💡Okay, so I have this problem here: I need to find the symmetric center of the function ( y = sinleft(2x - frac{pi}{6}right) ). The options are given as A through D, with coordinates involving pi over 12, pi over 3, pi over 6, and negative pi over 6, all with y-coordinate zero. Hmm, interesting.First, I remember that the sine function is symmetric about its midline. The standard sine function ( y = sin(x) ) has a midline at y=0 and is symmetric about every point of the form ( (kpi, 0) ) where k is an integer. So, these points are the centers of symmetry for the sine curve.But in this case, the function is transformed. It's ( sin(2x - frac{pi}{6}) ). So, I need to figure out how this transformation affects the symmetric centers.Let me recall that the general form of a sine function is ( y = sin(Bx + C) ). The period of such a function is ( frac{2pi}{|B|} ), and the phase shift is ( -frac{C}{B} ). So, in this case, B is 2 and C is ( -frac{pi}{6} ). Therefore, the phase shift is ( -frac{-pi/6}{2} = frac{pi}{12} ). So, the graph is shifted to the right by pi over 12.But how does this affect the symmetric centers? Well, the original sine function has symmetric centers at every multiple of pi, right? So, if the function is shifted, those centers should also shift accordingly.Let me think. If the original function ( y = sin(x) ) has symmetric centers at ( (kpi, 0) ), then after a horizontal shift, those centers should move by the same amount. Since the phase shift here is ( frac{pi}{12} ) to the right, each symmetric center should also shift to the right by ( frac{pi}{12} ).But wait, the period is also changed. The original period is ( 2pi ), but with B=2, the period becomes ( pi ). So, the distance between consecutive symmetric centers is now ( frac{pi}{2} ) instead of pi, because the symmetric centers occur every half-period.Wait, let me clarify. For the standard sine function, the symmetric centers are at every pi interval because the sine function is symmetric about those points. So, if the period is halved, the distance between symmetric centers would also be halved. So, instead of every pi, they would be every pi over 2.But hold on, is that correct? Let me think again. The standard sine function has a period of ( 2pi ), and it's symmetric about every point ( (kpi, 0) ). So, the distance between these centers is pi, which is half the period. So, if the period is now pi, the distance between symmetric centers should be half of that, which is pi over 2.Therefore, the symmetric centers for the transformed function should be at ( left(frac{pi}{12} + frac{kpi}{2}, 0right) ) where k is an integer. So, for k=0, it's ( frac{pi}{12} ), for k=1, it's ( frac{pi}{12} + frac{pi}{2} = frac{7pi}{12} ), and so on.Looking at the options given, A is ( left(frac{pi}{12}, 0right) ), which is the first symmetric center. B is ( left(frac{pi}{3}, 0right) ), which is approximately 1.047, while pi over 12 is about 0.2618. C is ( left(frac{pi}{6}, 0right) ), which is about 0.5236, and D is ( left(-frac{pi}{6}, 0right) ), which is negative.So, the first symmetric center is at pi over 12, which is option A. So, I think the answer is A.But let me verify this another way. Maybe by using the formula for the center of symmetry.I remember that for a function ( y = sin(Bx + C) ), the centers of symmetry can be found by solving for x when the argument of the sine function is an integer multiple of pi. So, setting ( Bx + C = kpi ), solving for x gives ( x = frac{kpi - C}{B} ).In this case, ( B = 2 ) and ( C = -frac{pi}{6} ). So, substituting, we get ( x = frac{kpi - (-pi/6)}{2} = frac{kpi + pi/6}{2} = frac{kpi}{2} + frac{pi}{12} ).So, the symmetric centers are at ( x = frac{kpi}{2} + frac{pi}{12} ), y=0. So, for k=0, x is pi over 12, which is option A. For k=1, it's pi over 2 plus pi over 12, which is 7 pi over 12, which isn't one of the options. So, the primary symmetric center is at pi over 12, which is option A.Therefore, I'm confident that the correct answer is A.**Final Answer**boxed{A}

💡Alright, so I've got this triangle problem here. It says that in a triangle with sides of lengths a, b, and c, the equation (a + b + c)(a + b - c) equals 4ab. And I need to find the angle opposite the side of length c, in degrees. Hmm, okay, let me try to figure this out step by step.First, I remember that in triangles, there are some important formulas, like the Law of Cosines, which relates the lengths of the sides to the cosine of one of its angles. Maybe that will come in handy here. But before jumping into that, let me look at the given equation and see if I can simplify it or manipulate it to find a relationship between the sides.So, the equation is (a + b + c)(a + b - c) = 4ab. Hmm, that looks a bit complicated, but maybe if I expand the left side, it'll become clearer. Let me try that.Expanding (a + b + c)(a + b - c). I can think of this as ( (a + b) + c ) times ( (a + b) - c ). That's similar to the difference of squares formula, where (x + y)(x - y) = x² - y². So, applying that here, x is (a + b) and y is c. So, it should be (a + b)² - c².Let me write that out:(a + b + c)(a + b - c) = (a + b)² - c².Okay, so expanding (a + b)², that's a² + 2ab + b². So, putting it all together, the left side becomes a² + 2ab + b² - c².So, the equation now is:a² + 2ab + b² - c² = 4ab.Hmm, let me rearrange this equation to group like terms. I'll subtract 4ab from both sides to get:a² + 2ab + b² - c² - 4ab = 0.Simplifying the terms with ab:2ab - 4ab is -2ab. So, the equation becomes:a² - 2ab + b² - c² = 0.Hmm, that looks like a quadratic in terms of a and b. Wait, a² - 2ab + b² is actually (a - b)². Let me check that:(a - b)² = a² - 2ab + b². Yes, that's correct. So, substituting that in, the equation becomes:(a - b)² - c² = 0.So, (a - b)² = c².Taking square roots on both sides, we get:|a - b| = c.But since a, b, and c are lengths of sides of a triangle, they must be positive. So, |a - b| = c implies that either a - b = c or b - a = c. But in a triangle, the length of any side must be less than the sum of the other two sides. So, if a - b = c, then a = b + c. But then, according to the triangle inequality, a must be less than b + c, but here a equals b + c, which would make the triangle degenerate, meaning it would be a straight line, not a proper triangle. Similarly, if b - a = c, then b = a + c, which again violates the triangle inequality because b would be equal to a + c, making it a straight line.Wait, that doesn't make sense. Did I make a mistake somewhere? Let me double-check my steps.Starting from the beginning: (a + b + c)(a + b - c) = 4ab.Expanding the left side: (a + b + c)(a + b - c) = (a + b)² - c² = a² + 2ab + b² - c².Setting that equal to 4ab: a² + 2ab + b² - c² = 4ab.Subtracting 4ab from both sides: a² - 2ab + b² - c² = 0.Which is (a - b)² - c² = 0, so (a - b)² = c², leading to |a - b| = c.Hmm, so this suggests that c is equal to the absolute difference of a and b. But in a triangle, the sum of any two sides must be greater than the third side. So, if c = |a - b|, then either a = b + c or b = a + c, which would mean that one side is equal to the sum of the other two, which is not possible in a valid triangle. That would make the triangle degenerate, meaning it collapses into a straight line, which isn't a triangle at all.This seems contradictory. Maybe I made a wrong assumption somewhere. Let me think again.Wait, perhaps I should approach this differently. Maybe instead of expanding the left side, I can relate it to the Law of Cosines. The Law of Cosines states that c² = a² + b² - 2ab cos C, where C is the angle opposite side c. Maybe I can use this to substitute into the equation.Let me write down the Law of Cosines:c² = a² + b² - 2ab cos C.From the equation we derived earlier, (a - b)² = c², which is a² - 2ab + b² = c².So, substituting c² from the Law of Cosines into this equation:a² - 2ab + b² = a² + b² - 2ab cos C.Let me write that out:a² - 2ab + b² = a² + b² - 2ab cos C.Hmm, let's subtract a² + b² from both sides:-2ab = -2ab cos C.Dividing both sides by -2ab (assuming a and b are not zero, which they can't be in a triangle):1 = cos C.So, cos C = 1.What angle has a cosine of 1? Well, cos 0° = 1. So, angle C is 0 degrees.Wait, but an angle of 0 degrees in a triangle would mean that the two sides adjacent to that angle are overlapping, making the triangle degenerate again. So, that doesn't make sense either.Hmm, so both approaches are leading me to a degenerate triangle, which isn't possible. Did I make a mistake in my calculations?Let me go back to the beginning. The given equation is (a + b + c)(a + b - c) = 4ab.I expanded it to a² + 2ab + b² - c² = 4ab.Then, subtracting 4ab from both sides: a² - 2ab + b² - c² = 0.Which is (a - b)² - c² = 0.So, (a - b)² = c².Which leads to |a - b| = c.But in a triangle, the sides must satisfy the triangle inequality: a + b > c, a + c > b, and b + c > a.If c = |a - b|, then without loss of generality, assume a ≥ b, so c = a - b.Then, a + b > c would be a + b > a - b, which simplifies to 2b > 0, which is always true since b is a positive length.Similarly, a + c > b would be a + (a - b) > b, which simplifies to 2a - b > b, so 2a > 2b, or a > b, which is consistent with our assumption.And b + c > a would be b + (a - b) > a, which simplifies to a > a, which is not true. So, this is a contradiction.Therefore, c cannot be equal to |a - b| because it violates the triangle inequality. So, my earlier conclusion that angle C is 0 degrees must be incorrect because it leads to a degenerate triangle, which isn't valid.Wait, maybe I made a mistake in the algebra when I substituted the Law of Cosines. Let me check that again.We had:(a - b)² = c², so c² = a² - 2ab + b².From the Law of Cosines, c² = a² + b² - 2ab cos C.So, setting them equal:a² - 2ab + b² = a² + b² - 2ab cos C.Subtracting a² + b² from both sides:-2ab = -2ab cos C.Dividing both sides by -2ab:1 = cos C.So, cos C = 1, which implies C = 0°, but as we saw, that's not possible in a valid triangle.Hmm, this is confusing. Maybe the problem is designed in such a way that the triangle is degenerate, but that seems unlikely. Perhaps I need to reconsider my approach.Wait, maybe I should consider that the given equation might not necessarily lead to c = |a - b|, but perhaps there's another way to interpret it. Let me think.Starting again with (a + b + c)(a + b - c) = 4ab.Let me denote s = a + b + c, which is the perimeter of the triangle. Then, the equation becomes s * (s - 2c) = 4ab.But I'm not sure if that helps directly. Alternatively, maybe I can express c in terms of a and b.From the equation:(a + b + c)(a + b - c) = 4ab.Let me denote (a + b) as x for simplicity. Then, the equation becomes (x + c)(x - c) = 4ab, which is x² - c² = 4ab.But x = a + b, so x² = (a + b)² = a² + 2ab + b².So, substituting back:a² + 2ab + b² - c² = 4ab.Rearranging:a² + b² - c² = 2ab.Hmm, that's interesting. So, a² + b² - c² = 2ab.Wait, from the Law of Cosines, we have:c² = a² + b² - 2ab cos C.So, rearranging that:a² + b² - c² = 2ab cos C.But from our earlier equation, a² + b² - c² = 2ab.So, setting them equal:2ab cos C = 2ab.Dividing both sides by 2ab (assuming a and b are not zero):cos C = 1.Again, this leads to C = 0°, which is problematic as before.Wait a minute, maybe I'm misapplying the Law of Cosines. Let me double-check that.Law of Cosines states that c² = a² + b² - 2ab cos C. So, solving for cos C:cos C = (a² + b² - c²) / (2ab).From our equation, we have a² + b² - c² = 2ab.So, substituting into cos C:cos C = (2ab) / (2ab) = 1.So, cos C = 1, which means C = 0°, as before.But as we saw earlier, this would make the triangle degenerate, which isn't possible. So, is there a mistake in the problem statement? Or perhaps I'm missing something.Wait, maybe the problem is designed to have a degenerate triangle, but that seems unusual. Alternatively, perhaps I'm misinterpreting the equation.Let me check the original equation again: (a + b + c)(a + b - c) = 4ab.Is there another way to interpret this? Maybe using Heron's formula? Heron's formula relates the area of a triangle to its sides, but I'm not sure if that's directly applicable here.Alternatively, maybe I can express c in terms of a and b and then find the angle C.From the equation:(a + b + c)(a + b - c) = 4ab.Let me solve for c.Let me denote (a + b + c) as S and (a + b - c) as D. Then, S * D = 4ab.But S = a + b + c and D = a + b - c.Multiplying them: (a + b + c)(a + b - c) = (a + b)² - c² = 4ab.As before, this leads to a² + 2ab + b² - c² = 4ab.Rearranging: a² + b² - c² = 2ab.From the Law of Cosines: a² + b² - c² = 2ab cos C.So, 2ab cos C = 2ab.Dividing both sides by 2ab: cos C = 1.Thus, C = 0°, which is the same conclusion as before.But again, this suggests a degenerate triangle, which isn't possible. So, perhaps the problem is designed to have a degenerate triangle, but that seems odd. Alternatively, maybe I'm missing a step or making an incorrect assumption.Wait, perhaps the problem is in the way I'm interpreting the equation. Let me check the original equation again: (a + b + c)(a + b - c) = 4ab.Is there a possibility that I've misapplied the expansion? Let me double-check that.Expanding (a + b + c)(a + b - c):First, multiply a by each term in the second parenthesis: a*(a) + a*(b) - a*(c) = a² + ab - ac.Then, multiply b by each term: b*(a) + b*(b) - b*(c) = ab + b² - bc.Then, multiply c by each term: c*(a) + c*(b) - c*(c) = ac + bc - c².Now, adding all these together:a² + ab - ac + ab + b² - bc + ac + bc - c².Simplify term by term:a² + ab - ac + ab + b² - bc + ac + bc - c².Combine like terms:a² + (ab + ab) + ( -ac + ac) + b² + ( -bc + bc) - c².Simplify:a² + 2ab + 0 + b² + 0 - c².So, indeed, it's a² + 2ab + b² - c².So, my expansion was correct.Therefore, the equation is a² + 2ab + b² - c² = 4ab.Rearranging gives a² + b² - c² = 2ab.From the Law of Cosines, a² + b² - c² = 2ab cos C.So, 2ab cos C = 2ab.Dividing both sides by 2ab: cos C = 1.Thus, C = 0°, which is the same conclusion as before.But as we saw earlier, this leads to a degenerate triangle, which isn't possible. So, perhaps the problem is designed to have a degenerate triangle, but that seems unlikely. Alternatively, maybe I'm missing something in the problem statement.Wait, perhaps the problem is in three-dimensional space, but the problem doesn't specify that. It just says "a triangle," which is typically assumed to be in a plane.Alternatively, maybe the problem is designed to have a specific type of triangle, like a right triangle, but in this case, it's leading to a degenerate triangle.Wait, let me think differently. Maybe instead of assuming that a, b, and c are the sides of a triangle, perhaps they are vectors or something else. But the problem states "a triangle with sides of lengths a, b, and c," so they must be the lengths of the sides.Alternatively, maybe I'm misapplying the Law of Cosines. Let me double-check that.Law of Cosines: c² = a² + b² - 2ab cos C.Yes, that's correct. So, solving for cos C:cos C = (a² + b² - c²) / (2ab).From our equation, a² + b² - c² = 2ab.So, cos C = (2ab) / (2ab) = 1.Thus, C = 0°, as before.But again, this suggests a degenerate triangle, which isn't possible. So, perhaps the problem is designed to have a degenerate triangle, but that seems odd.Alternatively, maybe I made a mistake in the initial steps. Let me try a different approach.Let me consider specific values for a and b to see what happens. Suppose a = b = 1. Then, let's see what c would be.From the equation: (1 + 1 + c)(1 + 1 - c) = 4*1*1.So, (2 + c)(2 - c) = 4.Expanding: 4 - c² = 4.Subtracting 4 from both sides: -c² = 0.So, c² = 0, which implies c = 0.But c = 0 would mean that the triangle has sides 1, 1, and 0, which is degenerate. So, again, we get a degenerate triangle.Alternatively, let's try a different approach. Maybe using the Law of Sines. But I don't see an immediate way to connect that with the given equation.Wait, perhaps I can express c in terms of a and b from the given equation and then use the Law of Cosines.From the equation: (a + b + c)(a + b - c) = 4ab.Let me solve for c.Let me denote (a + b + c) as S and (a + b - c) as D. Then, S * D = 4ab.But S = a + b + c and D = a + b - c.Multiplying them: (a + b + c)(a + b - c) = (a + b)² - c² = 4ab.As before, this leads to a² + 2ab + b² - c² = 4ab.Rearranging: a² + b² - c² = 2ab.From the Law of Cosines: a² + b² - c² = 2ab cos C.So, 2ab cos C = 2ab.Dividing both sides by 2ab: cos C = 1.Thus, C = 0°, as before.But again, this leads to a degenerate triangle. So, perhaps the problem is designed to have a degenerate triangle, but that seems unlikely.Alternatively, maybe the problem is designed to have a specific type of triangle, like a right triangle, but in this case, it's leading to a degenerate triangle.Wait, perhaps I'm overcomplicating this. Maybe the problem is designed to have angle C as 0°, even though it's degenerate. But in that case, it's not a valid triangle.Alternatively, maybe I made a mistake in the initial steps. Let me try a different approach.Let me consider the given equation: (a + b + c)(a + b - c) = 4ab.Let me divide both sides by ab:[(a + b + c)(a + b - c)] / ab = 4.Let me express this as [(a + b + c)/b] * [(a + b - c)/a] = 4.But I'm not sure if that helps.Alternatively, let me express c in terms of a and b.From the equation: (a + b + c)(a + b - c) = 4ab.Let me denote (a + b) as x. Then, the equation becomes (x + c)(x - c) = 4ab.Which is x² - c² = 4ab.But x = a + b, so x² = (a + b)² = a² + 2ab + b².Thus, a² + 2ab + b² - c² = 4ab.Rearranging: a² + b² - c² = 2ab.From the Law of Cosines: a² + b² - c² = 2ab cos C.So, 2ab cos C = 2ab.Dividing both sides by 2ab: cos C = 1.Thus, C = 0°, as before.But again, this leads to a degenerate triangle. So, perhaps the problem is designed to have a degenerate triangle, but that seems odd.Alternatively, maybe the problem is designed to have a specific type of triangle, like a right triangle, but in this case, it's leading to a degenerate triangle.Wait, perhaps I should consider that the problem might have a typo or is designed to have a degenerate triangle. But assuming it's a valid triangle, perhaps I'm missing something.Wait, let me think about the problem again. The equation is (a + b + c)(a + b - c) = 4ab.Let me consider the case where a = b. Let's see what happens.If a = b, then the equation becomes (a + a + c)(a + a - c) = 4a².Simplifying: (2a + c)(2a - c) = 4a².Expanding: 4a² - c² = 4a².Subtracting 4a² from both sides: -c² = 0.So, c² = 0, which implies c = 0. Again, a degenerate triangle.Hmm, so even when a = b, we get c = 0, which is degenerate.Alternatively, let's try a different approach. Maybe using the semi-perimeter.Let me denote s = (a + b + c)/2, the semi-perimeter.Then, the given equation is (2s)(2s - 2c) = 4ab.Simplifying: 4s(s - c) = 4ab.Dividing both sides by 4: s(s - c) = ab.But I'm not sure if that helps directly. Let me recall that Heron's formula for the area is sqrt[s(s - a)(s - b)(s - c)]. But I don't see an immediate connection.Alternatively, maybe I can express s - c in terms of a and b.s - c = (a + b + c)/2 - c = (a + b - c)/2.So, s(s - c) = [(a + b + c)/2] * [(a + b - c)/2] = [(a + b + c)(a + b - c)] / 4.But from the given equation, (a + b + c)(a + b - c) = 4ab.So, s(s - c) = 4ab / 4 = ab.Thus, s(s - c) = ab.But I'm not sure if that helps me find angle C.Alternatively, maybe I can use the formula for the area in terms of sides and angle.The area of a triangle can also be expressed as (1/2)ab sin C.But I don't see how that connects to the given equation.Wait, perhaps I can relate s(s - c) to the area.From Heron's formula: Area = sqrt[s(s - a)(s - b)(s - c)].But I don't see a direct way to connect this to the given equation.Alternatively, maybe I can express s - a and s - b in terms of the given equation.But I'm not sure.Wait, perhaps I can consider the ratio of the given equation to ab.From the given equation: (a + b + c)(a + b - c) = 4ab.Dividing both sides by ab: [(a + b + c)(a + b - c)] / ab = 4.Let me express this as [(a + b + c)/b] * [(a + b - c)/a] = 4.But I'm not sure if that helps.Alternatively, let me express (a + b + c)/b as (a/b + 1 + c/b) and (a + b - c)/a as (1 + b/a - c/a).But this seems complicated.Alternatively, maybe I can let a/b = k, so a = kb.Then, express everything in terms of k and b.Let me try that.Let a = kb, where k is a positive real number.Then, the equation becomes:(kb + b + c)(kb + b - c) = 4kb * b.Simplifying:(b(k + 1) + c)(b(k + 1) - c) = 4k b².Expanding the left side:[b(k + 1)]² - c² = 4k b².Which is b²(k + 1)² - c² = 4k b².Rearranging:b²(k + 1)² - 4k b² = c².Factor out b²:b²[(k + 1)² - 4k] = c².Simplify the expression inside the brackets:(k + 1)² - 4k = k² + 2k + 1 - 4k = k² - 2k + 1 = (k - 1)².So, c² = b²(k - 1)².Taking square roots:c = b|k - 1|.Since c is a length, it must be positive, so c = b|k - 1|.But from the triangle inequality, we have:a + b > c.Since a = kb, this becomes:kb + b > c.Which is b(k + 1) > c.But c = b|k - 1|.So, b(k + 1) > b|k - 1|.Dividing both sides by b (since b > 0):k + 1 > |k - 1|.This inequality must hold for the triangle to be valid.Let me consider two cases based on the value of k.Case 1: k ≥ 1.Then, |k - 1| = k - 1.So, the inequality becomes:k + 1 > k - 1.Simplifying:k + 1 > k - 1.Subtracting k from both sides:1 > -1.Which is always true.Case 2: k < 1.Then, |k - 1| = 1 - k.So, the inequality becomes:k + 1 > 1 - k.Simplifying:k + 1 > 1 - k.Subtracting 1 from both sides:k > -k.Adding k to both sides:2k > 0.Since k is a positive real number (as a and b are lengths), this is always true.So, in both cases, the inequality holds, meaning that c = b|k - 1| is valid for any positive k.But wait, from earlier, we saw that c = |a - b|, which would imply that c = |kb - b| = b|k - 1|, which matches what we have here.So, c = b|k - 1|.Now, let's use the Law of Cosines to find angle C.From the Law of Cosines:c² = a² + b² - 2ab cos C.Substituting a = kb and c = b|k - 1|:(b|k - 1|)² = (kb)² + b² - 2(kb)(b) cos C.Simplifying:b²(k - 1)² = k²b² + b² - 2k b² cos C.Divide both sides by b² (since b ≠ 0):(k - 1)² = k² + 1 - 2k cos C.Expanding the left side:k² - 2k + 1 = k² + 1 - 2k cos C.Subtracting k² + 1 from both sides:-2k = -2k cos C.Dividing both sides by -2k (assuming k ≠ 0):1 = cos C.Thus, cos C = 1, which implies C = 0°, as before.But again, this leads to a degenerate triangle, which isn't possible. So, perhaps the problem is designed to have a degenerate triangle, but that seems unlikely.Alternatively, maybe I'm missing something in the problem statement or misapplying the formulas.Wait, perhaps the problem is designed to have a specific type of triangle, like a right triangle, but in this case, it's leading to a degenerate triangle.Alternatively, maybe the problem is designed to have a specific angle, like 90°, but our calculations show it's 0°, which is conflicting.Wait, perhaps I made a mistake in the substitution. Let me double-check.From the Law of Cosines:c² = a² + b² - 2ab cos C.We have c = b|k - 1|, a = kb.So, c² = b²(k - 1)².Substituting into the Law of Cosines:b²(k - 1)² = (kb)² + b² - 2(kb)(b) cos C.Simplifying:b²(k² - 2k + 1) = k²b² + b² - 2k b² cos C.Dividing by b²:k² - 2k + 1 = k² + 1 - 2k cos C.Subtracting k² + 1 from both sides:-2k = -2k cos C.Dividing by -2k:1 = cos C.Thus, C = 0°, as before.So, it seems that no matter how I approach it, I end up with angle C being 0°, which suggests a degenerate triangle. But since the problem states it's a triangle, perhaps it's a trick question, and the angle is indeed 0°, even though it's degenerate.Alternatively, maybe the problem is designed to have a specific angle, and I'm misapplying the formulas.Wait, perhaps I should consider that the given equation might not necessarily lead to c = |a - b|, but perhaps there's another way to interpret it. Let me think.Wait, another approach: Let me consider the given equation as a quadratic in c.From (a + b + c)(a + b - c) = 4ab.Expanding: a² + 2ab + b² - c² = 4ab.Rearranging: -c² + a² + 2ab + b² - 4ab = 0.Simplifying: -c² + a² - 2ab + b² = 0.Which is: c² = a² - 2ab + b².So, c² = (a - b)².Thus, c = |a - b|.But as we saw earlier, this leads to a degenerate triangle.Therefore, the only solution is that the triangle is degenerate, with angle C being 0°.But since the problem states it's a triangle, perhaps it's a trick question, and the answer is indeed 0°, even though it's degenerate.Alternatively, perhaps the problem is designed to have a specific angle, and I'm missing something.Wait, perhaps I should consider that the given equation might have another solution. Let me think.From c² = (a - b)², we have c = |a - b|.But in a triangle, c must be less than a + b and greater than |a - b|.But here, c = |a - b|, which is the lower bound, making the triangle degenerate.Therefore, the only possible solution is a degenerate triangle with angle C = 0°.Thus, despite the contradiction with the triangle inequality, the angle opposite side c is 0°.So, after all this, I think the answer is 0°, even though it's a degenerate triangle.

💡To determine the weight of the new person, let's follow these steps:1. **Calculate the total weight increase for the group:** The average weight increases by 3.7 kg for 12 people. [ text{Total weight increase} = 12 times 3.7 , text{kg} = 44.4 , text{kg} ]2. **Determine the weight of the new person:** The new person replaces someone who weighed 57.3 kg. Therefore, the weight of the new person is the previous weight plus the total weight increase. [ W = 57.3 , text{kg} + 44.4 , text{kg} = 101.7 , text{kg} ]

💡Alright, so I have this problem here: Suppose that ( a ) and ( b ) are positive integers such that ( (a + bi)^2 = 3 + 4i ). I need to find ( a + bi ). Hmm, okay, let me think about how to approach this.First, I remember that when you square a complex number, you can expand it using the formula ( (a + bi)^2 ). Let me try expanding that. So, ( (a + bi)^2 = a^2 + 2abi + (bi)^2 ). Wait, ( (bi)^2 ) is ( b^2 i^2 ), and since ( i^2 = -1 ), that becomes ( -b^2 ). So, putting it all together, ( (a + bi)^2 = a^2 - b^2 + 2abi ).Now, the problem says this equals ( 3 + 4i ). So, I can set the real parts equal and the imaginary parts equal. That gives me two equations:1. Real part: ( a^2 - b^2 = 3 )2. Imaginary part: ( 2ab = 4 )Okay, so from the imaginary part, ( 2ab = 4 ), which simplifies to ( ab = 2 ). Since ( a ) and ( b ) are positive integers, I can list the possible pairs of ( (a, b) ) that satisfy this equation. The pairs would be ( (1, 2) ) and ( (2, 1) ) because ( 1 times 2 = 2 ) and ( 2 times 1 = 2 ).Now, I need to check which of these pairs also satisfies the real part equation ( a^2 - b^2 = 3 ).Let's try ( (a, b) = (1, 2) ):( 1^2 - 2^2 = 1 - 4 = -3 ). Hmm, that's not equal to 3. So, this pair doesn't work.Next, let's try ( (a, b) = (2, 1) ):( 2^2 - 1^2 = 4 - 1 = 3 ). Oh, that works! So, ( a = 2 ) and ( b = 1 ) satisfy both equations.Therefore, the complex number ( a + bi ) is ( 2 + i ).Wait, let me double-check to make sure I didn't make any mistakes. If I square ( 2 + i ), I should get ( 3 + 4i ).Calculating ( (2 + i)^2 ):( 2^2 + 2 times 2 times i + i^2 = 4 + 4i + (-1) = 3 + 4i ). Yep, that's correct.So, I think I've got the right answer here. It's always good to verify my work to avoid any mistakes, especially since I'm still getting comfortable with complex numbers.In summary, by expanding ( (a + bi)^2 ) and equating the real and imaginary parts to those of ( 3 + 4i ), I was able to set up a system of equations. Solving these equations led me to the values of ( a ) and ( b ), and confirming by squaring the result ensured that my answer is correct.

💡Alright, let's tackle this problem step by step. So, we have a student who multiplied 84 by a repeating decimal, specifically (1.overline{cd}), where (c) and (d) are digits. However, the student mistakenly thought it was (1.cd) instead of the repeating decimal. After recalculating, he found out that his answer was 0.6 less than the correct answer. We need to find the two-digit number (cd).First, let's understand the difference between (1.overline{cd}) and (1.cd). The notation (1.overline{cd}) means that the digits (cd) repeat indefinitely, so it's a repeating decimal. On the other hand, (1.cd) is a finite decimal, meaning it stops after two decimal places.Let's denote the repeating decimal (1.overline{cd}) as (x). So, (x = 1.overline{cd}). Similarly, the finite decimal (1.cd) can be written as (1 + frac{cd}{100}), where (cd) is the two-digit number formed by digits (c) and (d).The student calculated (84 times 1.cd) instead of (84 times 1.overline{cd}), and the difference between the correct answer and his answer was 0.6. So, mathematically, this can be written as:[84 times 1.overline{cd} - 84 times 1.cd = 0.6]We can factor out the 84:[84 times (1.overline{cd} - 1.cd) = 0.6]Now, let's find the difference between the repeating decimal and the finite decimal:[1.overline{cd} - 1.cd = 0.overline{cd}]This is because subtracting the finite decimal (1.cd) from the repeating decimal (1.overline{cd}) leaves us with the repeating part (0.overline{cd}).So, substituting back into our equation:[84 times 0.overline{cd} = 0.6]Let's denote (0.overline{cd}) as (y). Therefore:[84y = 0.6]Solving for (y):[y = frac{0.6}{84} = frac{6}{840} = frac{1}{140}]So, (0.overline{cd} = frac{1}{140}).Now, we need to express (frac{1}{140}) as a repeating decimal. Let's perform the division:[frac{1}{140} = 0.00714285714285...]Wait, that's a bit messy. Maybe there's a better way to find (cd). Since (0.overline{cd}) is a repeating decimal with two digits, it can be expressed as:[0.overline{cd} = frac{cd}{99}]So, we have:[frac{cd}{99} = frac{1}{140}]Cross-multiplying:[cd = frac{99}{140}]Calculating that:[cd = frac{99}{140} approx 0.7071]Hmm, that's approximately 0.7071, which is close to 0.71. But (cd) should be a two-digit integer. So, rounding 0.7071 to the nearest hundredth gives us 0.71, which corresponds to the two-digit number 71.Let me verify this. If (cd = 71), then:[0.overline{71} = frac{71}{99} approx 0.717171...]But earlier, we had (0.overline{cd} = frac{1}{140} approx 0.00714285...). Wait, that doesn't seem to match. Did I make a mistake somewhere?Let me go back. We had:[0.overline{cd} = frac{1}{140}]But (0.overline{cd}) is actually equal to (frac{cd}{99}). So:[frac{cd}{99} = frac{1}{140}]Cross-multiplying:[cd = frac{99}{140} approx 0.7071]But (cd) should be an integer between 00 and 99. So, 0.7071 is approximately 0.71, which would correspond to 71. However, when we plug 71 back into the equation, it doesn't exactly equal (frac{1}{140}). There's a discrepancy here.Maybe I need to approach this differently. Let's consider that (0.overline{cd}) is a repeating decimal, so it's equal to (frac{cd}{99}). Therefore:[frac{cd}{99} = frac{1}{140}]Multiplying both sides by 99:[cd = frac{99}{140}]Calculating this:[cd = 0.7071...]Again, we get approximately 0.7071, which is roughly 0.71. Since (cd) must be an integer, the closest two-digit number is 71. Even though it's not exact, it's the closest possible integer. Let's check if this makes sense in the context of the problem.If (cd = 71), then:[1.overline{71} = 1.717171...][1.71 = 1.71]Calculating the difference:[1.717171... - 1.71 = 0.007171...]Multiplying this difference by 84:[84 times 0.007171... approx 0.6]Yes, that checks out. So, even though (cd = 71) isn't exact in the fraction sense, it's the closest two-digit integer that satisfies the condition given in the problem. Therefore, the two-digit number (cd) is 71.

💡Alright, so I have this problem about a function f(x) that's both odd and periodic. The minimum positive period is π, and for x in [0, π/2], f(x) is given as -4 sin(2x). I need to find f(8π/3). Hmm, okay, let's break this down step by step.First off, since f(x) is periodic with period π, that means f(x + π) = f(x) for all x. So, if I have a value like 8π/3, which is bigger than π, I can subtract multiples of π to bring it within a range where I know the function's definition. Let me calculate how many times π fits into 8π/3.So, 8π/3 divided by π is just 8/3, which is approximately 2.666. That means I can subtract 2π from 8π/3 to get it within one period. Let's do that:8π/3 - 2π = 8π/3 - 6π/3 = 2π/3.Okay, so f(8π/3) is the same as f(2π/3). Now, 2π/3 is still bigger than π/2, which is the upper limit of the interval where f(x) is defined as -4 sin(2x). So, I need to figure out what f(2π/3) is.Since f(x) is an odd function, that means f(-x) = -f(x). Also, because it's periodic with period π, f(x + π) = f(x). Maybe I can use these properties to express f(2π/3) in terms of f at some point within [0, π/2], where I know the function's definition.Let me think: 2π/3 is equal to π - π/3. So, f(2π/3) = f(π - π/3). Since the period is π, f(π - π/3) = f(-π/3). But f is odd, so f(-π/3) = -f(π/3). Therefore, f(2π/3) = -f(π/3).Now, π/3 is within the interval [0, π/2], right? Because π/2 is approximately 1.5708 and π/3 is about 1.0472. So, I can use the given definition of f(x) on [0, π/2] to find f(π/3).Given that f(x) = -4 sin(2x) for x in [0, π/2], let's compute f(π/3):f(π/3) = -4 sin(2*(π/3)) = -4 sin(2π/3).I know that sin(2π/3) is sin(π - π/3) which is sin(π/3) = √3/2. So, sin(2π/3) = √3/2.Therefore, f(π/3) = -4*(√3/2) = -2√3.But earlier, I found that f(2π/3) = -f(π/3). So, plugging in the value:f(2π/3) = -(-2√3) = 2√3.Therefore, f(8π/3) = f(2π/3) = 2√3.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from f(8π/3), I subtracted 2π to get 2π/3. Then, I expressed 2π/3 as π - π/3, used the periodicity to get f(-π/3), then used the odd function property to get -f(π/3). Then, I calculated f(π/3) using the given function and found it to be -2√3, so f(2π/3) is 2√3. That seems correct.Just to be thorough, let me verify the periodicity and the odd function properties again. If f(x + π) = f(x), then f(8π/3) = f(8π/3 - 2π) = f(2π/3). Then, since 2π/3 is greater than π/2, I need to use the properties of the function to express it in terms of a value within [0, π/2]. Using the fact that f is odd and periodic, I can relate f(2π/3) to f(-π/3), which is -f(π/3). That seems solid.Calculating f(π/3) as -4 sin(2π/3) is correct because 2*(π/3) is 2π/3, and sin(2π/3) is indeed √3/2. Multiplying that by -4 gives -2√3, so f(π/3) is -2√3. Then, f(2π/3) is -f(π/3) = 2√3. Therefore, f(8π/3) is 2√3.I think that's all correct. I don't see any mistakes in my reasoning. So, the final answer should be 2√3.**Final Answer**The value of ( fleft(frac{8pi}{3}right) ) is boxed{2sqrt{3}}.