, B = π/6.Compute P:sin(A + π/6) = sin(π/2 + π/6) = sin(2π/3) = √3 / 2.So, P = 4*(√3 / 2) + 2 = 2√3 + 2 ≈ 5.464.Wait, but earlier when A = π/3, P = 6.Wait, let me compute when A = π/3.A = π/3, B = 2π/3 - π/3 = π/3.So, A = B = π/3, C = π/3.So, it's an equilateral triangle.Therefore, a = b = c = 2.So, perimeter P = 6.Yes, that makes sense.Similarly, when A approaches 0, a approaches 0, but since a must be positive, P approaches 4.But in reality, a can't be zero, so P can't be exactly 4.Therefore, the range of P is (4, 6].So, the perimeter can be as close to 4 as desired (but greater than 4) and up to 6.Therefore, the range is (4, 6].But let me confirm with another angle.Let me take A = π/4.Then, A + π/6 = π/4 + π/6 = (3π/12 + 2π/12) = 5π/12.sin(5π/12) = sin(75 degrees) = (√6 + √2)/4 ≈ 0.9659.So, P = 4*0.9659 + 2 ≈ 3.8636 + 2 ≈ 5.8636.Which is less than 6, as expected.Wait, but when A = π/3, P = 6.So, the maximum perimeter is 6, achieved when the triangle is equilateral.The minimum perimeter approaches 4 when one side approaches 0, but since sides must be positive, the perimeter can't actually be 4.Therefore, the range is (4, 6].So, for part (2), the perimeter of triangle ABC is in the interval (4, 6].But wait, in the earlier calculation, when A approaches 0, P approaches 4, but when A approaches 2π/3, P also approaches 4.But in reality, when A approaches 2π/3, angle B approaches 0, which is not possible, so P approaches 4 from above.Similarly, when A approaches 0, angle B approaches 2π/3, which is allowed, but side a approaches 0, which is not possible.Therefore, the perimeter can approach 4 but can't actually reach it.Hence, the range is (4, 6].So, summarizing:Part (1): Area = √3 / 2.Part (2): Perimeter ∈ (4, 6].I think that's the solution.

answer：Okay, so I have this equation: ((2x + y)^2 = 2017 + x^2). I need to find how many integer solutions exist for (x) and (y). Hmm, let me think about how to approach this.First, I should probably expand the left side of the equation to see if I can simplify it. Let's do that:[(2x + y)^2 = 4x^2 + 4xy + y^2]So now the equation becomes:[4x^2 + 4xy + y^2 = 2017 + x^2]Hmm, okay. Let me subtract (x^2) from both sides to get all terms on one side:[4x^2 + 4xy + y^2 - x^2 = 2017]Simplifying that, I get:[3x^2 + 4xy + y^2 = 2017]Alright, so now I have a quadratic equation in terms of (x) and (y). I need to find integer solutions for this. Maybe I can treat this as a quadratic in (y) and see if I can find integer values.Let me rearrange the equation to solve for (y):[y^2 + 4xy + 3x^2 - 2017 = 0]This is a quadratic in (y), so I can use the quadratic formula to solve for (y):[y = frac{-4x pm sqrt{(4x)^2 - 4 cdot 1 cdot (3x^2 - 2017)}}{2}]Simplifying inside the square root:[sqrt{16x^2 - 4(3x^2 - 2017)} = sqrt{16x^2 - 12x^2 + 8068} = sqrt{4x^2 + 8068}]So, the equation for (y) becomes:[y = frac{-4x pm sqrt{4x^2 + 8068}}{2}]Simplify further by factoring out a 2 from the numerator:[y = -2x pm frac{sqrt{4x^2 + 8068}}{2}]Hmm, for (y) to be an integer, the term (frac{sqrt{4x^2 + 8068}}{2}) must also be an integer. Let me denote:[sqrt{4x^2 + 8068} = 2k]Where (k) is an integer. Then, squaring both sides:[4x^2 + 8068 = 4k^2]Divide both sides by 4:[x^2 + 2017 = k^2]So, we have:[k^2 - x^2 = 2017]This is a difference of squares, which factors as:[(k - x)(k + x) = 2017]Now, 2017 is a prime number, right? Let me check: 2017 divided by primes less than its square root (which is around 44.9). It doesn't divide by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, or 43. So yes, 2017 is prime.Since 2017 is prime, the only positive integer factor pairs are (1, 2017) and (-1, -2017). So, we have two cases:1. (k - x = 1) and (k + x = 2017)2. (k - x = -1) and (k + x = -2017)Let's solve each case.**Case 1:**[k - x = 1 k + x = 2017]Adding both equations:[2k = 2018 implies k = 1009]Subtracting the first equation from the second:[2x = 2016 implies x = 1008]So, (x = 1008) and (k = 1009). Now, let's find (y):From earlier, we had:[y = -2x pm k]So,[y = -2(1008) pm 1009 = -2016 pm 1009]Calculating both possibilities:1. (y = -2016 + 1009 = -1007)2. (y = -2016 - 1009 = -3025)So, two solutions from this case: ((1008, -1007)) and ((1008, -3025)).**Case 2:**[k - x = -1 k + x = -2017]Adding both equations:[2k = -2018 implies k = -1009]Subtracting the first equation from the second:[2x = -2016 implies x = -1008]So, (x = -1008) and (k = -1009). Now, let's find (y):[y = -2x pm k = -2(-1008) pm (-1009) = 2016 pm (-1009)]Calculating both possibilities:1. (y = 2016 - 1009 = 1007)2. (y = 2016 + (-1009) = 1007)Wait, that gives the same value for both. Hmm, maybe I made a mistake.Wait, no, actually:[y = -2x pm k = 2016 pm (-1009)]So,1. (y = 2016 + (-1009) = 1007)2. (y = 2016 - (-1009) = 2016 + 1009 = 3025)Ah, that's better. So, two solutions from this case: ((-1008, 1007)) and ((-1008, 3025)).So, in total, we have four solutions:1. ((1008, -1007))2. ((1008, -3025))3. ((-1008, 1007))4. ((-1008, 3025))Let me double-check these solutions to make sure they satisfy the original equation.**Checking Solution 1: ((1008, -1007))**Compute ((2x + y)^2):[2(1008) + (-1007) = 2016 - 1007 = 1009][(1009)^2 = 1018081]Compute (2017 + x^2):[2017 + (1008)^2 = 2017 + 1016064 = 1018081]They match.**Checking Solution 2: ((1008, -3025))**Compute ((2x + y)^2):[2(1008) + (-3025) = 2016 - 3025 = -1009][(-1009)^2 = 1018081]Compute (2017 + x^2):Same as above, 1018081. They match.**Checking Solution 3: ((-1008, 1007))**Compute ((2x + y)^2):[2(-1008) + 1007 = -2016 + 1007 = -1009][(-1009)^2 = 1018081]Compute (2017 + x^2):[2017 + (-1008)^2 = 2017 + 1016064 = 1018081]They match.**Checking Solution 4: ((-1008, 3025))**Compute ((2x + y)^2):[2(-1008) + 3025 = -2016 + 3025 = 1009][(1009)^2 = 1018081]Compute (2017 + x^2):Same as above, 1018081. They match.Okay, all four solutions satisfy the original equation. So, it looks like there are four integer solutions.I think that's all the possible solutions because 2017 is prime, so the only factor pairs are (1, 2017) and (-1, -2017). Therefore, we don't have any other cases to consider.**Final Answer**The number of integer solutions is boxed{4}.